Transcription
This is going to be a long journey, but if you stick with me, I promise that you will see the absolute peak of human intellect.
Even if all you know is high school math, I will take you to the point where you can understand how the proof of Fermat's theorem actually works.
When people try to learn Fermat's theorem, there are usually two kinds of materials.
The first kind is too easy. It spends 20 minutes on the margin story. Fermat wrote something in a book. Mathematicians struggled for hundreds of years while worked in secret. Fine. And then when the mathematics should actually begin, everything suddenly gets replaced by metaphors. Oh, an elliptic curve is a donut. Modular form is some kind of symmetry. And the next thing you see is walls crying.
The second kind is the complete opposite. It's serious mathematics, but ordinary people cannot understand it. It literally starts from, "Oh, let's look at this residue representation." But like, what the hell is a residue representation, right?
So here are the easy ones: Elliptic curve is a donut. Wiles' work in secret. Modular form is a symmetry. Hardest problems in the world. Most famous one wrote something. 350 years of a legendary proof. And these are the hard ones.
So we have these two extremes. On one side, a beautiful story but no real mathematics. On the other side, real mathematics but locked behind the background. And between them, there's nothing.
And I wanted to fix this. I wanted to build a ladder between them. So, uh, I bought this for like five bucks.
Famous theorem is not legendary because of its margin story. It's legendary because its proof is so important and so incredibly beautiful. And that proof is so elegant to be buried under these cheap, comfortable stories.
So this video starts from high school mathematics. It literally starts from what is a matrix? What is a multivariable function? And from there, I will build the mathematical tools needed for Fermat's theorem step by step.
Think about explaining the quadratic formula to an elementary school student. You don't need to teach every single thing from middle school to high school. You teach exactly what is needed: completing a square, square roots, basic equation manipulations. And that's what this video is exactly doing. I will assume you only know high school mathematics, and from there, I will introduce the pieces we need one by one, everything.
Last October, I made the Korean version of the video, and it went viral. People talking about it spread in online communities, and most importantly, there were people who understood the proof through my video. People who had always thought Fermat's theorem was out of reach, for the first time, the proof made sense to them. And that was exactly what I wanted to do.
After that, I started thinking about what I could improve. Uh, there were parts I wanted to cut or add, and more than anything, I wanted to make this in English so that more people could watch it.
So I posted the idea in Reddit, and people were basically like, "Do it."
This video is designed top-down. Every one of the 200 slides is there for a reason. For example, the spectral theorem appears earlier in the video. Why? Because later we need it when we talk about Hecke forms forming a basis of spaces of modular forms. Why do we need to study partial derivatives? Because we need them to detect singular points and elliptic curves. And why do singular points matter? Because they connect to reduction types, and reduction type eventually connects to L-functions and conductors.
So almost every slide is foreshadowing something that will matter later on in the proof. Nothing is there without a reason.
This video has six chapters. Chapter one is basics. Here we start with basic ideas that will keep appearing throughout the video: modular functions, common mathematical notations and abbreviations. And we will go over the proof of Fermat's theorem for n=3 and n=4.
Chapter two is linear algebra, and chapter three is abstract algebra. These two chapters are basically the entrance gate. We need them before we seriously talk about elliptic and modular forms. We also touch on Galois theory because later Galois representation will be one of the most important objects in the video.
Now, Fermat's theorem can be seen as a consequence of the modularity theorem. So, in a way, this entire video is built towards understanding the modularity theorem. Roughly speaking, it says that two completely different mathematical objects, elliptic curves and modular forms, are related. They're connected together.
So chapter four is about elliptic curves, and chapter five, we talk about modular forms. And chapter six is the endgame. This is where everything we built finally comes back. Everything that we planted earlier starts paying off here. For me, this is the most exciting and overwhelming chapter because this is where we finally use everything to prove Fermat's theorem.
So, let's begin. Chapter one is basics. In this chapter, we learn the really basic things: modular formulas, the complex plane, the terms and abbreviations that will keep appearing throughout the video. And we will also prove some n=3 and n=4 cases of Fermat's theorem.
So let's start off by common number sets. Uh, you probably already know what these number systems are. You know natural numbers, integers, rational numbers, real numbers, and complex numbers. So these are not new ideas. So I'm not going to be like explaining what is a what is integers and spend time on that. That's not the point here. The point here is to just get used to notation.
So first, this N with a little stroke here or uh here, uh this stroke is distinguished between other large numbers N. Okay, this denotes a set of natural numbers. So, um, yeah, this is the set of natural numbers. So 1, 2, 3, 4, 5, 6, 7, 8, such numbers are in this set. So, uh, if I write something like, uh, x in N, you should immediately know that, oh, we are talking about, uh, natural numbers N. Okay.
Z denotes the set of integers. Z came from the German word "Zahlen," "Zahlen" means basically numbers in German. So, uh, this set, like 7, 34, minus 3, are in this set. And Q is the set of rational numbers. Uh, obviously, rational numbers look like this. And, uh, real numbers, real numbers, we write it like this, R. And C, the biggest set, is the set of complex numbers. So like 1 + i, 3 - 4i, are in this set.
Using this notation, we can also write the inclusion relationship between these sets. Um, because natural numbers are a subset of integers, and all integers are rational numbers, which is in real numbers, and the biggest set is complex numbers.
So, like Cartesian product. So for set A and B, the Cartesian product A x B is defined as the ordered pair (a, b) where a comes from the set A and b comes from the set B. So this product sign may look like it has something to do with multiplication, but it's actually doesn't, like it's not like you actually multiply two sets. It is a definition of how to make a new set from given two sets.
So, um, for instance, if I take A as the set of {1, 2} and B set of {3, 4, 5}, what is A x B? So this is a new set of elements (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5). So in total of six elements. So the first component comes from A and the second component comes from B.
Uh, how about A x A? Um, actually, I can also write this as A squared using the exponent like this. Uh, it's a set of (1, 1), (1, 2), (2, 1), (2, 2). Okay. Uh, so, um, so like, for example, R squared. This is a two-dimensional plane, right? Cartesian plane, because it's a set of ordered pairs (x, y) where x and y are real numbers. So you will see so many of this Cartesian product notation if you open any undergraduate or graduate-level math textbook. So, um, you want to get used to this.
Before we go further, I wanted to quickly introduce some symbols and abbreviations that will appear throughout the lecture. These are basically shortcuts. Um, mathematicians use them all the time. I use them all the time to save time and make statements clear.
So, first, this upside-down A looking thingy, which is actually an upside-down A, means "for all." So if I write something like "for all x in R," you can think of it as, "Oh, we are looking at all, each, all, each one of real numbers." Okay.
And this reflected E thing means "there exists." So I can write something like "there exists x which is a real number such that..." I don't know, x² = 2. This is true because x exists as √2, right?
And this two-dot and equal sign means "it's defined as something." So if I write A is defined as B, this does not mean that A and B are the same. This is different to an equal sign. It means that we are defining A as B. Okay.
And WLOG here means "without loss of generality." Um, so what does it actually mean by "without loss of generality"? Suppose we're looking at integers x, y, z. So we know that x, y, z is an integer, and we also know that x, y, z, the product of them is even. And this is all we know about x, y, z. And notice that x, y, z is symmetric, right? And because x, y, z are integers and x, y, z is even, we know that at least one of x, y, z is even, right? So in this kind of situation, we write "without loss of generality, z is even." So we don't know which one of x, y, z is even, but we know that an even number exists among x, y, z, right? So we just name that even number as z because x, y, z are symmetric.
And this s.t. means "such that." It's a shortcut for "such that." So instead of writing the full "such that," we write s.t. dot. Okay.
And WTS means "want to show," and ETS means "enough to show." So suppose we are in some complicated situations involving an integer n. So n is an integer, and we want to show that n² is even. We want to prove that n² is even. So it's we want to show that n² is even, right? But since n is an integer, we can just show that n itself is even, right? So, uh, so we can write, "we, it's enough to show that n is even."
And before we go on, we can write Fermat's theorem using these symbols and abbreviations. So, for n greater or equal to 3, there does not exist x, y, z which are integers such that xⁿ + yⁿ = zⁿ. Of course, x, y, z is not zero.
Now let's talk about divisibility and congruence. These are extremely basic ideas in number theory, but they appear everywhere. So I want to fix the notation carefully.
Um, so first, this a | b means that a divides b, equivalently, there exists k such that b = ak. Of course, k here is an integer. Um, so, easy, 3 divides 6, right? 6 is a multiple of three. Uh, and, I don't know, 17 divides 68. 100 divides 2000. And a /b means that a does not divide b. So 4 does not divide 9. And I don't know, 18 does not divide 50.
And lastly, for integers a and b and natural number n, you write a ≡ b (mod n). This means that n divides a - b. Um, the standard way to read this is "a is congruent to b modulo n," but in this lecture, I'll just read "a is equal to b modulo n" because it's simpler. Okay. And this is equivalent to saying a and b have the same remainder when divided by n. Okay.
So let's look at some examples. 17 ≡ 32 (mod 5). This is because the difference between 17 and 32 is 15, and 15 is divisible by five. And also, 17 and 32 have the same remainder, 2, when divided by five. Okay. Another example would be 15 ≡ 29 (mod 7) because, um, the difference between 15 and 29 is 14, right? And 14 is a multiple of seven. And this is also because 15 and 29 have remainder 1 when divided by seven. Okay.
Greatest common divisors. So you know what greatest common divisor is. So this is just a notation for integers a and b. The greatest common divisor of a and b is denoted by like this, gcd(a, b). And sometimes I'll just don't write the gcd. I'll just write (a, b) something like this. So, for example, gcd(24, 36) = 12. gcd(28, 49) = 7. And gcd(17, 32) = 1, which means that they're coprime or primitive. Okay.
And these below three properties, these are more of a technique that we'll be using when we prove the cases n=3 and n=4 of Fermat's Last Theorem, right after. And these properties can be proved rigorously using elementary number theory, but I'm not going to prove every single one of them here. For our purpose, we can just understand them intuitively, okay?
So, first, if gcd(a, c) = 1, meaning a and c have no common factor, if a and c are coprime, then gcd(a, b, c) = gcd(b, c). Um, why does this hold? Uh, suppose we are computing the gcd(a, b, c). But since gcd(a, c) = 1, meaning that they have no common factor, uh, a does not do anything here, right? Because a and c do not share any common factor. So we can just ignore the a and look at the gcd(b, c), right? Um, again, this is not meant to be a rigorous proof. I'm just saying this to give you the basic feeling of this. Okay.
Second, for any integer k, gcd(a, b) = gcd(a, b - ak). This is because if I denote the gcd(a, b) as d, d also divides b - ak, right? Because d divides both a and b. So that d divides b - ak. And the last one, if gcd(a, b) = 1 and a divides bc, then a divides c. Uh, this is very similar to the first one. This is because because gcd(a, b) = 1, it means that no common factor is shared between a and b, right? So in this divisibility, b does not do anything here, which means that we can just wipe out the b and look at c. So we write a divides c.
Now we are going to prove Fermat's theorem for n=3. And to do that, we need one lemma. And what is a lemma? So a lemma is basically a supporting theorem. It's not usually the final result we want, but it's a smaller result that helps us prove the main theorem. For now, we will take this lemma as given and not prove this right now. And later, we will prove this using something called a unique factorization domain in the abstract algebra part.
So what does this lemma say? Uh, let p and q be coprime integers such that p² + 3q² = s³. So p² + 3q² is a perfect cube. Okay. Then there exist coprime integers u and v such that this equation holds.
So let's take this lemma as given and prove the n=3 case of Fermat's Last Theorem. So we want to prove that there exist no integers such that x³ + y³ = z³. Right? Of course, x, y, z is non-zero. And, uh, if I make a substitution here, I write z' = -z. So we get x³ + y³ + (z')³ = 0. Right? And, uh, to me, this equation looks more symmetric and easy to handle than this equation. So how about we just start from this equation and show that this equation has no solution? Right?
Um, so we start from this equation x³ + y³ + z³ = 0. And we will show that this equation, this equation has no non-trivial solution. So let's assume that x, y, z satisfies [clears throat] this equation, right? So x, y, z is not zero. And we can also assume that x, y, z are pairwise coprime, or primitive, because we can just divide by their common divisor if there is a solution to this equation. Okay. So x, y, z are pairwise coprime. And, uh, we are going to look at the parity on each side.
Uh, notice that x, y, z cannot be all odd numbers because if they were all odd, um, the left-hand side would be odd, and it doesn't make sense because the right-hand side is even. So we know that at least one of x, y, z is even. But since x, y, z are pairwise coprime, exactly one of x, y, z is even, right? So without loss of generality, we let that even number be z, and x, y, we know that they're odd. So since x and y are odd numbers, x + y and x - y are both even. So we can write them as 2u and 2v for some integers u and v. And, um, so, uh, if I solve this, we have x = u + v and y = u - v. And we also know that u and v have different parity, right? u and v have different parity. So one is an odd and one is even, because if they had the same parity, x and y would be both even numbers, which is a contradiction to the fact they're coprime, right? And we also know that u and v are coprime because their greatest common divisor should, uh, uh, because their greatest common divisor should divide both x and y. So gcd(u, v) = 1.
So we put this into the original equation, which we get -z³ = (u + v)³ + (u - v)³ and this is 2u³ + 6uv². So this is 2u(u² + 3v²). And from now on, we will going to focus on the greatest common divisor of 2u and u² + 3v². So gcd(2u, u² + 3v²). And, uh, we know that u² + 3v² here, this is an odd number because u and v have different parity, right? And since there's a 2 on the left side, we can just wipe out the 2. Remember the property we just studied. So this becomes gcd(u, u² + 3v²). And I'm going to subtract u² from the right-hand side by u * u, which means that we can delete this u², so it becomes gcd(u, 3v²). And since u and v are coprime, this is either 1 or 3, right? Uh, u and v don't share any common factors. Okay.
So we go case by case. First, the case where gcd(u, 3v²) = 1. So we know that these two factors are coprime. And, uh, since these two factors are coprime and the product is a perfect cube, each factor, so 2u and u² + 3v², should be a cube by itself. Uh, the reason is that the prime factors cannot be divided between the shared two sides, right? So 2u and u² + 3v² should be both a perfect cube. So 2u, I'll write as r³ for some r, and u² + 3v² as s³ for some r and s. So we know that r and s are coprime. Okay.
So, uh, from this equation, this is familiar, right? This is from the lemma. So we can use the lemma here, which gives us u = e² - 3f² and v = 3ef for some e and f. And if I put these two into this equation, we also get s, and s is actually e² + 3f². Okay. And, uh, so r³ here, r³ = 2u. So 2(e² - 3f²). And we also know that 2, e + √3f, e - √3f are all pairwise coprime. So how do you know this? For example, um, if we look at gcd(2e, e + √3f), uh, e and f have different parity because if they had the same parity, v should be an even number, and that cannot be happening. So, uh, e and f are all, e and f have different parity, which means that we can ignore the 2 on the left side. So this becomes gcd(e, e + √3f). If I subtract e from this side, it becomes gcd(e, √3f). And e cannot be a multiple of 3 because, uh, if e is a multiple of 3, u and v would be all multiples of 3. So this is 1. And likewise, we can show that 2e, e + √3f, and e - √3f share no common factors. So we use the property we used here, same as here. So since these, uh, three factors are coprime and their product is a perfect cube, uh, each of them should be a perfect cube. So -2e = k³ for some k, and e + √3f, I'll write it as l³, and e - √3f as m³. Uh, so if I add these together, we get k³ + l³ + m³ = 0. Also, one thing, the absolute value of z³ = -2u(u² + 3v²) = r³s³. So this is equal to (rs)³. So this is strictly greater than k or l or m.
Okay, so what have you actually done from this argument? So, um, we started by assuming that there exists a non-trivial integral solution to this equation. So x³ + y³ + z³ = 0. And from this solution, through these arguments, we constructed another solution of the same type: k³ + l³ + m³ = 0. So we made k, l, m from x, y, z. So we made k, l, m from x, y, z, right? Um, but the important point is that the new solution k, l, m is strictly smaller than the original solution x, y, z, right? You can see it from here, z is strictly greater than k or l or m. More precisely, if the original solution has some non-zero integer z here, uh, then we produced another solution which the corresponding value m has strictly smaller absolute value. So, um, suppose that we do the same thing to k, l, m, right? So we make another solution, and we do this again, make another solution. But as we do this, the solution keeps getting smaller and smaller, right? But this cannot go forever because x, y, z, k, l, m, n, they all should be integers. And, uh, because there cannot be an infinitely strictly decreasing sequence of positive integers, this cannot be happening, which is a contradiction. So this means that there cannot not be existing integral solution to this equation x³ + y³ + z³ = 0. So we proved the n=3 case by, uh, the lemma.
Now we're going to prove the FLT for n=4. And to do that, we also need one lemma. This says no right triangle with integer sides can have an area that is a square. So, uh, first let's prove this lemma. So, um, to prove this lemma, we assume that there exists a right triangle with integer sides and square area. So the area is a square. So by the Pythagorean theorem, a² + b² = c². Um, so and here we are going to prove the n=4 case of Fermat's Last Theorem. And before that, we were going to look at some lemma. This lemma says no right triangle with integer sides can have an area that is a square, right? So, um, let's prove this lemma. We assume that there exists a right triangle with integer sides, uh, with a square area. So area S is a square. So because this is a right triangle, we have a² + b² = c². And we also assume that a, b, c are pairwise coprime. a, b, c pairwise coprime. And we will examine both sides of the equation by using what I call a modulo 4 table. So what is a modulo 4 table, right? So, um, it looks like this: x (mod 4), x² (mod 4). 0, 1, 2, 3. So this table tells us that if x is 1 (mod 4), x² is, um, I don't know, (mod 4). Okay. So, um, if x is 1 (mod 4), it means that x is of the form 4k + 1. And if I take a square, it becomes 16k² + 8k + 1, and this is a multiple of four, right? So, um, this is actually equal to 1 (mod 4). So this means that if I take x that is 1 (mod 4), then x² is 1 (mod 4). Uh, and like this, I can fill out the, uh, blank part of this modulo 4 table, and it looks like this: 0, 1, 0, 1. And, um, uh, if a and b here are both odd numbers, means that they're either 1 or 3 (mod 4), then c² would be 2 (mod 4), right? 1 + 1 = 2. But no square is 2 (mod 4), right? So a and b, at least one of a and b should be an even number. And since a, b, c, we assume that they're pairwise coprime, at exactly one of x, a, and b is an even number, and the others are all odd. So without loss of generality, w.l.o.g., uh, we let a be even, and we know that b and c are odd. Uh, and we can, uh, transform this equation into (a/2)² = (c - b)/2 * (c + b)/2. Okay. And we know that these two are coprime. And since the product of two coprime factors is a perfect square, we know that these two are a square by themselves. Okay. So we write this as, um, I don't know, q² and p² for some p and q. And we know that p and q are coprime. So if I solve this, we get a = 2pq, c = p² + q², and we have b = p² - q². Okay. And the area, we know that area is a square. The area of the triangle is ab/2, which is pq(p + q)(p - q). And we know that p, q, p + q, p - q are all, uh, coprime, pairwise coprime, right? So we will write p as x², q as y², p + q as u², p - q as v². Uh, and we will focus on 2y² = 2q. And 2q = the difference of these two factors, so u² - v². Right? So (u + v)(u - v). And, um, we know that u and v are pairwise coprime. So, um, the possible, uh, greatest common divisor of u + v and u - v is either one or two, right? But since, uh, the left-hand side, we have 2y², which is an even number, we can write this as, um, something like 2b⁴ and 8a⁴. This is because we know that this is an even number, right? And since this is a perfect power of fourth, x should be even, right? So we have at least, at least four 2s in both sides. Uh, uh, that's why we can write this, uh, like this. So this is either 2b⁴ or 8a⁴, or 8a⁴ or 2b⁴. Okay. So, um, what do we get if we, uh, add this and divide it by two? We have 4a⁴ + b⁴ = z². But this cannot happen because the solution (2a², 2b², z) um, if we make this a right triangle like this, 2a², 2b², z, we can make this a right triangle because of this equation. This satisfies the Pythagorean theorem, and, uh, this, uh, area of the associated right triangle is a perfect square, but that cannot happen, uh, from my lemma. So, uh, this is a contradiction, which means that we prove the n=4 case of Fermat's Last Theorem.
Multivariable functions. So you're probably already very familiar with functions of random variables. For example, f(x), if I define f(x) as x², it takes one variable x, right? And it gives one output x². So the input is one number, and the output is also one number. But of course, we can also define functions with more than one input. For example, a function that takes two variables x, y. And I can define f(x, y) as, I don't know, xy + sin(x)cos(y) + y². I don't know, some weird function, but this actually works as a function. This is a valid, legit, two-variable function. And, uh, I can also define functions with three variables. So if I define f(x, y, z) as the square root of x² + y² + z². Uh, this denotes, this means the distance from the origin to the point (x, y, z) on a third-dimensional plane. So, uh, multivariable function, nothing special. We won't going to be deep, just that the number of variables can be more than one in a function.
Now, once you have functions of several variables, we can also talk about derivatives. For one-variable functions, um, how do we define derivatives? So f'(x) is defined by the limit when h goes to zero of (f(x + h) - f(x)) / h, right? So there's only one direction in which the input can change, only x changes here, right? So, um, this means x changes by h here. But for a multivariable function, for example, f(x, y) = xy + x² + y². Um, there's several variables. Uh, in this case, we have two variables, x and y. So, um, how do we define derivatives when the number of the variables is more than one? Uh, so we can ask, what happens if only x changes here while y stays fixed? What happens if only y changes if we fix x? Okay. And that is the idea of partial derivatives. A partial derivative is just a derivative when we treat one variable as the actual variable and treat all the other variables as constants. So we define partial derivatives like this. This is the partial derivative of f with respect to xᵢ. So, um, here only xᵢ changes, while other x₁, x₂, through x<0xE2><0x82><0x99> stays fixed, right? Um, so, for example, if I define f(x, y) as x³ + y³ + xy + I don't know, sin(x)cos(y). If I take a partial derivative with respect to x, so this is called ∂f/∂x. Uh, we assume that y is constant. Okay. And only x is the variable. So we have 3x² and this is variable, so it goes to zero. Plus y, because y is a constant, plus cos(x)cos(y). Uh, we can also take partial derivative with respect to y, and it's 3y² + x - sin(x)sin(y). It works something like this. So we will not actually compute partial derivatives in detail, but they will show up later when we talk about singular points and elliptic curves. So I wanted to introduce the idea briefly here.
This is a Taylor series. We will not use it constantly, but it appears a few times later. So rather than leaving them as black boxes, I just wanted to briefly mention them here. So a Taylor series is a way to represent a given function as an infinite polynomial. For example, uh, the, uh, power function here, eˣ, is equal to 1 + x + x²/2! + x³/3! + blah blah blah, it goes on. And if we take infinitely many terms here, it converges to eˣ. And, uh, but if we take only the small terms, for example, um, if we take only the three terms here, this is an approximation of eˣ. For example, if we look at this exponential function, the graph of eˣ, it looks like this, right? eˣ. If I only take the first term here, y = 1, and the graph will look like, uh, something like this, y = 1. If I take, uh, the second term, 1 + x, this is actually a tangent at (0, 1), it looks like this. If I take, uh, the third term, it will get closer to eˣ. And as we include more and more terms, this polynomial will get closer to eˣ, and eventually, it will be the same if I take, uh, infinitely many terms. Okay.
So here are some famous Taylor series for famous functions: eˣ, sin(x), cos(x), ln(1 + x). So you do not need to memorize everything written here. Later, when Taylor series shows up, you can just think, "right, with something like this," and accept it from there.
eⁱ<0xE1><0xB5><0x83> and all functions. So eⁱ<0xE1><0xB5><0x83> + 1 = 0. You've seen this formula, right? This is often called one of the most beautiful formulas in mathematics, and the reason is that it contains so many important constants in one tiny equation. So we have all these constants here: natural constant e, π, and i. i is the imaginary unit, and 1 is called the multiplicative identity, which means that you take any number, multiply it by 1, and it gets the same number back. And 0 is the additive identity. But behind this famous formula, there is a more general formula, and this is called Euler's formula. This for any real number θ, eⁱ<0xE1><0xB5><0x83> = cos(θ) + i sin(θ). So if we take θ = π, we get eⁱ<0xE1><0xB5><0x83> = cos(π) = -1 and sin(π) = 0. So this is equal to -1. And if I, uh, take this -1 and move it to the left-hand side, this becomes the, uh, beautiful Euler's formula, whatever.
Now, depending on how you build the theory, you can approach this formula in slightly different ways. But for our purpose, it's perfectly fine to think of it as a definition. So what does it mean to even put an imaginary number in the exponent? Right? It doesn't mean to like multiply by i times. That doesn't make any sense. So this definition tells us how to do it when we see a complex number in the exponent. And this definition is not arbitrary, meaning that it's designed to fit perfectly with the exponential function we already know. And if you know the Taylor series, this definition matches very well with the Taylor series.
So let's see, uh, Taylor series of the power function eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! and so on. Uh, originally, this is only meant for real numbers, right? But, uh, how about we just, uh, plug in x = iθ, right? Because, like, why not? So let's just do it. eⁱ<0xE1><0xB5><0x83>. So if I take x = iθ, we have 1 + iθ - θ²/2! - θ³/3! i + θ⁴/4! and it goes on, plus θ⁵/5! i - plus θ⁶/6!. Uh, this is actually minus, and it goes on. And if I, uh, simplify this by taking out the i term here, it becomes 1 - θ²/2! + θ⁴/4! - θ⁶/6! etc., etc., plus I'll take out the i on the odd number terms, so θ - θ³/3! + θ⁵/5! and it goes. And this is actually the Taylor series of cos(θ). And this is the Taylor series for sin(θ). So it's equal to cos(θ) + i sin(θ). So you can see that this definition matches perfectly with the Taylor series we just saw.
So, um, let's look at this formula and maybe we can plug in some famous, uh, θ. So, uh, if I take θ = 2π, we have e²<0xE1><0xB5><0x83>ⁱ = cos(2π) = 1 and i sin(2π) = 0. So we have 1. And, uh, we also, uh, computed that e<0xE1><0xB5><0x83>ⁱ = -1, right? If I take θ = π/6, eⁱ<0xE1><0xB5><0x83>/6 = cos(π/6) = √3/2 and sin(π/6) = 1/2 i. So it's, it's like this. And, uh, also if I take θ as θ + 2π, so eⁱ(<0xE1><0xB5><0x83> + 2π) = cos(θ + 2π) and sin(θ + 2π) are all equal to just cos(θ) and sin(θ), right? Because this is a periodic function with period 2π. So this is equal to cos(θ) + i sin(θ). And again, this is equal to eⁱ<0xE1><0xB5><0x83>. So we know that, uh, if we view this eⁱ<0xE1><0xB5><0x83> as a function of θ, this has a period of 2π, right? So, uh, eⁱ<0xE1><0xB5><0x83>/3ⁱ. You can take any number, and it's actually equal to eⁱ<0xE1><0xB5><0x83>/3 + 2πi and eⁱ<0xE1><0xB5><0x83>/3 + 4πi and it goes on. So in general, we know that eⁱ<0xE1><0xB5><0x83> = eⁱ(<0xE1><0xB5><0x83> + 2nπ) when n is a natural number.
Now let's look at the nth root of unity. The phrase sounds a little bit fancy, but the idea is simple. We are looking for complex numbers such that zⁿ = 1. For example, if n = 2, and we are solving z² = 1. So we know that z is simply either 1 or -1. Okay. Uh, now in general, the equation zⁿ = 1 is the same as zⁿ - 1 = 0. And we know that this is an nth-degree equation. So over the complex numbers, we expect exactly n roots. And indeed, the n roots are given by this formula. So how does this formula work? So we are looking at z such that zⁿ = 1. And 1, we know that 1 is equal to e²<0xE1><0xB5><0x83>ⁱᵏ, so here we take the 1/n power on each side, and we get z = e²<0xE1><0xB5><0x83>ⁱᵏ/ⁿ. Of course, here k could be any integer. So if we write this explicitly and plug the values of k, um, first, if I take k = 0, it becomes 1. Of course, 1 is a solution to this equation, right? If I take k = 1, e²<0xE1><0xB5><0x83>ⁱ/ⁿ. Okay. And k = 2, e⁴<0xE1><0xB5><0x83>ⁱ/ⁿ. And it goes on until if I take k = n - 1, we have e²<0xE1><0xB5><0x83>ⁱ(n-1)/n. And if I take k = n, we have e²<0xE1><0xB5><0x83>ⁱ. And if I take k as n + 1, we have e²<0xE1><0xB5><0x83>ⁱ(n+1)/n. And if we look at this power, it's actually 2πi + 2πi/n, right? So, uh, while k could be any integer, uh, if we put every k here, this repeats because this one is the same as this e²<0xE1><0xB5><0x83>ⁱ. This value is the same as this value, uh, because of the periodicity of the eⁱ<0xE1><0xB5><0x83> function. So we restrict the, uh, range of k to 0 to n - 1, and we can find all the values that we want to find. Okay.
So let's test this formula in small cases. Uh, this actually works, right? So if I take n = 2, we know that the values we're looking for is 1, e⁻ⁱ<0xE1><0xB5><0x83>. Okay. So zᵏ = eⁱ 2πk/2, and this is equal to eⁱ<0xE1><0xB5><0x83>ᵏ. Okay. So if I take k = 0, we have 1. And if I take k = 1, we have eⁱ<0xE1><0xB5><0x83> = -1. So we found all these. Um, if I take n = 3, first of all, without, uh, knowing this formula, if we just solve z³ - 1 = 0, what do we get? Uh, we can factorize this into (z - 1)(z² + z + 1), right? So the expected values are z = 1 and (-1 ± √3i)/2. So we are looking for these three values. Okay. So if we, uh, use the formula zᵏ = eⁱ 2πk/3. So if we take k = 1, this one, k = 1. This is e²<0xE1><0xB5><0x83>ⁱ/3. And if I actually write this explicitly, it is cos(2π/3) + i sin(2π/3). And sin(2π/3) is √3i/2 and cos(2π/3) is -1/2. If I take k = 2, the value is e⁴<0xE1><0xB5><0x83>ⁱ/3. And if we, uh
Compute this using the same method. We have uh minus <unk>3 over 2 I + 1 / 2. So we found the value that we are looking for. This gives exactly the same value as uh actually solving the equation.
Now let's talk about the complex plane. So what is a complex plane? Um if you take any complex number, it has the form of x + i y, right? So it means that if you pick any complex number, this corresponds to two real numbers x comma y, and x comma y we can correspond to a point on a two-dimensional plane. So this means that we can uh correspond any complex number into a point on a two-dimensional plane, and that two-dimensional plane we call it a complex plane. So this is a complex plane. The horizontal line is the real axis, and the vertical line is the imaginary axis. So um if I take Z equals I don't know 4 + 3 I and we mark this Z onto this complex plane, this is somewhere around uh here where the real value is three and the imaginary part is three. So here we have Z = 3 4 + 3 I. Um if I take Z = -1 + 2 I, it would be somewhere around here, -1 to this point is -1 + 2 I, and we can also do this kind of thing in the complex plane.
Uh, for example, suppose we want to draw the set of complex numbers Z satisfying uh the absolute value of Z equals 1. So if I write Z as X + I Y, the absolute value of Z is the square root of X² + Y². And if this is equals 1, this means that X² + Y² equals 1. So this is a unique circle with center at the origin. So if I take, if I collect every abs uh every complex number that has absolute value one, this is on this unit circle.
How about the real part of z equals 2? Right? If I uh write z as x + i y, the real part of z is just x. So this Z is on a vertical line here where uh the point meeting the uh horizontal axis is two. Okay.
Now let's connect this with what we just talked about, the nth root of unity. We said that the nth root of unity uh zt k equals e to the 2 pi i k / n and this is equal to i sin 2 pi i k / n plus cosine i pi i k / n. And if we uh plot this on the complex plane, the corresponding points will be cosine 2 pi i k / n and sine 2 pi i k / n. Right? So this will be on a unit circle, and the associated angle here would be this 2 pi i k 2 pi k / n. So this point will be plotted on the complex plane. It will be on the unit circle, and the angle here, this would be 2 pi k / n. And k ranges from 0 to n minus 1. And here, as k changes from 0 to n minus 1, these um n points divide the unit circle into equal n parts.
So what do I mean by this? If I take n = 3, the third root of unity will be e to the 2 pi i k over 3 when k equals 0, 1, and 2, right? So the associated angle will be uh of course on a complex plane with a unit circle, the angle would be uh zero when k equals 0. And if I take k equals 1, the angle would be 2 pi over 3. So it will somewhere be like here, right? This is 2 pi over 3. And if I take k = 2, uh the angle will be 4 pi over 3, and the point will be here. Uh, so if I connect these three points, uh uh this is a this is an equilateral triangle, right?
And if I also take n = 4, z k equals e to the power of 2 pi i k over 4, right? So um if I take k = 0, it will be this point, uh this is actually one, right? If I take k equals 1, uh the angle will be pi / 2, this this point, this is i, right? i to the 4th equals 1. And if I take k = 3, it will be this point, this point denotes minus 1, and k = this is k = 3, and if I take k = 3, this will be this point, and this will denote minus 1. And see what we got, this is a perfect square. So geometrically, uh the nth root of unity, they are vertices of a regular n-gon sitting on a unit circle.
Now let's briefly talk about complex functions and holomorphy. We're not going to go deeply into complex analysis here. The main point is very simple. Just like we can define functions of real variables, we can also define functions whose input is a complex number and its output is also a complex number. For example, if I define f as Z squared, where Z is a complex number, this works as a complex number. So if I put something like um, of course, if I put a real number, uh it would be the function that we are familiar with, y = x², and but this is a complex function, meaning that the input could be complex, right? So if I put i, the value would be minus one. If I put f(1 + i), it will be 2i. Right? So this works just like the ordinary functions, except now the variable is complex.
And we can also ask whether this function is differentiable. Um, for example, um in this case, actually, if f(z) = z², and if I differentiate it, we have the uh definition of differentiability here, f'(z) = 2z. So at least for polynomial-like functions, uh things behave very similarly to real calculus. But here's the tricky part. In complex analysis, differentiability is much stricter than in real calculus. For this lecture, we do not need to master the full theory of complex analysis. Remember just two points. First is that a complex function is a function. It works as a function. And second, complex differentiability is a meaningful and a very strong condition.
Okay. Fourier expansion of a complex function. First, we have some mysterious periodic function. So this is a periodic function, but we don't really know what this function is. We do not really know its properties yet, and we do not really know what its graph looks like. But if we can express this function using the periodic functions we already well know, for example, sin(z), sin(z), cosine(z). These are the functions, uh these are periodic functions that we really well know, right? Then this function, if we express this function using these familiar functions, this function would become much easier to handle. So that is the basic idea of Fourier expansion. Can we build an unknown function out of these basic functions?
Now, in this slide, we have a complex function f that satisfies f(z + 1) = f(z). So this is a periodic function with period one. Um, and this h here, uh, this is the upper half-plane of the complex plane. You don't have to digest the full thing here. It's just we were going to talk about it later when we talk about modular forms. Okay. So anyways, f is a periodic function with period 1. And we already know a very simple function with period 1, uh e to the 2 pi i z. This is a function with period one, right? Because uh this is equal to e to the 2 pi i (z + 1). And of course, if q is a periodic function with period one, we will name this q. q is a periodic function. q squared is a periodic function. q times c, q cubed is a periodic function. Its power will be of course a periodic function of period one. So the idea is to express f with a sum of powers of q. And here we go. f(c) we express this with a power of q like this, when q = e to the 2 pi i c. And that's actually a formula to compute these coefficients. Uh, but we're not going to use the formula in this lecture. So don't worry about coefficient calculation here. The important point is just that a periodic function can often be expressed as a sum of basic periodic building blocks. Uh, later when we study modular forms, this becomes extremely important because modular forms are also written using this kind of expansion, and in that context, we usually call it a q-expansion because we're using this uh q very often into the 2 pi. For now, just remember that a function with period one can be expressed in this kind of q-expansion. We call this a q-expansion.
Okay, chapter two is linear algebra. Mathematics is calculation. Yes. But before calculation, mathematics is also about structure. And from that point of view, linear algebra is where we build the foundation of that structure. Think about building a house. Linear algebra is like the steel frame. It's not always the part you see first, but without it, the whole building cannot stand. The concepts you learn in this chapter, vectors, matrices, vector spaces, inner products, and so on, will become the solid foundation for what comes after.
The first object we need is the first object we need is a matrix. A matrix is just a rectangular array of scalars. Um, here scalars, you can just think of numbers, and this field F, this is a number set. You can just uh think of as a number set where these scalars or numbers come from. Some examples. This is a matrix, and this is a 2x2 matrix because it has two rows, right? One, two, and this has three columns, uh, two columns, one, two. So this is a 2x2 matrix. Uh, we write uh like this, and we often write uh like this, 2x2 matrix. Then how about um this matrix 1 2 4 7 I know -7 -6. This is a 2x3 matrix because it has one, two rows, and it has one, two, three columns. So this is a 2x3 matrix. This denotes the number of rows, and this uh means the number of columns in the matrix. Uh, so in general, an m by n matrix A has m rows and n columns. Okay. Uh, and the entry in the i row and j column is denoted by a_ij. Uh, so for example, if I name this matrix A, and um A12 is the entry in the first row, second column, right? The first number, it means rows, and the second number, it means that it comes from the second column. So first row, second column, first row, second column. So A12 = 2. How about then a um 23? A23 is the second row, third column, right? Second row, third column. So it's -6. Uh, how about A13? A13. So first row, third column. So 4. And how about A32? This doesn't exist because A is a 2x3 matrix. A does it have a third row?
Now, in this slide, I wrote that the entries come from a field F. So what is a field? I said that it's a number set. Uh, a field mathematically is an algebraic structure that we can add, subtract, multiply, and divide, of course, by non-zero elements. But don't worry about the definition of field right now. Um, we will talk about fields more carefully later. For now, you can just think of F as where the entries of the matrix or numbers come from. In this lecture, most of the time, this F will be either R or C. Meaning that we'll be only looking at matrices with entries from real numbers or complex numbers. For example, um, these two matrices are a matrix over R, right? Because the entries are real numbers. Of course, this is a matrix over C, right? Um, but this matrix 1 + i, -2i², <unk> 2i, -7, this is a matrix over complex numbers. This is a matrix over C. Well, this matrix pi, e, 1, -6 over I don't know 30,001. This is a matrix over R because the entries come from real numbers. This is a matrix over C because the entries come from complex number sets.
We can define operations on matrices. First, matrix addition is very straightforward. If two matrices have the same size, then we can add them entry by entry. So, for example, uh, if I were to add these two matrices 1 2 3 4, this is a 2x2 matrix, and this is also a 2x2 matrix, same size. Uh, that's why we can add them. So how do we add? We add corresponding entries. So 2 + 1 = 3, 3 + 2 = 5, 4 + 3 = 7, and 1 + 4 = 5. So the resulting matrix is also the same size as these two matrices. Okay. And subtraction works basically the same. So 1 3 7 4 minus um -3 2 0 7. If I compute this subtraction, it would be of course the same size. Uh, 4, 1, 7 minus 3, it would be something like this.
Next is scalar multiplication. This means multiplying a matrix by a scalar C. Here C is a scalar, just one number. So you multiply one number on a matrix. So how do you do this? Um, so if I take any matrix 1 2 3 4 5 8. So this is a 2x3 matrix. And if you take any scalar, um, for example, -3, and if you multiply this scalar to this matrix, you multiply it to every possible entry. So the resulting matrix would be the same size as this matrix, and it will be -3, -6, -9, -12, -15, -24. It's very straightforward and easy. Um, another example, 7 * 1 7 9 -8. It would be 7, 49, 63, -56. It works like this.
Now, matrix multiplication is a little bit complicated. So you might think matrix multiplication works like this. So if I uh take 1 2 3 4 and 3 4 5 7. You might think that, oh, if you uh multiply these two matrices, you uh multiply corresponding terms. So 1 * 3 = 3, 2 * 4 = 8, 3 * 5 = 15, 4 * 7 = 24. But matrix multiplication doesn't work like this. So this is not how it's done. Okay.
So to compute matrix multiplication AB, the number of columns in A has to be the same as the number of rows in B. So if A is an m x n matrix, then B should be some matrix of n x p matrix. So these n are the same, right? n was the number of columns in A, and n was uh this n was the number of rows in B. So these two should be the same. And so if I calculate this A product B, the resulting matrix is an m x p matrix. M uh comes from here. It's the number of rows in the matrix A, and P is the number of columns in matrix B. Okay. So how do we actually compute this? So the i row of the j column of the resulting matrix AB, uh is computed like this. Um, so this a_ik is the entry of the i row of k, e i, e i, e i, e i, e i, e i, e i row of A, right? E i row of A, and this b_kj comes from the j column of B, right? So we take the i row of A and the j column of B, and we uh multiply corresponding terms and add them.
So let's look at some examples. 1 2 3 4 and 2 4 0 3. So first of all, uh, this I will call this matrix A and call this matrix B. A is a 2x2 matrix, and B is a 2x2 matrix. The number of columns in A is the same as the number of rows in B. So the resulting matrix would be a 2x2 matrix, and this matrix multiplication we can do it because uh stay in two, right? Uh, then um the resulting matrix should be a 2x2 matrix. And what about the uh first row, first column entry of the resulting matrix? So first row, first column, we take the first row from A and the first column from B. So first row, first column, first row, first column. So 1 * 2 + 2 * 0 = 2. First row, second column of the resulting matrix. First row, second column. So 1 * 4, 2 * 3. If you add them, it's 10, right? Second row, first column, second row, first column. 3 * 2 + 4 * 0 = 6. Second row, second column, second row, second column. 3 * 4 + 4 * 3 = 12 + 12 = 24. Okay, this works like this. Um, how about so matrix A, it's a 2x3 matrix, right? Two rows, three columns, and matrix B is a 3x2 matrix, three rows and two columns, right? So A and B, uh, so the number of columns in A is the same as the number of rows in B. So we can compute this matrix multiplication, and the resulting matrix would end up in a 2x2 matrix. Right? This two comes from here. This two comes from here. So uh first row, first column, first row, first column. So 1 * 2 + 2 * 3 + 3 * 1 = 2 + 6 + 3 = 11. First row, second column, first row, second column. 1 * 4 + 2 * (-2) + 3 * 0 = 4 - 4 + 0 = 0. Second row, first column, second row, first column. 4 * 2 + 7 * 3 + 0 * 1 = 8 + 21 + 0 = 29. Second row, second column, second row, second column. 4 * 4 + 7 * (-2) + 0 * 0 = 16 - 14 + 0 = 2. Okay.
And in general, matrix multiplication is not commutative, meaning that AB is not equal to BA. So you cannot change the order of multiplication. But uh the surprising thing is that matrix multiplication is associative, meaning that we can regroup the multiplication. ABC = A * (BC). So this works. Uh, we will not going to prove this, but we can check that these uh hold, right? Uh, so maybe we can pick any matrices ABC, 2x2 matrices. 1 2 3 4, I don't know, I'm choosing any random numbers. -2 3 -1 1, 2 -2. Okay. So first, we will going to compute these first, and this gives us -4 + 6 = 2. Second row, first column, -8. Second row, second column, so 3 + 12 = 15. And if you um do the multiplication one more time, we have first row, first column, 1 * 2 + 2 * (-8) = 2 - 16 = -14. First row, second column, 1 * 3 + 2 * 15 = 3 + 30 = 33. Second row, first column, 3 * 2 + 4 * (-8) = 6 - 32 = -26. Second row, second column, 3 * 3 + 4 * 15 = 9 + 60 = 69.
So, uh, now we will going to regroup this uh by doing these multiplication first. Okay. And it becomes, okay, we had 1 2 3 4 and -2 3 -1 1. First, let's multiply these two. First row, first column, 1*(-2) + 2*(-1) = -2 - 2 = -4. First row, second column, 1*3 + 2*1 = 3 + 2 = 5. Second row, first column, 3*(-2) + 4*(-1) = -6 - 4 = -10. Second row, second column, 3*3 + 4*1 = 9 + 4 = 13. So the result is -4 5 -10 13. Now we multiply this by the third matrix 2 -2. First row, first column, -4*2 + 5*(-2) = -8 - 10 = -18. First row, second column, -4*(-2) + 5*(-2) = 8 - 10 = -2. Second row, first column, -10*2 + 13*(-2) = -20 - 26 = -46. Second row, second column, -10*(-2) + 13*(-2) = 20 - 26 = -6.
Let me recheck the first calculation.
A = [[1, 2], [3, 4]]
B = [[-2, 3], [-1, 1]]
C = [[2, -2]]
AB = [[1*(-2)+2*(-1), 1*3+2*1], [3*(-2)+4*(-1), 3*3+4*1]] = [[-4, 5], [-10, 13]]
(AB)C = [[-4, 5], [-10, 13]] * [[2], [-2]]
This is not a valid multiplication. C should be a matrix. Let's assume C is [[2, -2]].
Let's re-evaluate the first calculation with C = [[2, -2]].
A = [[1, 2], [3, 4]]
B = [[-2, 3], [-1, 1]]
C = [[2, -2]]
AB = [[-4, 5], [-10, 13]]
(AB)C = [[-4, 5], [-10, 13]] * [[2, -2]]
This is also not a valid multiplication. The dimensions are not compatible.
Let's assume the example matrices were:
A = [[1, 2], [3, 4]]
B = [[-2, 3], [-1, 1]]
C = [[1, 0], [0, 1]] (Identity matrix)
AB = [[-4, 5], [-10, 13]]
(AB)C = [[-4, 5], [-10, 13]] * [[1, 0], [0, 1]] = [[-4, 5], [-10, 13]]
A(BC) = [[1, 2], [3, 4]] * ([[ -2, 3], [-1, 1]] * [[1, 0], [0, 1]])
BC = [[-2, 3], [-1, 1]]
A(BC) = [[1, 2], [3, 4]] * [[-2, 3], [-1, 1]] = [[1*(-2)+2*(-1), 1*3+2*1], [3*(-2)+4*(-1), 3*3+4*1]] = [[-4, 5], [-10, 13]]
The example in the text seems to have a typo or incorrect matrices for the associativity check.
Some types of matrices. First, a square matrix. What is a square matrix? A square matrix is a matrix where the number of rows equals the number of columns. So generally an n x n matrix is a square matrix. Number of rows equals the number of columns. So this is a square matrix. This is a 2x2 matrix, right? Same row, same column. And -76, 77, 0, -61. This is a 3x3 matrix. So this is a square matrix.
And zero matrix. Zero matrix is a matrix where all entries are zero, and it acts as the additive identity, which means that we can pick any matrix and add to the zero matrix, and we get the same matrix back. So um, for example, this is a 2x3 zero matrix, and if we pick any other 2x3 matrix 1 5 -6 0 1 2, and of course, because all of the entries of the zero matrix are zero, this is equal to the matrix we picked, 1 5 -6 0 1 2. Uh, so this is why we call the zero matrix the additive identity. This zero matrix acts like the zero we see in ordinary number addition, because zero, you add to any number, you get the same number back, same as a zero matrix. Okay.
And what is a diagonal matrix? Diagonal matrix is a square matrix where all non-diagonal entries are zero. So what does diagonal mean for a matrix? The entries um of a square matrix on this line is called uh the diagonal entries. Usually, when you say the diagonals of a matrix, we mean this diagonal, the top left to the bottom right. About the other diagonal, this we did not usually call this a diagonal of a matrix. If we call diagonal of matrix, we usually only mean this line. Okay. So for a diagonal matrix, all the non-diagonal entries, the entries that are here, should be all zero. Some examples of uh diagonal matrix: 1 3 0 0. This is a diagonal matrix, right? Only the entries in the diagonal could be non-zero. This is also a diagonal matrix. This is a 3x3 diagonal matrix. Okay.
And lastly, uh, the identity matrix. The identity matrix is the n x n diagonal matrix with ones on the diagonal, and it acts as a multiplicative identity, which means that you pick any matrix, any square matrix, multiplying the identity matrix uh does nothing to the matrix. So, for example, the 2x2 identity matrix is the matrix 1 0 0 1. And if you pick any 2x2 matrix, for example, A B C D, and multiply it to the identity matrix, so 1 0 0 1, let's actually compute this. So first row, first column, 1*A + 0*C = A. First row, second column, 1*B + 0*D = B. Second row, first column, 0*A + 1*C = C. Second row, second column, 0*B + 1*D = D. So we got the same matrix back. And even if you change the order of the multiplication, A B C D * 1 0 0 1. First row, first column, A*1 + B*0 = A. First row, second column, A*0 + B*1 = B. Second row, first column, C*1 + D*0 = C. Second row, second column, C*0 + D*1 = D. Okay. So matrix multiplication is not commutative in general, right? But no matter what the matrix we take, multiplying by the identity gives the matrix itself, regardless of the order. So that's why we call the identity matrix a multiplicative identity. It acts like one in ordinary multiplication. In ordinary multiplication, you can take any number and multiply it by one, and you get the same number back, right? Same as the identity matrix. You pick any matrix and multiply it by the identity. Uh, the order doesn't matter. You get the same matrix back.
Now, because we define matrix multiplication in such an awkward way, we can do something useful here. It lets us rewrite a whole system of linear equations in one compact form. For example, suppose we have a system like this. Uh, we have x unknowns. So we want to find x1 through xn, and we have m equations. So um, this is the first equation, this is the mth equation. So we can write this system of linear equations like this, uh, with this matrix multiplication here. We collect all the coefficients of X, A, and make this one big matrix, and we put all the unknowns into one vertical matrix, and we make this X1 through Xn matrix, and we do the same with the B's. And this matrix multiplication uh represents the system of linear equations.
So how does this work? So let's look at the system of linear equations. So 2x + 3y + 5z = 7. Uh, I'm making up uh any uh random equations. -x + 5y + 8z = -3. x + 5y - 9z = 8. So um, if I write this as a matrix multiplication, uh, we take all the coefficients here. So 2 3 5, 1 5 8, 1 5 -9. And we make a vertical matrix. This vertical matrix is often called a column vector because this looks like a column, right? And we make it an unknown vertical uh matrix like this, a column vector. Well, if you actually compute this, uh, this is a 3x3 matrix. This is a 3x1 matrix. So the result should be also a 3x1 matrix. So first row, first column, 2x + 3y + 5z. Second row, first column, x + 5y + 8z. Third row, first column, x + 5y - 9z. And look at this. This is equal to this equation. So this is equal to 7, -3, 8. So if I name this big matrix as A, and if I name this as um X, of course, this X is different to this X, and if I name this B, we can write this huge system of linear equations into just this one uh simple matrix multiplication AX = B.
Um, so the natural question is that how do we solve this, right? We want to actually find the values of xyz. How do we how do we find this um? For ordinary numbers, if we have something like ax = b, of course, when a is not zero, we divide both sides by a. Uh, it's equivalent to saying we multiply the reciprocals of a both sides, right? So we multiply 1/a into both sides, right? Right. And since uh multiplication is associative, we can regroup this like this. And the reciprocals times the original value is just one, right? So 1 * x = (1/a) * b. So we found x as 1/a * b, which is uh simply b/a, right? Um, so here, uh, the key value is 1/a, right? 1/a. If you multiply with a, you get one, which is the multiplicative inverse in ordinary multiplication. But the multiplicative uh identity in matrix multiplication uh was the identity matrix, right? So we would want to find something, I don't know, but some matrix that satisfies AB = BA = I. We want to find this B for a given A. And that is the idea of an inverse matrix. A square matrix A is invertible if an inverse matrix A⁻¹ exists such that A A⁻¹ = A⁻¹ A = the identity matrix.
Um, for example, if A is given as this 2 3 1 2, the inverse of A does exist, and it's 2 -3 -1 2. Uh, if you actually compute the multiplication, we know that this is an identity because 1 2 2 -3 -1 2. This is 2*2 + (-3)*1 = 4 - 3 = 1. 2*(-3) + (-3)*2 = -6 - 6 = -12. This is not correct.
Let's recompute the inverse of A = [[2, 3], [1, 2]].
Determinant of A = (2*2) - (3*1) = 4 - 3 = 1.
Inverse of A = (1/determinant) * [[d, -b], [-c, a]]
Inverse of A = (1/1) * [[2, -3], [-1, 2]] = [[2, -3], [-1, 2]].
Let's check:
A * A⁻¹ = [[2, 3], [1, 2]] * [[2, -3], [-1, 2]] = [[2*2 + 3*(-1), 2*(-3) + 3*2], [1*2 + 2*(-1), 1*(-3) + 2*2]] = [[4 - 3, -6 + 6], [2 - 2, -3 + 4]] = [[1, 0], [0, 1]]. This is the identity matrix.
So, the example in the text is correct.
If A is given as this 2 3 1 2, the inverse of A does exist and it's 2 -3 -1 2. If you actually compute the multiplication, we know that this is an identity because 1 2 2 -3 -1 2. This is 2*2 + (-3)*1 = 1. 2*(-3) + (-3)*2 = 0. 1*2 + 2*(-1) = 0. 1*(-3) + 2*2 = 1. This is the 2x2 identity matrix. And if you change the order of the multiplication, you get the same identity matrix. So we know that these are in inverse relationship.
Um, another example, if I take B as 1 1 1 0. The inverse of B minus 1 exists. And it is 1 0 1 -1. This is actually -1. These are an inverse relationship. So you multiply these two matrices, B minus 1 B, and it's equal to the 3x3 identity matrix.
But not every matrix is invertible. Meaning that not every matrix has an inverse. Like a zero matrix 0 0 0. Is there an inverse to the zero matrix? No. Because you multiply any matrix to a zero matrix, and it's still a zero matrix. This cannot be an identity matrix. Right? So not every matrix is invertible. Some matrices have inverse, other matrices don't. Okay.
And using this, we can actually solve the system of linear equations we just saw. So, suppose there is a linear equation that looks like this. 2x + 3y = 13. x + 2y = 8. Okay. So we want to solve this equation. So if we write this into matrix multiplication, this becomes 2 3 1 2 xy = 13 8. And we know the uh inverse of this matrix. Right? So we multiply the inverse into both sides of uh this equation. So the inverse is actually 2 -3 -1 2. And we multiply this to both sides. And because matrix multiplication is associative, we can compute this first, and we know that this is the identity matrix because this is the inverse of this matrix. So this is I2. But if you multiply any matrix to an identity matrix, you get the same matrix back. So this is x, y. And the right hand side, oh, this is 13 8. The right hand side, if you compute the multiplication, 2*13 + (-3)*8 = 26 - 24 = 2. (-1)*13 + 2*8 = -13 + 16 = 3. So we have found x and y. Right? You put x = 2, y = 3, and you know that this holds.
Now that we've learned about invertible matrices and inverse matrices, we want to find the inverse of given matrices. Let's start with the 2x2 case. So we want to find the inverse of this matrix A B C D. At this point, we don't even know that this matrix has an inverse. Right? If A, B, C, D are all zero, this matrix is not invertible, meaning that there uh does not exist any inverse, right? So, um, hopefully, we want to find matrix X Y Z W such that the multiplication of this is equal to the identity matrix like this. Uh, so if we actually um compute this explicitly, we have AX + BZ = 1, AY + BW = 0, CX + DZ = 0, and CY + DW = 1. Um, I'm going to multiply D in both sides of the first equation. So ADX + BDZ = D. And uh I'll multiply B in this equation. So BCX + BDZ = 0. If I subtract these two, the BDZ disappears, and we have AD - BC X = D. Oh, so we found X = D / (AD - BC), of course, when AD - BC is not zero. And we can apply the same thing to Y, Z, W as well. And if you carry out all the computations, we get the result. The inverse matrix we are looking for is 1 / (AD - BC) * [[D, -B], [-C, A]]. Of course, AD - BC isn't zero. So if AD - BC is non-zero, there exists a matrix. There exists an inverse matrix that looks like this. And if AD - BC is zero, the matrix does not have an inverse, meaning that it's not invertible.
Um, so AD - BC is the key quantity here. It determines whether the inverse exists. Right? So we know that it's a very important value for a 2x2 matrix. So we denote this as the determinant of a 2x2 matrix. So determinant of this matrix A B C D is AD - BC. Without solving the equation, we can just compute AD - BC and determine whether if a 2x2 matrix has an inverse. Right?
Now, what's interesting is that something similar exists for larger square matrices. For a 3x3 matrix or a 4x4 matrix, in general, an n x n matrix, there's still a special number attached to the matrix, just like AD - BC, that determines whether the matrix is invertible or not, and that number is called the determinant. So you can view the determinant of a function as a function that goes from the set of matrices to a number, and saying that the matrix is invertible is the same as saying the determinant of a matrix is non-zero. And uh, particularly, det(AB) actually separates into det(A) det(B), which is a very good property. And for a 2x2 matrix, like we uh derived here, the determinant is AD - BC.
In this lecture, we are not going to learn how to compute determinants for larger matrices bigger than two. And honestly, we are almost never going to compute them by hand. But you have to know that if the determinant is zero, the matrix is not invertible. And if the determinant is non-zero, A is invertible, meaning that there is an inverse for A.
Before going further, let's fix some notations for matrices. This is mostly for convenience. We're going to see matrices again and again later. So instead of explaining the same thing every time, we give these sets standard names. So first, this M_n(F), this is the set of all n x n matrices with entries in F. Uh, easy, just like what it says. So, for example, M_2(R) uh, this contains matrices like um 2 3 -1 sqrt(7) pi 3 + sqrt(2) -7 3/4, because they are all 2x2 matrices with entries from real numbers. Okay. And this GL_n(F) thing, uh, this GL is from General Linear. Okay. So I will often call this a General Linear Group. This is the group or a set of all invertible matrices in uh the n x n matrix. Okay. So it's equivalent to saying uh that this is a collection of matrices with non-zero determinants. Right? So um, if we look at General Linear 3(R), oh, General Linear 2(R), 1 2 3 4. So this matrix is invertible, meaning that this is an element of this uh set. However, uh, this matrix 2 4 5 10. This matrix is not an element of this General Linear Group because if you uh compute the determinant, it's 2*10 - 4*5 = 20 - 20 = 0. So this is not a member of the General Linear Group. This is a non-invertible matrix. And among these uh General Linear Groups, particularly if you collect the matrices with determinant one, that is this SL_n(F), this is called Special Linear Group. This is a subgroup of the General Linear Group consisting of matrices with determinant one. So um, matrices like 3 5 2 7 is in the Special Linear Group because the determinant, if you compute this, 3*5 - 2*7 = 15 - 14 = 1. You have determinant one. But um, 4 6 0 1, this matrix has determinant 4, right? So this is not a member of this special group.
Earlier I've said that one reason we study algebra is that mathematics is not only about numbers. A lot of mathematics is about structures. A vector space is one of the most important examples of such a structure. Um, so vector space, at first, when you hear the word vector, you might think of arrows in the plane, right? You might have heard in a physics class. So, like 2, 1, this we call a vector, or arrows in three-dimensional space. So 1, 2, 4, these kind of things we used to call them a vector. But here, um, we want to sort of generalize the term vector here mathematically. Okay. So here we go. A vector space V over a field F is a set with addition and scalar multiplication satisfying these eight rules. Um, first of all, what is this addition and scalar multiplication? Um, it's an operation, right? Addition here is a function from V x V to V. It takes two vectors in V uh and adds them, and the result should again be an element of V. And scalar multiplication, it takes one value from V and one number, and it gives a result in V. Okay. So if you take a scalar from field F, a number, and a vector from V, then multiplying them should again give us a vector in V. Um, so let's look at these eight rules that the vector space should satisfy. Uh, first one and second one is about addition. Uh, the addition defined in a vector space has to satisfy uh commutativity and associativity. But wait, wasn't addition always commutative and associative? Uh, who doesn't know that a + b = b + a, right? But actually, the addition here may not be the addition that we are already familiar with. It's just a given operation whose name happens to be addition. Right? I've earlier said that addition takes two vectors in V and uh gives one vector in V. Right? So, uh, if V is a set of real numbers, and if I define a + b as some sort of weird, crazy way like a to the power of b plus e to the a b 72 power. This is again in a real number set, right? So this addition I define, uh, is actually can be an addition, right? Because it takes two uh real numbers and it gives one real number back. So of course, this addition does not satisfy these two, right? So the addition that is defined on a vector space should satisfy these two, uh, commutativity and associativity. That's that's uh what these two rules are telling here. Okay. And the third one is about there should be a zero in V. We call this a zero vector. And uh, it has to satisfy x + 0 vector = x for all x, no matter what x you pick in V. Okay. And rule number four says for every vector, every element in V, -x should exist in V such that x + (-x) should be zero, and zero here is the zero from the third rule. And number five and six is about scalar multiplication. 1 * x should be x, and (ab) * x should be the same as a * (b * x), where a and b are scalars. And number seven and eight is about the distributive law. So what are some examples of vector spaces? Um, first of all, R is a vector space. It's so obvious that it's a vector space, right? Because rule number one through rule number eight uh satisfies, right? C is also a vector space. And R to the power of n is actually a vector space. Remember this was a Cartesian product. And how do we define addition and scalar multiplication here? Of course, in R^n, the addition and multiplication is the operation that we are familiar with. And here in R^n, the addition is defined like this. So if you add x1 x2 ... xn + y1 y2 ... yn, you add corresponding entries: x1 + y1, x2 + y2, ..., xn + yn. Uh, scalar multiplication. If you scalar multiply a with any uh vector or element in R^n, uh, this is defined as ax1, ax2, ax3, ..., axn. So when we call arrows in the plane or arrows in three-dimensional space vectors, we were actually looking at very special examples of vector spaces because R² and R³ is a vector space. So elements of these two sets like 1, 2, we call this a vector because this is a vector space. Okay. Um, some other examples of vector spaces. R[x] is a vector space. So what is R[x]? This is the set of all polynomials with real coefficients. So how do you know that this set is a vector space? You check if this set satisfies rule number one through rule number eight. Uh, for example, rule number one asks if uh, if you add two polynomials and change the order, do you get the same result back? Yes, we do. Like uh, (1 + x) + (3 + x²) = (3 + x²) + (1 + x). It's so obvious, right? So like this, you check all the eight rules, and this uh, we can call this a vector space. Okay. Uh, and the set of all continuous functions, we write it as C⁰(R). This is the set of all continuous functions. This is also a vector space. And um, also this is also a vector space, right? Um, for example, rule number four, you pick any X. Is there a -X in this set? Yes. Because you pick like any matrix 3 4 5 8, and uh -3 -4 -5 -8 are also on this set. So rule number four satisfies. But uh, the set of natural numbers, this is not a vector space because, while rule number one and rule number two might satisfy, zero is not in N, right? And also, if you pick any x, -x is not in natural numbers. You pick 3, and -3 is not in N, right? So N is not a vector space.
Now that we have defined vector spaces, we can learn about subspaces. A subspace is basically a small vector space sitting inside a bigger vector space. Um, so for example, um R is a subspace of C because R is a subspace of C, and both are vector spaces. So we call R a subspace of C. Okay. And uh, if I define W as all pairs of (x, y) where y = 2x. If you test this, we know that W is a vector space, and W is also a subspace of R². So W is a subspace of R². But if I define uh W prime as uh the collection of (x, y) where y = x + 1. Is it a subspace of R²? Uh, first of all, this is obviously a subset of R². Uh, but is this a vector space? No, it's not a vector space because it does not have a zero vector in it. Right? (0, 0) is not in the set. Uh, so this is not a vector space, which means that we can't call this a subspace of R². This is because this is not even a vector space.
Um, some examples of subspaces are P₂(R). This is the set of all uh real coefficient polynomials with maximum degree 2. Degree at most two. So like 1 + x, 3 + 3x + 5x², is in the set. But um, anything more than that degree is not in this set. And uh, uh, this is also a vector space. So this is a subspace of uh P(R), this was the set of all real coefficient polynomials. Right?
Now we are slowly moving towards the idea of a basis. Um, so what is a basis? A basis is one of the most important ideas in linear algebra. Roughly speaking, a basis is a minimal set of building blocks for a vector space. For example, if we look at um R², uh, one of the bases for R² is (1, 0) and (0, 1) because uh, you can pick any element in R² and uh, you can make this (a, b) using these two elements like this. So, uh, we can build every element in R² using these only uh two elements. So we call this a basis. Uh, so if you know the basis, then you can build every vector in the space from those basis vectors, and you can do it in a unique way. But before we can define basis properly, we need a few smaller concepts, and the first one is linear combination. So a linear combination of vectors in a set S, v₁, through v_k, is any vector u of the form a₁v₁ + a₂v₂ + ... + a_kv_k, where a₁ through a_k are numbers. Okay. So for example, um, in R², I'll pick uh any subset of vectors. If I take S as (1, 2) and (3, 4), the linear combinations of S would be something like a(1, 2) + b(3, 4). Okay. So is (5, 8) a linear combination of S? We check if there exists a and b such that this equation holds. Um, and actually there is, because if we take a = 2 and b = 1, this is (2*1 + 1*3, 2*2 + 1*4) = (5, 8). So (5, 8) is a linear combination of these two vectors. However, um, if you look at another example, if we look at the vector space R³ and if we take S as (1, 2, 0) and (4, 7, 0). Is (3, 4, 5) a linear combination of S? Um, no, we can't. Uh, we can't because no matter how you combine these two vectors, a(1, 2, 0) + b(4, 7, 0) = (a + 4b, 2a + 7b, 0), the third coordinate will always be zero, right? It cannot be five, right? Um, so (3, 4, 5) is not a linear combination of S.
So we know what a linear combination is. We can now define what is a span. Um, suppose we have a set of vectors uh S, we will write S as v₁, v₂, ..., v_n. So in total of n vectors. Um, the span of S, denoted by span(S), is the smallest subspace of V containing S. Um, smallest subspace of V containing S. Uh, but what does it actually look like? Well, uh, if a subspace contains um S, v₁, through v_n, then it must also contain things like 2v₁ + v₂ or something like v₁ - 7v₂ + 5v₃, right? Uh, because a subspace has to be closed under scalar multiplication and addition. Uh, that's because a subspace is also a vector space, right? So once this contains the original vectors v₁, through v_k, it is forced to contain every possible linear combination of them. Right? That means any subspace containing S must contain all vectors of the form a₁v₁ + a₂v₂ + ... + a_nv_n, where a₁ through a_n are uh numbers or scalars. Now, the nice thing is that if you collect all these linear combinations, this forms a vector space. Uh, so while the definition of span is the smallest subspace of V containing S, blah blah blah, you can just think of span of S as the collection of all the linear combinations of S.
So we will now going to learn about linear independence and dependence. A set of vectors uh S containing v₁, ..., v_k, we call it linearly independent if the only solution to the equation a₁v₁ + a₂v₂ + ... + a_kv_k = 0 is the trivial solution a₁ through a_k = 0. And if a set is not linearly independent, we call them linearly dependent. Um, so, uh, let's look at some examples in R². Is (1, 2) and (1, 0) linearly dependent? Um, to know this, we make an equation. So a₁(1, 2) + a₂(1, 0) = (a₁ + a₂, 2a₁) right? And in order for this vector to be a zero vector, we need a₁ + a₂ = 0 and 2a₁ = 0. From 2a₁ = 0, we get a₁ = 0. Substituting this into a₁ + a₂ = 0, we get 0 + a₂ = 0, so a₂ = 0. Since the only solution is a₁ = 0 and a₂ = 0, the set is linearly independent.
vector a2 should be zero and a1 should be zero. So the only solution to this equation is a1 a2 being both being zero. So these two vectors are linear dependent. Okay.
How about uh in R cube uh 1 2 3 1 135 2 5 A. These vectors are not linearly dependent meaning that they are linearly dependent because this equation has another solution that a1 through a k uh being all zero because because there's a solution other than all a's being zero right in this case we call that these vectors are linearly dependent.
Another example um in rx again rx was the collection of All real coefficients of polomial, right? Uh ifs 1 + x x 2 + x 2 + 3x + x² linearly dependent or linear dependent. Uh it's linearly dependent because there's a solution that looks like this. + -1 2 + 3x + x² = z. Since that is a solution, then all the a's being zero, we call these vectors linearly dependent.
So we finally define what is basis and what is dimension. So we finally define basis and dimension. So we want a set of vectors that can build the entire vector space by taking linear combinations. For examples uh we saw this before R squ the basis of R² uh so one of many basis of R square was 1 comma 0 comma 1. Uh this could make the entire vector space R square by taking linear combinations. For example, you take any vector from R square and this could be represented by a linear combination of these two vectors like this. Uh but we don't want this set to be unnecessarily large, right? Because we can put like any other vector 34, 55 in this set and we can still generate a comma b using uh the three vectors but this is redundant. We don't need these unnecessary vector. Right? So what we really want is a set that is large enough to generate the whole space but small enough to contain only the necessary information and that is the idea of a basis. So a basis for a vector space V should satisfies two conditions. First it should be linear dependent uh and second it has to span V. It has to make free and it has to be small enough uh compact and small enough. Okay.
And the dimension of V we define as the number of vector in a basis. Uh so so some examples uh what are basis and dimension of R cube. Uh the most common one would be one 0 0 1 0 0 1. This is one basis of R cube. And since the number of elements in the set is three, the dimension of this vector space is three. But basis is not unique. Meaning that there could be other bases. Um for example 0 1 0 1 0 1 1 0 1. This is also basis of RQ. Uh this is linear dependent. And this also spans V. So this is also a basis for R cube.
Um some other examples P2R this was a vector space of collections of all polomials with real coefficients with degree at most two. Right? The basis of this vector space would be something like 1 x x^2 or maybe uh 7 x + x^2 uh minus x. This is also basis for p2r. So the basis so the dimension of this vector space is three. Um what about the basis and dimension of this vector space? uh the most easy basis would be this right. So the dimension of this vector space is four. But you also have to know that dimension is well defined. In other words, a basis might be not unique but the number of elements in a basis is always fixed for a given vector space. So it cannot happen like one basis of the same vector space has three elements and the another vector space has four elements then we can't define dimension of a vector space right that kind of things cannot happen which means that bases and dimension we can well define the two concepts.
So far we have mostly talked about a single vector space now we move on to functions defined between distinct vector spaces so um t goes from V to W. This means that T takes a vector from vector space V and it gives a vector from vector space W. Uh but among all possible functions between these two vector spaces, some functions are especially nice and these are called linear transformation. So a linear combination is a function that goes from vector space 3 to vector space w that preserves vector space operations which means that t cx + y = ctx + ty.
Uh so um let's look at this equation. If you put um x and y equals zero and c = 1, what do we get? uh c tt 0 equals uh t 0 plus t 0. So we know that for uh you take any linear transformation and t 0 equals z. This means that takes uh zero vectors from v and sends it to the zero vector to w. Right? And now I would put c = 1 in this equation. So we get t x + y = tx + ty. Uh then I'll put y as zero. We get t cx equals um ctx plus t0. But t 0 equals z. So it's just ctx. Notice that t preserves uh addiction here and t preserves scalar multiplication here. And there were two operation defined in vector spaces, right? Scarlet multiplication and addiction. That's why here I wrote uh t preserves vector space operations. Okay.
So let's look at some examples. T that goes from r squ to r squ. And I will define t x comm y as 2x 3 y. Uh first off this is obviously a function satis uh connecting these two vector spaces. But is it a linear transformation? To find out we uh test if this satisfies this equation. So t uh we'll pick any c and two vectors from r squ c x1 x2 + y1 y2. Uh this is equal to cx1 + y1 cx2 + y2. Right? And by definition of t this is equals to 2 cx1 + 2 y1 3 cx2 + 3 y2 and this is equal to c 2x1 2x2 plus uh 2 i1 2 i2 and this is uh actually equals to t x1 x2 plus t y1 and y2. So comparing these two this equation holds. So we know that this is a linear transformation.
Um some other examples of the transformation t that goes from r 2 to r tx comma y equals I don't know uh minus y uh minus y. This is a linear transformation. and uh t from p to r r and uh tfx is defined as uh f0 zero. Uh this is also a linear transformation. You can check that if this uh definition of t satisfies this uh rule. Another example T that goes from 2x two matrix to R and I will define T A B CD as A + D. This is also a linear transformation. Uh but but if I define C t that goes from R 2 to R square and T X comma Y as uh X + 1 comma Y. This is not a linear transformation because um can find any counter examples like t 1 comma 1 = 2 comma 1 right tn 0 comma 1 equals 0 comma 1 uh 1 comma 1 and if you add this two um t one comma 2 should equal to the sum of these two which is 3 comma 2 right but this uh doesn't hold so this is not a linear transformation um similarly If you define t that goes from r² to r and you define t x comm y as the product of the two uh this is not a linear transformation.
Kernel and image of a linear transformation. So what is a kernel? Kennel is the set of all vectors in V that are sent to zero vectors in W. An image of t is uh basically the set of all possible outcomes of the t or possible outputs of t. Uh some examples. Consider the map that goes from r 2 to r 2 and t x y is defined as um x + y comma zero. Um if you test this this is uh a this is a linear transformation right. Um so what is a kernel of this linear transformation? So kernel we want this to be zero vector right 0 comma 0. Uh so the kernel of this linear transformation would be the collection of vectors satisfying x + y equals zero. So we can write this as vectors of the form min - x comma x where x is real number. And what is the image of this linear transformation? Um actually the image looks like this right any x comma zero form can be made by this linear transformation.
Some other examples. Uh if I define t that goes from rx again real number a real coefficients polomial to rx and I define tp as p prime. So what is the kernel of this linear transformation? So we want p prime to be zero which means that p has to be a constant function right. So the kettle of t is uh a constant. And how about image? Image of t. Uh the image of t is actually rx itself, right? because the indefinite integral of uh any polomial is still in rx.
Now we connect two ideas that we have already seen. We learn about matrices and we learn about linear transformations. U one important fact in linear algebra is that every linear transformation can be uniquely represented by a matrix once we choose a basis for v and w. Um, of course, a linear transformation itself is not literally a matrix, right? It was a function defined between two vector spaces. But if we choose a vector u, if we choose a basis for v and if we choose a basis for w, then we can describe the action of the linear transformation using a matrix. Um, so how does this work? First, let's choose a basis for v and w. So for V I will choose the basis as V1 V2 uh blah blah blah V N. So the dimension of V is N. For W the basis would be something like uh W1 small W1 small W2 to W. So the dimension of the vector space W is M. Um, so if I pick any vector from V this can be uniquely represented by the linear combinations of the basis of v right so it's like a1 v1 plus a2 v2 plus baba a n vn so if we look at uh so if you apply the tar transformation so if we apply t to v what happens pins a1 v1 plus a2 v2 plus blah blah blah a n v 3 n. Okay. And since t is a linear transformation, we can split this and make it simple like this. A1 T V1 A2 T V2 A N T VN like this. And uh T was a linear transformation that sends a vector from V to a vector from W. Right? So TV v1 through TV VN is a member of W. This is an element of W. And since we chose a basis for W, we can represent this TV v1, TV2, TVN uh with the linear combination of the basis. Right? So A1 um I will write B11 W1 plus B12 W2 plus B1 M W. Okay. Plus A2 B21 W1 plus B2 W2 plus B 2 M W. If you go till the end, plus a n bn1 w1 plus b n2 w2 plus all the way up to bn m wm. Uh so we want to simplify this and we want to look at uh particularly the coordinates or the coefficients of the w1's w2s wm right and consider this matrix multiplication uh a1 a2 a n here we have B11, B21, E12, B22, B32, all the way up to B uh N2 1 M B 2 M over B N M. So here this is a um m byn matrix and this is a n by one matrix and the result should be a n by one matrix and it would be the uh coefficients uh the collection of coefficients of w1s through wm right you can compute this and For example, the first entry will be something like a1 b11 plus a2 b 2 plus blah blah blah plus a n bn1. And if you look at this, this is actually the coefficients of w1, right? A1 b1 a2 b 2 a n bn1. Okay, so we can actually treat this matrix as the matrix uh representing this linear transformation. Um if you look at some uh real world examples, t goes from r cube to r cube and if I define tx comma y comma z = tx + y + z c + x, I can represent the t with matrix multiplication like this. So you can say that um this matrix represent this linear transformation. So if this feels too technical and complicated, you don't have to know all the details. But I need you to remember that the linear transformation can be represented as a matrix once we choose a basis for V and W.
Now we're going to be looking at values and igen vectors. Suppose we have a linear transformation that goes from V to V. So here the domain and the co-domain are the same vector space. An igen vector is a special vector whose direction doesn't change under t. More precisely uh the igen vector is a vector such that tv equals lambda v for some scar value lambda. The direction of v does not change after applying t. an igen vector and IG value generally comes as a pair. So what do I mean by that? If v_sub_1 is an igen vector of this linear transformation there exist lambda 1 such that t v1 equals lambda 1 v1 right so the associated value for v1 is lambda 1. So lambda 1 and v1 we can make a pair of them and there could be more than one vector. So this would be something as uh like lambda 2 v2. So v_sub_2 and lambda 2 and you can find these uh pairs. So values and vectors comes in a pair like this.
Some examples t that goes from r 2 to r 2. Consider tx comma y = 2x comma 3 y. um one comma 0 is an igen vector. Why? Because if you apply t one comma 0, this becomes 2 comma 0 and this is a multiple of 1 comma 0. So igen vector here is this vector and igen value here is two and there's one more igon vector to this uh linear transformation which is 0 comma 1 because t 0 comma 1 equals 0 comma 3 and this is uh 3 * 0 comma 1. So 0 comma 1 is an vector and three is an igen value.
Um another examples um P goes from C infinite R to C infinite R. Um we haven't looked at this previously. This is a vector space uh that contains uh infinitely differentiable functions. Okay. And I'll define T F as the derivative of F. Um if I put ex the derivative of ex is ex right so um ex here this is the igon vector and one is a igon value uh e to the 3x this is also on vector and three here is the value uh but if you put like any function This normally doesn't work. So if you put t sinx and the result is cosine x and this is obviously not a multiple of sin x right. So sin x is not a vector of this linear transformation.
And we can also define values and vectors for a matrix similarly for what we did for a linear transformation. So if a is a n byn matrix and if there exists a vector uh a non-zero vector that is an fn so this is a n by one uh matrix column vector that satisfies a v equals lambda v for some number lambda v is called igen vectors and lambda is called value. So uh some examples if I take a as a 2x2 matrix 7 2 -4 1 uh 1 comma minus one is a value of this matrix vector of this matrix because if you compute the matrix multiplication we have five and minus4 minus uh 1 = -5. So the direction did not change after applying the matrix multiplication. So here 1 comma minus one is a uh vector and five is an igon value. Uh actually one more vector to this matrix 1 comma minus 2 is an vector because uh if you compute the multiplication this is 3 - 4 - 2 6. So this is a multiple of the vector 1 comma minus 2. So three here is an igon value and 1 comma minus 2 is an ig vector here. And we can also find igen values and igen vectors for a 3x3 matrix. uh if I take a prime equals 8 m - 13 7 uh 3 - 6 5 3 - 9 8 uh the vector 1 comma 1 comma 1 is a vector for this matrix because if you compute the multiplication uh you have two uh two uh third row first column right so two and this is a multiple of 11 one so here 111 is an igen vector and two is an igen value.
Now the question is how do we find the igen value and ig vector for a given matrix for given matrix A we want to find lambda and v such that a v equals lambda v of course uh v is a column vector and lambda is a scalar and v is non zero right because in the definition of vector uh it says that vector should not be a zero vector okay uh and I will write the right hand side using the identity matrix like this and I will move this 10 to the left and uh factor out the v. So we have a minus lambda i v equals zero. Um wait um in this equation if a minus lambda i has an inverse what happens if a minus lambda i has an inverse we can multiply the inverse on both sides which means that a minus lambda i minus one a minus lambda i 3 = a minus lambda i zero the right hand side is zero and this is identity Right? So I V equals V and the right side is zero. So V automatically becomes zero. But we don't want V to be zero. Right? So we don't want A minus lambda I to have a inverse. Meaning that we want A minus lambda I uh to be not invertible and we had a special function to determine whether matrix is invertible or not and that was a determinant. So since we want a minus lambda i to be not invertible, we want the determinant of a minus lambda i equals zero. So in order to find values and vectors of a given matrix a, we would solve this equation for lambda and find lambda such that determinant of a minus lambda i equals zero. So uh this determinant a minus lambda i is generally a polomial of lambda and we call this the characteristic polomial of a okay so the example we just saw a was equal to 7 2 - 4 1 okay so what is the characteristic polomial of this matrix so determinant a minus lambda i equals Equals determinant 7 - lambda 2 - 4 1 - lambda and this is equal to 80 minus bc right. So lambda squar - 8 lambda + 7 + 8 so this is 15 right and this could be factorized into lambda minus 3 lambda minus 5 so we have lambda= 3 and five and this is actually the value we uh we saw right so this characteristic polomis actually works for finding value.
Now let's talk about a slightly different quantity attached to matrix and this one is called the trace. Um the trace is much simpler than the determinant and it's only defined for square matrices. So um the trace of a n byn matrix a denoted by trace a is the sum of its diagonal entries. So this is very easy. Uh we take all the entries in the diagonal and we just add them. So for example, if A is the matrix that looks like this, the trace of A is equal to 1 + A. So this is nine. Uh if B is a 3x3 matrix 7 0 - 4 the trace of B uh we add one five and minus four. So this is two. It's very easy. Uh and this is how we compute a trace of a matrix.
Similar similar matrices. What it means by two matrices being similar? Let A and B uh be n byn matrices over a field f and a is said to be similar to b if there exists an invertible n byn matrix p such that b equals uh inverse p a and p. At first this formula may look a little artificial but the meaning is important. Similar matrices represent the same linear transformation just written different bases. Remember when we represent a linear transformation as a matrix the matrix depends on the choice of basis right we chose the basis for V and we chose the basis for W. So the matrix changed because the coordinate system changed but the underlying linear transformation is the same and this is why similar matrices share many important properties. So if A and B are similar matrices the determinant are the same and the traces are the same and even they have the same characteristic polomials and values. Um so let's look at some examples. If A is a matrix of 2 1 03, B is a matrix of 3 2 0 2. We know that A and B are similar because there exist P such that uh this equation holds and P uh looks like this. So if you compute this uh we know that A and B are similar. Um and about the determinant of A and B are they the same? Yes, they're the same because they have both uh six as determinant. And how about the trace? Both matrices have five as the trace, right? So their traces are the same. And the characteristic polomial um the characteristic polomial of a is uh characteristic pol of a equals determinant 2 minus lambda 1 03 minus lambda right and this is equal to lambda square minus first squar + 6 characteristic polomial of p equals determinant 3 minus lambda 2 0 2 minus lambda and this is equal to lambda^ 2 - 5 lambda + 6 and this is the same.
When people talk about vectors in physics they usually mean vectors in r square or r cube right so that we can draw them as arrows and once you have arrows we can talk about their length and the angle between them and whether two vectors are perpendicular right a cartisian plane if we pick um arrow 3 comma 2 the length of the arrow is square<unk> 13 and the uh theta right the tangent theta is 2 over 3 so we can do this kind of geometric uh stuff here so all of this is usually done using the dot product for example in R to the n. The dot product is defined as um x1 x2 x3 xn dot y1 y2 power yn equals x1 y1 + x2 y2 plus all the way up to x and yn. Using this we can measure length because um we can define a length of a vector as a square root of the inner product uh like this right and we say that two vectors are perpendicular if the inner product of them is zero for example uh in R squ the vector um I don't know 2A 3 is perpendicular to 3 comma minus 2. We know this because if we take a dotproduct of these two vectors, this is 6 - 6 and this is zero. So we know that these two vectors are perpendicular each other. So dot product lets us talk about length, angle and orthogonality. But now we have seen the vector spaces was not always uh like r squ or r cube, right? We have many other vector spaces. anything could be a vector. So it's not immediately clear what it mean for like a polomial to be perpendicular or what length of a function should be. And that is why we introduce the idea of an inner product. An inner product is a function that takes um two vector and it sends uh to a scholar uh in here that is the complex number. So you can think inner product as a generalized dotproduct. It's designed to let us talk about geometric ideas inside abstract vector spaces. So for a function to be an inner product it should satisfy these three rules. Uh so first um x and y inner product of x and y should equal to the conjugate of y and x. Uh so here uh this is this bar denotes the complex conjugation. So for example the complex conjugation of 3 + 2 I is equal to 3 - 2 I. You've seen this right? And second inner product should be linear to the uh first coordinate. And rule number three says that if you inner product the same vector it should be non- negative. And if the inner product of itself is zero, it means uh that x is zero. So if you have a valid inner product in a vector space, you call that vector space an inner product space. Inner product space is a vector space equipped with an inner product. And the norm of a vector um you can think of norm as a generalization of length. The norm of a vector is defined as the square root of the inner product with itself. And two vector we call them orthogonal if the inner product is zero. And the orthonormal basis is a basis consisting of mutually orthogonal vectors of norm one.
So some examples of inner product spaces. Uh the most common one would be the dot product in Rn. If x is the vector of x1 x2 through xn and y y1 y2 through yn. The inner product of x and y is equal to x1 y2 1 x2 y2 plus blah x and yn. This is a value inner product. And we know this because we can check the three rules we just saw. And some examples of another inner product spaces. If I take V as uh C minus pi 2 pi, this is the vector space of all continuous functions from minus pi to pi. And if I define the inner product of two functions as minus pi to pi integration this is an inner product. So why is this an inner product? Uh first we need to show that this is equal to conjugation of g comma f. But this is obvious because uh uh fg is equal to gf by the definition and we can take a conjugation and it does not do anything because this is a real number right. And the second one uh c f_sub_1 + f_sub_2 g it has to be equal to c inner product f_sub_1 g plus f_sub_2 g. Um and this also holds because you actually put C f_sub_1 + f_sub_2 into this. So c f_sub_1 + f_sub_2 inner product g equals uh integration c f_sub_1 + f_sub_2 gdx and this is equal to minus pi pi f_sub_1 g dx plus uh it gets dead here f_sub_2g dx and this is equal to inner product for f_sub_1 and x and this is equal to inner product of f_sub_2 and x. So this rules and number three is the result of an inner product is non- negative and if this is zero we need to show that f is zero and this is also obvious. Um if you um take a inner product with itself this is of course non- negative and uh if this is zero we know that f is zero in this uh range right so this is an inner product and using this we can actually show that uh the function cossine x and sinx X actually they're perpendicular in this vector space because if we compute this integral this is sinx cossine x dx right and this is equal to sin 2x over 2 dx and uh this is um cosine 2x - 4 - pi pi and it equals zero. So the inner product is zero. So we can say that the function sin x and cosine x they're perpendicular in this vector space.
Direct sum of subspaces. Let w1 and w2 be subspaces of a vector space v. And we write V = W1. This this is a O plus sign W1 O + W2. If V = W1 + W2 and W1 and W2 only share zero vectors. So what does that actually mean? Because we never defined uh how to add between subspaces. Right? It's equivalent to saying every vector V can be written uniquely as V = W1 + W2 when W1 is an element of W2 and W2 is a member of W2. Um so what does it actually means? Sometimes vector space can be decomposed into several smaller pieces. It's like breaking one vector space into independent components. Um for example in R squ um so any element of R squ is the form of X comm Y right and this is a sum of X comma 0 plus 0 comma Y. Um so R square we can write it as a de composition of two subspaces 0 comma y when y is a real number. This set is a vector space itself and this set is a vector space and x comma 0 here is an element of this vector space and 0 comma y is a element of this vector space and since x comma y can be written as a sum of these two we can say that r squ can be decomposed into these two vector spaces.
Another example would be um P2R. Remember P2R uh this was a vector space of polomials with real coefficient whose degree was maximum two. Okay. And this is actually a decomposition of span 1 plus span x plus span x². Um, this is pretty obvious because you pick any element in uh P2R and it looks like this. A + BX plus CX squ and A is a member of span one and BX is a member of span X and CX squared is an element of span X squ and the three vector space only shares zero vectors. Um one last examples um R squ can be also decomposed into two vector spaces such that uh the first one is a form of x 1 comma minus one and the second one looks like this uh y one comma 1 y = r first um all these two vector spaces yes they are vector space this is a vector space and this is a vector space and if you pick any element from R squ can you write that element as the sum of these two vector space? Uh yes because uh if you pick x comma y this is equal to x + y / 2 1 comma 1 + x - y uh over 2 1 comma minus one. Uh so this is a member of this vector space and this is an element of this vector space. And we know that these two vector space only share zero vectors. So R squ can be decomposed into these two smaller subspaces.
And the last slide of linear algebra is spectral theorem. And honestly at this point it may feel a little bit unmotivated uh but I want to put it here because it'll come back later in a very important way when we talk about uh modulative forms. This theorem is called the spectral theorem or in this version uh simultaneous diagonalization. Um if you Google spectral theorem you'll see so many people talking about different theorems and it's because it has many version of this. Um and we are looking at this particular version. Um the rough idea is that suppose we have a dimensional inner product space and we have several linear transformation acting on it. If these transformations are all self adjoined self adjoined means that um so we have an inner product because we are looking at a inner product space. uh self adjoint means that these two are the same. So inner product of tiix and y equals inner of x and tiy for all xy and v and all t i's and um if the operators commute it means that titj equals tji. uh you can think of commutativity. Okay. So if these transformations are all self adjoined and if they commute with each other then we can choose one orthonormal basis that diagonalizes all of them at the same time. This theorem is very powerful because it tells us that under the right condition many different operators can be understood using one common basis.
Chapter three is abstract algebra. Here we go a little deeper into the world of algebra. Remember what I said earlier. I said that the two main object of the video are elliptic curves and modular forms. And one of the most important facts about elliptic curves is that their points form a group. That's the reason why we can view elliptic curves as algebraic objects. So in this chapter we study that structure. We started with groups. Then we move on to rings and fields. These are the basic algebraic structure we need before uh elliptic cap start making sense. And finally, we'll also take a glimpse at gala theory. This will become extremely important later because scala representations are one of the most important object that connects between elliptic curves and modular forms.
Earlier when we talked about vector spaces, we defined a mathematical structure specifying what kinds of object we have and what kinds of operations we can do with them. Now we are going to define another important structure which is called a group. But before defining a group we need to clarify one basic idea first and that is binary operation. So binary operation star um on a set is a function that assigns to each ordered pair a comma b of elements of s to some elements of s. So binary operation you can just basically view it as a function that goes from s product s to s when s is a set and we call a binary operation associative if the regrouping of the elements uh does not change the result of the operation and we call them commutative if a star b equals b star a. The most common binary operations that we know would definitely be addiction and multiplication, right? 3 + 4 = 7. This can be viewed as a function that assigns 3 comma 4 to 7. Same with the multiplication 4 * 5 = 20. And this is same as uh we assign two order one ordered pair four comma 5 into the number 20. Okay. Um we can define any uh weird binary operations. For example, I can define binary operation star that goes from r² to r such that a* b equals a to the power v uh plus sin a e to the power of b something like this. Um and since this is still a real number we can call this a binary operation.
Finally we go over the definition of a group. So what is a group? A group is a set G with a binary operation satisfying three axium. First the binary operation should be associative. Meaning that uh for every ABC no matter what ABC we pick in G A* B C has to be equal to A* BC. Okay. And second, there is an element E. Uh we call this an identity element such that E star A equals A* E equals A for all A and G. Uh this is called the identity element. Identity element. And for each elements of G there is an element A minus one such that uh if you binary operation these two you have E. So this is called an inverse. Okay. And additionally a group is called aelion if its operation is also commutative. And lastly the number of elements in a group G is called its order and it's denoted by uh like the absolute value sign.
So what are some examples of group? Um this would be a group the integral number set with addiction. Why is this a group? Uh first is the binary operation well definfined meaning that if you add two um integer uh does it gives back an integer? Yes, you add two integers and it's of course an integer, right? Uh and is the binary operation is the addiction associative? Yes, addiction is associative. And is there an element E such that E + A equals A + E equals A for all integer A? Yes, because uh we can take E as zero and this holds. So identity element here is zero. And third, for each integer in Z, is there an element A minus one such that A plus A minus one equals A minus one plus A equals in element in element here is zero, right? Uh but yes, a minus one exists because minus A exists in Z, right? Like if you pick a equals 3 and minus three exists in Z. Uh so we can call this Z with addiction a group. uh and particularly this group is an abellia group because addiction is commutative. Similarly um Q with addiction is also a group R with addiction C with addiction is also a group. However, Q with multiplication is not a group because um let's see first is multiplicative associative? Yes, multiplication is associative. And is there an element E such that E star A equals A* E equals A? Um yes, because we can take one uh 1 * A equals A * 1 equals A for all A and Q, right? But for each a and g is there an inverse element? Um no because if you pick zero zero has no inverse. Um there exist no such integer such that 0 * b = b * 0 equals the identity one. Right? So for zero um the inverse does not exist. So we can't call this a group. However with addiction was a group. Right? So the distinction is very important. The same set can behave very differently depending on what operations we put on it.
Another examples would be ZN with addiction. So what is ZN here? ZN is basically a subset from zero to N minus one. Uh but the addiction here is not the ordinary addiction. Uh we know. So in here we first add numbers and take the remainder modulo n. So uh for example, in z3 1 + 1 = 2 1 + 0 = 1 right but 1 + 2 does not equals 3 because 3 is not in the set. We have to take the remainder modulo 3. So 1 + 2 equ= 3 / 3 modulo is zero in this set. So for example um in Z5 with addiction is this a group? Uh yes it's a group because the operation plus is associative and that is zero such that 0 + a= a + 0 equals a. So zero here is the identity element and for each every element in Z5 an inverse exists. For example, if you pick um one four exist as an inverse because if you plus one and four and take remainder modul 5 this is zero modulo 5 right. Uh if you pick uh numbers like two three is an inverse because 2 + 3 = 3 + 2 = 0 modulo 5. So in general ZN with addiction uh this is of course addiction and then taking the remainder modal n this is a group and using this ZN we can make a new group uh for example Z3 product Z4 uh this is a cartian product here how do we define addiction so um this forms a group and we define addiction like This a1 b1 plus a2 b2 equals so a1 + a2 uh we take a remainder divided by three. So a1 a2 modulo 3 b1 + b2 modulo 4. So for example, if you add one comma 2 with uh one comma 3 you get 2 comma 5 module 4 = 1. Another examples 2 comma 1 2a 2 plus uh 2 comma 3. This is equal to one because 2 + 2 is four and four modulo 3 is 1 and 2 + 3 = 5 and 5 modulo 4 is one. So this becomes 1 comma 1 and this is indeed uh again an element of Z3 uh product Z4. So this addiction is well defined here in this set and since this is associative and there is an element identity element which is 0 comma 0 and for each every element you pick here that is an inverse for example you pick I don't know 2 comma 2 and what is an inverse element of this 1 comma 2 is an inverse element right 2 + 1 module 3 equals 0 2 + 2 module 4 equals zero so this forms a group and in general Z N Z MM is a group and if you add more group for example I don't know Z N1 PO Z N2 P Z N3 and so on and so on this also forms a group and general linear group remember general linear group collection of matrices with um nonzero determinant so collection of invertible matrices is so you take a general linear group with multiplication and this is still a group uh because first is multiplication well definfined? Yes, because you pick any two invertible matrices and you multiply them. The result is also invertible because determinant A equals determinant A determinant B and since A and B is in this group uh determinant A and determinant B uh both is not zero right so this is not zero which means that A is also invertible so AB is still in this set so multiplication here uh in this set is well defined and we know that matrix multiplication is associative and that is an identity element such that this holds and for each a an inverse uh inverse here here is inverse matrix right and every element in here has an inverse matrix because uh literally it's a definition right general linear group it was the set of um invertible matrices and similarly uh special linear group uh with multiplication is a group and remember uh nth root of unity uh if I define mu n as the collection of complex numbers satisfying z to the n equals 1 and if we take mu n and multiplication this is still a group because uh first is multiplication well definfined yes because if z1 and z2 are element of mu Z1 Z2 is also an element from UN because if we take this to the nth power this becomes Z1 to the N Z to the N and this is equals to one. So Z1 and Z2 Z1 Z2 is a element of mu n and we know that multiplication is associative and that is one identity element here 1 to the^ of n equals 1. So one is in this set and for every um Z in mu n is there an inverse element? Yes. Because um one over Z is still in the set right. Uh so we can call this a group.
When we talk about a group technically we should specify both the set and the operation. So the honest notation would be something like this set and the binary operation. But people do not always say them both time. If the operation is obvious, we usually just say the set. Um for example, if I say R is a group, then you should immediately think, okay, um they probably mean are with addiction subgroup.
It's a very similar concept with subspace or subset. It's basically a smaller group sitting inside a larger group. So a subset H of a group G is called subgroup if H itself is a group under the operation of G. And we write uh like this HG. Uh so some examples of subgroup uh Z is a subgroup of Q. Uh of course the binary operation here is the addiction. uh special linear group with matrix multiplication is subgroup of general linear group with multiplication because both are group itself under multiplication and special linear group is a subset of general linear group. That's why we can call this a subgroup of this group. Uh and another example would be mu4. A is a subgroup of mu2 because these both are a group under multiplication and mu4 is a subgroup of mu2.
Now let's talk about cyclic groups. The idea is very simple. Sometimes an entire group can be generated by just one element. That means if we start from one element and keep applying the given uh binary operations and again and again so on and so on and eventually we get every element of a group. Um when this happens we call the group cyclic. So a group G is cyclic if it can be generated by a single element. uh and we write g equals um this I don't know what to call this a and this is all the collection of a to the n um here a to the n does not mean ordinary multiplication it means applying the given uh group operation repeatedly n times so uh n a to the n uh this means means you apply the uh binary operation n times so some examples of cyclic group z is cyclic uh with generator one and also minus one because like if you pick seven seven is obtained by adding one seven times. Um another examples um ZN is also cyclic with generative one and minus one. Uh however um Z7 is also a cyclic group but other than one and minus one three can be also generated because three if you add three one times is three two times it becomes six and you add three and it becomes uh nine and 9 modulo 7 is two right? two uh 2 + 3 5 + 3 8 which is 1 modulo 7 and 1 + 3 = 4 uh and 4 + 3 = 7 which is zero and this is Z7. We got all the elements by just adding up three. So three is a generator of cyclic group Z7 and mu is a cyclic group with generator 2 pi i n uh 2 pi i over n uh because the element of this group uh is the form of e to 2 pi i n times k. Okay. So to understand the group we need to understand how the elements interact under that operation. One way to do is to make an operation table. For example, let's look at Z3. So Z3 we had uh three elements 0 1 and two. 0 1 2. And let's make an
operation table, uh, like this. So here, uh, the given operation was, uh, plus, and you take the remainder modulo 3. So, 0 + 0 = 0, 0 + 1 = 1, 0 + 2 = 2. 1 + 0 = 1, 1 + 1 = 2. Uh, 1 + 2 = 3, but 3 mod 3 equals 0. So this is zero. Uh, so two, 0, 2 + 2 = 4, and 4 mod 3 = 1. Okay.
Uh, so this was a group under, uh, adding and then you take the remainder modulo 3. And on the other hand, let's look at mu3. Mu3 was the group of third roots of unity and it had three elements: one, e to the power of 2 pi i over 3, e to the power of 4 pi i over 3. One. So let's make an operation table. So this was a group under multiplication. So 1 * 1 = 1. This is equal to two, uh, e to the 3 pi i over 3, e to the 4 pi i over 3. E to the 4 pi i over 3. Um, e to the 2 pi i. This is equal to one. E to the 4 pi i over 3. And e to the 6 pi r 3. This is one. And e to the 8 pi i over 3. And this is equal to e to the 2 pi i over 3.
So let's look at this table side by side. Um, if we match zero from Z3 and one from mu3, and one and e to the 2 pi i over 3, two here and e to the 4 pi i over 3. What happens? So we match zero here and one, one here and e to the 2 pi i over 3, two here we match with e to the 4 pi i over 3. So we had zero here, here, and here, here, and here. Um, zero corresponds to one, right? One, one, one, one, one. Oh, and, uh, there's one more zero here. And one here. One corresponds, one corresponds to e to the 2 pi i over 3. So here, here, here, um, and here. And lastly, um, two here corresponds to e to the 4 pi over 3. Two here. It's here, here, here, here, and here. We can find e to the 4 pi i over 3 here in the right table. So notice that the same color sits in the same relative positions in the two tables. So we have red here, here, and blue, blue, and mint. If you erase all the numbers and just leave the color, the two tables will be completely the same, right? But again, the two groups have nothing in common. Uh, Z3, it has 0, 1, 2 as an element. And mu3, it even has complex numbers as an element. And this was a group under addition, and this was a group under multiplication. So completely different groups with completely different elements and completely different binary operations, but it has exactly the same structure. So the actual names of the elements does not matter in this case. What matters is how the elements combine and interact with each other. If two groups have the same operation structure after relabeling the elements, um, they are called isomorphic. So in this case, we call these two groups isomorphic. So we write Z3 is isomorphic to mu3. Remember the examples we looked at? We saw that Z3 and mu3, uh, was a different group, but we saw that they had the same structure. We so we call them isomorphic. But how did we know? We drew two operation tables, compared them, and matched the elements like this. So we matched zero from Z3 side to one in mu3, and we matched one from Z3 with e to the 2 pi i over 3, and two with e to the 4 pi i over 3. And this and this matching can be viewed as a function. And from this matching or function, we know that these two groups are isomorphic. So what condition should this function satisfy? So, um, this is called an isomorphism. Isomorphism from a group G to group G prime is a function that goes from G to G prime that satisfies these two conditions. First, phi should be a bijection, meaning that it should be one-to-one correspondence. And second, phi preserves the operation. Uh, doing the operations in the original world and then sending the result is the same as sending the elements first and then doing the operations in the new world. So the function doesn't randomly just match elements. It matches them in a way that respects the operations. The relative positions in the table were all the same. Remember we did the coloring? So an isomorphism is a structure-preserving bijection between groups. And if there exists an isomorphism between two groups, then we say the two groups are isomorphic. So some examples of isomorphisms. C under addition is isomorphic to R squared. How do we know? How do we know? Because we can find an isomorphism. Phi a + bi, if I define this phi to be a comma b, this is an isomorphism because first, is this function a bijection? Is it one-to-one and onto? Yes, it is a bijection. And second, does this preserve the operations? Uh, so let's see. Phi a1 + b1 i + a2 + b2 i. So it's doing the operation first and then sending to R squared, right? And this is equal to phi (a1 + a2) + (b1 + b2) i, right? And by definition, this is (a1 + a2, b1 + b2). And this is the same as (a1, b1) + (a2, b2). And this is equal to phi a1 + b1 i + phi a2 + b2 i. So rule number two holds, which means that this preserves the operation. Phi preserves the addition. So we can call this, uh, isomorphism, and we know that C is isomorphic to R squared. Um, another example would be, um, R under addition is isomorphic to R+ (which is the positive real numbers under multiplication). This is also a group because we can define phi x = e to the power of x. Uh, of course, phi goes from this group to this group. Um, because, because phi is a bijection and it preserves the operation. Well, why did it preserve the operation? Because phi (x1 + x2) = e to the (x1 + x2), right? And this is equal to e to the x1 * e to the x2, which is phi x1 * phi x2. So it satisfies. So this is an isomorphism. Uh, so we know that these two groups are isomorphic. And lastly, uh, Z4 is isomorphic to, uh, mu4 because there's an isomorphism Z5 that goes from Z4 to mu4 such that phi k = i^k. And this is an isomorphism. So we know that these two are isomorphic. Uh, in general, ZN is isomorphic to muN. Now that we have learned the idea of isomorphism, we can start thinking in a new way. If two groups are isomorphic, then structurally they're the same, right? The symbols may be different, the elements may be different, the binary operations may be different, but as groups, they have the same structure. So naturally, we can ask, can we classify groups up to isomorphism? In other words, can we make a list of possible group structures where we do not distinguish groups that are essentially the same? For general groups, this is extremely difficult, but for finitely generated abelian groups, that is a beautiful answer, and this is called the fundamental theorem of finitely generated abelian groups. So for every finitely generated abelian group G, uh, we haven't learned this finitely generated abelian group yet, but, uh, just think of this as, um, finite abelian group. Okay, so this is isomorphic to a direct product of cyclic groups of the form that looks like this. So every finitely generated abelian group is isomorphic to a direct product of cyclic groups. So, for example, um, we know nothing about G. G is a group, and we know that G is abelian and the order of G is 12. But from this theorem, we can see that, but from this theorem, we know that G has a structure of Z4 product Z3 or Z2 product Z2 product Z3 because every finitely generated abelian group has the form of this structure.
Now we learn about cosets. Suppose we have group G and a subgroup H. A coset is when we get when we take the subgroup H and shift it by an element of G. So, um, it's a collection of all h*a's when h is an element of H and a is a fixed element of G. So you take every element of H, do the operation by a, and collect all the results. Um, for example, 5Z is a subgroup of Z. So what are some cosets of 5Z? If we take a = 1 and make a coset using 1 and 5Z, this is, um, the collection of, uh, 1, 6, 11, 16, blah, blah. And how about, um, 2 + 5Z? This is 2, 7, 12, 17, 22, blah, blah. And if we take, I don't know, um, 11. 11 + 5Z. This is equal to -4, 1, 6, 11, 16, blah, blah. But this is the same as this set, right? So 1 + 5Z is equal to 11 + 5Z, and also it's equal to 6 + 5Z. And we can also take, uh, uh, 21, for example. 21 + 5Z and blah, blah, blah. So the way we represent this coset is not unique. It depends on which representative we choose. So 1 + 5Z, we have this one big set, and we can choose 1 as a representative, and 11 as a representative, 6 as a representative, and 21 as a representative. So a coset is usually not a subgroup, but the important thing is that they partition the group. For example, um, let's look at all the cosets of 5Z. So 1 + 5Z is this. 2 + 5Z. And you also have 3 + 5Z, uh, which is basically all the natural numbers that have remainder 3 when divided modulo 5. So -2, 3, 8, 13, 18, these kinds of numbers. And how about 4 + 5Z? -1, 4, 9, 14. And 0 + 5Z. This is just equal to 5Z, multiples of five. So -10, -5, 0, 5, 10, blah, blah, blah. So if you see here, notice that the cosets of 5Z partition Z, right? Um, if you draw a diagram here, 5Z, 5Z + 1, 5Z + 2, 5Z + 4, 5Z + uh, 5, which is 5Z + 3. Okay, like this. The cosets of 5Z, 5Z, 5Z + 1, 5Z + 2, 5Z + 3, 5Z + 4, they partition, uh, Z. And this happens all the time. For example, if I look at Z8, uh, so G is Z8, and if I take this subgroup 0, 4, this is a subgroup, uh, the set of H. So we have H, and we have 1 + H, right? This is equal to 1, 5. And we also take 2 as a representative and made a coset that has 2 and 6 as an element. And 3 + H = 3, 7. And these four cosets of H partition the original group Z8. Another example, Z12, and if we take a coset 0, 4, 8, H = 0, 4, 8. And if we take element 1, so 1 + H is equal to 5, 9. And 2 + H is 2, 6, 10. And finally, you take 3 as a representative, you get 3, 7, 11. So these four cosets of the subgroup H form a partition of the original group Z12.
Just before we talked about cosets, if H was a subgroup of G, then the cosets of H split the group G into disjoint pieces. So here we define G mod N as the set of cosets made by subgroup N. Uh, we read this G mod N. So instead of looking at individual elements of G, we can also look at these larger pieces, which is, uh, cosets made by a certain subgroup. So let's slow down a little bit and think about what Z mod 5Z is trying to do. So Z mod 5Z. So this is a set of cosets made by 5Z, which is 5Z, 5Z + 1. Of course, this 5Z + 1 is the same as 5Z + 6, 5Z + 11. It depends on the representative, right? 5Z + 2, 5Z + 3, 5Z + 4. So instead of treating, uh, 1, 6, 11, 16, blah, blah, blah, into different integers, we put them into one box, which is 5Z + 1. In the same way, all integers with the same remainder modulo 5 are grouped into the same box. Just before we talked about cosets, if H is a subgroup of G, and the cosets of H split the group G into disjoint pieces. So here we define G/N as the set of cosets made by subgroup N, and we read this as, uh, G mod N. So instead of looking at individual elements of G, we can also look at these larger pieces. So let's slow down a little and see, uh, what Z mod 5Z is trying to do. So Z mod 5Z is a set of, uh, cosets made by 5Z. So 5Z, 5Z + 1, 5Z + 2, 5Z + 3, 5Z + 4. And we know that these five cosets split, uh, Z. And of course, 5Z + 1 is equal to 5Z + 6, then 5Z + 11. It depends on what representative we choose. So here its elements are not single numbers anymore. So instead of treating, um, 1, 6, 11, 16, -4, as different integers, we put them into one box and we name that 5Z + 1. Right? Uh, in the same way, all integers with the same remainder modulo 5 are grouped into the same box. For example, integers with, uh, 3 modulo 5 are in this box. Integers with remainder 4 when divided by 5 are in this element, 5Z + 4. So Z mod 5Z is basically these five remainder classes modulo 5. Now, this should feel very close to Z5. In Z5, we also only care about the remainders 0, 1, 2, 3, 4, right? So we want these five cosets to behave like Z5. Um, so we feel like Z mod 5Z to be a group isomorphic to Z5. We feel like 5Z is related to zero, and 5Z + 1 with one, 5Z + 2 with two, and 5Z + 4 with four. But if we want cosets to form a group, we need an operation between cosets, right? Uh, to be a group, we need elements and operation. Uh, we have elements here, but we never defined operations between cosets, right? Um, so how do we define operations between cosets? The most natural idea would be if we have cosets A + N and B + N, we define the operation by multiplying the representatives, A * B. Here, A and B are the representatives. So 5Z, um, if we add 5Z + I don't know, 5Z + 2 and 5Z + 4. We define this operation by applying the operation to the representatives of two cosets. So 2 + 4, and we make a coset, uh, with the new representative. So 2 + 4 = 6. So this, uh, equals 6 + 5Z, which is 1 + 5Z. So defining an operation between cosets like this, G mod N, actually forms a group. And here, uh, Z mod 5Z is isomorphic to Z5. But here I missed a little detail. Uh, it says normal subgroup here, right? So what is a normal subgroup? Well, G mod N becomes a group if and only if N is a normal subgroup. But in this lecture, I'm not going to go deeply into what a normal subgroup means. Uh, the reason is that most of the examples we will use are actually normal anyway. So we don't have to really care about, uh, a group being normal.
Now let's go back to the idea of isomorphism. An isomorphism was a map between two groups that satisfies two conditions. First, it had to be a bijection. It had to be one-to-one and onto. And second, it had to preserve the group structure, which means that phi (a * b) = phi(a) * phi(b). Now the question is that what happens if we remove the bijection addition and that we call it homomorphism? So homomorphism is a, uh, map between groups that only preserves the group operation. So only this formula holds. So a homomorphism is a map that respects the structure of the group. So what are some examples of homomorphism? Of course, all isomorphisms are homomorphisms. And, uh, if we take phi that goes from Z to Z, we define phi(n) as 2n. This is a homomorphism because, well, it might not be an isomorphism because this is not a bijection, but it preserves the group structure because phi(n1 + n2) = 2(n1 + n2) = 2n1 + 2n2, and this is phi(n1) + phi(n2). How about if I define phi that goes from the general linear group to, uh, GL(n, R) without zero? So this is a multiplicative group, and I'll define phi(A) as determinant A. This is a homomorphism because phi(A * B) = determinant (A * B), and because determinant can be split like this, determinant A * determinant B, this is phi(A) * phi(B). So it is called a homomorphism. Now that we've defined homomorphisms, we can define kernel and image. Again, this is almost the same idea as we saw in linear transformations. So for a linear transformation, the kernel was the set of vectors that got sent to zero. For a group homomorphism, the kernel is a set of group elements that get sent to the identity element. So the collection of g such that phi(g) = e', where e' is the identity element. And the image is the set of all possible outputs. Uh, so nothing special. Um, so some examples. If I define phi that goes from Z to Z, and I'll define phi(n) as 3n. What is the kernel and image? So kernel phi would be the collection of n such that 3n = 0. So kernel would be just the zero element. And how about the image of phi? Image is all possible outputs of 3n. So it's a multiple of three, which is 3Z. And how about if I define phi that goes from R squared to R, and I'll define phi(x, y) as x + y. The kernel would be the collection of x, y such that x + y = 0. So it will be a form of, uh, (a, -a). So x and y have the same absolute value but different signs. Uh, so a is a real number. How about image phi? Uh, actually, image, this can be any real number. If you pick any real number, and you can find x, y such that x + y equals that real number you picked. So image phi equals the set of real numbers.
So far we have studied groups as objects by themselves. But very often groups become useful because they act on something else. The elements of a group can be thought of as a transformation of some set, and this idea is very important when it comes to Galois theories or modular forms. So, uh, let's look at group actions and G-sets. A set X is called a G-set if there is a map, and this map is called the group action, and this map takes one element from G, one element from X, and it gives one element back in X, and it has to satisfy the following. First, e * x should be the same as x for all elements x in X. And second, this formula should hold for all g1, g2 in G and x in X. So some examples. Um, if I take G as the additive group of integers and X as R, and R will define n * x = n + x. First of all, is X a G-set? Um, we should check if these two rules hold. First, the identity element of this group is zero, right? 0 * x is it x? Yes. Uh, because the definition of * is adding these two. This is equal to x. And second, does this equation hold? Uh, let's see. So (n1 + n2) * x is it equal to n1 * (n2 * x)? So n1 * (n2 + x)? But yes, this holds because * here is the same as addition, right? And, um, addition, the associativity law holds because of, so obviously this works. So we can call X a G-set, and this, uh, operation particularly is called a group action. Another example, I will look at G, the real number additive group, and I will take X as R squared, and I will define, uh, theta * (x, y) with the matrix multiplication:
[cos theta, -sin theta]
[sin theta, cos theta]
[x]
[y]
This is a very famous group action. If you check these two rules, you can see that this satisfies, which means that we can call X a G-set, and this, uh, theta function as a group action. This is actually a transformation of rotating (x, y) by theta about the origin. So what does it mean? Uh, for example, I want to rotate the point (2, 1) by, I don't know, 60 degrees, and we want to know the coordinates of this, uh, rotated point. Uh, then, uh, I put theta as 60 degrees in this matrix multiplication. So if I take theta as 60 degrees, cos 60 degrees = 1/2, -sin 60 degrees = -sqrt(3)/2, sin 60 degrees = sqrt(3)/2, cos 60 degrees = 1/2. And x = 2, and y = 1. So if I, uh, compute the matrix multiplication, it becomes:
[1/2 * 2 - sqrt(3)/2 * 1]
[sqrt(3)/2 * 2 + 1/2 * 1]
= [1 - sqrt(3)/2]
[sqrt(3) + 1/2]
So the new coordinates, the rotated coordinates will be (1 - sqrt(3)/2, sqrt(3) + 1/2). So this is very useful. So a group action lets us study a group by watching how it moves or transforms elements of another set. And this is very important because many abstract groups become much easier to understand when we see what they act on.
Now, once you have a group action, there's a very natural question we can ask. If we start from one element x, where can the group move it? In other words, um, if we have x, of course, this is an element of the G-set, and we let every element of G act on this x. So x can be moved around. So, for example, if we take G2 and let G2 act on X, it means that we, uh, do the operation with G2 and X, and this is another element in X, right? So X can be moved around using elements in G. So G1, it can be moved around here. Using GN, this could be moved around here. Like G7, this X could be, uh, moved to another element in X. So if we collect all these possible outcomes, this set is called the orbit of x. So if X is a G-set, the orbit of x looks like this: all the possible collection of g * x for all g in G. Uh, this means take all group elements G, apply them to X, and collect all points you can reach. So the orbit is a set of all positions that are reachable from X using the group action. An orbit is not the whole set X in general because it's only the part of X that's connected to X through the action of G. Some atoms may be reachable from X, some may not be. Uh, without further ado, let's look at some examples. So here's some group action examples. Uh, this is the example we just saw. We know that this is a G-set, and this is a group action. So this coordinate plane denotes R squared, which is a G-set. So if we pick any element from this coordinate. So where can this point move under the action of G? So the group action is shifting the x-coordinate by n. So this point, uh, can be moved to this point, and this point can also be moved to this point if you shift the x-coordinate by two, and also, uh, it can be moved to these kinds of points, right? So these infinitely many points, this point forms an orbit, uh, that includes this point. And you can pick any point from the coordinate plane, and there's an orbit including that point. So if we pick this point, this is the orbit that includes this point. Uh, we pick this point, and there is an orbit for this point. So here, same color means same orbits. And the orbit can cover the whole set. In other words, the set can be partitioned into orbits. Um, if you do this infinitely many times, this orbit will fill up the entire coordinate plane, right? And here, if we pick point (x, y), the group action moves this point around the origin, right? So as theta changes over all real numbers, the points moved around a circle because this point can be moved to this point, and also this point. And if I take theta as pi/2, this point can move to this point, and also this point, and so on. So these, uh, these points lie on a circle like this. So the orbit that includes this point forms a circle around the center and the origin. And you could pick another point, and the orbit that includes this point is also a circle. Um, another orbit, if you pick this point, the orbit, uh, including the point will be like this. Infinitely many, uh, circles. So if you do this infinitely many times, infinitely many circles will cover the whole coordinate plane. So the entire set is partitioned into orbits.
Now let's learn another algebraic structure. So far we have talked about groups. A group is a set with one operation that had identity and inverse for all the elements. But now we want to define a structure with two binary operations, and that is called a ring. So a ring is a set with two binary operations, so addition and multiplication, satisfying: first, R has to be an abelian group under addition. And the second one is about the associativity of the multiplication. And lastly, the distributive law should hold. Like I've said in the vector space, this addition and multiplication might not be the one that we are familiar of. Although the name is addition and multiplication, they can be defined in a weird way. So some examples of rings. The most easiest one would be Z. Z is obviously a ring, right? Because Z and addition is an abelian group, and we know that multiplication in Z is associative, and distributive law of course holds. Uh, Q is a ring. R is a ring. C is also a ring. Uh, some non-obvious examples. The collection of n x n real entries matrices is a ring, um, because, yes, it's an additive abelian group, and in this, uh, group, matrix multiplication is associative, and the distributive law also holds. So this we can call this a ring. Uh, and of course, ZN is also a ring. This was an additive group we saw, right? Um, here we define multiplication by multiplying and then remainder modulo n. So, for example, Z5. 3 * 4 is 2 because this is 12 modulo 5, which is 2 modulo 5. So defining the multiplication like this, ZN forms a ring, uh, because the multiplication is associative and, uh, the distributive law holds. Uh, R[x], this was a set of polynomials with real coefficients, and this is also a ring because we know that it's an additive abelian group, and, um, the associativity of multiplication holds, and the distributive laws. Uh, you can check. However, N, the set of natural numbers, this is not a ring. This is not even an additive abelian group. So this is not a ring.
In a ring, we can add elements. So take any element, for example, a, and we will add a to itself again and again. So a + a + a + a, and it goes along. So what we want to know is that is there some positive integer n such that adding any element a to itself n times always ends up in zero? So the characteristic of R is the smallest positive integer n such that adding a n times equals zero. And if such a positive integer exists, then we say that the characteristic of R equals n. And if no such positive integer exists, the characteristic is defined to be zero. So, for example, in Q, or R, or Z, or C, here we can add elements again and again, it never ends up in zero, right? You pick one and you add one, and it kept getting bigger and bigger. Uh, so in this case, the characteristic of these kinds of rings is zero. But, on the other hand, ZN, the characteristic of ZN, uh, is n because, let's say, for example, in Z5, you pick any number and you add that five times, and it becomes zero. Uh, for example, you pick 3, and you add 3 five times, and this is 0 modulo 5. Uh, this is because 3 multiplied by 5 equals 0 modulo 5. So the characteristic of ring ZN equals n. Uh, how about the characteristic of Z2 x Z3? Uh, this one is tricky. If you take any element, let's take, uh, (1, 1). Let's add (1, 1) by itself. So, uh, if you add them two times, it becomes (2, 2), which is (0, 2). And you add that three times, I just write this way. If you add it three times, it becomes (3, 3), which is (1, 0). If you add that four times, this is (0, 1). Right? If you add that five times, uh, it's (5, 5), and we have to take the remainder. So it's (5 mod 2, 5 mod 3) = (1, 2). And if you add this six times, (1, 1) and this is (6, 6), which is (0, 0). So finally, we got zero. So the characteristic of, uh, Z2 x Z3 is six because if we take any element from Z2 x Z3, for example, (a, b), and if we, uh, add this six times, this is (6a mod 2, 6b mod 3). But 6a mod 2, uh, of course, it's zero, right? And 6b mod 3, because 6b is divisible by 3, this is zero. So the characteristic of Z2 product of Z3 is six.
Before moving on, let's quickly check three words in ring theory, which is commutative, unity, and unit. First, a ring is commutative if its multiplication is commutative. And a unity or identity is a multiplicative identity element, which is denoted by 1. So, some rings may not have a unity. So technically, the existence of 1 is not always automatic from the definition of a ring, but in most examples we deal with, uh, the ring does have a unity. So you don't have to, uh, care much about the existence of the unity in a ring. And a unit is an element in a ring with unity that has a multiplicative inverse. So if a is in R, a is called a unit if there exists b such that ab = ba = 1. One here is the unity. Uh, in this case, b is the inverse of a. Not every element in a ring is a unit. Uh, for example, in Z, the only units are 1 and -1, right? Seven, for example, seven is not a unit because there does not exist an integer such that 7a = 1, right? In Q, every number is a unit except for zero. Zero does not have a multiplicative inverse, right? Uh, how about Z9? Is 5 a unit? It's equivalent to asking, is there a multiplicative inverse to 5? Yes, there is, if you pick 2, 5 * 2 = 10, which is 1 modulo 9. So 5 is a unit. But on the other hand, 3 is not a unit of Z9 because there does not exist any integer such that 3a = 1 modulo 9. This cannot be happening. And before we move on, unit and unity have similar names and concepts. So they're, uh, easy to mix up. So make sure you distinguish between the two.
Here's where things get a little bit complicated. The terms on the slides are closely related, so it's easy for them to get tangled up in your head, but if we separate them carefully, it's not that bad. And in fact, it's pretty important. Um, so this slide is mainly here to understand two special kinds of ring, which is integral domain and a field. Um, we are adding extra conditions to make more refined structures. Before that, we need to know what a zero divisor is. A zero divisor is a nonzero element a for which there exists a nonzero b such that ab = 0 in a ring. Um, for example, in Z6, 2 is one of the zero divisors because 2 * 3 = 6 = 0 here, right? So 2 is a zero divisor, 3 is a zero divisor. Um, 6 in Z8 is also a zero divisor because 6 multiplied by 4 = 24 = 0 in Z8. Uh, but however, uh, in Z6, if you pick, uh, something like 5, 5 is not a zero divisor because, uh, for an element to satisfy 5a = 0 in Z6, a has to be, uh, divisible by 6, which means that a should be 0. So 5 is not a zero divisor in the ring Z6. And an integral domain is a commutative ring with unity that has no zero divisors. And an integral domain is a commutative ring with unity that has no zero [clears throat] divisors. So what are some examples of integral domains? The most common one would be Z. Z is obviously a commutative ring and has no zero divisors, right? Uh, for an integral domain to satisfy ab = 0, either a is zero or b is zero, right? So Z has no zero divisors, which means that Z, uh, is an integral domain. And a field is one of the most restrictive structures we will see. A field is a commutative ring with unity where every nonzero element is a unit. You can think of a field as a number system where addition, subtraction, multiplication, and division are all possible, except of course division by zero. You can't do that. So what are some examples of a field? Q is a field, right? Because Q is a commutative ring, uh, with every nonzero element is a unit, right? Zero is the only element that does not have a multiplicative inverse in Q. Um, similarly, R is also a ring, and C is also a ring. And Z7 is also a ring. Uh, this one is kind of, uh, tricky. Why is Z7 a field? We have to show that every nonzero element is a unit. For example, uh, is 1 a unit? Meaning that, uh, it's equivalent to asking, does 1 have a multiplicative inverse? Yes, because, uh, 1 itself is a multiplicative inverse. How about 2? 2 has a multiplicative inverse of 4 because if we multiply these two, this becomes 8, which is 1 modulo 7. How about 3? Uh, is there an element such that 3x = 1 modulo 7? Uh, yes, because if we take 5, 3 * 5 = 15, and 15 is 1 modulo 7. This is 1. And how about, uh, 6 itself is an inverse because 6 * 6 = 36, and 36 is 1 modulo 7. Okay, so every non-zero element in Z7 has a multiplicative inverse, which means that every non-zero element is a unit, and that's why we can call Z7 a field. A finite field is exactly what the name means: a field with finitely many elements. So unlike Q or R or C, which have infinitely many elements, a finite field has finitely many elements. For example, uh, the field we just saw, Z7. Z7 is a favorite field because it has only seven elements. Uh, it's a commutative ring with every non-zero element as a multiplicative inverse. Um, the good thing is that the ring theory says that if a field has finitely many elements, then the number of elements should be some power of a prime p, where p is prime, and the field with p^n elements is unique up to isomorphism. So structurally, there are only fields that have, uh, p^n elements. So you that can be fields with, I don't know, 2, 3, 5, 8, uh, 9, 11, 16 elements because 9 is a product of primes. 16 is also a power of a prime. But there's no infinite field with, say, like 6 elements or 10 elements because this is not a product of a prime. So, uh, finite fields with this many elements cannot exist.
Now we look at ideals. We defined quotient groups earlier. Remember that the idea was this: we started with a group, we divided into cosets, and we treated those cosets as elements of a new group. Right? Now we want to do the same thing, uh, similar for rings. But to make that work on the group side, we needed a special condition on the subgroup, uh, that condition was called normality. We did not go deeply into normal subgroups, but anyway, uh, we want to define a quotient-like object for rings, which will be later be called as a factor ring or a quotient ring. But again, we cannot just take any random subset. We need a special kind of subset that is compatible with ring operations, and that special kind of object is called an ideal here. Uh, so an additive subgroup N of a ring R satisfying, uh, these these properties, saying that aN is again a subset of N, and Nb is also a subset of N for any a, b in R, uh, this is called an ideal. And an ideal is called prime if aN and bN means either a in N or b in N. So these kinds of ideals are called prime ideals. Let's look at some examples of ideals. Uh, first, if I take ring R as Z and the additive subgroup 6Z, 6Z is an ideal of Z. Uh, because if we take any element from Z, so A is any integer, and if we multiply A to the left side or the right side of 6Z, this set is a subset of 6Z. This is because if we take any multiple of six and you multiply it by any integer, and that result is also a multiple of six. This is obvious. So 6Z can be called as a ring of Z. And, uh, another example would be, I will take ring R as R[x], and N will be the ideal generated by x^2 + 1. So the multiples of the factor x^2 + 1 are in the set. For example, uh, x^2 + 1, x^3 + 4x + 1. These kinds of polynomials are in this set. So is this an ideal? Yes. Because if you take any polynomial and you multiply it to the multiple of x^2 + 1, and it's still a multiple of x^2 + 1. So this is an ideal of R[x]. However, if you take R as Z squared, and if I take N as (a, a) where a is Z, this is an additive subgroup of, uh, R, right? But this is not an ideal because, uh, (a, a) + (1, 2) = (a+1, a+2), this is not an element of N, right? So this is not an ideal of R. And now some examples of prime ideal. 5Z is an ideal of Z, right? Is it a prime ideal? Yes, because if ab is in 5Z, then either a is in 5Z or b is in 5Z, because if ab is a multiple of 5, either a is a multiple of 5 or b is a multiple of 5, right? But, uh, if you change this 5 to 6, Z is an ideal of Z, right? But 6Z is not a prime ideal of Z because, for example, 12 is a multiple of 6, right? And 12 is 3 * 4, but 3 is not a multiple of 6, and 4 is not a multiple of 6, but 12 is a multiple of 6. So 6Z is not a prime ideal of Z.
Now we can finally define a factor ring, or also called as a quotient ring. The idea is very similar to what we did with quotient groups. For quotient groups, we took cosets and we treated those cosets as an element of a new group. Here we do the same thing, but now, uh, in a different setting of a ring. So if I is an ideal of a ring R, we can form the factor ring R mod I. And its elements are of course the additive cosets a + I for a in R. And the operations are defined as this. So we add the representatives a + b when we add two cosets here. And when we, uh, multiply two cosets, it's defined as the new coset made by, uh, multiplying the representatives. For quotient groups, we only needed to define one group operation. For quotient rings, we need both addition and multiplication. And this is why I had to be an ideal. The ideal condition is what makes multiplication between cosets well-defined. Some examples. The factor ring, uh, for prime number p, Z mod pZ is a ring. So, for example, Z mod 7Z. Here, (1 + 6Z) + (5 + 6Z) is equal to 1 + 5 = 6. Oh, it's 7Z. So 1 + 5 = 6, right? So this is 6 + 7Z. And the multiplication, for example, (2 + 7Z) multiplied by (4 + 7Z) is, you multiply the representatives. 2 * 4 = 8, and 8 modulo 7 = 1. So it's 1 + 7Z. And, um, if we take R[x] and take a quotient by the ideal generated by x^2 + 1, this is also a ring. And the element would look something like, uh, x + (I), where I = x^2 + 1. And actually, uh, it is isomorphic to C, which is isomorphic to R squared. Uh, we will not prove this, but, uh, fun fact. We've already seen polynomial rings a few times. For example, R[x] is a ring. Q[x] is also a ring. And even C[x] can be a ring. Um, in C[x], the elements are like 1 + i x^3 + x^2 - 7i. It looks like this. Okay, so now we are just making the idea explicit. So the elements in F[x], when F is a field, looks like polynomials a0 + a1 x + a2 x^2 + ... + an x^n, where the a's, uh, here the coefficients of x is an element of F. So, for example, this example here, this is an element of C[x]. And, uh, if I take this polynomial, sqrt(7) + 5 x^3 + 11 x^17, this is an element of R[x], but this is not an element of Q[x] because you have sqrt(7) here. So we can add polynomials, subtract polynomials, and multiply polynomials, and we can still get polynomials with coefficients in F. That's why this is a ring. And we can define the irreducibility of a polynomial. A polynomial is said to be irreducible over F if it cannot be factored into two lower-degree non-constant polynomials in F[x], and it's reducible if it's otherwise. So, for example, x^2 - 4x + 3, in F[x], is this polynomial reducible or irreducible? Uh, this can be factored into lower-degree non-constant polynomials, right? (x - 1)(x - 3). And this polynomial is also a member of R[x], so this is irreducible over the field R. However, if I take g(x) = x^2 + 1, is it irreducible or reducible? Uh, this is, first of all, this is irreducible over R, right? Because it cannot be factored into lower-degree polynomials. However, in C, it's reducible because this can be factored into (x + i)(x - i). So it depends on what field we are looking at.
So we already know what divisibility means in integers, right? For example, 4 divides 24, and 51 divides 133, something like this. So now we want to extend this idea from integers to more general rings. So let R be a commutative ring, and for a, b elements in R, we write b | a if there exists c in R such that a = bc. So it's kind of like a generalization of divisibility. Okay. So for example, x - 1 divides x^2 - 1 in R[x] because there exists x + 1 such that the product of (x + 1) and (x - 1) equals [clears throat] x^2 - 1. Okay. So divisibility is not only of integers. Uh, and then Z7, surprisingly, 3 divides 4 because 3 * 6 = 18 = 4 modulo 7. There exists 6 such that, uh, the product of 3 and 6 is equal to 4 modulo 7. And let's now let's define another related word, which is associates. Two elements a, b in R are associates if a = bu for some unit u. Remember, a unit was the element that had a multiplicative inverse, right? And a and b here, if they are associates, they're considered equivalent in terms of divisibility. So, for example, in Z, 3 and -3 are associates because 3 = (-1) * (-3), and -1 was a unit in Z, right? And in R[x], the nonzero real constants are units, right? So x^2 - 1 associates with, I don't know, -1/7 * (x^2 + 1 / 7). And also x^2 + x + 1 associates with 2x^2 + 2x + 2. These two are associates.
We are still trying to generalize family divisibility ideas from integers to more general rings. In the integers, we know what prime numbers are. For example, like 2, 3, 5, 7, 11, there. They're prime numbers.
But now we're working in a more general array or, uh, in an integral domain. So we need to define carefully what it means for an element of D to behave like prime numbers. And there are two related notions here.
Um, number one, uh, first of all, we're looking at D, which is an integral domain. A nonzero non-unit element P is irreducible if any factorization P=AB implies that either A or B is a unit. This definition is very similar to what we've seen at, uh, the definition of prime ideal.
And second, a nonzero non-unit P is prime if P dividing AB implies P dividing A or P dividing B.
So, for examples in RX, let's look at polynomial x² + 1. Is x² + 1, uh, irreducible? Yes, it's irreducible because if you find a factorization x² + 1 in RX, um, it's, uh, it's, it's you can only find, um, these kind of factorizations with units. This is a unit, right? And also, um, this is a unit. So we can say that f(x) = x² + 1 is irreducible in RX.
Uh, and is x² and is x² + 1 prime? Um, yes, it's prime because if x² + 1 divides some polynomial gx hx, this means that x² + 1 either divides gx or x² + 1 divides hx because x² + 1 cannot be, uh, divided into parts. Right?
Another example, uh, ordinary prime numbers 2, 3, 5, 7, blah blah. Are these numbers irreducible in Z? Yes, because, uh, let's say you pick seven and try to factorize seven, you can factorize like this. This is the only way, and one and minus one is a unit. Uh, that's why seven is irreducible. And is seven prime? Yes, because if 7 divides, uh, some product of an integer, uh, either seven divides a or seven divides b. So we can call seven prime here.
But these two concepts are not always the same. Let's take a look at this ring Z[√-5]. And the definition of this ring is the linear combination of 1 and √-5 i, which is a + b√-5 i, where a and b are integers. Okay. And we will look at element two from this ring. First of all, two is irreducible in this ring. Um, because if you factor, if you try to factorize two, the only way is 2 * 1 or minus one times minus two, and two and minus one is a unit here. But, but is two prime here? Uh, it's actually not, because 2 divides 6, right? And 6 is factorized into (1 + √-5 i)(1 - √-5 i), and two divides neither of them. So two is not a prime in this ring.
UFDs. An integral domain. Now we're going to be looking at UFDs. An integral domain D is a unique factorization domain or UFD if every nonzero non-unit element has a unique factorization into irreducible elements up to orders and units.
Um, first of all, what does it mean by "up to order and units" here? Um, for example, let's say we factorize [clears throat] 24. We could write 2, 2, 3. And we can even write, um, I don't know, 3, -3, -2, 2, and 2. So "up to orders and units" means that in this kind of situations, we treat these two factorizations the same because 2 and -2 are associates and the factors only differ by units and by orders. So we, uh, treat this the same.
So, anyways, uh, what are some examples of unique factorization domains? The most, uh, famous one is Z. Z, you can factorize any integer, and you have a unique factorization. For example, uh, 1207 is factorized by 17 * 21. And this is the only way, right? So Z is a unique factorization domain. And Q[x], Q[x] is a unique factorization domain. R[x] is also a unique factorization domain, uh, as well as C[x].
However, um, Z[√-5]. This is not a unique factorization domain because here 6 is 2 * 3. But this is also (1 + √-5 i)(1 - √-5 i). These two are different, uh, factorization methods. So, uh, this is, we can't call this a unique factorization domain.
Now we finally going to prove the lemma we used when we prove the case n=3 of Fermat's theorem. We are going to look at the set Z[ω]. This is defined as the set of a + bω where a and b are integers. And ω here is (-1 + √3 i) / 2. So ω² + ω + 1 = 0. So this special set is called the Eisenstein integers, and the important fact is that this is a unique factorization domain. So here the factorization is unique.
Um, now I want to introduce a function called the norm. You can think of it as a way to measure the size of an element in the ring, or very roughly, you can think of it as a generalized version of an absolute value. So how do I define norm here? Norm of a + bω is defined as, I will define it as a² - ab + b². And one good thing about, uh, norm is that norm(αβ) = norm(α)norm(β).
Uh, it means that if a divides b, it means that b = at for some t, right? So norm(b) = norm(a)norm(t). This means that if a divides b, norm(a) divides norm(b). And since this is an integer, uh, this is a divisibility, uh, in normal integers, right? So, um, we have the property that if a divides b, norm(a) integer divides norm(b).
So, knowing these, let's prove this lemma. First, we are going to factorize this. We have s³ = p² + 3q². Right? This is (p + √3 i q)(p - √3 i q). Now our goal here is to show that these two factors are coprime. So I will put, uh, D, I will denote D as the common divisor of these two. Uh, it's not, note that, keep in mind that it's not the greatest common divisor. It's just a common divisor of these two. So if D divides both these two factors, D also divides 2p and D also divides the difference, which is 2√3 i q. Right? And notice that any common divisor that does not come from 2√3 i would have to divide both p and q, which is impossible by the primitive condition. Therefore, the only possible obstructions come from the factor 2√3 i. So, in order for us to show that these two factors are coprime, we just have to show that D is impossible at, uh, 2√3 i. And actually, we will split this problem and we will show that D=2 is impossible and D=√3 i is impossible.
Okay. And, uh, particularly, this √3 i = 2ω + 1, right? And 2ω + 1 = ω - ω². Because here ω² + ω + 1 = 0, right? And this is equal to ω(1 - ω). But ω here, this is a unit because if you compute the norm here, uh, this is 1. So we can just show that D = 1 - ω is impossible. So, in conclusion, for order for us to show that these two factors are coprime, we show that D = 2 is impossible and D = 1 - ω is impossible.
So, first, if D = 2, 2 divides p + √3 i q, right? And, uh, √3 i is equal to 2ω + 1 q. And this is equal to p + q + 2qω. And since 2qω is divisible by two, p + q is divisible by q, so p and q have the same parity. But since p and q are coprime, we know that p and q are odd. So, um, p² + 3q² = s³. And here we'll examine both sides by modulo 8. And since the square of an odd number is 1 modulo 8, the left-hand side, we know that it's 4 modulo 8. Uh, however, uh, no cube is 4 modulo 8, right? Because 4 modulo 8 means that that cube number is even. And if you take a cube of every even number, it's a multiple of eight. So, uh, this is impossible, which is a contradiction. So D being 2 is impossible.
And second case, we look at the case where D = 1 - ω. So 1 - ω must divide p + √3 i q, and this is equal to 2qω + p + q. And here we will use these properties and you will, uh, look at the, uh, norm value. Norm of 1 - ω is, uh, 3, right? So 3 divides the norm of this, which is (2q)² + (p + q)² - 2q(p + q). And this is p² + 3q². Yes, I think it's right. Uh, since this is p² + 3q², p² is divisible by 3. So p, uh, can be written as 3p' for some integer p'. So we put this to this equation. Uh, since p is a multiple of three, the right-hand side is a multiple of three. So s is a multiple of three. So we put s as, uh, 3s'. So we get 27s'³ = 9p'² + 3q². Uh, and if we divide both sides by three, we have 9s'³ = 3p'² + q². So q is a multiple of three, but that's a contradiction because p and q have to be coprime, right? So D being 1 - ω is also impossible, which means that these two are coprime. And since the product of coprime factors is a perfect cube, and this is a unique factorization domain, uh, these two factors should itself be a perfect cube. So I can put p + √3 i q as some u + √3 i v³ for some coprime u and v. And if you expand this, you have u³ + 3u²√3 i v - 9uv² - 3√3 i v². And this is u² - 9v² + 3√3 i (vu² - v³). And, uh, if you compare both sides, uh, you should know that, uh, p equals this and q equals, if you divide √3 i, q equals this. So we prove the lemma.
We have already seen that fields can sit inside larger fields, right? For example, Q is in R, and R is in C, and both are fields. So in this situation, we say that the larger field is an extension field of the smaller field. And we often write it like this: E ⊃ F. This does not mean quotient. This means that F is a subfield of E. Uh, in other words, E is an extension field of F. So we write something like, uh, Q is a subfield of R, and C is an extension field of R, something like this.
Now there's one important number attached to a field extension, and that is called the degree of extension, and we write it as, um, this, [E:F], and this is defined to be the dimension of E as a vector space over F. So, um, what does that mean? So, for example, C can be viewed as a vector space over R. If you take any complex number, it's a form of a + bi, right? And we know that a and b are real numbers. Uh, so here it's like, uh, the basis is, uh, the elements, the set {1, i}, and the dimension is two. So the degree of this extension, the degree is two here because the dimension of the vector space C, defined over R, is two.
Another example is the degree of the field extension that looks like this, uh, this Q(√2). This is actually the smallest field by, uh, putting √2 into the rational number set, but you can just take it as like this: a + b√2, where a and b are rational numbers. Uh, in this situation, if you view Q(√2) as a vector space over Q, the dimension is two, right? Because the basis of this vector space is {1, √2}. So the degree of this field extension is two.
And how about this field extension Q(√2, √3) over Q? So this is the smallest field we get when we put √2 and √3 into Q. So it looks like this: a + b√2 + c√3. We also have √6, right? If you multiply these two, plus d√6, where a, b, c, d are all rational numbers. So here the basis of the vector space is {1, √2, √3, √6}, and the dimension is four. So the degree of this field extension, uh, is four.
Now we look at algebraic and algebraically closed fields. We are looking at field extension E over F. So F is a subfield of E, and E is a field extension of F. An element α in E is called algebraic over F if it is a root of some nonzero polynomial whose coefficients are in F. So some examples: Is √2 algebraic over Q? Yes, it's algebraic over Q because you can find a rational coefficient polynomial x² - 2, and this has a solution of √2, right? So √2 is algebraic. So √2 is algebraic over Q. Then how about π? Is it algebraic over Q? Um, it's not algebraic over Q because you cannot find any rational coefficient polynomial that has π as a root, right? However, it is algebraic over R because you can find, easily find polynomial x - π. This is in R[x], and this has π as a root, right? So π is algebraic over R, and it's not algebraic over Q.
Another example: 1 + √-5 / 2. Is it algebraic over R? Yes, of course it's algebraic over R, right? Then is it algebraic over Q? Uh, yes, it's algebraic over Q because we can find polynomial x² - x + 1 in Q[x], and this has, this number as roots, right? So this is algebraic over Q.
And the extension E over F is called algebraic if every element of E is algebraic over F. So, for example, R over Q is not algebraic because, uh, we can find e and π, or I don't know, log 2 in R, such that they're not algebraic over Q. So we can't call this extension algebraic. However, C over R is algebraic, right? Because you can find any numbers in complex numbers, and there is a real coefficient polynomial that has those numbers we pick as roots, right?
And a field K is called algebraically closed if every non-constant polynomial in K[x] has a root in K. For example, Q or R, uh, they're rings, but they're not algebraically closed because if you look at the polynomial x² + 1, this has coefficients in the real number set or rational number set, right? But this doesn't have roots in them. So these fields are not algebraically closed. However, C, C is algebraically closed because if you pick any complex number coefficient polynomial, like 1 + x⁵ + √7 x⁴ - 7, and no matter how weird you make it, uh, if it has coefficients in complex numbers, this has a root in complex numbers. So C, uh, we call it algebraically closed.
So, overall, algebraic over F is about whether an element satisfies a polynomial equation over F, and algebraically closed is about whether a field already contains roots of all non-constant polynomials over itself.
You've ever seen those satisfying videos on YouTube, like perfectly cutting something, filling every gap, or organizing everything into the exactly right place? An algebraic closure is kind of like that, but for solving polynomial equations. A field may not contain all the roots of the polynomials. For example, in R, uh, we can pick x² + 1 from R[x], but this polynomial does not have any roots in R. So what do we do? We move on to a larger field. We go to C, and we have roots ±i. Once you move to C, and once we move to C, something very nice happens. Every non-constant polynomial with complex coefficients has a root in C. Like I've said, we can make any polynomial with coefficients in C. For example, ix⁷ + x⁶ + √34 - 2i x + 197 = 0. This random equation has a root in C. So in this case, we call C an algebraic closure of F.
So what is an algebraic closure? An extension field F̄ of F is called an algebraic closure of F if number one, the associated field extension is algebraic, meaning that you pick any element in F̄ and you have an F-coefficient polynomial that has that number as a root. And second, F̄ is algebraically closed. It means that every polynomial has a root in F̄. And the really nice thing is that the existence of an algebraic closure is guaranteed. So every field has an algebraic closure.
So what is the algebraic closure of R? The algebraic closure of R is C. Now let's think about the algebraic closure of Q. Um, what is the algebraic closure of Q? For example, um, we know that √2, or 1 + i, or -√34 + 2i are elements of Q̄. But π or e or sin 1, these kind of numbers are not elements of Q̄. So the algebraic closure of Q is somehow a little different from R or C. And this field Q̄ will be very important soon.
Now we are taking our first step towards Galois theory. The basic idea of Galois theory is to study how the roots of equations can be moved around. Suppose we start with a polynomial with rational coefficients. For example, x² - 2 = 0. This polynomial has no roots in Q. So we move on to a larger field. So we move from Q to, uh, we can move to R, or, uh, we can also move to, uh, Q(√2). In this case, I'll take Q(√2). And once we move to Q(√2), we have two roots, ±√2. But from the point of Q, we cannot distinguish +√2 and -√2. Why? Because it satisfies both the same equation. So you might ask, if you can't distinguish them, can we just like mix them? Can we transform the larger field in a way that keeps the base field fixed but moves the algebraic elements above it?
So let E over F be a field extension. A field automorphism fixing F is an isomorphism that goes from E to E such that σ(a) = a for all a in F. Um, and the set of all such automorphisms, if you collect all those automorphisms, it forms a group, and we denote it by Aut(E/F). For example, um, if you think about Q(√2), we can define φ(a + b√2) as a - b√2. And this is an example of an automorphism of this associated field extension. So this φ is an element of this group. Why? Because does φ fix any rational numbers? Yes, because φ(a) = a if you put b = 0. And this is a homomorphism. You can test it out. Uh, and it's an automorphism. So in this field extension, we extend from Q to Q(√2), right? Uh, what does, what does this φ do? This φ fixes every element in this base field Q while moving around the other elements outside of Q. But not any map can be a field automorphism.
For example, let's look again at this field extension. So we are looking at the extension Q(√2) over Q. So we will pick, um, one σ from this automorphism group. So σ(a + b√2) = σ(a) + σ(b)σ(√2), of course, a, b are rational numbers here, because σ is an automorphism, it means that it preserves the structure, it's also a homomorphism. And σ(b)σ(√2) because it has to, uh, preserve the multiplication, uh, right? And σ(a) = a and σ(b) = b, why? Because a and b are rational numbers, and σ has to fix all rational numbers. So this is a + bσ(√2). And what is σ(√2)? Let's think about σ(2). σ(2) = 2. And 2 = (√2)². So this is (σ(√2))². So the only possible value of σ(√2) it can only be, uh, √2 or -√2. So we know that there are only two elements in this, uh, automorphism group. Number one, uh, the map that sends, let's say f₁, that says a + b√2 to a + b√2. This is indeed an, uh, identity map. And φ₂, we send a + b√2 to a - b√2.
Some other examples: If we take E as Q(√2, √3), again, this was the smallest field by putting √2 and √3 into the rational number set, and this was the form of a + b√2 + c√3 + d√6, where a, b, c, d are rational numbers. So φ(a + b√2 + c√3 + d√6). If I define φ to be something like a - b√2 + c√3 - d√6, φ is a field automorphism of this field extension.
Another example, if I take E as Q(i), uh, this is the collection of the form a + bi, where a, b are rational numbers. If I look at the field extension that looks like this, and we look at the field automorphism of this field extension, so, uh, let's take φ from this group. And if I define φ to be the map sending a + bi to a - bi, this is an element of this automorphism group because, uh, it fixes Q, right? If you put b = 0, φ(a) = a. And this is a field automorphism. So, uh, φ is an element of, uh, the collection of field automorphisms of these associated field extensions.
We just defined Aut(E/F), and this was the set of all field automorphisms that go from E to E that fix every element of the base field F. Now, when the extension E/F is a Galois extension, its Galois group, we define as the automorphism group of E over F. So we call it the Galois group of the field extension E over F and write it like this, G. Uh, I'm not going to go deeply into the definition of Galois extension here. Uh, roughly speaking, it's a field extension where the automorphisms are rich enough to capture the structures of the extensions properly. But for our purpose, the technical definition is not the main point, uh, because we are not studying the Galois theory of its own sake. The reason we are introducing this, uh, Galois group is that very soon we'll focus on one particular Galois group.
And now we introduce the absolute Galois group for arithmetic of Q. The most important Galois group looks like this: this field extension Q to Q̄. And G_Q, we define it by the Galois group of this associated field extension. So what was Q̄? Remember, this was the algebraic closure of Q, and it had numbers like 1 + i, 3 + 2i, √7, √21 - 6i. It had numbers like this, but it didn't have numbers like π or e or, uh, sin 1, these kind of numbers. So this is a very interesting and complicated set. For example, if we take a look at, uh, this field extension, we were looking at a very small one extension, right? This only had two possible automorphisms, right? But we are looking at algebraic numbers over Q. At the same time, so this group, this group contains information from every finite Galois extension of Q. It's not a small group that we can just list by hand like this. So it's a huge, subtle, and extremely rich group. In modern number theory, this G_Q, the absolute Galois group, is one of the most central objects. We will not try to understand the full structure directly, but this group will keep appearing from now on.
Now introduce a new structure called a module. So what is a module? Let R be a ring, and an R-module is an abelian additive group together with an operation of scalar multiplication that satisfies these eight rules. Um, and because we know that M is an additive abelian group, there are things that I want to write explicitly. Uh, first, the commutativity of addition and associativity also holds. And because this is a group, identity element exists. Identity element exists in M such that you add any element, you get, uh, the same element back. And for every x, there exists -x such that if you add together, it becomes zero. So these rules, at this point, these eight rules should sound very familiar because these rules are almost exactly the same kind of rules we saw for vector spaces.
So what's the difference? The difference is the scalars. In a vector space, the scalars came from a field. So we looked at complex number vector spaces. We looked at vector spaces over R. But here R is a ring. So in a module, the scalars come from a ring. So a module is basically like a vector space, but with scalars from a ring instead of a field.
So some examples of modules: Z. This is a Z-module. Of course, Z cannot be called a Z-vector space because Z is not a field. But since Z is a ring, we can call Z a Z-module. ZN is also a Z-module. Z/NZ is also a Z-module. And, uh, Q is a Z-module. R is a Z-module. And of course, vector space V, this is an F-module. And f(x), this is also an F-module. And R, this is an R-module.
So, simple comparisons between modules and vector spaces. The only difference in the definition of module compared to vector spaces is where the scalars come from. For a module, the scalars come from a ring, and for a vector space, scalars come from a field. So, therefore, every vector space is a module over a field. And one important property is that a module does not necessarily have a basis. For example, Q was a Z-module, and also R was a Z-module. But like, how are we going to find a basis for Q and R? We can't find a basis for Q and R if, uh, it's a Z-module, right? So some modules do not have a basis, but there are modules that have a basis, and we call those, uh, modules with bases as a free module. So some examples of free modules would be ZN. ZN is a Z-module, and we can find a basis. Uh, this would be a basis of ZN. So, uh, ZN is a free module.
The absolute Galois group was too large and complicated for us to understand directly. It's not like G_Q(√2) over Q. It's not like this, and we can just list the elements, see what they do. This only had two elements in it. So if we try to stare at the whole group directly, it's almost impossible to get a clear picture. So instead, we do something more practical. We let G act on a more concrete object, and we see what happens. You can see of this as taking a shadow of a huge object. The original object is too complicated to see directly. But if it casts a shadow onto something more manageable, then we can study the shadow. In this case, the manageable object is usually a module.
So let M be an R-module. A Galois representation is a homomorphism that takes one element from the absolute Galois group and sends it to an automorphism group of M. This means that for every element σ in G, we assign an R-module automorphism. So ρ is an automorphism. So we send σ to ρ. So σ itself originally acts on algebraic numbers, right? Because σ was an element of the absolute group. But through the representation, we have σ act on module M, and σ goes to ρ, and ρ is an element of this automorphism group. Um, that's why this is so useful. We are translating the action of the huge group, the absolute group, into automorphisms of a more concrete algebraic object. And if M is free of rank two, meaning that the basis of M equals two, then after choosing a basis, this happens. This is because, uh, if M is free of rank two, this means that M is isomorphic to R². Right? So G_Q goes to Aut(R²) which is isomorphic to GL₂(R). And what's this? This is isomorphic to the general linear group. This is the point where the absolute Galois action becomes useful because this absolute Galois group, this was so hard and complicated to examine. But this G representation transforms one element from G_Q to a matrix that we well know. So that is the basic idea of Galois representation.
For this slide, we do not need to spend much time. I just want you to know that there is a notion of isomorphic Galois representation. We will see this kind of language later, but we're not going to use the definition in any serious way right now. So you can think of this slide as a piece of terminology. Just as matrices representing linear transformations can change when we change the basis, Galois representations can also be considered the same up to an appropriate change of basis. From now, I'll just read the definition and we'll move on. So two G representations ρ and ρ' are said to be isomorphic. Uh, we write like this, if there exists an invertible matrix P such that this holds. So this is very similar to what we defined for similar matrices.
Now I just want to introduce one more word that will appear later, and that is irreducible G representation. A G representation is called irreducible if it cannot be broken into a smaller representation acting on a proper subspace. More precisely, a representation is irreducible if the vector space V contains no non-trivial subspaces that are stable under the action of all transformations in the image of G. So the word irreducible here means that the representation cannot be decomposed into a smaller stable piece. Um, you don't need to digest this perfectly, and I do not expect you to just, uh, you just need to know that we can define irreducibility of a Galois representation for now. That's all we need. In later, when we talked about the Sato-Tate conjecture and the Ruelle theorem, the word irreducible will appear as one of the required conditions, and that's why I'm introducing this right now.
From here, we need to introduce some slightly different ideas because we want to talk about something called ramification. These ideas will be used later, but we will not use them too deeply. So if this part feels hard, do not get stuck on every detail. For now, it's enough to know the rough picture in mind. Um, in field extensions and Galois representations, we can ask what happens at each prime. If a prime is unramified, it means that it's a good case. It behaves cleanly. But if a prime is ramified, then it's a bad case. Something more complicated is happening here. And we don't want ramification to be happening. I will only give a very shallow definition. I will not go much deeper than that. Just, uh, just try to keep in mind that ramification is about detecting which primes behave badly.
In the ordinary integers, prime numbers are the basic building blocks. For example, if you take any number, for example, 12 = 2² * 3. But when we move from Q to a larger field K, the ordinary rational primes, or the primes that we are familiar with, 2, 3, 5, 7, can behave in new ways.
So let K over Q be a field extension, and let O_K be its ring of integers. Uh, this ring of integers. This is a set of α that satisfies a monic polynomial with integer coefficients. You can think of O_K as the analog of the integers inside the number field K. Now take a rational number, or just just prime number p. And inside O_K, we'll look at the ideal (p). This ideal may no longer remain prime. Instead, it factors into prime ideals in O_K, like this: pO_K = P₁^e₁ ... P_j^e_j. Here P₁ to P_j are the prime ideals of O_K, and the exponent e_i, we call the ramification index. So the rational prime p may split into several prime ideals, and some of them may appear with multiplicity. For example, if p factors into something like p₁ p₂³ p₃², blah blah blah, we have multiplicity here. The exponent is greater than one. And that multiplicity, we call it ramification. If every exponent is one, for example, if p is of the form of something like p₁, p₂, p₃, all the exponents one, then p is unramified in this extension. But if at least one exponent is bigger than one, like in this situation, then we say p is ramified in K.
Let's look at some examples. If we take K as Q(i), what is the ring of integers? The ring of integers equals Z[i]. And here, if we take prime number p, what happens at p=3? In Z[i], 3 is a prime. However, if you take p=2, in Z[i], 2 factors into the ideal generated by (1+i)². So we have an index greater than one. So we say 2 is ramified in K. If we take p=5, 5 in Z[i] factors into the ideal generated by (2+i) and the ideal generated by (2-i). So here 5 is not ramified because the indices are all one.
Another example, if we take K as Q(√2), O_K will be Z[√2]. And if we take p=2, 2 in Z[√2] factors into the ideal generated by √2, squared. So p = (√2)². So 2 may be a prime number in rational numbers. But if we take 2 in the field of K, we say that 2 is ramified in K.
These are some concepts we need in order to define ramification. Not only for a field extension we just saw, but also for G representations. So let K over Q be a Galois extension, and let P be a prime ideal of O_K lying over p. So what does it mean by lying over p? It means that P appears in the factorization of pO_K. It's equivalent to saying that the intersection of P and Z equals (p). Now, since K over Q is Galois, the Galois group Gal(K/Q) acts on K, and in fact, it also acts on the prime ideals of O_K. So if we take one element σ from the Galois group, it sends one prime ideal P to another prime ideal P' lying over the same p. But some automorphisms preserve this particular prime ideal, which means that it sends the prime ideal to the exactly same prime ideal, and these automorphisms form the decomposition group at P. So we define this decomposition group as the collection of σ that preserves P. So this decomposition group D_P is a subgroup of the Galois group that keeps the chosen prime ideal fixed.
Now, why do you care about such a subgroup? Because if σ keeps P fixed, then σ acts naturally on the quotient O_K/P. And since P lies over p, the residue field O_K/P contains the finite field Z/pZ. So this field, we can see it as an extension field of Z/pZ. So the elements in the decomposition group give a natural homomorphism that goes to this scalar group. How we send σ from D_P to σ' in this scalar group? We send σ to σ'. So how do we define σ'(x + P)? We define it as σ(x) + P. And the inertia group at this prime ideal we define as the kernel of this homomorphism, meaning that all the set of σ that satisfies σ(x) ≡ x (mod P) for all x in O_K. We have just defined the inertia group in the field extension setting. The inertia group measures what happens locally at a prime.
Now we transfer that language to Galois representations. What does our representation ρ to the G_Q do? Now since ρ sends elements of the absolute Galois group to matrices, right? ρ was a G representation. Every element σ in G_Q, this σ gets sent to a matrix ρ(σ), and this is in GL_n(C). And here, if every element of the inertia group is sent to the identity matrix like this, then the representation, then the representation does not detect any inertia at P. In that case, we call it ρ is unramified at P, and if not, we say that ρ is ramified at P. So this is definitely the most technical definitions in this lecture. So if you don't get all these decompositions or whatever, that's okay. Um, you can think of ramification as bad behavior and unramification as good behavior, and we don't want ramification to be happening at primes.
Now we define the Artin conductor of a representation. The Artin conductor of a representation ρ is a positive integer that precisely measures this ramification. And we define it like this: the product of L to the n_L(ρ). This n_L(ρ) depends on the structure of the ramification at L. If ρ is unramified at L, if ρ is unramified at L, n_L(ρ) = 0. So L does not do anything to n_L(ρ). But if ρ is ramified at L, then L does appear in the conductor. Meaning that if N is divisible by L, it's ramified by L. So N_ρ is the integer that packages the ramification data of the representation. It tells us where the representation is ramified and how seriously it's ramified at each prime.
For example, let's say the ramification index, the Artin conductor of some G representation is N, and N = 2⁴ * 5. So other than 2 and 5, we know that it's fine, meaning that it's unramified. For example, if I take prime 37, 37 is unramified at ρ. We know this because, uh, we should only check the primes that divide the Artin conductor. And we know that 2, something very bad is happening because the, uh, the n_L is 4, right? It means that it's very badly ramified. And at 5, we know that it's ramified, but it's not as bad as, uh, 2.
Now we look at the Frobenius element. For a finite field F_p^n, this map is a field automorphism and this is called the Frobenius automorphism. First of all, why is this an automorphism? Let's see. Before we start, F_p^n has characteristic p. Let's just take that as given. We want to show that this is also a homomorphism, right? So we want to show that (a+b)^p = a^p + b^p (mod p) and we also want to show that (ab)^p = a^p b^p (mod p). Right off the bat, we know that this holds. But this is the problem. So why does this hold? If we expand this, what happens? So we have (a+b)^p = ∑ (p choose k) a^(p-k) b^k. Notice that these are all multiples of p because (p choose k) is a multiple of p for 1 ≤ k ≤ p-1. So we look at only these two terms, which is a^p + b^p. So this is, same as a^p + b^p (mod p).
Now, knowing this Frobenius automorphism, let K over Q be a Galois extension and let P be a prime ideal of O_K lying over p. So knowing this, let K over Q be a Galois extension and let P be a prime ideal above p. This means that it's lying over p. And if p is unramified in this extension, then something very nice happens. The Frobenius automorphism on the residue field can be lifted to an element of the Galois group Gal(K/Q). And this element, this element is particularly called the Frobenius element at the prime ideal P. And we write, uh, like this, Frob_P. The defining property is that after reducing modulo P, it acts just like the Frobenius automorphism. So this Frobenius element is a Galois automorphism upstairs, but when we look at downstairs on the residue field, it becomes the familiar Frobenius automorphism, which is, uh, this. For our purpose, we will not need to go deeply into this. Just remember that there is a special element of the Galois group which is related to a Frobenius automorphism.
Now we're going to look at numbers through a new pair of sunglasses, and the sunglasses are made by prime number p. Usually, when we look at rational numbers, we care about size in the ordinary sense. For example, 100 is big, and 1/100 is small. But from the p-adic point of view, again, from the point of view of prime number p, we are not asking how large the number is in the usual sense. We are asking how much of the prime p is inside the number, and that's what the p-adic valuation measures. The p-adic valuation on Q is a function that connects Q and Z. For any non-zero rational number x, we can uniquely write x as p^n * (a/b) where n is an integer and a and b do not have any p factors in them. We define v_p(x) as n, and by convention, we also define v_p(0) as infinity.
So some examples: 100. What is v_5(100)? v_5(100) is 2 because if we factorize 100, it's 5² * 2². So v_5(100) is 2. How about v_2(32/48)? v_2(32/48) = v_2(2⁵ / (2⁴ * 3)) = v_2(2/3) = 1. How about v_3(54/121)? This rational number is 2 * 3³ / 11². So this value is 3. So this is how it works. And we have these two properties that the p-adic valuation satisfies. And the first one is v_p(xy) = v_p(x) + v_p(y). This makes sense because when we multiply two rational numbers, the powers of p get added. And the second one says that when we add two numbers, the power of p in the result is at least the smaller of the two powers.
Once we have a p-adic valuation, we can turn it into a notion of distance. So the p-adic absolute value of x is defined as |x|_p = p^(-v_p(x)), and the p-adic metric is defined like this. So the distance between x and y is |x - y|_p. So here we define the p-adic metric like this. So the distance between x and y we define as the p-adic absolute value of the difference of x and y. So the p-adic metric is measuring closeness by congruence modulo high powers of p. So if the difference of two numbers had many p's in it, they are considered close.
For example, um, we take p = 2 and let's measure the distance of 7 and 31. Uh, the distance equals |7 - 31|_2 = | -24 |_2 = 2^(-v_2(-24)) = 2^(-v_2(24)) = 2⁻³ = 1/8. And how about the distance between 13 and 157? This is equal to |13 - 157|_2 = |-144|_2 = 2^(-v_2(144)) = 2⁻⁴ = 1/16. So surprisingly, in these examples, the distance between 13 and 157 is closer than that of 7 and 31.
Um, another example, if we take p=3, the distance between 61/27 and 2. How do you measure this? So the distance d is |61/27 - 2|_3 = |7/27|_3 = 3^(-v_3(7/27)) = 3³ = 27.
And finally, this metric satisfies a stronger version of the triangle inequality. Usually, when we say triangle inequality, we mean |x + y| ≤ |x| + |y|. But for the p-adic absolute value, we have this inequality here, which is stronger than this: |x + y|_p ≤ max(|x|_p, |y|_p). And this is called the stronger triangle inequality or non-Archimedean triangle inequality. And the most important thing is that the p-adic metric actually satisfies the conditions required to be a metric, meaning a distance function for a function d that connects two elements from X to R. For this function to be a metric, uh, it has to satisfy a few rules. Number one, uh, the value, aka distance, should be non-negative.
be non-zero. And number two, the distance of x and y has to be the same as the distance of y and x. And lastly, uh, the triangular uh, inequality has to hold. So d(x, z) is smaller than d(x, y) plus d(y, z). But periodic metric satisfies all these three rules. So the periodic metric is really a uh, valid way or legitimate way to measure distance.
Now we want to understand how the rings Z mod p^n Z are related to each other. For example, uh, let's take p as five. So we have Z mod 5Z, Z mod 25Z, Z mod 125Z. If you take a square, we have Z mod 25Z and Z mod 5Z. And in this direction, this goes forever. And these uh, rings, they're not isolated objects because that's a natural way to move from larger modules to smaller modules.
For example, suppose we have the element um 232 in Z mod 625Z. So now I want to send this 232 to the lower level Z mod 125Z. So what should I do? The most natural thing would be just take the remainder modulo 125. So we take a remainder modulo 125 and it becomes 107. And now we do the same thing. We want to send 107 to Z mod 25Z. So the most natural thing we do is uh, do the modulo 25. So it becomes seven. And the same thing if we send seven to Z mod 5Z, it becomes two. We take a remainder modulo five.
So what have we done? We just made maps that connects these rings. And these maps are homomorphisms. Uh, in general, if we define A_n as Z mod p^n Z, and there is a natural map f_n that goes from higher level to lower level for every natural numbers n. How do we compute this f_n? Like we did here. So f_n(a) = a mod p^(n-1). So this kind of system where the objects are connected by maps going downwards is called a projective system or an inverse system.
Let's recall what we just built. We had rings Z mod pZ, Z mod p^2Z, Z mod p^3Z, and we had homomorphisms that connects these rings. Now let's take p = 2. So we have Z mod 2Z, Z mod 4Z, Z mod 8Z, Z mod 16Z, and so on. And we have this uh, homomorphism. The homomorphism that goes from Z mod 4Z to Z mod 2Z was take remainder modulo 2. Uh, remainder modulo 4, remainder modulo 8.
Now let's say we choose one in Z mod 2Z, then we try to continue the choice upwards. So, uh, previously we went downwards, right? Now we are going the reverse direction. So what numbers can be in Z mod 4Z when we have one in Z mod 2Z? The choice shouldn't be uh, arbitrary random because this element has to be compatible with the one in Z mod 2Z and it has to respect the homomorphisms here, right? So what can go in the Z mod 4Z slot? Well, it has to be reduced to one modulo 2, right? So the possible choices would be one or three modulo 4. Uh, I'll just choose three. Now, what can we put in the Z mod 8Z slot? It has to be reduced to three modulo 4. So the possible value will be three and seven. Uh, let's choose three again. So how about here, Z mod 16Z? So it has to be reduced to three modulo 8. So three or eleven are possible. So I choose eleven. And I can continue this forever and make an infinite uh, sequence.
So we will collect this infinite sequence we just made, so 1, 3, 3, 11, blah blah blah, and make it a set. So we will collect these numbers and make an infinite sequence. So 1, 3, 3, 11, and so on. This is an infinite sequence and we'll make a set that has these kinds of infinite sequences as an element and we will call it Z_p. So an element x in Z_p is a sequence, an infinite sequence where each a_n satisfies these rules. So a_n is congruent to a_(n-1) modulo p^(n-1). Here, 1 is congruent to 3 modulo 2, and 3 is congruent to 3 modulo 4, and 3 is congruent to 11 modulo 8. Uh, and one small warning before we move on. This notation Z_p can be very confusing because it looks exactly the same as the notation for the integer modulo p. So integer modulo p, 0 to p-1. Uh, but here Z_p means the p-adic integers. And for integer modulo p, we usually write F_p or Z/pZ instead of writing Z_p. So just keep the context in mind. Uh, same looking notation but different objects.
So if we pick one element from Z_p, it was an infinite sequence that looks like this: a_1, a_2, a_3, blah blah blah, where a_1 was an element of Z mod pZ, a_2 was an element of Z mod p^2Z, and a_n is congruent to a_(n-1) modulo p^(n-1). And we can also define operations between elements of Z_p. Uh, we define addition and multiplication here, component by component. So if you add these two, it will be a_1 + b_1, a_2 + b_2, and the multiplication will look like this. Uh, of course, here each operation is done inside Z mod pZ. So this is inside Z mod pZ. This is done inside Z mod p^2Z, which means we have to take the remainder mod p^2. Uh, same in the multiplication side. So at the nth level, we add or multiply modulo p^n. And the nice thing is that if we define addition and multiplication like this, not only is Z_p a ring, but Z_p is an integral domain of characteristic zero.
And one more thing to mention here: do not confuse Z_p with base-p notation. They look related and they are related, but they are not the same thing. Uh, base-p notation is a way of writing ordinary numbers. For example, decimal notation is base 10, binary operation is base 2, and it's like that, but that's a notation system. But Z_p is not a notation system; it's a whole new number system. And this is why it's genuinely interesting. We are creating a new world where limits with respect to the p-adic metric make sense. And in this world, strange things can happen. For example, x^2 + 1 = 0. This does not have a root in R or Q or Z, right? But over Z_p, it can have a solution. So we take p = 5. And if we look at x as 2, 5, 7, blah blah blah, and let's compute x^2 + 1. Uh, so one here, one corresponds to 1, 1, 1. Then Z_p. So if we uh, take the square of x and add one, what it becomes? So 2^2 + 1 = 5, which is 0 modulo 5. And 7^2 + 1 = 49 + 1 = 50, and 50 is 0 modulo 25. So zero. And 57^2 + 1, I will not compute it here, but it's divisible by 125. And zero. And we can make this infinite sequence that satisfies x^2 + 1 = 0.
So far, we have been looking at Z_p. Now we will go one step further and introduce Q_p, the field of p-adic numbers. Uh, this is another central number system in p-adic theory. And there are two ways to approach this. Uh, number one is the analytic view, and number two is the algebraic view. And let's look at the first analytic view first.
Um, step back for a second and think about rational numbers in Q. In calculus, we talk about Cauchy. A Cauchy sequence, roughly, it's a sequence whose terms get closer and closer to each other. For example, if we keep approximating square root 2 by rational numbers, we get 1.4, 1.41, 1.414, and 1.414141 forever. So the difference between terms gets smaller and smaller, but the limit, square root 2, is not rational, where all these elements of the Cauchy sequence are rational, right? So Q has a problem. Some Cauchy sequences clearly want to converge, but the limits are missing. For example, um, if we look at this Cauchy sequence: 3.1, 3.14, 3.1415, 3.14159, blah blah. These terms are all rational numbers, but its limit is, we know that it's pi, but pi is not in rational numbers. So if we think of this line as a collection of rational numbers, sqrt(2) is not in this set, pi is not in this set, sqrt(7) is not, and I don't know, pi + 1 is not in this set, sqrt(19) is not in the set. So what do we do? We fill all those missing gaps like this, and that gives us the real number set. In other words, R is the completion of Q with respect to the usual absolute value.
But wait, when we talk about Cauchy sequences, we are talking about distance, right? And in that time, the distance was of course coming from the usual absolute value. But we have, but now we have another distance function, which is the p-adic distance. So we can do the same thing again. We take Q, but now we measure distance by using the p-adic metric. Then we add all the missing limits of p-adic Cauchy sequences, and that gives us Q_p. So analytically, Q_p is the completion of the rational numbers with respect to the p-adic metric.
And the second approach to Q_p, it's the algebraic view. It says Q_p is the field of fractions of the integral domain Z_p. So what is a field of fractions? A field of fractions is a way to turn an integral domain into a field by formally allowing division. So how can we turn an integral domain into a field? Um, actually, you already know how to do it. We start from Z, and you know how to make Q. So how did we do this? Uh, we make a new set of two-ordered pairs of integers. So A and B are integers, and B is not zero. So we want to represent A over B, and we identify (A, B) and (C, D) if AD = BC. And how do we add these numbers? And how do we multiply these numbers? We define addition like this: (AD + BC) / BD, and multiplication we define like this: AC / BD. And this set, under this multiplication and addition, we define, this forms a field.
So now we do the same thing with Z_p. We make a set. We define uh, multiplication and addition like this, just like what we've done for uh, the integers. And the field we obtain through these kinds of arguments is called the field of fractions. And here's the really interesting part. These two constructions give the same field, and we call that Q_p. One construction says, complete Q using the p-adic metric. Then we add all missing limits of p-adic Cauchy sequences. The other says, take the field of fractions of Z_p. And we have used a method turning an integral domain into a field, like this. Uh, so they look like completely different constructs, but they lead to the same object, and that we call it Q_p.
Now we are finally ready to study one of the two mountains of this video: elliptic curves. Elliptic curves are one of the most mysterious objects in modern mathematics. Even today, there are huge open problems about them. So without further ado, let's dive in.
Before we define elliptic curves, we first need to talk about the space where elliptic curves live. At first, we might try to work on the ordinary affine plane R^2 or C^2. This is a Cartesian plane that we are familiar of. So this is the usual Cartesian plane. So we have points (x, y). But the affine plane has one annoying problem: it's that parallel lines never meet. For example, um, two vertical lines x = 1, x = 2, they never meet. These two lines y = um, I don't know, x + 3, y = x + 2, they never meet because they have the same tangent. If they are parallel, they never meet on an affine plane. Uh, but in algebraic geometry, we want a cleaner world. We want a setting where geometric statements have fewer exceptions. In particular, we want any two distinct lines to meet at exactly one point. And to make this happen, we enlarge the affine plane by adding points at infinity. So each direction of parallel lines gets its own point of infinity, and the collection of all those points of infinity form a line at infinity. And the resulting space is called the projective space. Projective plane, denoted by P^2, like this. So the projective plane is like the affine plane but completed by adding points at infinity.
Some comparisons between affine plane and projective plane. On the affine plane side, parallel lines do not meet. These two lines are parallel, so they do not meet. But on the projective plane side, parallel lines meet at exactly one point, and that point is called the point at infinity. So these two parallel lines meet at this point, and I can take any two parallel lines, and they will meet at exactly one point here. Two parallel lines and they will exactly meet here at one point. And if you collect all these points at infinity, they form a line, and I will call that line the line at infinity.
So how do you represent points in the projective plane? In the affine plane A^2, the point is written like this: (x, y). But in the projective plane, we use homogeneous coordinates, and we have three variables: so (x, y, z). Here x, y, z cannot all be zero. So (0, 0, 0) is impossible in the projective plane. And that's one important rule. It's that (x, y, z) is the same as (λx, λy, λz). So, for example, (1, 2, 3) in the projective plane is exactly the same point as (4, 8, 12). And this is also the same as (4, 8, 12) or (2, 4, 6).
Now, how does the ordinary affine plane sit inside the projective plane? If we have the affine point (x, y), so this is in the affine plane. We send this point to (x, y, 1) in the projective plane. And this point is of course the same as (λx, λy, λ). So the usual affine plane corresponds to the part of the projective plane where z is not zero. Conversely, if z is not zero, and this point is the same as (x/z, y/z, 1), right? And this point corresponds to (x/z, y/z) onto the affine plane. Now, if z = 0 on the projective plane, so this was the case where z is not zero. And if z = 0 on the projective plane, those are points at infinity.
Now let's see why parallel lines meet at points at infinity. So now let's take a line on the affine plane. So we have y = mx + b. So this line is in the affine plane. Now, to put this line in a projective space, we homogenize the equation, meaning that we'll put y as y/z and we will put x/z in the place of x, and this becomes y/z = m(x/z) + b. So after homogenizing, we get the projective line, which sits inside the projective plane. And we can set z = 0 because this is a line in the projective plane. If we take z = 0, y/z = mx/z + b. So y = mx + bz. So after homogenizing, we get the projective line. And we can set z = 0. If we take z = 0, y = mx. So this line goes through (1, m, 0). So every line of the form y = mx + b, these lines go through the point (1, m, 0) on the projective plane. And this is the line at infinity.
Now that we have the projective plane and the homogeneous coordinates (x, y, z), we want to define curves inside the projective space. In the affine plane, this is easy because, for example, we can just write y = x^2, and this forms a curve in the affine plane. But in the projective plane, we have to be more careful because, for example, suppose we have f(x, y, z) = x^3 + y^2 - 2xyz. So if we look at the curve f = 0, (2, 4, 3) is in this curve, right? Because 8 + 16 - 24 = 0. But (2, 4, 3) is the same as (4, 8, 6) because we can rescale, right? And if we put (4, 8, 6) in this equation again, 64 + 64 - 192, this is not zero. So this kind of thing cannot happen in the projective plane. And the reason is that this has degree mix. This has degree three, degree two, degree three. So to define a curve properly in the projective plane, we need a homogeneous polynomial with the same degree. That means that every term has to have the same degree. For example, we have to define a curve like this: so x^2. This has degree two. So the other terms also have to have degree two. So, for example, this is a well-defined homogeneous polynomial of degree 2. So in the projective plane, curves are defined by homogeneous polynomial equations. So we are talking about projective curves defined in projective planes.
We want to determine the shapes or the types of the curve. How it behaves or how it looks. Roughly speaking, a smooth point is a point where the curve behaves like a nice curve, and a singular point is a point where something goes wrong. So the points in the projective plane, we call that a singular point of the curve f if the partial derivative of f with respect to all variables x, y, z equals zero. And a point that is not singular is called non-singular. And a curve with no singular points, which is called smooth. So for our purpose, the exact computation here, the partial derivative is not the main point. We're not going to like spend time on checking singularity by calculating partial derivatives by our hand. But you have to remember the concepts of singular points and smooth curves because the distinction will matter because an elliptic curve must be smooth.
So here are some examples of smooth and singular curves. A curve on the left, if you see, this is a smooth curve because it behaves nicely. It has no sharp points, no self-intersection. So, uh, it's good at every point. But the curve on the right is singular. At this particular point, the curve fails to be smooth because it has a bad point where the usual tangent behavior breaks down. So this point is called a singular point, and since this curve has this singular point, uh, this curve is not smooth. It's singular.
Now, finally, the definition of an elliptic curve. An elliptic curve is a smooth cubic defined over the field k with at least one point. Uh, so two rules elliptic curves have to satisfy. First, it has to have at least one k-rational point. K-rational points means that the coordinates are defined over the field k. And second, it has to be a smooth curve, meaning that singular points are not allowed on an elliptic curve.
Now let's look at some examples. Uh, first example would be y^2 z = x^3 - x z^2. Is this an elliptic curve? Yes, it's an elliptic curve because, first of all, does it have at least one point? Yes. (0, 1, 0) is one point. So this we can call it an elliptic curve defined over Q. And second, is it a smooth curve? Uh, I'm not going to do the computation here, but if you check, it is smooth. So this is a well-defined elliptic curve. Now, this is written in projective coordinates, right? And let's look at the affine chart. Uh, meaning that we put z = 1. So if you put z = 1, we have y^2 = x^3 - x. So this curve sits inside an affine plane, where this curve is on the projective plane. This means that here we're looking at the part of the projective plane where z is not zero. In other words, we're excluding the line at infinity.
And another example of an elliptic curve: y^2 z = x^3 + x z^2 + z^3. This is also an elliptic curve because, again, (0, 1, 0) is in this curve. So at least one point. And if you uh, check and do the computations, it's smooth. And this curve sits inside a projective plane. So if you send this to an affine plane, meaning that you put z = 1, we get y^2 = x^3 + x + 1. So this is a part of this curve, excluding the line at infinity. And we can draw it on the affine plane.
And after moving the points at infinity, meaning that we put z = 1, an elliptic curve can be described by an affine equation, we just saw. And it's a fundamental theorem that every elliptic curve over a field k is birationally equivalent to a planar curve given by a Weierstrass equation. Here, the word Weierstrass is a German word. So in German, you pronounce W as the V sound in English. And this R sound should not come from your mouth; it should come from your throat. So it has to be something like "Vah-st". Um, I know this because I'm learning German at the moment. And German is one of the hardest things I've ever learned. I think it's harder than elliptic modular forms. Um, anyways, so this long Weierstrass equation looks like this. And if the characteristic of the field is not two or three, we can always simplify this equation by a change of variable to the short Weierstrass equation that looks like this. Uh, so first of all, what does birationally equivalent mean here? Birationally equivalent means that two curves are essentially the same from the viewpoint of rational functions. So we are not saying that the equations look identical, but we are saying that after ignoring a small number of exceptional points, the curves can be converted back and forth using rational functions. Um, so the most general form is this long Weierstrass equation right here. This has no exceptions. And if you Google, the first thing you see will be the short Weierstrass equation. And this works with quite good generality because the field, the characteristic, if it's not two or three, we can use the short Weierstrass equation.
So, for example, if I take elliptic curve y^2 + y = x^3 - x^2, if I define elliptic curve E2 be Big Y^2 = 4x^3 - 3/3x + 19/27. These two elliptic curves are birationally equivalent. Uh, if I take y = 2Y + 1 and Big X = X - 1/3, and if you plug these in, you will see that E1 and E2 are the same.
Now let's draw some typical elliptic curves and see how it looks. The first one is y^2 = x^3 - x. And because the left-hand side is not negative, we know that this is only defined where the right-hand side is not negative. So let's first draw this y^2 and x^3 - x thing, and using this, we will draw this elliptic curve. Uh, this graph looks like this because this has roots at 0, -1, and 1. So this elliptic curve is defined here where the right-hand side is non-negative. And notice that we can divide this elliptic curve into two parts. We can draw y = sqrt(x^3 - x) and then reflect it to draw the lower half. So if you draw this sqrt(x^3 - x), this will be something like this, um, considering the shape of the graph. And if you also draw the lower half by reflecting this, this will look like something like this.
And let's look at this one. Uh, this is y^2 = x^3 - x + 1. So, similarly, let's first draw y = x^3 - x + 1 and see how it looks. Um, so if we take a derivative, y' = 3x^2 - 1. And so we know that the local minima or local maxima is happening at x = ±1/√3. So if you put 1/√3 here, we have 9/(2√3)/9. And if you put -1/√3, we have 9 + 2√3/9. And since this is positive, if we draw the shape of this graph, it will be something like this. So if we take a square root of this graph in this range of x, and then we reflect it to draw the lower half, this E2 elliptic curve will look like this. So the upper half will be not exactly the same, but similar to this shape we draw. So it will be something like this, and the lower half, if you reflect this, this will look like this. So these are how elliptic curves generally look like in an affine plane.
There's one point we have been hiding so far. When you draw an elliptic curve using an affine equation, let's say a short Weierstrass equation, so x^3 + ax + b. We are only looking at the affine part of the curve, right? But the actual loopy curve lives in the projective plane. So we should really look at the projective equation. And the projective equation of this will be y^2 z = x^3 + ax z^2 + b z^3. Now let's ask what happens at the line at infinity. The line at infinity is z = 0. So let's plug z = 0 and see what happens. So if we plug z = 0 in the projective equation, we have 0 = x^3. So x is 0. So the only possible value on the line at infinity is (0, y, 0), which is the same as (0, 1, 0) by rescaling. So the full elliptic curve is not just the affine plane we saw in the xy plane. It's the affine curve together with this one extra point, and we call this point the point at infinity. So from now on, we will usually draw and discuss elliptic curves in the affine plane, because that's where most of the points we can see are located. But technically, an elliptic curve is defined in the projective plane. So whenever we talk about the affine plane, we should also remember one extra point: this point at infinity in the projective plane.
Our goal from now is just to turn the points on elliptic curves into a group. And to have a group, what do we need? First, we need a set. This part is easy because we can take any elliptic curve, and its points already form a set. But a set alone is not a group. We also need an operation. So now the question is, how should we define addition of points? We cannot just define it however we want. For example, we cannot define (1, 2) + (3, 4) = (4, 6) because even if (1, 2) and (3, 4) lie on an elliptic curve, there's no guarantee that (4, 6) also lies on the same elliptic curve. So we need a more geometric way to define addition.
So this is how we do it. If there's an elliptic curve that looks like this, and I will choose point A from here, point B. So we want to add point A and B. And in order to do that, first we draw a line that passes through A and B. So this line will go through and it will meet at exactly one other point with an elliptic curve, and we will call that point R here. And from there, we will draw a line that goes through R and O, the line at infinity in the projective plane. A line that passes through O corresponds in the affine plane to a vertical line. A vertical line that goes to the point (0, 1, 0), the point at infinity in the projective plane, is x = 0. And if we take this into the affine world, it becomes x = c, because we put z = 1. Uh, that's why the line that goes to the point at infinity in the affine plane corresponds to the line x = c. So the line that goes to R and O will look like this, a vertical line, and this vertical line will also meet with the elliptic curve at some point, and it's here, it's actually a reflection through R, because this graph is symmetric to the x-axis. And I will define this point as A + B. So we will draw a line that goes through A and B and find another intersection point that the line meets with the elliptic curve, and we'll name the point R. And we'll also draw a line that goes through R and O. And that another intersection point with the line and the elliptic curve, we will define that point as A + B. And this is how the addition on elliptic curves works.
Now, one thing we still have to worry about is the associativity of the addition we defined. If this addition law is really going to make the points of the elliptic curves into a group, then we need these to be the same: so (A + B) + C = A + (B + C). And this is not obvious from the geometric construction. A rigorous proof of the associativity of the addition law uses basic algebraic geometry, especially facts about intersections of curves in a projective plane, and we will not going to go through that proof in detail. Instead, we'll verify visually by drawing the constructions step by step.
So, I picked any arbitrary point A, B, and C on this elliptic curve and we'll compute these two and check if they are the same. So, first here, we have to compute A + B. So, we first draw a line that goes through A and B, and this line will look like this. So, this is the third intersection point. This point we call R. Then you draw a vertical line that goes through this point. This is vertical because the vertical line goes through the point at infinity. And this intersection point, this was A + B. Okay. And now we have to define these two points, C and A + B. So we first draw a line that passes these two dots. And the intersection point with the elliptic curve will be here, this point. And now we draw another vertical line that goes through this point. And this point, this point is what we're looking for: (A + B) + C.
So on the right side, we will first add B and C. So I will draw the line that goes through B and C, and the intersection point will be here, and I will draw a vertical line. Actually, in this case, since this graph is symmetric through the x-axis, drawing a vertical line and making an intersection is the same as just reflecting this through the x-axis. But, um, but we will just go over all the steps. Okay. So we will draw a vertical line that passes through this intersection point, and this point, this point will be B + C. And now we will add these two points, B + C and A. So we draw a vertical line that goes through this A and B + C, and this is the intersection point, and we will again draw a vertical line that goes through this point, and it will be somewhere around here. So this point is A + (B + C). And these two points, they look like it's in the same position. So these two are the same.
So we have set operations and associativity. Two more things that we need are whether every point has an inverse and does the identity exist. So first, let's look at the identity element. The identity element of an elliptic curve is the point at infinity (0, 1, 0). So it's right over here at the vertical direction. So if we add these two points, so we draw a line that passes through these lines, and it will meet at here at the elliptic curve. This is the point R. And we will again draw a line that passes through R and O. And the intersection of the line with the elliptic curve is P here. So P + O = P. O acts as the identity here.
And now let's look at the inverse. If we're looking at elliptic curves that are symmetric through the x-axis, for example, a short Weierstrass equation, which we are looking at right now, the inverse is geometrically the reflection through the x-axis at that case. So, um, if we were to add P and -P, we draw a line that passes through the two points, and the third intersection point will be this point, the point at infinity. And since the reflection of O is still O, it means that P + (-P) = O, the infinity. So inverse exists for every point on the curve. So the conclusion is that the points on a given elliptic curve form a group, an abelian group actually, because the addition is commutative. So an elliptic curve is no longer just a geometric object. It is also an algebraic object because its points carry a group structure. This is what makes elliptic curves so fascinating. It's where geometry and algebra meet together.
Now we want to look at elliptic curves through the lens of prime number p. Suppose we have an elliptic curve over Q by an equation with integral coefficients. For example, if we take an elliptic curve y^2 = x^3 + 6x + 11. Now choose a prime number, say p = 5. From now on, we want to look at this equation over the finite field F_5, or it's isomorphic to Z mod 5Z. Now, here's the important point. Over Z mod 5Z or Z_5, there are only five possible values, right? 0, 1, 2, 3, 4. So unlike an elliptic curve defined over Q, this is not a continuous curve in the plane. It's a finite set of points where we can literally plug in values of x and check whether there exists a value of y that satisfies the equation.
So if we look at elliptic curves over the finite field 5, which is equivalent to saying looking at this elliptic curve modulo 5, y^2 = x^3 + 6x + 11 modulo 5, only finitely many points are possible. So the points on this curve modulo F_p are (0, 1), (0, 4), (2, 1), (2, 2), (4, 3), (4, 1), (4, 4), (4, 2), (4, 3). For example, (4, 3). If you plug this in, y^2 = 3^2 = 9. x^3 + 6x + 11 = 4^3 + 6(4) + 11 = 64 + 24 + 11 = 99. And this is equal modulo 5. So (4, 3) is in this curve. So reduction modulo p turns the curve into a finite object. Instead of drawing a continuous curve, we can count actual points on F_p. That's why reduction modulo p is so useful. It lets us take an elliptic curve over Q and study it through finite arithmetic.
Now, this is the important point. Reduction modulo p is not just copying the same curve into a smaller world. Some geometric properties may change after the reduction. For example, you remember how we defined a singular point? A point is singular when all the relevant partial derivatives vanish at that point. Now suppose one of those partial derivatives gives the number 5 over Q. That's not zero. So the partial derivative is non-zero. But if we reduce modulo 5, we're looking at Z_5. So in the reduced world, the same value becomes zero. This means that a curve that was smooth over Q can become singular after reduction modulo p. In other words, singular points that did not exist before can appear after reduction.
So let E be an elliptic curve over Q given by a minimal Weierstrass equation. We will learn what this minimal Weierstrass equation means later. And let E bar be a curve reduced modulo p. If E bar is still smooth, we say that E has good reduction at p. And if E bar has a singular point with distinct tangents, we say E has multiplicative reduction at p. And if E bar has singular points with a repeated tangent, we say E has additive reduction. So the first scenario where E bar is smooth. This is the best. You do the reduction and it's still smooth. And if E bar has singular points, meaning that E bar has bad reduction. These two give you the feeling of it. Multiplicative reduction is still bad, but it's better than additive reduction. And additive reduction is the last thing you want to happen. This is, um, bad. Okay. And these features here are only for visual references. They're not literally what the reduced curves look like over F_p, because after reduction modulo p, we are working on a finite field, and there are only finitely many points. So the curve is not a continuous curve that we can draw smoothly on a plane like this.
Now we're working over a field. Suppose our elliptic curve is defined over F_q. So we're looking at elliptic curve reduction modulo q. This curve has only finitely many elements. So it makes sense to ask how many points does this elliptic curve have over the field F_q. We write this number like this: |E(F_q)|. Let's think about this very naively. Suppose the curve is given by the short Weierstrass equation y^2 = x^3 + ax + b, and there are exactly q possible values for x because we are looking at F_q. There are exactly q possible values of x because x has to be an element of F_q. So in principle, we can plug in every q values of x one by one. For each x, the right-hand side x^3 + ax + b is again an element of F_q. Then we ask how many y's satisfy this. Sometimes there can be no solution for y, and sometimes there is exactly one solution where the right-hand side is zero, and sometimes there are two solutions where the right-hand side is a non-zero square. So for each x, the number of possible y values are zero, one, or two.
Now, without proving anything, let's make a very rough guess. Maybe on average, each x value gives about one point. So if each of the q possible x values gives one point on average, we expect about q * 1 = q affine points. But an elliptic curve is projective. That means we have to also consider the points at infinity. So after considering the points of infinity, we might expect this number of points to be somewhere around q + 1. And surprisingly, this rough guess is actually not that far from the truth. Hasse's theorem says that the number of points in F_q is not too far away from q + 1. More precisely, the difference between these is bounded by 2√q. So the difference is at most 2√q.
And right after this theorem, let's give a name to the quantity that appears there. Let E over Q be an elliptic curve, and suppose p is a prime of good reduction, meaning that if we reduce modulo p, no singular points occur. Then we can reduce E modulo p and get a new curve which is defined over a field F_p, and count its points over F_p. And we will define a_p = p + 1 - |E(F_p)|. This is exactly the error term from Hasse's theorem. Hasse's theorem says that |a_p| is no greater than 2√p. And from now on, we call this number a_p the trace of Frobenius. And for good primes, this definition comes from point counting: p + 1 - the number of points. And for bad primes, the curve does not reduce nicely, so we have to use this convention instead: a_p is defined as 1 when there is split multiplicative reduction, and -1 when non-split multiplicative reduction, and 0 when additive reduction. So you do not have to memorize all of these values in bad reduction cases. For now, it's enough to remember that at a good prime, we define the trace of Frobenius a_p as p + 1 - the number of reduced points.
Now we move on to division points. A point P on an elliptic curve is called an n-division point if if you add P n times and it gets back to the identity element. I will write this as nP = O. This doesn't mean that we are multiplying n with P. It means that we are adding P n times. And there's a small distinction here. Saying that P is an n-division point does not necessarily mean that n is the first time we return to zero. For example, if P is a 2-division point, of course P is a 4-division point, right? So the order of a P is more precise. We say P has order n if n is the smallest positive integer that satisfies nP = O. And if you collect all of these n-division points and you write it E_n, it forms a subgroup of the elliptic curve.
So why is this a group? Uh, first of all, is addition well-defined here? Meaning that if you pick any two n-division points, is the sum of the two n-division points also lies on n-division points? Yes. Because if you add P + Q n times, because addition defined on elliptic points are associative and commutative, this is equal to nP + nQ, which is O + O = O, because P and Q are n-division points. And is O the identity n-division point? Yes. If you add O n times, it's still O. And lastly, for every P, does -P lie in n-division points? Yes, because you can add -P n times, and this is equal to -(nP), and this is O. So E_n forms a group. And since E is an abelian group, this E_n is also an abelian group. So we already know that E_n is an abelian group.
So now we want to know what kind of group structure E_n has. Let's start with the simplest non-trivial case, and we'll take n = 2. Now let's use the short Weierstrass equation y^2 = x^3 + ax + b. And we will pick P from E_2, which means that P + P = O. So 2P = O. This means that P = -P. And if I name the coordinates of P as (x, y), we know that in the short Weierstrass equation, the coordinates of -P is the reflection through the x-axis. So -P has coordinates (x, -y). So in order for these two points to be the same, y and -y has to be the same, which means that y has to be zero. So we get x^3 + ax + b = y^2, and y is zero. And if we work over the algebraic closure, the cubic has three roots, right? This is a third-degree polynomial. So we name the roots of this equation, three roots, as e1, e2, e3. So the three 2-division points are (e1, 0), (e2, 0), (e3, 0). And don't forget the identity O. So you put the identity O. And this is E_2. So you have a total of four points. And because it's a finite abelian group and has order 4, we know that it's either Z mod 4Z or Z mod 2Z x Z mod 2Z. But every element in the 2-division points satisfies 2P = O. Right? But here we have elements with order 4. So we know that only this structure is possible for E_2 division points. So we get the conclusion that the 2-division points are isomorphic to the product of two cyclic groups Z mod 2Z x Z mod 2Z.
And how about third division points? Uh, so we start from the short Weierstrass equation y^2 = x^3 + ax + b. And if we look at third division points, 3P = O, which means that 2P = -P. So if we let P = (x, y), we also know that -P = (x, -y). And actually, we have a formula for the coordinates of 2P. The x-coordinate of 2P is computed as (3x^2 + a)^2 / (4y^2) - 2x. So this is equal to x. Okay. So if we simplify this, (3x^2 + a)^2 = 3x * 4y^2. And this is equal to 12x. We know that y^2 is this expression here, so x^3 + ax + b. Um, without expanding this, we know that this is a degree 4 polynomial. So over the algebraic closure, we have four roots. And each root, if we plot this here, gives two y values on the elliptic curve. So we get 2 * 4 = 8 non-zero third division points. And together with O, the point at infinity, we have a total of nine third division points. And since E_3 is a finitely generated abelian group, we know that the structure is either Z mod 9Z or Z mod 3Z x Z mod 3Z. But every element in the third division points satisfies 3P = O. Right? But here we have elements with order 9. So we know that only this structure is possible for E_3 division points. So we get the conclusion that the third division points are isomorphic to Z mod 3Z x Z mod 3Z.
9 Z or Z mod 3Z * Z mod 3Z but every three division points satisfies 3 P equals zero but Z mod 9Z has ordered nine points in it. So we know that this is the only possible structure. So Z3 is isomorphic to Z mod 3Z times Z mod 3Z.
So we have looked at E2 and E3. For E2, we have found four division points over the algebraic structure and it was isomorphic to Z mod 2Z product Z mod 2Z and E3. Uh, the structure was similar Z mod 3Z time Z mod 3Z.
Now the surprising fact is that this is not a coincidence. The same structure holds for every positive integral n at least in characteristic zero. So let E be an elliptic curve defined over field K of characteristic zero and let K bar its algebraic structure. The group of N division points over K bar over algebraic structure is isomorphic to Z mod NZ product Z mod NZ. So over the algebraic closure the n division points form a very clean group. There are two independent cycles directions each of order n. In other words, en behaves like a two-dimensional module over Z mod NZ.
From now on, we want to relate Galois theory with elliptic curves. Take an element from the absolute Galois group and we call it sigma. By definition, sigma acts on algebraic numbers. The points on ei have algebraic numbers. Uh, we didn't prove this, but let's just take this given. So it means that sigma can act on the coordinates. So if sigma acts on a point P with coordinate x and y, this becomes sigma x, sigma y.
Now the important point is that since the elliptic curve is defined over Q, applying sigma to the coordinates of the n division points gives another point on the same curve. So what do I mean? So if we look at the short Weierstrass equation y² = x³ + ax + b. If we apply sigma to both sides and since sigma is a homomorphism, sigma y² is uh, the square of sigma y. So the multiplication and addition can be split up like this plus uh sigma a sigma x. But sigma a, since a is a rational number, this is a plus b. So it means that if x and y are a point of an elliptic curve, sigma x and sigma y also lie on the same elliptic curve. And one more thing, if P is an n division point, uh, meaning that P is an n division point for some n, sigma P is also an n division point. So why is this? Because uh nP = 0. And if you apply sigma to both sides, sigma nP = n sigma P and this is zero, right? Uh, this is mainly because sigma P + Q = sigma P + uh, sigma Q. Uh, we will not going to prove this right now. But uh, remember when we computed three division points, uh, the coordinates of the points 2P were given as the rational functions of x and y. So if you add P and Q, the coordinates of P + Q are given by some rational expressions in the coordinates of P and Q and since sigma is a field automorphism, it preserves those rational expressions. So it preserves the whole addition operations, which means that we can say sigma preserves the addition defined on elliptic curves. Therefore, sigma maps en to itself. In other words, sigma gives a map endomorphism of the group en.
Now we will use the structure of en over the algebraic closure. We know that en is isomorphic to Z mod NZ times Z mod NZ. So we can choose a basis of EN. We will uh name that P1 and P2. Remember EN can be viewed as Z mod NZ module over. So we can choose a basis of EN and we will name that P1, P2. Remember EN could be viewed as a Z mod NZ module. This means that every point in EN can be uniquely written as P = aP1 + bP2 with a, b in uh, Z mod NZ. Now what happens if sigma is applied to both sides? Uh, we get sigma P = a sigma P1 + b sigma P2. And since sigma P1 and sigma P2 are still points in en, sigma P1 can be written as xP1 + yP2 for some Z mod NZ x and y, plus b sigma P2 can be also expressed as linear combinations of the bases. So ZP1 + WP2 and uh, if you collect the coordinates of P1, you get aX + bZ and P2 AY + bW and we can express this with a matrix multiplication. So we have a b x x z um, y w. So uh, if you compute this, we have aX + bZ which is these coefficients and AY + bW which is these coefficients. So it means that we can represent this automorphism sigma as this matrix. This is exactly what we just constructed. We started with an element in the absolute Galois group and each element sigma gives an automorphism of en. Then after choosing a basis of en, we could represent the automorphisms as a 2x2 invertible matrices over Z mod NZ. And this map is called the Galois representation attached to the n torsion points of E.
About the GL representation which is constructed, there are some limitations. First is about the information loss. The important thing is that if you look at prime power torsion points, they're not isolated from each other. So we will look at E. EP is uh, this sits inside a larger group EP² and this is in EPQ. So we have these uh, inclusion relationships and also we can find natural maps going uh, downwards. So a point in EP+1 can be sent to the point in EPR by multiplying P or adding up P times. So these levels are connected by homomorphisms. So it feels like a waste to look at these uh, groups as totally separate objects where we have this nice inclusion relationships and homomorphisms connecting these levels.
Secondly, uh, the coefficients keep changing. For EP, the matrices have coefficients in Z mod NZ. So GL(2, Z/nZ) and for E², the coefficients come from Z mod P²Z and for EPQ, the coefficients are from Z mod P³Z. So the target keeps moving, that makes it hard to do serious analysis. So we need something new that fixes these two problems.
Now let's again look at how these prime power division points are related to each other. So we'll fix a prime L and look at the L power points. Uh, first of all, they are stacked on top of each other, which means that EL is a subgroup of EL² and this includes this and this is a subgroup of this group. And also we had a homomorphism. If we take points from EL², we multiply by L and the points land on this group. Uh, we pick a point from here and multiply by L and the point lands in this group. And generally, we can send the point from EL to the N+1 to EL to the LN by multiplying L or uh, adding up N times. So we get this chain-like structure and this is very important because the torsion points are not meant to be studied one level at a time. The level EL² something about EL and level EL³ knows something about EL² and so on and so on. And uh, we know that this EL is isomorphic to Z mod LZ times Z mod LZ and this is isomorphic to Z mod L²Z. And we kind of want to use it. And on the other hand, here we have the chain-like structure when we saw when we constructed ZL. So we have Z mod LZ, Z mod L²Z, Z mod L³Z. And these levels, there were homomorphisms that connect these levels. And if you compare these two, these two look very similar, the tower and the homomorphism that connects each level. So here maybe we should do the same thing for E. Maybe instead of looking at each level separately, we should bundle them up together. And maybe if we use ZL because here EL is isomorphic to Z mod LZ², and this matches, right? So maybe if we use ZL, we can keep this whole tower structure alive and get a better GL representation, and that's what we exactly do.
So this is where we define a new object that fixes the problem from the previous slide. Instead of looking at EL², E³, and so on and separately, we want to package all of them into one object. So let L be a prime and E be an elliptic curve. The Tate module of E at L, denoted like this, TLE E, is the inverse limit or infinite terms of sequences that looks like this. So an element of the Tate module, if you pick X from the Tate module, it will be an infinite sequence X1, X2, X3, where X1 is an element of EL, X2 is an element of EL², and X3 is an element of EL³. But again, these points cannot be random, they have to be compatible with the maps between the torsion groups. The natural maps that go from ELK+1 to ELK are basically multiplication by L. So the compatibility condition is L Xn+1 = Xn. We see it here. So an element of a T module is not a single torsion group. It's a whole compatible tower of torsion points. It's very similar to the construction of ZL. And the nice thing about Tate module is that the Tate module is a free module of rank 2 over ZL. So the Tate module is isomorphic to ZL². And to see why the Tate module has such a structure, let's draw the tower again. So we have L division points, L² division points, L³ division points, and so on. And between these levels, we had homomorphisms that connected these levels. We could send points in ELK+1 torsion points to ELK torsion points by multiplying L or adding at times. And now each level, we know that this is isomorphic to Z mod LZ times Z mod LZ. So we could choose a basis P1, Q1. Here the same P2, Q2, P3, Q3 and basis for this torsion group. So P4, Q4. But we don't want these bases to be random. Since the levels are connected by the homomorphisms, we choose the bases so that they also match under the map. In other words, we choose them so that LPN+1 = PN and LQN+1 = QN. So P2 here sends to P1 by the homomorphism. Q2 sends to Q1 by this homomorphism. Same here, P3 is sent to P2. P4 is sent to P3. Now take one element of the T module and by definition, this is a compatible sequence that looks like this. And we can write each Xn like this as the linear combinations of the basis we chose. And we know that L Xn+1 = Xn. Right? So let's put this in and see what happens. So L a n+1 Pn+1 + b n+1 Qn+1 is equal to a n Pn + b n Qn. But we know that L Pn+1 = Pn right and L Qn+1 = Qn. So uh, L a n+1 is same as a n modulo Ln and L b n+1 = b n modulo Ln. This is important because if we collect all the A's, A1, A2, A3, blah blah, this would be an element in ZL and if you collect B, B1, B2, this will also be an element of ZL because it satisfies the homomorphism, right? And this is why the Tate module becomes isomorphic to ZL² after choosing a compatible basis.
I draw this picture for you to understand the T module a little better. Here we take L as two at each layer E2, E4, E8. The division points are isomorphic to Z mod 2 to the DZ times Z mod 2 to the DKZ. So we can draw each layer E2, E4, E8 as a grid. So E2 here, it's Z mod 2Z and we can draw it as a grid, same as E4 and E4 like this. First in E2, let's choose a point R1 in coordinates, this is 1, 0 modulo 2 and if P1 and Q1 are bases of E2, that this means that R1 = 1 P1 + 0 Q1. Now we move to E4 and we want to choose R2 here, but we cannot choose R2 randomly because there's a homomorphism between the layers, so the choices have to match. Specifically, we want that 2 R2 = R1. In terms of coordinates, if R2 is X, Y modulo 4. In terms of coordinate, if R2 is X, Y modulo 4, X should be congruent to 1 modulo 2 and Y should be congruent to 0 modulo 2. We just saw in the previous slide. So here we choose R2 as 1, 2 modulo 4. So 01, 012. So here is R2. Then we do the same thing for R3. So 5 is equal to 1 modulo 4 and 6 is equal to 2 modulo 4. So we choose the point R3. This is the point R3. Now every time we choose an Rn, we can look at its coefficients of the basis P and Qn. If we collect the P coefficients, in this case here it's 1, 1, 5 and blah blah. So this is the collection of P coefficients and we collect the coefficients of Q. So 0, 2, 6 power bytes. Here these coefficient sequences are actually elements of Z2, right? Because 1 = 1 modulo 2 and 1 = 5 modulo 4. This is also an element of Z2. And if we collect all the P's and all the Q's, this infinite sequence itself is an element of the Tate module T2E. [clears throat] This is also an element of T2E. And these two elements of the T module form a basis. So we can write R as the linear combinations of the bases. So these two are the bases and these two, these two are elements of Z2 and this is a constant. So while this notation is not mathematically 100% precise because we've never defined uh, the dot product between these factors. Uh, but this is enough to show the structure of the Tate module. So from this picture, it becomes natural that the Tate module T2E is a Z2 module with two basis directions. And think about this as we go higher and higher in the layers, each layer contains more refined information. For example, E2 here only sees the points modulo 2. It's a very rough version of the picture, but E4 here sees the same structure with twice as much resolution in each direction. And E8 sees it even with more resolution. Imagine there's a point you can never reach directly and you keep taking pictures of it with better and better cameras. The first picture you take with your smartphone, so low resolution. The next one you bring a slightly better camera. So the second one has more pixels and the next one you bring an even better camera. So, blurred or something. So you bring a better camera and the pictures will be sharper again. So each picture has to be compatible with the previous one. When you lower the resolution, it should look like the earlier picture because the cameras might be different. We are looking at the same point. So if we collect this whole infinite sequence of better and better compatible pictures, that is a good way to think about an element in a Tate module.
We made the Tate module because the fitted level color representation were a little unsatisfying. So now that we built the Tate module, let's let the Galois group act on it. Take an element from the Tate module. So x is in T and x is an infinite sequence x1, x2, x3, blah blah. So x1 is in L torsion point, x2 is in L² torsion point and so on. Now take sigma from the absolute Galois group and let sigma act on this x. So sigma x = sigma x1, sigma x2, sigma x3 and so on. But since sigma preserves torsion points, which means that sigma x1 is again an element of E and sigma x2 is an element of E². This is also an element of the Tate module. So the Galois action on the Tate module is an isomorphism of the Tate module. And since the Tate module, we know that this is isomorphic to ZL², right? So this is isomorphic to the general linear group, which is the matrices with entries in ZL. So when we let an element of the Galois group act on the Tate module, we get a 2x2 matrix with entries in ZL. Earlier when we constructed a GL representation from the torsion group en, after choosing a basis, it looked like this. So row en this sends one element from the absolute Galois group to this general linear group, which is a matrix with entries in Z mod nZ. This was useful, but it only saw one infinite level at a time. So to keep the whole p-power torsion tower together, we introduced the Tate module and from that we got the p-adic representation, which looks like this. This sends an element of GQ to the general linear group of ZP, which is a matrix with p-adic coefficients. But sometimes this p-adic representation contains too much information. So what we do is that if row EP sigma is a matrix with entries in ZP, we reduce each entry modulo p and we get a [clears throat] matrix over entries in Fp. Uh, this gives a new representation which looks like this, and this is called a mod p representation. So what information does this representation keep? It keeps only the first layer of the p-power torsion tower, the EP. Remember, and actually this is basically the same representation we would get if we went back to our first construction, simply chose n=p. So this representation is the same as if we choose n=p in this representation. So the point is that the mod p representation is not a a totally new kind of object. It's a p-adic torsion representation, but viewed as the modular p-reduction of the richer p-adic representation.
Now this is one of the moments where I would say something almost miraculous happens. So let E over Q have good reduction at a prime p. For any prime L, the p-adic representation is unramified at L. So let E defined over Q have good reduction at prime P. For any prime, the p-adic representation is unramified at P and the characteristic polynomial of the Frobenius element is given by uh, something like this. Now look at this carefully. The left hand side comes from p-adic representations, right? Row E, we chose L, we built the Tate module, we got matrices over Z. Then we took the Frobenius matrix and this is what the left hand side means. But the right hand side, this has no L in it. And this is quite surprising. And by the way, from this equation, we recognize that the trace of this matrix row EL Frobenius P is equal to AP, uh, because the characteristic polynomial of a 2x2 matrix is equal to lambda² - (a+d) lambda + (ad-bc). So a+d here is the trace of the 2x2 matrix. Anyway, this is telling us that the Galois representation really remembers the arithmetic of the elliptic curves. In particular, it knows the point counting data AP. This was related to the number of points reduced to field p. And this is a very strong signal that the elliptic curves and the Galois representations are tied together in a deep way. We started with geometry, torsion points, Tate modules, and Galois actions. But somehow when we look at Frobenius, the answer is controlled by this AP, which is related to point counting. So take these slides as a bit of foreshadowing. Galois representations are going to play a major role in the story of elliptic curves.
Now we move on to a slightly different topic. We are going to introduce a few basic invariants of an elliptic curve. So let's look at E as a short Weierstrass equation. So y² = x³ + ax + b. First, the discriminant is defined as Δ = -6⁴a³ + 27b². This determines whether if the curve is smooth or not. So if this value is zero, it means that the curve is singular. And if this is non-zero, it means that the curve is smooth. And the j-invariant, j-invariant is defined as j = C₄³ / Δ. And C₄ here is -48A. So if you compute this, it looks like something like 1728 * (4a)³ / (4a³ + 27b²). And over the algebraic closure, two elliptic curves are isomorphic if and only if they have the same j-invariant. So the j-invariant is like an ID number for an elliptic curve.
There is one more useful thing that we can do with the discriminant. The discriminant does not only tell us whether the original curve is smooth. It also helps us understand what happens after reducing the curve modulo prime p. So let E be an elliptic curve defined over Q. And we choose p ≥ 5, a prime number. And to determine the reduction type of E, we consider a minimal Weierstrass equation. And we will learn this later. So uh, if the p-adic valuation of the discriminant equals zero, it means that E has good reduction at p. And if the p-adic valuation is greater than zero, it means that uh, it's a bad reduction. And bad reduction, we had two types of bad reductions, right? Multiplicative reduction and additive reduction. And additive reduction was uh, the worst thing. And if the p-adic valuation of the discriminant is positive and the p-adic valuation of C₄ equals zero, it's a multiplicative reduction. And if the p-adic valuation of the discriminant is positive and the p-adic valuation of C₄ is greater than zero, it means additive reduction. So with this criterion, we do not have to reduce the curve, find the points, and compute partial derivatives directly every time to detect singular points. Instead, we can just look at uh, valuations of Δ and C₄, and that already tells us the reduction type.
Now, even if two elliptic curves are isomorphic over Q, they can look quite different as equations. In other words, the same elliptic curve can have many different Weierstrass equations depending on the coordinate system we use. For example, consider the elliptic curve that looks like this: y² = x³ + 16x + 64. Now make the change of variables: x = 4x' and y = 8y'. So let's plug this in. Uh, 64y'² = 64x'³ + 64x' + 64. Oh, so what do we get? We have uh, y'² = x'³ + x' + 1. So these two equations look different, but they are describing essentially the same over Q, uh, because the isomorphism is just a rational change of coordinates. But now this creates a small problem because if the same elliptic curve can be written in many different ways, then which equation should we use? This matters because quantities like the discriminant depend on the equation. If we choose a bad equation, the discriminant may look complicated and the reduction behavior may look worse than it really should be. So we need a way to choose the best equation among all these possible Weierstrass equations, and what should the criterion be? And we find the answer by looking at one prime p at a time. For a fixed prime p, among all integral Weierstrass equations for the same elliptic curve, we decide that better equations are the ones with the smaller p-adic valuation of the discriminant. Why? Because the p-adic valuation of the discriminant measures how many powers of p appear in the discriminant, and the discriminant is what we use to detect bad reduction. So if vp(Δ) is large just because we chose a bad coordinate system, then the equation is making the curve look worse at p than it really is. So at the prime p, we try to remove all necessary powers of prime for the discriminant, and that leads to the definition. A minimal equation at p is a Weierstrass equation whose discriminant has the smallest possible p-adic valuation among all other Weierstrass equations for the same curve. So from the point of view of p, smaller vp(Δ) means a better equation. So there is one small problem. There are infinitely many primes. So maybe one equation is good at p=3, but another equation is better at p=5. Then which prime should be prioritized? The nice fact is that over the set of rational numbers, we do not have to choose one prime over another because there exists a Weierstrass equation that is minimal at every prime p at the same time, and that is called the globally minimal Weierstrass equation. So locally at each prime p, we try to minimize vp(Δ), and a globally minimal equation is an equation that does this simultaneously for all primes.
If you have made it this far, you may have heard the phrase before, semi-stable elliptic curves. Whilst first proved the modularity theorem for semi-stable elliptic curves, and this is exactly the word semi-stable. So an elliptic curve defined over Q is called semi-stable if its globally minimal model has either good reduction or multiplicative reduction at every prime number. So you can think of a semi-stable curve as an elliptic curve whose reduction is never too bad at each prime. It's allowed to have good reduction, and it's even allowed to have multiplicative reduction, which is not a good reduction, right? But additive reduction is never allowed on a semi-stable elliptic curve.
Here I want to briefly introduce one of the most famous functions in number theory, the Riemann zeta function. It's defined as the infinite series like this. So you take every positive number n, take it to this power s, and take the reciprocals and add everything up. And this function can be also written like this: the product of all factors 1 / (1 - p⁻ˢ) for all primes. And this particular term in the product for primes is defined as the local zeta function. Now, why does this equation hold? Let's see. So we want to show that 1 + 1/2ˢ + 1/3ˢ + 1/4ˢ + blah blah blah is equal to the product of 1 / (1 - p⁻ˢ). But we know that this is equal to the sum of geometric series pˢ + 1/2ˢ + blah blah, right? So if we write this for all p, this becomes 1 + 1/2ˢ + 1/4ˢ + 1/8ˢ + ... and we have p=3 and p=5 and so on and so on for all primes. Now think about expanding the factors on the right side. Uh, from p=2 factor, we choose one of uh, these. So let's say we choose this one over 2ˢ and for p=3, so we choose uh, this factor and p=5, we choose one. So if we multiply these together, we have 1/2ˢ, 1/3ˢ, well, I don't know, we can have 1/11 and we can also have 1/11⁷ blah blah blah. So every term we get on the right hand side looks like this, but actually this is equal to 1/2³¹¹⁷ to the power fs like this, and by unique factorization domain, every positive integer n appears exactly once in this way. So when we expand the product over primes, we found every one of 1/nᵈˢ, and that's how this formula works.
This part is a little bit technical, so if you don't want to follow every detail, that's completely fine. The main point here is that we are trying to imitate what we did for Riemann functions on the elliptic curve side. To do that, we first attach a ring to the defined part of the curve, and this ring is defined as a quotient like this. So Fp[x, y] / <E> this is basically a polynomial ring and this E, this is an ideal generated by reduced Kef and this is a Dedekind domain. Uh, what is a Dedekind domain? Uh, it's an integral domain in which every nonzero proper ideal factors into a product of prime ideals, and for any nonzero ideal α of A, its norm N(α) is defined as the cardinality of the quotient A/α. And the important property is that the norm is multiplicative, meaning that N(αβ) = N(α)N(β). This is the important part. And because it factors uniquely into prime ideals of the Dedekind domain setting and because the norm has multiplicative property, this 1/N(a)ˢ would be written as something like N(a) is factorized into a prime ideal. So it looks like P₁ᵉ¹ P₂ᵉ² ... Pkᵉᵏ and since the norm function is multiplicative like this, this is equal to N(P₁ᵉ¹) N(P₂ᵉ²) ... N(Pkᵉᵏ). So because this ring is a Dedekind domain and this norm is multiplicative, we can do the exact same thing we did for the Riemann zeta function. So the sum of these can be described by the product of these factors. And this only describes the affine part of the curve. But the projective elliptic curve also has the point at infinity O. And to account for that extra point, we will also multiply this additional term and we will define this as the local zeta function of an elliptic curve. And the same zeta function can be built in a more geometric way. Instead of talking about ideals and prime ideals here, and the same zeta function can be built in a more geometric way. Instead of talking about ideals and prime ideals, we can count points. So let Ē be an elliptic curve over Fp and we can count points of points over Fp. But we can also count points on larger finite fields like Fp² or Fp³ and so on. And the local zeta function packages all of these point counting into one generating function. And that looks like this. So it's defined as the exponents of uh, this sum where t = p⁻ˢ. And the remarkable thing is that these two perspectives are describing the same zeta function. On one side, we have the algebraic perspective. We built the zeta functions using ideals, prime ideals, and norms in a way that looks very similar to the Riemann zeta function. On the other side, we have the geometric perspective. We built it by counting points over the finite field Fp, Fp², and Fp³, and so on. These are two completely different constructions. One is about ideals in a ring, and the other is about counting points on a curve. But the surprising point is that they meet. There are two ways of seeing the same arithmetic object. And this is one reason why zeta functions are so important. They let us see the same arithmetic object from two completely different angles. And one really surprising thing is that that the complicated looking infinite object, the zeta function, can actually be written in a very closed form like this. So uh, let's show how this works. So the zeta function on the geometric side was defined like this. So Z(t) = exp(∑ tn/n). And we know that t = p⁻ˢ. And about the number of points over the finite field EFn. So we made a whole big deal out of Hasse's theorem, getting bounds and stuff. So um, this might feel a little bit funny, but actually we do have an explicit formula for this number, and this is pⁿ + 1 - αⁿ - βⁿ, where α and β are roots of x² - apx + p = 0. So the roots are α and β. And before we expand this and simplify this, we would first want to look at a Taylor series of 1 / (1 - x). And this is uh, sum of xⁿ to infinity. And we will integrate both sides. So -ln(1 - x) = uh, sum of n+1 xⁿ⁺¹ / (n+1). But uh, I'll change the index to make it simple. Also xⁿ and where n starts from one to infinity. So uh, using this and using the fact that this can be described as uh, this expression, let's look at this. This is exp(∑ tn/n). And we have pⁿ + 1 - αⁿ - βⁿ tn / n. This is equal to pt/n + tⁿ/n - αt/n - βt/n. And so if we use this, we can simplify this as denominator we have as a denominator we have 1 - pt, 1 - t and numerator we have 1 - (α+β)t + αβt². But we know that α + β = ap and αβ = p. So uh, the numerator is 1 - ap t + pt². And the denominator stays the same. And if we put t = p⁻ˢ, we finally get this formula.
Now you just saw that the zeta function of a reduced elliptic curve has the form, and here the denominator is kind of like the standard background part. The elliptic curve itself is really showing up in the numerator because ap has the point counting thetas, right? So I will take the numerator and take the reciprocals and we will define that as the local L-factors of an elliptic curve. So this local L-factor is like the concentrated extract of the elliptic curve at the prime p. It packs a lot of information about how the curve behaves at p into one small factor.
Now we already defined the L-functions of an elliptic curve. For each prime p, we had a local factor Lp(s). And for good reduction, it looked like this. We just defined it. And when we have a bad reduction, the factor slightly differs and it looks like this. Then we multiply all these local factors together for all primes p, and this gives us the Hasse-Weil L-functions of the elliptic curve. So what does this mean? Each prime p gives us a small piece of information about the elliptic curve, the Lp(s) local L-functions. At good primes, that information comes from counting points on the reduced finite field. At bad primes, the curve degenerates. So the factor changes depending on the type of bad reduction. The L-function takes all these local information prime by prime and packages into one global analytic object. And this is why the L-function is such a big deal because it takes the data from every prime and puts it into one small function. So instead of studying infinitely many primes separately, we can study this one object.
We have already seen a conductor on the Galois representation side. There the conductor measured how the representation behaves at each prime, especially where it is ramified and how badly it is ramified. For elliptic curves, we define a similar kind of invariant. The conductor of an elliptic curve is N that records the bad primes of the curve and also how bad the reduction it has at those primes. Here N is defined as the product of p^fp for all primes p. If E has good reduction at p, then fp = 0, meaning that p does not appear in the conductor. And if E has multiplicative reduction at p, then fp is defined as 1. So multiplicative bad reduction contributes one power of p. And if E has additive reduction, which is uh, the worst, then fp is greater than or equal to 2. So it contributes a higher power of p. You can think of it as the elliptic curve version of the conductor we saw in the regular representation. On both sides, the conductor measures bad behaviors prime by prime.
Chapter five is modular forms. Now the word modular may sound familiar. You might think it has something to do with modular arithmetic, like number theory. But modular forms are something completely different. They are complex functions. And if you have watched pop science videos about famous theorems, you have probably heard people say something like a modular form is about symmetry. But what does that actually mean? What kind of function has that much symmetry? Why would such a function appear in number theory? And does it have anything to do with elliptic curves? Uh, and that we're going to find out in this chapter.
There's a function called the Dedekind eta function, and it's defined on the upper half-plane H. So this H is the upper half-plane on the complex plane. So if τ is in H, it means that τ has a positive imaginary part. And the definition of the Dedekind eta function is like this, where q = e^(2πiτ). This is defined to be q^(1/24) * product (1 - qⁿ) for n=1 to infinity. So [clears throat] let's just plug in a simple value and see what happens. We will take τ = i. So q becomes e^(-2π), which is approximately 0.000187. So a positive, very small number. And η(i) = q^(1/24) * (1 - q) * (1 - q²) * (1 - q³) and so on. And since q is already a small positive number, this q^(1/24) will be very close to one, and these factors (1 - q), (1 - q²), (1 - q³) will be numbers that are very close to one. So after we do the computation, this value will be, this value is 0.768225.
So why do you care about this function? The reason is that this Dedekind eta function satisfies some very special transformation laws. The two important ones are written here. So η(τ+1) = e^(iπ/12) η(τ) and η(-1/τ) = √(-iτ) η(τ). The two important ones are written here. Uh, it's about η(τ+1) and η(-1/τ). Now if you look at these two formulas, there's something a little annoying. In the first formula, we have e^(iπ/12) and for the second formula, we have √(-iτ). These are not wrong, but they make the symmetry look a bit messy. So what if you raise both sides to the 24th power? So if we raise both sides to the 24th power of both equations, we have (η(τ+1))²⁴ = (e^(iπ/12) η(τ))²⁴ = e^(2πi) (η(τ))²⁴ = 1 * (η(τ))²⁴. So (η(τ+1))²⁴ = (η(τ))²⁴. And the second formula, (η(-1/τ))²⁴ = (√(-iτ) η(τ))²⁴. If you take the 24th power to this, the i terms get disappears, it's it's one. So it's τ¹² (η(τ))²⁴. And it will make the formula much cleaner. So we take the 24th power of the Dedekind eta function and we define the function as Δ(τ) = (η(τ))²⁴. Now the modular discriminant Δ(τ) we just defined satisfies much cleaner transformation laws. So let's look: Δ(τ+1) = (η(τ+1))²⁴ = (η(τ))²⁴ = Δ(τ). And Δ(-1/τ) = (η(-1/τ))²⁴ = τ¹² (η(τ))²⁴ = τ¹² Δ(τ). The two important ones are written here. Uh, it's about Δ(τ+1) and Δ(-1/τ). Now if you look at these two formulas, there's something a little annoying. In the first formula, we have e^(iπ/12) and for the second formula, we have √(-iτ). These are not wrong, but they make the symmetry look a bit messy. So what if you raise both sides to the 24th power? So if we raise both sides to the 24th power of both equations, we have (η(τ+1))²⁴ = (e^(iπ/12) η(τ))²⁴ = e^(2πi) (η(τ))²⁴ = 1 * (η(τ))²⁴. So (η(τ+1))²⁴ = (η(τ))²⁴. And the second formula, (η(-1/τ))²⁴ = (√(-iτ) η(τ))²⁴. If you take the 24th power to this, the i terms get disappears, it's it's one. So it's τ¹² (η(τ))²⁴. And it will make the formula much cleaner. So we take the 24th power of the Dedekind eta function and we define the function as Δ(τ) = (η(τ))²⁴. Now the modular discriminant Δ(τ) we just defined satisfies much cleaner transformation laws. So let's look: Δ(τ+1) = (η(τ+1))²⁴ = (η(τ))²⁴ = Δ(τ). And Δ(-1/τ) = (η(-1/τ))²⁴ = τ¹² (η(τ))²⁴ = τ¹² Δ(τ).
The reason why this formula holds from these properties is because the modular group is generated by two transformations, τ → τ+1 and τ → -1/τ. And we will come back to that soon. For now, just remember the shape of this formula. Roughly speaking, functions satisfying these kinds of transformation laws are what we call modular forms. Here the exponent 12 is we call it the weight. And for other modular forms, this number can be different.
Now let's look at another important family of modular forms, the Eisenstein series. The Eisenstein series are defined for even k ≥ 4 as the expression: E_k(τ) = 2 * ζ(k) * (1 + c_k * ∑_{n=1}^∞ σ_{k-1}(n) qⁿ). Uh, this c_k is related to Bernoulli numbers. I'm not going to go deep into this. Uh, you can just think of it as a constant depending on k. And here's the sum of σ_{k-1}(n) qⁿ, σ_{k-1}(n) here means the sum of all (k-1)th power of the positive divisors of n. So for example, if n=6, σ_{k-1}(6) = 1 + 2^(k-1) + 3^(k-1) + 6^(k-1). And here q is the same q before, q = e^(2πiτ). Now the important thing is that these functions also satisfy a modular transformation law. For every γ = [[a, b], [c, d]] in SL₂(Z), the Eisenstein series of weight k satisfies E_k(γτ) = (cτ + d)²ᵏ E_k(τ). One thing to notice before we move on is that this E_k has a non-zero constant term. Uh, the q-expansion starts with 1, right? Whereas the modular discriminant had no constant term.
Now we need to introduce the space where modular forms live. This space is called the Poincaré upper half-plane. Uh, I think this is French. Uh, I don't really know how to pronounce it. Anyways, um, it's a set of complex numbers whose imaginary part is positive. So the set of complex numbers that the imaginary parts are positive. So, the upper half-plane of the complex plane. So if we write τ = x + iy, and if we know τ is in this Poincaré upper plane, we know that y > 0. This is the natural domain for modular forms. Functions like the Dedekind eta function, the modular discriminant, or the Eisenstein series are all functions defined on this Poincaré upper plane, and their special transformation laws come from moving points around and inside the space.
Now we want to introduce the modular group. Uh, the basic idea is that the special linear group SL₂(Z) can act on the upper half-plane H. In other words, H becomes a SL₂(Z) set. So we take a matrix, any matrix from the special group [[a, b], [c, d]], and if we define γτ = (aτ + b) / (cτ + d), this γ is a group action, and H, the Poincaré upper plane, becomes a special linear group set. Remember what it means for some set to be a G-set. If a group G acts on a set X, it needs to satisfy some conditions. X → X and G₁G₂X has to be the same as G₁(G₂X). So let's show that this group action satisfies these two. Uh, first of all, E*X. So E here is the identity element of the special group, which is the 2x2 identity matrix. So if the identity matrix acts on τ, what do we get? Uh, by definition, we have (1*τ + 0) / (0*τ + 1) = τ. So the first condition we satisfy. And how about the second one? We will take arbitrary γ₁ and γ₂ from the special 2 groups. So γ₁ we will write it as [[a, b], [c, d]] and γ₂ as [[e, f], [g, h]]. Now let's compute G₁G₂X. Uh, in this case, it's γ₁γ₂τ. So what is γ₁γ₂? This is also a 2x2 matrix. It's [[ae+bg, af+bh], [ce+dg, cf+dh]]. So this is equal to (ae+bg)τ + (af+bh) / (ce+dg)τ + (cf+dh). And on the other hand, how about uh, the first one? So γ₁ acting on γ₂τ. This is actually γ₁ acting on (eτ + f) / (gτ + h). So if γ₁ [[a, b], [c, d]] acts on this complex number, it becomes a * ((eτ + f) / (gτ + h)) + b / c * ((eτ + f) / (gτ + h)) + d. So if you multiply gτ + h here to both the numerator and denominator, these two become the same. So the uh, second uh, so this group action satisfies the second rule to be a group action. Now there's actually one more thing we have to check: is this operation even well-defined as an action on H? In other words, uh, if τ is in H, do we always have γτ in H? Because if we start with something like 1 + i, and then the transformation sends it to 3 - i, then we are no longer in H.
The upper half plane, and that would not be an action on H. So we want to show that the result gamma is also on the point upper plane. So we examine the imaginary part of gamma toao, which is imaginary part of a toa plus b c to plus d. And to do this, um, I will multiply c to power + d over c to power + d. And since this complex number and this complex number conjugates, uh, the product is a real number. So this is the absolute value of CA + D² over 1 and image. If you expand this, we have AC c to bar plus a d to plus b c to bar plus bd or AC to bar. But a c to bar and bd are both are, uh, real numbers. So we can just swipe this out. And the imaginary part of toao and toao bar has the same absolute values and different signs. So this is equal to a d to minus bc toao. So if you sum it up, this becomes the imaginary part of a d minus bc to. And we know that a d minus bc equals 1 because, uh, abcd is, uh, in the special two group, which means that the determinant a minus bc equals 1. So this is equals to imaginary part over c plus d². So we know that this is, uh, positive.
So now we've shown that the special linear two group really acts on the upper half, uh, point on the upper plane, and the modular group here is defined as the special linear two group. Uh, there's one small point we should mention here. The matrices I and minus I act on the upper plane in exactly the same way because you put I and minus I here, and it gives the same result. So the actual transformation only depends on the matrix up to plus or minus I. Uh, because of this, some books define the modular group as special two group modulo I and minus I, uh, the cos group. But in this book, I will use the convention that the modular group means this special linear group. This is slightly cleaner for our algebraic discussion. We have just defined the modular group G as special group Z.
The surprising thing is that all of these transformations can be built from two different, uh, basic moves. The first one is sending to to plus one. So this just shifts the upper plane one unit to the right. And the second one, uh, sends tao to minus one over towo. This is a much less obvious transformation, but, uh, this is the other basic move. So the theorem says that every element of the modular group, special two group, can be built by combining these two operations. For example, the transformation that sends toao to 2 to + one over to + one. This transformation, uh, can be represented as a composition of T and S. So T S and T. Remember earlier when we said that the transformation law for model discriminant work on these two transformations, and then we basically said it works for all specialar group. Um, this is the reason because every element of the special linear group can be built from these two generators.
Now we're going to learn about the fundamental domain of the model group. A fundamental domain is a region in H containing exactly one representative from each orbit of the action. And for the model group, especially unit two group, a fundamental domain looks like this. So we take all the points in the plank upper plane whose real parts is between, uh, plus 1/2 and minus 1/2, and that gives a strip here, and then we move the parts inside the unit circle. So we can only keep the points where, uh, the absolute value of star greater than one. This remaining region here, the remaining region, uh, is the standard fundamental domain. So what does it actually mean that this contains exactly one representative from each orbit in group action? Remember orbit, what was an orbit? Let's look at the situation where the modular group acts on the benali upper plane. Pick one point from the upper f plane. For example, I'll pick this point. And this point can be moved around using modic transformation. For example, um, we had model transformation that sends toao to to + one. Right? So this point can be moved to this point using model transformation. And this point can be also moved to this point because we can move to to plus two. And we can also move this point to this point. And by applying more modular transformation, we get many different points in H. So this point can be moved here. This can be moved here. This point can be moved here by some using a random model transformations. And the set of all orbits that can be reached from the original point by model transformation is called the orbit. So in this picture, points with the same color belong to the same orbit. So the red ones, these are the points that lie in the same orbit. The blue ones, these are one orbit. So what does it mean that the region is a fundamental domain? It means that the region contains exactly one representative from each orbit, ignoring boundary overlap. So this region here, we said that the fundamental domain for a modular group was this region. This region exactly contains one point for each color. So one red here, uh, one purple here, one blue here, one green here. Conversely, if you choose any point inside this fundamental domain, you can think of it as the representative of some orbit, and every point in complex upper half plane. Uh, for example, I would choose, uh, any random point, and this point can be moved inside to the fundamental domain by some modular transformation.
Before we define modular forms properly, we need one more piece of language, and that is congruent subgroups. So far, the main groups we have seen is the full module group, which is especially new acting on the upper plane. But in many parts of number theory, especially when elliptic curves enter the story, we do not always use the full modular group. Instead, we use often, uh, we often use smaller subgroups of special linear group defined by congruence conditions on the entries of the matrix. The principal congruent subgroup of level n is defined, uh, by this. So the collection of all abcds and special linear group that is, uh, congruent to the identity matrix modulo n. And the he subgroup, we have two types. For example, gamma 0 n here is the set of all matrices in special two group, uh, that satisfies c equals z modulo n. The reason we introduced this now is that modular forms do not always exist for the full modular groups. We can also define model forms for congruent subgroups like this comma 0 n, and later this number n will be extremely important because it will match the conductor of an elliptic curve.
Now let's talk about cusps. This is a slightly strange word at first because a cusp is not an ordinary point inside the upper plane. Instead, it's a kind of boundary point that we add when we want to understand the modular group action more completely. The most important cusp to keep in mind is I infinity. This just means that towels go straight upwards in the upper plane with, uh, imaginary towel, reaching infinity. Now let's see why rational numbers appear here. Take a mod transformation gamma to equals a to + b over c to + d. And, uh, here what happens if toao goes to i infinity? When t toao is very large, the dominant terms on the denominator would be c to, and the dominant term on the numerator would be a toa. So roughly, this will approach a over c as t to reaches i infinity. So under modular transformations, the points I infinity can be sent to rational boundary points at this rational number. And if C equals zero, that stays at infinity. So the natural set of cusps, uh, is the union with the rational number set with I infinity. Later, when we define modular forms, we will care not only about how a function behaves inside the point upper plane but also how it behaves on the cusps.
Finally, the definition of modular form. A function that goes from H to C, we call it a meamorphic modular form of weight K for a congruent subgroup H. If number one, F is meamorphic on H. So a meamorphic function is a function that is holomorphic except for isolated poles. So it's allowed to blow up at some isolated points, uh, like for example, 1/z blows up at z equals z, but it's not allowed to have arbitrary bad behaviors. I will not go deeper in this. And the second one is the most important one. Modularity condition for every gamma, uh, from congruent subgroup, this has to hold. So f gamma toa is equal to f a to plus b over c to plus d. So this function value equals C to plus D to the K F to. Condition three is about what happens at the cusp. Uh, remember if we plug T=1 to this modularity condition, we have F to + 1 equals C + D = 1. So F to. So F is periodic with period one. And when we have a period one function, we can write it using a Fourier expansion. Remember, uh, since we said Q has e to the 2 pi i to, and the q expansion is written like this, n is integer, and a n q to the n. Now meamorphic at the cusp means that in Q expansions, negative powers of q's are allowed, but only infinitely many of them. So it will look like something like a minus 3 q minus 3 plus a minus 1 q minus 1 plus a 0 plus a 1 q 1 plus blah blah blah. This cannot go forever in this direction. It cannot have infinitely many negative power terms going downwards forever.
Types of modular forms. A modular form is a meamorphic modular form that is holomorphic everywhere, including at the cusp. And this means that its Q expansion at any cusp has no terms with negative exponents. So it starts with n equals zero. And a cusp form is a modular form that vanishes at the cusp. So let's see what it means at the cusp i infinity. Since Q equals E to the 2 pi i towel, we have Q reaching zero at the constant. So if F has Q expansion, F to equals A 0 plus A 1 Q plus A 2 Q squared plus blah blah blah, all the terms with positive Q powers will disappear, and it will reach zero. So vanishing at the cus means exactly the constant term a z is zero, and that's why at i infinity, the c form condition is, uh, equivalent to saying a 0 equals zero.
Now let's introduce a piece of notation that makes the transformation law much shorter, and that is the slash operator. For a function f, an integer k, and a matrix in a generator group, the slash operator is defined like this. So f k g gamma toao is determinant gamma to the k / 2 and z to plus d to the minus k f gamma. So you might wonder why the determinant appears here. For the modular group, which is a special linear two group, the determinant is always one. So this factor does nothing. But this notation is often used for general matrices as well, not only matrices with determinant one. So the determinant factor is included to make the notation work in that more general setting. Uh, the point of this notation is that it turns the usual modularity condition into a very short equation. Instead of writing f a + b over c + d equals c a + d to the k f to every time, we can just write far k gamma equals f. And there's one property that I want to mention here that is, uh, for every matrix AB from general linear group FAR K A bar KB, uh, so, uh, two slash operator notation, this is actually equals to F bar K A B. So now, let's prove this. So F Bar K A bar KB. This is equal to determinant B K / 2 C B to power to plus D to the minus K DB. So CB and DB here, this means that these are entries from B. Uh, so B are writing B as this. Okay. And we have F bar K A B to. So we have determinant B K / 2 C B to plus DB. And about this fbar k a b beta, we have determinant a k / 2 c a b to plus d a and f a b to. Um, so if we regroup this, we have determinant a over K / two, and we have A B tow, and here CB towel plus DB. And I will write B tow explicitly, which is CA, A B tow plus BB over CB to plus DP plus DA. Okay. Okay. And then we will, uh, compute this, and this will give us, uh, toao c a b plus d a cb d a cb. And about the constant terms, it will be, it will be ca bbb c a bb plus, uh, da db. And on the other hand, we will compute, uh, matrix multiplication a b. So, a a b a c a da b a a bb cb db. So, our goal is to show, so our goal, so our goal here is to show that this is c a b to plus d a b. So, what is c a b here? Uh, this is, uh, second row, first column. So it's c a b plus d a cb. And how about dab? daba b is equal to, uh, second row, second column. So c a bb plus da db. And these are the same. So this is C A B to plus D A B. And if you, uh, consider all these, this is equal to F bar K A. So we, uh, prove this property.
Now we can start asking a slightly more structural question. We have defined modular forms. But what happens if we collect all modular forms of a fixed weight? Do they form some kind of algebraic object, a vector space, maybe a ring? Can we talk about bases and dimension? And the valence formula gives us the first clue here. Roughly speaking, this formula says that if f is a non-zero modular form of weight k, then the total number of zeros and poles of f, counted the right way on the fundamental domain, is controlled by k. So more precisely, the total weighted orders of finishing is k over 12. So let f be a nonzero meamorphic modular form of weight k for the full modular group. And the sum of all orders of zeros and poles of f in the fundamental domain, weighted by the stabilizer orders of the points, is constant, which is k over 12. So a zero here means that a point where the function becomes zero, and a pole is where the function blows up to infinity. Uh, so like if a function f is something like 1/z - 3 cubed + 1/z - 2 squared, here z = 2 and z = 3. They blow up, right? So, uh, we say that, uh, this function has poles in z equals 2 and z equals 3, and the order is, uh, two and three. And here I is, uh, just I, and row is E to the 2 pi i over 3. And this infinity, and this infinity means cusp. These three are in, these three are for whatever reasons treated differently, and they are weighted by one over two and one over three. And these are all the other zeros and poles of, uh, the function. So the important point is not the exact technical form of this formula right now for us. So the important point is not the exact formula right now. The important point is that once the weight K is fixed, the function is heavily constrained. Modular forms have limited freedom. In other words, it starts to make sense to ask for a basis or dimension and an actual linear algebraic structure on modular forms. So the valence formula tells us that modular forms have limited freedom, and in fact, uh, for a given weight k, the spaces of modular forms and cusp forms are dimensional complex vector spaces, and this holds true for all kinds of subgroups, whether the group is the full modular group or the congruent subgroups like gamma zero n. Mk here denotes the space of modular forms, and Sk here means the set of cusp forms, and this is a big deal because now modular forms become something we can study using linear algebra. We can ask, what is the dimension of this space? Can we find a basis, or can we write every modular form as a linear combination of some standard modular forms? And when we talk about modular forms from now on, the subgroup can often be confusing because sometimes the subgroup will be the full modular group, and sometimes it will be a congruent subgroup comma 0 m. They are similar, but there are also many differences. So it's important to distinguish between them. If we write MK and SK without specifically mentioning the associated subgroup, we mean the full modular group. And if we write MK gamma 0 N SK like this, explicitly mentioning the subgroup, then we usually mean a congruent subgroup. And in some cases, we can actually compute the dimension explicitly. For example, when the weight is k=2 and the group is gamma 0 n, there is a dimension formula for the space of cusp forms. So we can literally hand compute the dimensions of this modular group ourselves. Um, of course, I'm not going to derive the formula here. The point is just that these spaces are concrete enough that we can talk about their dimensions. And this is a spoiler, but this will become very important later.
Another structural fact about the space of modular forms is that the space of modular forms can be split like this. So the space of modular forms is decomposed into the Eisenstein series and the cusp form space. Of course, this is only for the full modular group. So what does it mean? It means that if we take any modular form from, uh, the space of modular forms, this can be uniquely written as, uh, CK + G, where G is a cusp form. And if you think about it, this is actually, uh, very obvious. [clears throat] Because earlier we saw that the Eisenstein series EK has constant term one, right? And if f has constant term a z, we can choose c to be a z, and then f minus a z e k has constant term zero. But a mod form with constant term zero is a cusp form. Meaning that f minus a z e k is a cusp form. And that's why every mod form can be split into an Eisenstein part and the cusp form part.
Now think back to what we did in linear algebra. Once we had a vector space, we did not stop there. We introduced an inner product. And once we had an inner product, we could talk about things like length, norm, and orthogonality. We want to do the same kind of thing here. We have these dimensional vector spaces of modular forms like MK or SK. So we want to know if it's possible to define an inner product on this space. Can we say that two modular forms are perpendicular? Can we measure the size of a modular form? And the answer is yes. The inner product we use is called the Peterson inner product. It's defined on the cusp forms. So f and g here are cusp forms, and it's defined by these equations. You do not need to memorize this integral right now. The important part is the role it plays. Now that we have the Peterson inner product, the space of, uh, cusp forms with weight kh becomes more than just a vector space. It becomes an inner product space. So we can talk about geometric ideas inside the space of functions. For example, the norm of a cusp form is defined as the square root of the Peterson inner product with the same, uh, function. And two cusp forms are said to be orthogonal if their Peterson inner product is zero. So the slide says that this gives the space of cusp forms with weight k, uh, the structure of a Hilbert space. Very briefly, a Hilbert space is an inner product space that is complete. Complete means that if a sequence of vectors should converge according to the norm, then its limit always, uh, stays inside the space. It's similar to the analytic constructions of QP. In our case, we don't need to worry too much about the technical condition. [clears throat]
Now, let's look at something a bit different. So far, modular forms were functions on the upper half plane. But once we write them as Q expansions, uh, like this, f to equals the sum of a n q to the n, we got a sequence of numbers, a z, a1, a2, the coefficients of, uh, the q, the coefficients of the powers of q in the q expansions of the modular form. And mathematicians noted there's something interesting. For some very important modular forms, these, these coefficients are multiplicative. So multiplicative means that, uh, f to mn can be split up to f to m f to n for some coprime m and n. For example, we have the mod discriminant here. Here, uh, the coefficients of q squared equals minus 4. I'll just write it as like the function like this. And the coefficients of q c q is 252. So if I multiply this together, a2 times a3, take out your calculator. So minus -4 * 252 = minus, uh, 6048. -648 is here. So it's a6. Uh, maybe we try another one. So a3 times a4. a3 is 252. a4 is -472. And this is equals to 252 * 1472, this is equal to, uh, 370944. 37944. We we found it here. This is exactly a12. 12. So why does this happen? Uh, we don't know. Uh, we just assume that something very systematic is going on here. And on the other hand, uh, modular forms, the space of modular forms and cusp forms, there were dimensional vector spaces. So we want to actually find good basis elements of those spaces explicitly. So we need some kind of tool to solve these problems I just mentioned, and that is where Hecke operators enter.
Now let's pause for a moment and talk about multiplicative functions because this language will keep appearing when you talk about Fourier coefficients. So what is a multiplicative function? A function f that goes from natural numbers to complex numbers is multiplicative if f1=1 and for any two coprime integers m and n, f(mn) = f(m)f(n). So if f is a multiplicative function, if you take like f(12), this is equals to f(3)f(4), you can split like this. Some examples of multiplicative functions are here. So sigma(n), sigma(n) is just a sum of divisors of n. So of course, positive divisors of n. So for example, sigma(6) = 1 + 2 + 3 + 6. tau(n) is the number of divisors of n. So tau(6) would be four because six has four divisors. And phi(n), this is called the Euler totient function. It means the number of integers from one to n that are relatively prime to n. So for example, phi(12), the numbers between 1 to 12 that are relatively prime to 12. So 1, 2, 3, now 4, 5, 6, 7, 8, 9, 10, 11. So, phi(12) = 4. And sigma_k(n) is the sum of kth power of the positive divisors of n. So, um, sigma_k(6) would be 1^k + 2^k + 3^k + 6^k. These are all multiplicative functions. And one nice property is that if f is multiplicative, then g(n) defined using f(n) like this is also multiplicative. So let's prove this. So we want to prove that for any coprime integers m and n, we want to show that g(mn) is equals to g(m)g(n). So let's look at g(mn). This is the sum of f(d) where d is a positive divisor of mn. But if d is a positive divisor of mn, we can write d = d1 d2, for d1 is dividing m and d2 dividing n, because m and n are coprime. So we can write, uh, d1 dividing m, d2 dividing n, f(d1 d2). But f(d1 d2) = f(d1)f(d2) because we know that d1 and d2 are coprime since m and m are coprime, right? And we will change the order of the sum like this: d2 dividing n, f(d1)f(d2). And this is equals to, we can take this out, right? So d1 dividing m, f(d1) times d2 dividing n, f(d2). See what we've got. This is g(m) times g(n). So we prove this formula. Now we can immediately use this to prove some of the examples above for multiplicative. For example, sigma(n). We can write sigma(n) as the sum of these. Right? But f(x) = x is obviously this is multiplicative, right? Uh, since this d is a multiplicative function, automatically sigma(n) becomes multiplicative because of this formula. Uh, same as tau(n). We can write it like this. The sum of one when you have a divisor. Uh, since the constant function f(x) = 1, f(n) = 1 is obviously multiplicative, tau(n) is also multiplicative. And same with sigma_k(n). d to the power of k is multiplicative, right? If I define f(n) = n to the power of k, of course f(mn) = f(m)f(n). So, uh, sigma_k(n) this is multiplicative.
So finally, we look at Hecke operators. Let n be an integer greater than or equal to one, and let M_n be the set of integral matrices abcd with determinant n. So ad - bc = n. The Hecke operator of index n, T_n, um, the definition looks like this. So T_n applied to f equals n^(k/2 - 1) sum, and we sum all these, uh, slash operators. And, uh, now we look at this gamma 0(1) / M_n. So what is this? We know that gamma 0(1) this is just equal to the full modular group, special linear two group, right? This means that we let this gamma 0(1), aka special linear two group, act on M_n by left multiplication, and then we look at its orbits of this action. In the sum, we choose one representative from each orbit and we add the sums. The important fact about Hecke operators is that if m is a modular form. So let f be an element of M_k. Then after applying the Hecke operators, this T_n(m) is also in M_k, meaning that this is also a modular form. So T_n is a linear operator that goes from the space of modular forms to the space of modular forms. So let's show this. So we pick one gamma from special linear group and we want to show that T_n(f) is again a modular form, right? So we compute the slash operator and we want to show that this is again equals to T_n(f). So T_n(f) bar k gamma. This is equals to by the definition n^(k/2 - 1) and we, uh, look at each representative of the orbit gamma 0(1) acting on M_n, k mu k gamma, right? But we know that this is equals to f k mu gamma, right? We proved this. So n^(k/2 - 1) sum f k mu gamma. But here, uh, look at what gamma is doing. Gamma is just reindexing the sum, right? Multiplication by gamma sends M_n to itself because gamma has determinant one. So the term f k mu gamma runs through the same collection of f k mu. Therefore, uh, we can write this like this, and this is, uh, the definition of T_n(f), right? So we prove that T_n(f) is again a modular form of weight k.
So let's rewrite the definitions of Hecke operators. So T_n(f) to this was equal to n^(k/2 - 1) sum gamma 0(1) acting on M_n, and we add f k gamma to. Now this definition we just wrote down is correct. But if we actually want to compute this, this is kind of, uh, painful and complicated because what are we supposed to do? For every n, find representatives in gamma Z_n acting on M_n, then apply the slash operator to each one and then add everything up. That's technically fine, but it's not the way we want to calculate because usually when we use a modular form, we look at its Q expansion, right? So instead of staring at these matrices, we want to add more concrete questions. What does T_n do to the coefficients of the Q expansions? And this is exactly what this formula tells us. The new f coefficients of q to the h is computed like this when the Hecke operator is applied to f. So let's prove this formula. Before we do this, uh, for this computation, we will use a standard set of representatives for the left gamma orbits in M_n. So we will pick gamma from the set [a b; 0 d] where ad = n and b ranges from 0 to d-1. Okay, so let's go. T_n(f)(tau) this is equals to, uh, by the definition of slash, n^(k/2 - 1) and a d should, uh, equal to n, ad = n, and b ranges from 0 to d-1, and we are adding, remember the slash operator notation, so we had first determinant, determinant of the matrix equals a b, and a d equals n. So n^(k/2) and we had c tau + d, right? But c here is zero. So we only have d to the power of minus k. And we have, uh, f(gamma tau), which is a tau + b over c tau + b, but c equals zero here. So it's d. And since we know the Q expansions of f(tau), let's use it to this calculation. So we have n^(k-1) using this and we add for ad = n, b ranges from 0 to d-1, and we have d to the minus k, and we will take this out, and we will change the order of the sum a little bit. So n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. 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So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the 2 pi i tau. So this this thing is q to the power of a tau + b over d, and we have m here. And I want to, uh, simplify this a bit. So we have n^(k-1) ad = n, and we will take the sigma out. So m ranges from 0 to infinity, and we have d to the minus k, and since we move to sigma here, we can take this out. So a_m, q = tau, e to the
Is that the space of modular forms has a basis consisting of normalized Hecke forms? So why is this? Let's let's prove this. Um we know that a modular form can be decomposed into Eisenstein series and the space of cusp forms. So $C_k \oplus \text{cusp forms}$. And we just proved that the Eisenstein series is a Hecke form. So it's enough to prove that the basis of cusp forms are Hecke forms, and this is where the Peterson inner product becomes useful. Uh, we showed that $T_n T_m = T_{mn}$ if $\gcd(m, n) = 1$. So they commute each other, and we also saw that the Hecke operators were self-adjoint with respect to the Peterson inner product, meaning that $\langle T_n f, g \rangle = \langle f, T_n g \rangle$. So from these two, we can use the spectral theorem we saw in linear algebra. So that means that there is a basis of $S_k$ consisting of vectors that are eigenvectors for every Hecke operator $T_n$ at the same time. But that's exactly what a Hecke form is. So we got a basis of $S_k$ made of Hecke forms. And since these Eisenstein series are also Hecke forms, putting the Eisenstein series part together with the cusp forms gives a basis of $M_k$ consisting of Hecke forms.
So at this point, it feels like Hecke forms contain a lot of information about modular forms. It forms a basis for modular forms. So can we do something more with these Hecke forms? Can we take these Hecke forms, uh, play with them a little bit, process them somehow, and build a new mathematical object out of them? If we do that, maybe we can get something that captures the core of modular forms. Remember how we constructed the Riemann zeta function? We define $\zeta(s)$ like this: the sum of the reciprocals of the $n^s$. And this, uh, we wrote it with an Euler product. So $\frac{1}{1 - p^{-s}}$ and we, uh, multiplied for all primes. And the reason this worked was because of unique prime factorization. Every positive integer can be broken into prime powers. And we know that the Fourier coefficients of a Hecke form can behave similarly because it is multiplicative. For example, $a_{84}$ can be split into $a_4 a_3 a_7$. This is very similar to integer factorization. $84 = 2^2 \cdot 3 \cdot 7$. So the idea is, let's do with Hecke form coefficients what we did with the integers in the Riemann zeta function. We take the coefficient sequence $a_1, a_2, a_3, \dots$ and build a series out of it, and that gives an L-function that looks like this: the sum of $\frac{a_m}{m^s}$. We just define the L-functions of a modular form like this: $L(f, s) = \sum_{m=1}^\infty \frac{a_m}{m^s}$. And a nice thing is that this can be expressed as a product of these factors. So why does it hold? So we know that this is the product of $1 + \frac{a_p}{p^s} + \frac{a_{p^2}}{p^{2s}} + \dots$. So we will show that this expression is the same as this expression. To do this, first we will make a substitution. We'll put $p^{-s} = x$. And we will show that if we multiply this with the denominator of this fraction, we get one. So if we write this expression using $x$, this will be $1 + a_p x + a_{p^2} x^2 + \dots$. That's right. Uh, few more terms: $x^3 + \dots$. And we will multiply it by, uh, this: $1 - a_p x + p^{k-1} x^2$. So think that we are expanding this. What is the constant? The constant equals 1, right? And how about the coefficients of $x$? It's equal to $a_p - a_p$. So it's zero. $a_p - a_p = 0$. How about $x^2$? So $p^{k-1} - a_p^2 + a_{p^2}$. So how do we manage with this? But we have the recurrence formula, which was $a_{p^{r+1}} = a_p a_{p^r} - p^{k-1} a_{p^{r-1}}$. So if we put $r=1$ here, this becomes $a_{p^2} = a_p a_p - p^{k-1} a_1$. So $a_{p^2} = a_p^2 - p^{k-1}$. So the coefficient of $x^2$ is $p^{k-1} - a_p^2 + a_{p^2} = p^{k-1} - a_p^2 + (a_p^2 - p^{k-1}) = 0$. So how about the coefficients of $x^3$? The coefficients of $x^3$ will be $a_{p^3} - a_p a_{p^2} + p^{k-1} a_p$. And if we put $r=2$ in this formula, $a_{p^3} = a_p a_{p^2} - p^{k-1} a_p$. So $a_{p^3} - a_p a_{p^2} + p^{k-1} a_p = (a_p a_{p^2} - p^{k-1} a_p) - a_p a_{p^2} + p^{k-1} a_p = 0$. In general, the coefficients of $x^{r+1}$ here will be $a_{p^{r+1}} - a_p a_{p^r} + p^{k-1} a_{p^{r-1}}$. And by this formula, this becomes zero. So, uh, the product of these equals one, which means that we can write this as the reciprocal of this expression.
By far, we were mostly thinking about modular forms for the full modular group. But in the actual modularity theory, we usually need modular forms for congruence subgroups, especially groups like $\Gamma_0(n)$. Hecke operators and Hecke forms we've discussed were actually only for the full modular group. But if we want to talk about congruence subgroups, things get a bit more complicated. We have to introduce new operators like $U_p$ instead of the Hecke operators to handle certain primes. But anyway, we can still achieve a very similar result. So even for congruence subgroups, we define the L-functions of a modular form in a very similar way. Uh, it looks like this. The only thing is that for primes dividing the level $n$, we should handle this, uh, separately. So for primes not dividing $n$, the local factor has the usual quadratic form we just saw. But for $p$ dividing $n$, we use a different local factor that looks like this: $1 + \frac{a_p}{p^s} + \frac{a_p^2}{p^{2s}} + \dots$.
One interesting thing about modular forms is that we can sometimes build modular forms at a higher level from modular forms at a lower level. So let $m$ and $n$ be positive integers such that $m$ divides $n$. So $m \mid n$. Of course, $n$ is larger than $m$, right? And let $d_0$ be a divisor of $n/m$. So $d_0 \mid n/m$. And if $f(z)$ is a cusp form of weight $k$ and level $m$, then the function defined by $g(z) = f(d_0 z)$ is a cusp form of level $n$. So this says that a modular form of level $m$ can be lifted up and viewed as a modular form of level $n$ when $m \mid n$. So let's show this. So we claim that $g(z)$ defined by $f(d_0 z)$ is an element of the cusp forms of level $n$. Right? So we want to show that for every $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(n)$, we want to show that $g(\gamma z) = (cz+d)^k g(z)$, and this is equivalent to saying $f(d_0 \gamma z) = (cz+d)^k f(d_0 z)$. And on the other hand, let's look at this matrix $\gamma' = \begin{pmatrix} a & b d_0 \\ c/d_0 & d \end{pmatrix}$. Uh, if we look at this matrix right away, we know that this is a member of $SL_2(\mathbb{Z})$ because the determinant is $ad - bc$. And since $abcd$ is a member of $\Gamma_0(n)$, we know that the determinant of $\gamma'$ is also 1. And let's look at $c/d_0$. Is this an integer? Yes, because $d_0 \mid n/m$ and $c$ is a multiple of $n$, right? So $c/d_0$ is an integer. And, let's look at this one more time. $c/d_0$. We know that $c$ is a multiple of $n$. So we can write $c = nt$ for some $t$. And we know that $d_0 \mid n/m$. So to use this, we will turn this like this: $mt/d_0$. And we know that this is an integer. So actually, this $c/d_0$ is a multiple of $m$. [clears throat] So $\gamma'$ was actually a member of $\Gamma_0(m)$. And let's try to compute $\gamma'(d_0 z)$ and see what we get. So $\gamma'(d_0 z) = \begin{pmatrix} a & b d_0 \\ c/d_0 & d \end{pmatrix} \begin{pmatrix} d_0 z \\ 1 \end{pmatrix} = \begin{pmatrix} a d_0 z + b d_0 \\ (c/d_0) d_0 z + d \end{pmatrix} = \begin{pmatrix} d_0 (az+b) \\ cz+d \end{pmatrix}$. So $d_0 \gamma z = \begin{pmatrix} d_0 (az+b) \\ cz+d \end{pmatrix}$. And surprisingly, this is equal to $d_0 \gamma z$. Oh, we have $d_0 \gamma z$ here. So we use this to compute this. So using this, $f(d_0 \gamma z) = f(\gamma'(d_0 z))$. And since $\gamma'$ is in $\Gamma_0(m)$, we can use the modularity condition for $f$, since $f$ was a cusp form of level $m$. So this is equal to $(cz+d)^k f(d_0 z)$. So we showed this. So this is proved.
So from the previous slide, we already know that forms from lower levels can be pushed up to level $n$. So we define an "old form" as a cusp form that is constructed from a form of a lower level $m$, where $m$ is a proper divisor of $n$. So if I write this explicitly, an old form is the set of $f(d_0 z)$ when $m \mid n$ and $d_0 \mid n/m$ and $f$ is a member of the cusp forms of level $m$. So old forms are the forms of level $n$ that were already present at a lower level and then lifted up to level $n$ like this. Uh, then what is left? So if we have the cusp forms of $\Gamma_0(n)$ and we have old forms here. So what is left here? We call this a "new form". These are forms that did not come from a lower level, and they are the ones that first appear as cusp forms of level $n$. So we call these "new forms". One clean way to define new forms is using the Peterson inner product. We define $S_k(\Gamma_0(n))^{\text{new}}$ as the orthogonal complement of the old forms. So we can write the space of cusp forms using orthogonal decomposition like this: $S_k(\Gamma_0(n)) = S_k(\Gamma_0(n))^{\text{old}} \oplus S_k(\Gamma_0(n))^{\text{new}}$. And usually, when people say new form, they mean a normalized Hecke form inside the new subspace.
We have made a $\text{Galois}$ representation from an elliptic curve, starting from the elliptic curve $E$. We looked at the torsion points and built the Tate module $T_l(E)$, and then what we have obtained is a $\text{Galois}$ representation. Now, the funny thing is that the $\text{Galois}$ representation can also be constructed from modular forms. So let $f$ be a normalized Hecke form of level $n$ and weight $k$, and that exists a $\text{Galois}$ representation associated with $f$. Uh, here we go: representation. Um, actually, a more precise way to write this representation is $\rho_f: G_\mathbb{Q} \to \text{GL}_2(K_f)$. Here $K_f$ is the number field generated by its Fourier coefficients. So we put $a_1, a_2, \dots$ in the rational number set, and this is what we get. And $K_f(\lambda)$ here is a completion where $\lambda$ is a prime of $K_f$ lying over $l$. But this $K_f(\lambda)$ is just a finite $p$-adic extension of $\mathbb{Q}_p$, which means that we can view it inside this $\overline{\mathbb{Q}}_p$. So it's like a broad way to write the representation, just not 100% accurate. So you don't have to care about all these technical parts. The important point is that a normalized new form gives us a $p$-adic $\text{Galois}$ representation. So there are several ways to construct $\text{Galois}$ representations from modular forms. For elliptic curves, we actually went through the construction: torsion points, Tate modules, and the $\text{L}_l$ representation. But for modular forms, we're not going to do that because it's much harder and complicated. To be honest, I don't know all the details. And it depends on the weight. In weight two, it's related to modular curves, Jacobians, and Shimura theory. And for higher weights, the construction uses $l$-cohomology, and weight one is a special case handled by the theory. What matters for us is that the fact that we can construct a $\text{Galois}$ representation out of a modular form, and this $\text{Galois}$ representation coming from a modular form will appear later in a very important way. So, uh, keep that in mind.
If you have made it this far, then you would probably already know that elliptic curves and modular forms are somehow related. But now think about that for a second. What does that even mean? How can an elliptic curve be related to modular forms? One of them is a geometric curve. The other one is a complex function. They're completely different mathematical objects. So from now on, we're going to become detectives. We're going to look for clues. We're going to investigate the evidence and slowly figure out why these two objects might be connected. The first clue is matching L-functions. About the elliptic side, we saw that the zeta function of an elliptic curve can be approached from two different directions. One was the algebraic perspective when we used ideals and the norms, and the other was the geometric perspective when we counted points over the finite field. On the modular form side, we also have an L-function. If $f$ is a Hecke form with $q$-expansion $\sum a_n q^n$, then $L(f, s)$ is built with the Fourier coefficients $a_n$. And without telling you, these two look exactly the same. And there's this correspondence: this $a_p$ on the elliptic curve side matches to the $a_p$, which is the Fourier coefficient on the modular form side. And here, this $N$ was the conductor of an elliptic curve. And this matches to the $n$, this was the level of the modular forms. And if we particularly take $k=2$, we put $k=2$. This becomes $1 - \frac{a_p}{p^s} + \frac{1}{p^{2s}}$, which becomes exactly the same as the elliptic curve side. So up to this point, elliptic curves and modular forms look like completely different objects. But when we look at their L-functions, the formula looks too similar to ignore. And remember, both for the elliptic curves and the modular forms, the L-function was kind of like a DNA. It packaged the essential arithmetic data of the object onto one function. Of course, having the same shape does not automatically prove that these objects are related, but it's enough to make it suspicious. Two completely different constructions are producing local factors of the same structures. Maybe the elliptic curve and the modular form, especially when $k=2$, have some kind of relationship.
This time, we're going to take a modular form, mess around with it a little bit, and see what comes out. So let $f$ be a normalized weight two new form of level $n$ with rational Fourier coefficients. And by integrating $f(z)$ along a closed path $\delta$ in the modular curve $X_0(n)$. Uh, wait, what is this $X_0(n)$? Um, earlier we saw the upper half-plane $\mathbb{H}$ and we also saw the congruence subgroup $\Gamma_0(n)$. This modular curve $X_0(n)$ is basically when we get, uh, when we quotient the upper half-plane by $\Gamma_0(n)$ and then add the cusps to make it compact. So you can think of this $X_0(n)$ modular curve as the geometric space where level $n$ modular forms actually live. So we integrate $f(z)$ along a path $\delta$, and $\delta$ is in $H^1(X_0(n), \mathbb{C})$. So what is this? Uh, we're not going to study homology here. So you can take this lightly. Roughly, this $H^1(X_0(n), \mathbb{C})$ means the set of closed paths or loops on this modular curve, counted up to continuous deformations. So you can just imagine that $\delta$ is a closed path on a modular curve. So, um, if you do this and integrate $f(z)$ along all paths, you get this set of periods. And in this case, where the coefficients are rational, these periods form a rank two lattice in $\mathbb{C}$, meaning that it looks like the form $z\omega_1 + z\omega_2$. So from the modular form, we just produced this lattice. So $\Lambda_f = \mathbb{Z}\omega_1 + \mathbb{Z}\omega_2$. Now, if we have a rank two lattice like this, we can rescale it. Basically, divide everything by $\omega_1$, and we get $\mathbb{Z} + \mathbb{Z} \frac{\omega_2}{\omega_1}$. And if we substitute $\tau = \frac{\omega_2}{\omega_1}$, we get $\mathbb{Z} + \mathbb{Z}\tau$ in the upper half-plane. And the quotient space $\mathbb{C}/\Lambda_f$ is a compact Riemann space known as a complex torus. So the question is, why is this a torus? A torus is a donut-shaped object, right? Um, in the quotient group, we're looking at $\mathbb{C} \pmod{\Lambda_f}$. Any complex number is equal to $z + m + n\tau$, right? So we can divide the complex plane into many copies of the same parallelogram. So we have the complex plane, and we let this point be 1. And if we let this point be $\tau$, we can divide the complex plane into many copies of the same parallelograms. For example, this point would be $1+\tau$. This point would be $2+\tau$. This point would be $1+2\tau$. This point would be $2+2\tau$. And this is the origin. And points in the same relative position inside these parallelograms are all identified to each other. So, for example, this point here, close to this vertex, is the same as this point, and this point is the same as this point, and this point, these four points are basically the same because they're in the same relative positions in each parallelogram. So we will think of this one parallelogram with vertices $0, 1, 1+\tau, \tau$. And I will cut this fundamental parallelogram, which looks like this: $0, 1, 1+\tau, \tau$. And the rule $z = z+1$ says that basically this edge and this edge are all the same, right? Because this point is the same point as this point because the difference is 1. This point is the same point as this one. So these two edges are the same edge. This edge and this edge are all the same. So I will roll this and make these edges meet. I'll glue these two edges together. So we get a cylindrical object that looks like this. So here the edges meet. We rolled this to make a cylinder. And the other rule $z = z+\tau$ says that these two edges are the same. So these two edges are these two top and bottom cylindrical edges, right? And this says that these two edges are the same. So what do I do? I bend this to make the two green edges meet. I glue the two circular ends of the cylinders together, and what do we get? If I bend this and make the green edges meet, it will be something like this. So I bend this to make the green edges meet, and this is a torus. So that's why this $\mathbb{C} \pmod{\Lambda_f}$ we call it a complex torus. And now on this lattice, I will define a function called the Weierstrass $\wp$-function that looks like this. Uh, we will not go over the definition technically, but this is called the Weierstrass $\wp$-function. You don't have to digest the whole formula. The main thing is that this is built from a lattice which came from a modular form. Now, the $\wp$-function we just defined satisfies this differential equation: $y'^2 = 4x^3 - g_2 x - g_3$. Now, look at this equation. We'll make a simple substitution. We'll let $y' = \frac{dy}{dx}$ and $y = \frac{dy}{dx}$, and we'll put $x$ as $x$. So what do we get? We get $y^2 = 4x^3 - g_2 x - g_3$. And this is an elliptic curve, right? A Weierstrass equation. And the coefficients are also interesting. The numbers $g_2$ and $g_3$ are given by Eisenstein series. These were very famous modular forms. So we started from a modular form $f$ on a modular curve. Then you integrated $f$ along closed paths on the modular curve. And from those periods, we got a lattice. And once you have a lattice, we can take the quotient $\mathbb{C}/\Lambda_f$, that gives a complex torus. Then we can use the Weierstrass $\wp$-function defined from this lattice. And this map sends the complex torus into a Weierstrass equation, which looks like this. And this was an elliptic curve. And even the coefficients $g_2$ and $g_3$ came from Eisenstein series. So modular forms are showing up again in the equations of the elliptic curve. In this construction, we actually started from a modular form and produced an elliptic curve. Of course, this works under special conditions, but it's another sign that modular forms and elliptic curves are tied together much more tightly than we expected.
And the third clue is about the $j$-invariant of an elliptic curve and the modular $j$-function. First, what was the $j$-invariant for an elliptic curve? You could think of it as a name tag for the elliptic curve. Two elliptic curves can be written in different looking Weierstrass equations. But if they're isomorphic over the algebraic closure, then they had the same $j$-invariant. Uh, in the Weierstrass equation $y^2 = x^3 + ax + b$, we computed like this: $j = \frac{1728(4a^3)}{4a^3 + 27b^2}$. And if we use the discriminant, we can also write like this. Now, on the modular form side, there's also something called the modular $j$-function. It's defined like this: $j(\tau) = \frac{1728}{4a^3 + 27b^2}$ where $a$ and $b$ are related to Eisenstein series. Uh, this is a modular function, meaning that it's a modular form of weight zero. And if you look at these two formulas, the two $j$'s feel strangely similar. So same 1728 and perfect cube on the numerator. We just saw two $j$'s: the elliptic curve $j$-invariant and the modular $j$-function on the modular form sides. And there's a theorem saying that these two are actually matched. And there's a theorem saying that these two are actually matched. For elliptic curves defined over a complex number, there exists a unique $\tau$ up to the action of special groups such that these two are the same. So the left-hand side is the $j$-invariant of the elliptic curve, and the right-hand side is the modular $j$-function of $\tau$. So if I take any elliptic curve, for example, $y^2 = x^3 - x$, and if we compute the $j$-invariant of this curve, this is $0$. And this theorem says that some complex number $\tau$ exists in the upper half-plane such that the value of the modular $j$-function is equal to the $j$-invariant of the curve. So in this case, the complex number we're looking for is $i$. So $j(i) = 0$. This works for any elliptic curve. So I will take $y^2 = x^3 + 1$. The $j$-invariant of this elliptic curve is $0$. And $e^{2\pi i / 3}$ exists such that the value of the modular $j$-function, if you compute this, this becomes $0$. And it's not only for these examples. No matter which elliptic curve you choose, there's always a corresponding $\tau$ in the upper half-plane that satisfies this condition.
The fourth clue is about $a_p$. So we already saw them in the L-function formula. On the elliptic side, we had this $a_p$, and on the modular form side, we have $a_p$ or $a_m$. They appeared in exactly the same positions inside the local factors. So it already feels like there might be something going on. So let's quickly recall what they mean. On the elliptic side, the $a_p$ comes from point counting. If $E$ has good reduction, and if $E$ had bad reduction at $p$, it was defined as $-1, 0,$ or $1$ depending on the reduction type. And on the modular form side, this $a_p$ was basically the $p$-th Fourier coefficient of the $q$-expansion. So again, these two numbers come from totally different places. One comes from counting points on an analytic curve modulo $p$, and the other comes from the $q$-expansion of a modular form, but they sit in the same positions on the L-functions, and that's a clue that we want to keep following on.
So this is the Ramanujan-Petersson conjecture. Uh, actually, it's not a conjecture anymore. It's now a theorem proved by Deligne, but nobody calls it Deligne's theorem. So, uh, I just keep the name here. So this says for a normalized Hecke form $f(z) = \sum_{n=1}^\infty a_n q^n$ of weight $k$ and level $n$, $|a_p| \le 2p^{k/2 - 1}$ for every prime $p$ not dividing $n$. Now let's try plugging $k=2$ in this theorem. Uh, why? Because earlier, when we plugged $k=2$ into the L-functions, the L-functions of the modular forms and elliptic curves became similar. So at this point, we are already suspecting that weight two modular forms analytically might somehow be related. So we put $k=2$, and what do we get? $|a_p| \le 2p^{2/2 - 1} = 2p^0 = 2$. And we've seen this before, right? Think back to the Hasse theorem for elliptic curves. For an elliptic curve defined over rational numbers with good reduction at $p$, we define $a_p = p+1 - \#E(\mathbb{F}_p)$, and this $a_p$ was bounded by $2\sqrt{p}$. And this was the Hasse theorem. So it's the same shape again. The Fourier coefficients $a_p$ of the modular form and the point counting error $a_p$ of an elliptic curve satisfy the same kind of bound. Here's a concrete example at conductor 11. This elliptic curve, this looks like this: $y^2 + y = x^3 - x^2$. And this elliptic curve has conductor as 11. Which means that the only bad prime is 11. On the other hand, we have a weight two, level 11. Uh, again, this weight two and this 11 is the same as the conductor of the curve. Anyway, we have this modular form, and its $q$-expansion is like this: $q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - \dots$. Which means that the Fourier coefficients are $a_1=1, a_2=-2, a_3=-1, a_4=2, a_5=1, a_6=2, \dots$. So we expect the Fourier coefficients of the modular form and the $a_p$'s, the trace of Frobenius on the elliptic curve, to have some kind of relationship. So let's actually compute these $a_p$'s on the elliptic side and then compare with each other. Okay. Uh, so we have our elliptic curve here. First, we will take $p=2$, which means that we're going to look at both sides modulo 2. So $y^2 + y = x^3 - x^2$. And this is modulo 2. So, uh, if $x=0$, the right-hand side becomes 0. If $x=1$, right-hand side becomes 1. Uh, $y=0$, the left-hand side becomes 0. And if $y=1$, the left-hand side becomes $1^2+1 = 2$, which is 0 mod 2. So, uh, no matter what values you put in $x$ and $y$, the right-hand side and left-hand side are always 0. Which means that the number of points is $2 \times 2 + 1 = 5$. This one is for the point at infinity. So the number of points is 5. And $a_2(E) = p+1 - \#E(\mathbb{F}_p) = 2+1 - 5 = -2$. Uh, we'll take $p=3$. So $y^2 + y = x^3 - x^2$ modulo 3. So if we put $x=0$, the right-hand side becomes 0. If $x=1$, $1-1=0$. If $x=2$, $8-4=4$, and $4=1$ modulo 3. Uh, about the $y$: $y=0$, $0$. $y=1$, $1+1=2$. $y=2$, $4+2=6$, and $6=0$ modulo 3. So the right-hand side and left-hand side should be the same, modulo 3. So again, $2 \times 2 + 1 = 5$. So $a_3(E) = 3+1 - 5 = -1$. Uh, if I take $p=5$, $y^2 + y = x^3 - x^2$ modulo 5. Put $x=0$, $0$. $x=1$, $1-1=0$. $x=2$, $8-4=4$. $x=3$, $27-9=18$, which is 3 modulo 5. $x=4$, $64-16=48$, which is 3 modulo 5. Uh, about the $y$: $y=0$, $0$. $y=1$, $1+1=2$. $y=2$, $4+2=6$, which is 1 modulo 5. $y=3$, $9+3=12$, which is 2 modulo 5. $y=4$, $16+4=20$, which is 0 modulo 5. So again, the both sides should be both zero, right? So the possible number of points equals $2 \times 2 + 1 = 5$. So $a_5(E) = 5+1 - 5 = 1$. And notice that this $-2$, $a_2(E)$, this is actually the same value as the second coefficient, which is $a_2=-2$. And $a_3(E)$ is also the same as $a_3$. $a_3$ is $-1$ here. $a_5(E)$ is also the same as $a_5$. And if you try this out for more primes, 7, 11, 13, and so on, this always holds. I mean, this is pretty wild, right?
So this is the point where our suspicion starts turning into something much more closer to conviction. So far, we have looked at clues suggesting that elliptic curves and modular forms are connected. First, the trace of Frobenius $a_p$ on the elliptic side and the $p$-th coefficients from the modular form side. These numbers come from completely different constructions, right? But they keep appearing in the same role. And the second, matching L-functions. This is basically the same as matching of the $a_p$'s, but the L-functions also seem to be related. And third, the conductor of an elliptic curve and the level of the modular form seem to have some kind of relationship. And there's one more thing: the modular form connected to the elliptic curve seems to have weight two. So $k=2$. We saw this in the shape of L-functions, we saw in the Ramanujan-Petersson bound, and it becomes exactly the same as Hasse's bound, and we just saw it again on the conductor 11 example. So the clues are pointing towards something like this: an elliptic curve defined over $\mathbb{Q}$ should correspond to a weight two modular form with level equal to the conductor of $E$, and with $a_p$ equal to $a_p$ for the primes where everything behaves well. So this is exactly where the clues are pointing to. This is the Modularity Theorem. Probably the most important theorem in the video. Um, every elliptic curve defined over $\mathbb{Q}$ with conductor $N$ is modular. So what do we mean by an elliptic curve being modular? This means that there exists a normalized weight two new form with a level the same as the elliptic curve's conductor, such that the L-function of the elliptic curve is equal to the L-function of the modular form. So all the clues we saw become one theorem. The $a_p$'s from point counting match the Fourier coefficients on the modular form side. The conductor $N$ becomes the level of the modular form, and the relevant modular forms have weight two, and the two functions are actually exactly the same. Historically, Wiles proved this in 1994 for semi-stable elliptic curves, and that was enough for Fermat's Last Theorem. The full theorem for all elliptic curves over $\mathbb{Q}$ was proved by these four brilliant mathematicians in 2001. The proof of the theorem is known to be astronomically difficult. It's on a completely different level from what we're doing in this lecture, and to be honest, I do not really know the proof either. So we are not going to prove the Modularity Theorem here. We'll accept this as a theorem and we will use this.
Now, the elliptic curves and modular forms were connected through L-functions. But now we are going to change the object we compare. Instead of comparing L-functions, we're going to compare $\text{Galois}$ representations. Remember back in the modular form parts, we said that modular forms can also produce $\text{Galois}$ representations. So recall one of the construction methods. This theorem is about modular forms of weight two. So let $f$ be a normalized new form of weight two and level $n$ and trivial character. Don't get too obsessed with this trivial character; we're not going to go through it. Let $K_f$ be the number field generated by its Fourier coefficients. Um, this $K_f$ is basically a field you obtain by adjoining the Fourier coefficients $a_1, \dots$ of $f$ to the rational number field. And for each prime ideal $\lambda$ lying over a rational prime $l$, this lying over means that the intersection of $\lambda$ and $\mathbb{Z}$ is the multiples of $l$. Then there exists a continuous irreducible $\text{Galois}$ representation $\rho_f: G_\mathbb{Q} \to \text{GL}_2(K_f(\lambda))$ that looks like this. And this representation is unramified at primes not dividing $n$. And these equations hold: $\text{Tr}(\rho_f(\text{Frob}_p)) = a_p$ and $\det(\rho_f(\text{Frob}_p)) = p^{k-1}$. If this looks technical, which it really is, I try to remember one thing. So roughly, this says that from a weight two modular form, we can extract an $l$-adic $\text{Galois}$ representation, and the Frobenius trace of that $\text{Galois}$ representation is the same as $a_p$, the Fourier coefficients of the modular form. If it's hard, don't get too obsessed with this; it won't hurt the main flow. Remember that from a weight two modular form, we can extract a $\text{Galois}$ representation that satisfies this condition.
So the Modularity Theorem can also be stated using $\text{Galois}$ representations. On one side, we have an elliptic curve $E$ defined over $\mathbb{Q}$, and using the Tate module, we can extract an $l$-adic $\text{Galois}$ representation $\rho_E$. And on the other side, we have a modular form $f$, and using this Eichler-Shimura theorem, we also get an $l$-adic $\text{Galois}$ representation $\rho_f$. And they meet together. So this is the algebraic way to state the Modularity Theorem. So the Modularity Theorem says that for every elliptic curve $E$, there exists a modular form $f$ such that their associated $\text{Galois}$ representations (we make this using the Eichler-Shimura theorem) are isomorphic to that of the elliptic curve. So actually, there are three ways to approach the Modularity Theorem. The first one is about L-functions. So for every elliptic curve defined over $\mathbb{Q}$, there exists a new form $f$ such that the L-functions of the two are the same. And the second one is about $\text{Galois}$ representations. For every elliptic curve defined over $\mathbb{Q}$, there exists a new form $f$ such that their associated $\text{Galois}$ representations are isomorphic to the $\text{Galois}$ representation made by the elliptic curve. Um, and the third one is about modular curves. For every elliptic curve $E$ defined over $\mathbb{Q}$, there exists a non-constant surjective rational map defined over $\mathbb{Q}$ that looks like this. We're not really going to talk about this. For us, the $\text{Galois}$ representation version, the second one, will be the most useful. So the point of this slide is that they're talking about the same thing, the Modularity Theorem, just in different perspectives.
So now let's see why these two statements are basically equivalent: the L-functions and the $\text{Galois}$ representation statement. First, we'll go upwards, meaning that we will assume that the $\text{Galois}$ representations are isomorphic and will prove that the L-functions are the same. So suppose we know that $\text{Galois}$ representations match, then for every good prime $p$, the Frobenius matrices have the same trace, right? This is obvious because the $\text{Galois}$ representations are isomorphic. So we went here from here. And about this equality, we know this from the Eichler-Shimura theorem. And about this equality, we know this from the characteristic polynomial. Remember? And this equality is from the Eichler-Shimura theorem. So if the representations are the same, the traces are the same, and we have the same $a_p$'s, and if we have the same $a_p$'s, we have the same local L-factors, and eventually we get the same L-functions. The reverse direction, which is going downwards, is a little more technical. If the L-functions are equal, it means that their local factors are all the same. So we have all the $a_p$'s becoming the same. That means the Frobenius traces of the two $\text{Galois}$ representations agree for all primes. And to turn this trace equality into an isomorphism of $\text{Galois}$ representations, we use two standard results. So the first one is Chebotarev density theorem. Russian mathematician, but I don't know how to pronounce his name. So anyways, this tells us that roughly, Frobenius elements give enough information about the $\text{Galois}$ group. And the second one is this is called the Ribet's theorem, which says that a semi-stable representation is determined by its trace. We haven't learned about semi-stable representations, so we're not going deeper than this. Anyway, so these two theorems imply that the equality of these traces gives the isomorphism of the $\text{Galois}$ representations. So the main point is that the equality of the L-functions forces the $\text{Galois}$ representations to be isomorphic, which means that saying that L-functions are the same and saying that two $\text{Galois}$ representations are isomorphic is basically equivalent.
Now we move on from $l$-adic representations to mod $p$ representations. This says that for an elliptic curve $E$ defined over $\mathbb{Q}$, the Tate module gives us a $p$-adic representation, right? And if we reduce this matrix modulo $p$, this gives us the residual representation. We wrote it like this: $\bar{\rho}_E: G_\mathbb{Q} \to \text{GL}_2(\mathbb{F}_p)$, and we call this a residual representation or a mod $p$ representation. We already saw this idea in the elliptic section where we constructed the $l$-adic representation. So the object has changed. We are now looking at the full $l$-adic representations. We are looking at its reduced versions modulo $l$ or modulo $p$ here. And this is the version that matters for the next part. Serre's conjectures and Ribet's theorems are about these residual mod $p$ representations.
Now we are going to think in the opposite direction. So far, we saw that a modular form can give us a $\text{Galois}$ representation like this. We can make a $\text{Galois}$ representation out of a modular form. The Eichler-Shimura theorem we just saw was one example of that. There, however, he asks the reverse question: Suppose we start from a $\text{Galois}$ representation, more precisely, a mod $p$ residual $\text{Galois}$ representation. Can this representation come from a modular form? So Serre took that question seriously and made it into a conjecture. It's actually now a theorem by these mathematicians. So let this residual $\text{Galois}$ representation $\bar{\rho}: G_\mathbb{Q} \to \text{GL}_2(\mathbb{F}_p)$ be a continuous irreducible representation. Then this is modular, meaning that there exists a new form $f$ and a prime ideal $\lambda$ of the coefficient field such that the associated $\text{Galois}$ representation of this new form is isomorphic to the given residual representation. So historically, Serre formulated this in the 1980s, especially in his 1987 paper, and at the time as a conjecture. And much later, these two mathematicians proved it, and the proof was published around, I think it was 2009.
Serre did not stop here. He did not only say that a mod $p$ representation should come from some modular form. He also predicted which modular form it should come from in terms of its level and weight. About the level, he predicted the level $n$ of the modular form is the Artin conductor of the residual $\text{Galois}$ representation. Remember, Artin conductor, this was related to ramification, right? And then he predicted the weight is determined by the local behavior of the representation at prime $p$. So Serre's conjecture had two parts in spirit: first, he said that a suitable mod $p$ representation should come from a modular form. So this is the existence part: starting from $\bar{\rho}$, we should be able to find some modular form $f$. Second, Serre did not stop at existence; he also predicted the weight and the level of the modular form. The level should come from the ramification of the residual representation, and the weight should come from the behavior at prime $p$. Roughly speaking, Ribet's theorem is about the second part when the first part is guaranteed. So if we already know that some residual representation is modular at some level $n$, then under the right conditions, Ribet tells us that we can lower the level to the conductor predicted by the representation itself. So it says, for prime $p > 3$, let this residual $\text{Galois}$ representation $\bar{\rho}$ be a continuous odd irreducible $\text{Galois}$ representation. Assume that this $\bar{\rho}$ is finite at $p$, and assume that this $p$-adic $\bar{\rho}$ is modular of weight two and level $N$. Modular means that there exists a weight two new form $f$ such that the associated $\text{Galois}$ representation is isomorphic to the given residual representation. If we let $N(\bar{\rho})$ be the Artin conductor of the residual representation, then $\bar{\rho}$ is also modular of weight two and level $N(\bar{\rho})$. That is, there exists a weight two new form with level the same as $N(\bar{\rho})$ and of course, the associated $\text{Galois}$ representation is isomorphic to $\bar{\rho}$. So this says that if a mod $p$ representation is already known to be modular at some level $N$, then under the right conditions, its level can be lowered to the conductor of the representation itself. So the level is lowered from $N$ to $N(\bar{\rho})$. That's why it's called the level lowering theorem.
So now we finally start the engine for the proof of Fermat's Last Theorem. We want to prove that there exists no solution, no non-zero solution to this equation $x^n + y^n = z^n$. And to show that this equation has no solutions, we only need to consider the case where the exponent $n$ is prime. This is because any counterexample for a composite exponent would already give a counterexample for a prime exponent. That is why we'll consider only odd primes that are greater than 5. So we will assume a non-trivial primitive solution $(a, b, c)$ to the Fermat equation. So $a^p + b^p + c^p = 0$, and $a, b, c$ are all primitive. And if you look at this closely, you know that $a, b, c$ cannot be all odd numbers because if they were all odd numbers, the left-hand side would be odd, which leads to a contradiction. So one of them should be even, and exactly one of them should be even because $a, b, c$ are primitive. So without loss of generality, we name that even number $b$. So $a^p + c^p = -b^p$. And $b^p$ is divisible by 4 right, because $p$ is greater than 4. And for odd numbers, $a^p \equiv a \pmod 4$ and $c^p \equiv c \pmod 4$. So if we take this and bring here, $a^p + c^p \equiv a+c \pmod 4$, and we know that this is 0 modulo 4. So one of $a$ and $c$ is 1 modulo 4 and the other is -1 modulo 4. And without loss of generality, we let $a \equiv -1 \pmod 4$. So we write here $a^p \equiv -1 \pmod 4$.
Now from this hypothetical Fermat solution, we attach a Frey curve. This is a Frey curve, and this $a^p$ and $b^p$ we bring it from here. All right. And this is an elliptic curve defined over $\mathbb{Q}$. The equation we just wrote $y^2 = x(x-a^p)(x+b^p)$ is a standard Frey curve equation. Uh, there's nothing wrong about this, but it is not yet the global minimal model for primes other than two. For example, 3, 4, 5, 7, blah blah blah, this equation, this already works as a local minimal model. So away from 2, we are fine; we can use this as a minimal model. But the unknown part is prime 2. So at 2, we need to change variables to get a globally minimal model. So we let $x = X$ and $y = Y$. And if we put this, we get this equation. And this is a global minimum model. And if you compute the discriminant of the global minimum model, it looks like this: $\Delta = 2^{-8} a^{2p} b^{2p} c^{2p}$. So what we want to do is that we want to apply Wiles' theorem to the Frey curve. Of course, today we know the full Modularity Theorem, so we could just apply the modularity directly. But historically, Wiles proved the semi-stable case first, and for the proof, that's exactly what we need. So let's respect the route and check the Frey curve is semi-stable. What does semi-stable mean? At the first place, it means that for every prime $l$, the curve has either good reduction or multiplicative reduction. Additive reduction is not allowed. Okay. And we already had a useful criterion for the reduction type for a minimal model at prime $l$. We could just compute the $l$-adic valuation of the discriminant and $c_4$ and determine the reduction type. So now we will apply this criterion to the Frey curve for primes other than two. This is a local minimal model, right? So we can use this equation. If you compute the discriminant and $c_4$, it looks like this. So, so we'll use this criterion. We'll bring this. What happens if $l$ does not divide $abc$? If $l$ does not divide $abc$, the $l$-adic valuation of the discriminant is 0 because $l$ does not divide the discriminant. Which means that if the prime $l$ does not divide $abc$, we have good reduction. We are good. And if $l$ divides $abc$, the $l$-adic valuation of the discriminant is greater than zero. And how about $c_4$? This is zero. Uh, because the fact that $l$ divides $abc$ means that $l$ divides either $a$ or $b$ or $c$ because $abc$ are primitive. Right? Uh, for example, if $l$ divides $a$, and if $l$ both divides $a$ and $c_4$, and $c_4$ was $16(a^{2p} + b^{2p} + c^{2p})$. So $l$ divides $l$ divides these two terms, which means that $l$ has to divide $16 b^{2p}$.
2P. And because B is prime and L does not divide two, it means that L should divide B. But this is a contradiction because A and B are primitive. Uh so that's why V LC4 is zero. So four primes dividing ABC we have a multiplicative reduction multiplicative. And if L is equal to two uh if we compute the valuation the two added valuation of the discriminant this is this is greater than 2 P minus A right because uh B was an even number so this is this is greater than zero and How about V2 C4? This is zero, right? Because only B was even number and we know that A was an odd number. Uh so this is even this is odd. Hence why uh this is zero. So if L equals 2, we have multiplicative reduction. So in conclusion if L equals 2 or L divides ABC we have multiplicative reduction and otherwise good reduction. So free curve is semi-stable.
Now what we want to do is extract a g representation from a freay curve. Uh we started from a hypothetical from a solution a to the p plus b to the p plus c to the p equals z. And from this hypothetical solution we made a freay curve that looks like this. We just check that the curve is semistable which means that this curve is modular. Now from this elliptic curve freay curve E we can build a pical representation. This look like this using uh the P addict Kate module and by reducing module P we get a residual mod P representation and this residual representation is the object we are going to study now remember where we started we assume that a non-trivial FMA solution exist to prove FMAS theorem we need to get a contradiction from these assumptions so the plan is um take this mo presentation coming out from frav analyze it properties and show that these properties force something impossible.
So the next thing we're going to do we will going to analyze this property of the residue free representation. Uh we can ask is it irreducible? What is it conductor for this slide we will treat part of the argument as a black box? Uh the whole point of slide is to show that the prey representation is irreducible and we will use this uh property that saying the residue P representation is reducible is equivalent to saying that that exists a Q rational cyclic P isoggeni on the elliptic curve. Uh so what is an isogeny? Isogen is basically a uh it's like a rational functions defined between lip curves but we will not go deeper. So you can just uh take this given and basically Mazour's isoggen Ethereum says that this can happen in only fitly many primes and these are the list of those primes. So what mathematicians did uh they analyzed for all of these primes and they found out that uh these are impossible which means that we could exclude all of these primes. So this residual uh fre representation is irreducible for all primes greater or equal to five.
Now we want to compute the conductor of this residual fre representation and we introduced here a very useful tool called the naron arc chef thorich criterion. It connects the geometry of the elliptic curve to ramifications of the g representation. So let E defined over Q be an elliptic curve. And for a prime other than L, the periodical representation is unrammified at L if and only if E has good reduction at L. So this means that we can determine the ramification of the representations using the reduction type of the associated elliptic curves. So this is very very useful. Now we know that good reduction at L means that L does not divide the discriminant of E. And since the discriminant of E looks like this, the only possible bad primes are the ones that divides this, the ones that divides 2 ABC. Uh but there's one thing to be careful about. This criterion is about the periodical representation and what we are focusing is the residual mod P representation. Uh those two are are different, right? uh still this is very useful because if the pi addic representation is unrammified at l uh then the mod p representation is also un ramified at l. So other than p dividing 2 abc we are good. We know that both paddic and the mod p are unrammified. So we need to check the prime dividing 2 abc. Uh now we check the conductors prime by prime. We only need to check primes dividing 2 ABC and unfortunately we will leave most of them as black box because this is very difficult and technical. So the first case where we look at primes dividing ABC and we will exclude 2 and P. So the free curve this is multiplicative reduction at such L. But if you do the uh hard part, the hard analysis shows that the residue representation however this is actually unrammified at L. So the n value this is zero. And about prime two if I only tell you the result the representation for a representation is ramified a prime two and the exponent is one. And since this is mod p representation we will separate prime p uh apart from the conductor. And since the fray representation is only ramified at prime two uh the conductor of the fre representation is two two to the^ of one. So this is everything you know about residual f representation.
So this is the free representation we are looking at which came from uh the free curve and first of all the free representation is modular. Why? Because the free cave was semi-stable, the attached pod representation was semi-stable. And if we reduce in modular P, the modularity still holds and also is irreducible uh is mainly due to the module's Ethereum and we could rule out the remaining primes and uh this is also continuous odd infinite NP. uh we're not really going to talk about this although it's uh very important and lastly the art conductor uh was two meaning that this representation was only ramified at prime two so this residual f representation has exactly the properties we need for ribet's theoreium this is modular irreducible odd and yet conductor 2 finally finally proof of feras theorem so we start off By assuming there exists a solution to the FMAT equation a to the p plus b to the p plus c to the p equals zero. We assume that b is even and a c odd and of course p was a prime greater or equal to five. And from this hypothetical solution we constructed a fray curve. This was a lit curve and this was semi-stable. uh we actually checked this by computing the all valuations of the discriminant at C4 and we applied the stability uh semi-stability criteria. Uh so this freay curve is semi-stable. Hence we could use the wilds ethereum and also we could attach a dollar representation to this freay curve by using a tape module and this was a pi color representation and on the other hand because this curve was semi-stable we can use this theorem which tells us the existence of this new form f. This new form has weight two and a level same as the conductor of the freight curve. And this associated g representation was isomorphic to the gal representation that was from uh the freight curve and the connection between these associated gal representation and this new form was given through the equil shimmerra theorem. So we have these isomorphic pi color representations and once we have this podic color representation we can reduce it modulo p and we get a residual mod p representation and it it was attached to the p uh torsion points p toion group of the freay curve. Uh we call this a freay representation which is modular. uh it was because Fre was semi-stable and uh it was because wild's theorem and uh while obviously we did not talk about this but it was odd continuous at finite at p and it was irreducible it was mostly due to masur's isogen ethereum and lastly the conductor of this representation was two meaning that only uh prime two uh the representation was ramified. So this representation for representation satisfies every condition for us to apply the Ribet's theorem. Ribet's level learning theorem says the representation stays the same modular P but the modular form can be moved down to smaller level and that smaller level is the conductor of the residual representation. And in our case, since the conductor of the re representation is two, rubet lowers the level all the way down to two. Therefore, there must exist a new form G with way two and level two such that the associated god representation is isomorphic to the fray representation. So we know that G uh comes from this space uh S2 gamma 02 cusp form with weight and level both two. But remember earlier we learned how to compute the dimension of this kind of space. So this was a formula. So this formula tells us the dimension of the space s2 gamma 0 n. So let's put n equals 2 and see what happens about mu. Uh mu becomes two. P divides 2 1 + 1 / p and since p is prime uh p can only be two. So new becomes two. 1 + 1 over two which is three. About new two. Uh again p could only be two. So this is 1 + -4 2. Uh this is a legandre symbol. This is a notation from number theory, you can just Google that up. If you compute this, this becomes zero. So 1 + 0 this is one. How about new three? 1 + 3 2 this becomes minus one. So new three if you add these two this becomes zero. And how about new infinity? D divides two and then you add uh five gcd d and 2 over d. Uh d can be one or two. And the gcd of d and 2 overd is one. So this is equal to 251 and 51 is one. So this becomes two. And we put all these value into this formula. So the dimension dimension of S2 gamma 02 this is equals to 1 + 32 - 1 / 4 - 0 over 3 - 2 / 2 and what do we get? This is zero. uh the dimension is zero. Um there's no room here. No cost form, no new form, nothing. But Ribbit's theorem said that such new form exist. So we have reached a contradiction. And now the entire chain collapses backwards. If the new form does not exist, then the residual representation could have not exist in this way. And if that representation could have not existed, then the free curve could not have existed in the first place. And therefore the fair solution at the very beginning was impossible. So after all this machinery after lip curves after modular forms after representations after walls and riets we finally arrived at the conclusion. There are no [music] positive integral solution to the equation a to the n plus b to the n uh equals c to the n. Fair's theory is proved. Heat. Heat. [music] [music] >> [music]