 This lecture is part of an online graduate course on Galois theory, and will be about finite fields. What we're going to do is we're going to use the splitting fields that we discussed earlier in order to classify finite fields. In particular, we're going to use the following magic polynomial, x to the p to the n minus x, where p is prime, and n is an integer greater than or equal to 1. And the classification of finite fields we will get is that they all have order p to the n for some prime and some integer n, and for each prime power p to the n, there's a finite field that is as unique up to isomorphism. And you may remember that splitting fields, well, the splitting field of a polynomial is in some sense unique, but in some sense it wasn't really. There was a certain amount of ambiguity about what you mean by the splitting field, and we will find this problem turning up when we try and write down finite fields explicitly. Anyway, let's first of all do the easy stuff. So we just recall that fields have a characteristic, and the characteristic of a field is either 0 or it's a prime greater than 0. And the characteristic is just the kernel of the map from z to the field. Remember, the kernel must be a prime ideal, so it's either generous by 0 or by a prime. And if the field is characteristic 0, then it contains the rational number, so it can't possibly be finite. And if it is characteristic p, then it contains z modulo pz, which is the finite field of order p, sometimes denoted by fp. And now we notice that if f is finite, this means that it must have characteristic p for some p, so the finite field of order p is contained in f. This means f is a vector space over the finite field of order p, and if the dimension is equal to n, then this means the order of f is equal to p to the n. So it's easy to see the order of any finite field must be a power of some prime. Well now we've got to show that for any given prime power there's a unique field of that order. Let's start just by reviewing a bit of notation. So as usual, we'll use p to be a prime. This is the characteristic we're going to work over. And q is often used to mean a power of p. q is equal to p to the n, where n is a positive integer. The finite field of order q will be denoted by gfq, that stands for Galois field, because Galois was one of the first to investigate them. It's also denoted by fq, and of course it can also be denoted by gfp to the n, or fp to the n. So all of these denote, well as I said, we can't really talk about the finite field of order q, the usual ambiguous problem, so these are ways of denoting a finite field of order q if you're not worried too much about exactly which notation for it you're using. So let's first do existence. So I want to construct a finite field of order p to the n, and that's easy. I'm going to take the, well, not the a splitting field of x to the q minus x. So let's call the splitting field l, so the finite field of order p is contained in l. And now what I'm going to do is look at the roots of x to the q minus x. And what I'm going to show is the set of these roots is actually a field. To do this, all I've got to do is to show if it's closed under addition, multiplication, subtraction, and division. Well, multiplication is easy because of x to the q equals x and y to the q equals y, then obviously xy to the q is equal to xy. Addition is a little bit trickier. If x to the q is equal to x and y to the q is equal to y, then x plus y to the q is also equal to x plus y. And this follows because you remember that over a field of characteristic p, a plus b to the p is equal to a to the p plus b to the p because all the binomial coefficients p choose i equal to 0 for nought less than i less than p. And the same is true if p is a prime, is a power of q. So this is closed under addition and multiplication and therefore it's a field. We can also check that all the roots are distinct. So you remember a polynomial has distinct roots if it's co-prime to its derivative. So let's look at the polynomial x to the q minus x. Well, its derivative is qx to the q minus 1 minus 1, which is equal to minus 1 because q is equal to 0 over the finite field. So it's certainly co-prime to its derivative and all its roots are distinct, so it has q distinct roots. So the set of roots is the splitting field of x to the q minus x because all roots of x to the q minus x are in the set of roots and the set of roots is a field and it has order p to the n. So we found a field of order p to the n. By the way we can say a little bit more about the construction of this field. What is really going on is that we have a Frobenius endomorphism, that's right it is phi, where we write phi of a is equal to a to the p. Now if p equals nought in your ring, then phi of a b is phi of a times phi of b and phi of a plus b is phi of a plus phi of b. So phi is a homomorphism from your ring to itself. And if we look at the roots of x to the q minus x equals nought, this is equivalent to saying that phi to the n of x is equal to x because phi of x is just x to the power of p, so if you compose that n times this is just saying x to the q is equal to 0. So the roots of this are the fixed points of the endomorphism phi to the n. Well it's actually an isomorphism and this sort of explains why the roots of this are closed under addition and multiplication. It's because they're really the fixed points of an automorphism of something. So now we want to show uniqueness and to do this all we need to do is to show that any finite field of order q is a splitting field of x to the q minus x. So all we need to show that all elements a of the finite field are roots of x to the q minus x because this will immediately show that it's the splitting field of it because we found q different roots of it. So for a equals 0, this is trivial, a equals 0 is obviously a root of this. So a is not equal to 0, you have to think for a few seconds and you notice that a is in the group of non-zero elements of the field which is order q minus 1. And by Lagrange's theorem if a is an element of a group of order q minus 1 it means that a to the q minus 1 is equal to 1 and now we can just multiply it by a and we find a to the q is equal to a so in either case a is a root of this. So this shows that any two finite fields of order q are isomorphic because we showed that any two splitting fields are isomorphic. So this actually shows another way of thinking where this funny polynomial comes from. This polynomial is just the polynomial whose roots are exactly the elements of the finite field. Well next we have the following problem. The problem is to construct a finite field of order q equals p to the n. Well haven't I just done that? I mean I said we can construct the finite field as taking the splitting field but the problem is as we said talking about the splitting field is a little bit ambiguous. So suppose I take the splitting field of x to q minus x over fp. Well what on earth are its elements? How do we write down an element of this or how do we multiply them together? You'll see we can't it's ambiguous. So how do we actually write down an explicit finite field? What we do is we choose an irreducible polynomial f in fp of x of degree n and then we form a finite field gfq to be fp of x modulo f. So this will give us a field of order p to the n. So we actually have to find some irreducible polynomials. And let's just do this for p equals 2 and construct a few small finite fields. So let's recall what the irreducible polynomials over the field with two elements are. Well if you want to find primes there's a well-known method using the sieve of erotosthanes. You just write out all numbers in order and cross out the ones which aren't primes and you can do exactly the same thing with polynomials over a finite field. It's an easy exercise and if you do this for p equals 2 you find the following irreducible polynomials. So degree 1 we get those. Degree 2 the only one we get is this. Degree 3 we have that and xq plus x squared plus 1. Degree 4 we get x to the 4 plus x plus 1. x to the 4 plus x cube plus x squared plus x plus 1. And x to the 4 plus x cubed plus 1. So let's use these to construct some finite fields. Let's start with order 4 which is 2 squared. Well this is going to be f2 of x. We take polynomials over the finite field and we quotient out by a degree 2 irreducible polynomial and we don't have a whole lot of choice in this because there is only one irreducible polynomial of degree 2. So here's our finite field of order 4. Now it's usual to use the element omega instead of x for this. The reason is that x cubed is equal to 1 and it's traditional to use omega as a cube root of 1. So we usually write this as f2 of omega modulo omega squared plus omega plus 1. And the elements of the finite field of order 4 are therefore 0, 1, omega and omega plus 1. And you can add them using the fact that 2 omega is 0 and you can multiply them using the fact that omega squared is equal to omega plus 1. So this gives a completely explicit description of the finite field of order 4. So let's look at the finite field of order 8 which is 2 cubed. And we're going to take f2 of x modulo of polynomial and this time we've got a bit of choice. We could take one of these two polynomials and let's do x cubed plus x plus 1. Well this means that it's got 8 elements which can be represented by all polynomials to degree at most 2. So we can just write out the element 0, 1, x, x plus 1, x squared, x squared plus 1, x squared plus x, x squared plus x plus 1. So here are all the 8 elements of a field of order 8 and we can multiply them together by multiplying polynomials and then reducing using the fact that x cubed is x plus 1. Well that's not the only way we could construct a finite field of order 8 because we could instead use this polynomial x cubed plus x squared plus 1. So we could take f2, and I'm going to use y instead of x to avoid getting confused. So let's take f2 of y over y cubed plus y squared plus 1. So here's another finite field of order 8. And you see they don't actually look the same. The multiplication kind of looks different because y cubed is y squared plus 1, not y plus 1. However we know that these two must actually be isomorphic because there is only one finite field of order 8 up to isomorphism. So what's an isomorphism between them? And do you notice it's not actually immediately obvious how to get an isomorphism? Well it's not too difficult to find. For instance if we look at the element y plus 1 and cube it, we get y cubed plus y squared plus y plus 1. And now y cubed is y plus 1 so this is equal to y. So y plus 1 cubed plus y plus 1 plus 1 is equal to naught. So y satisfies, so y plus 1 satisfies the irreducible polynomial of x. So if we want an isomorphism between these two finite fields, we can just take x to y plus 1 and this will give you an isomorphism between them. So now let's have, you see this is getting to be a little bit of a problem. What we would really like is a standard ISO finite field. So what does this mean? Well ISO is the International Standards Organization and they do a lot of useful things like setting standards for sizes of screws or whatever. For instance one well-known one is trying to write dates. So if you try to write the January 2nd 2021, there are two ways of writing it. You can write it as 1-2-2-0-2-1 or you can write it as 2-1-2-0-2-1 depending on which country you're in. And this is really annoying because if you see this you're never really quite sure what the date is. So the International Standards Organization has come up with a standard way of writing dates which is 2-0-2-1-0-1-0-2. And now this is completely unambiguous because all the numbers occur with the most significant ones first. So thousands are more significant than hundreds, so two occurs before zero. But years are more significant than months and months are more significant than days and so on. And you also notice we don't miss out the zero here for much the same reason we don't miss out the zero here. It's just kind of silly. So this is a clearly better way of writing dates than either of these. So it's the canonical way to write dates. And what we want is something similar for finite fields. So we can ask what is the canonical finite field of order p to the n. We want to find the best possible one. And to do this we need to choose an irreducible polynomial over fp of degree n. And we need to choose the best possible one and it's sort of the same way this is the best possible way of writing dates. You can sort of think of it as follows. Maybe there are mathematicians on Alpha Centauri who are choosing the best possible irreducible polynomial. And we want to make sure that we're choosing the same one as they're choosing without consulting them so that I don't know whenever there's a galactic standards organization comes along it makes sure that we're using the right irreducible polynomial. So let's try and do this for the field of order 16. So let's try 2 to the 4. So there are three possible irreducible polynomials we could have. We could have x to the 4 plus x plus 1, x to the 4 plus x cubed plus x squared plus x plus 1, or we could have x to the 4 plus x cubed plus 1. And which of these is best? And let's write down some reasons. Well, what are the advantages of this one? Well, it's symmetric. If we change x to x to the minus 1, it sort of looks kind of similar. Another nice property it has is that its roots are the four primitive fifth roots of 1. So, you know, if we're choosing an irreducible degree 4 polynomial, we're choosing four special elements of the finite field. So we need to choose the most special four elements. And, you know, there's an obvious collection of four elements, which are the fifth roots of 1 in the field. Well, what about this one? Well, this is the first in lexicographic order. So the ISO committee to decide this, you know, there'll be someone called Aaron Ardvark or something who's very keen on lexicographic order, and he's going to vote for this polynomial here. Meanwhile, what about this polynomial? Well, I don't know, maybe there's some called Zach Zygmunt who really resents lexicographic order because it means his name is always lost on joint papers, and he doesn't like lexicographic order. So he points out that actually lexicographic order is merely a convention and you could equally well write the polynomial as 1 plus x cubed plus x to the 4. And this is kind of better because it allows you to apply the order to power series as well as polynomials. And he points out that in this sort of reversed order, this polynomial here comes first, so it's clearly the best polynomial to choose. So which of these is really the best one? And the answer is I really don't know. There doesn't seem to be any particularly good reason for choosing any one of these polynomials over the other two. And there just doesn't seem to be a nice canonical way of choosing a standard finite field. As I said, this is related to the problem that splitting fields aren't really unambiguously defined. You have this problem even for degree 2 polynomials over finite fields. So suppose you want to find an irreducible polynomial of degree 2, say. You might try x squared minus a. So you want to find a number a in a finite field of order p that's a that is not a square. And there are plenty of numbers in a finite field of order p that are not a square. In fact, about half of them are not squares. And it's pretty easy to find ones that aren't squares, but it's very difficult to do so in a really neat way. And you could say, well, let's just choose the smallest positive integer that's not a square. Well, yeah, that works, but it's not really satisfactory because if p is 3 mod 4, then it might be a bit difficult finding the smallest positive integer that's not a square. But there's a really obvious negative integer that's not a square, which is minus 1. So maybe you should choose minus 1 whenever p is 3 mod 4. And already you see it's getting a bit complicated and messy. That just doesn't seem to be a really neat way of doing this. So sorry, I'm afraid there isn't a standard finite field of order p to the m. Well, we notice that the polynomial x to the q minus x, the roots, are all elements of fq. And this allows us to do things like find the number of irreducible polynomials of given degree. So let's, instead of giving a theory about this, let's just solve the following problem. How many irreducible polynomials of degree 6 over f2? Well, it wouldn't be too difficult just to list them all, but what I want to do is to have a sort of lazy way of figuring this out without actually doing any calculation. So first of all, you notice that fp to the m is contained in fp to the n if and only if m divides n. This is fairly easy. So m must divide n because this is a vector space over this field. And conversely, if m does divide n, then you can construct this as being the set of roots of x to the p to the m minus x inside this field. So let's have a look at the field of order 64. So let me write it out. And altogether, it's got 64 elements. And inside it is the field of order 2 cubed. So here we have f2 cubed, which is 8 elements. And inside it, we also have the field f2 squared with 4 elements. And these overlap in the field f2 with 2 elements. And so there are 4 different sorts of elements inside the field with 64 elements. First of all, there are the 2 elements in the field of order 2. And these satisfy an irreducible polynomial of degree 1. Then we get these elements here. There are now 6 elements that are in the field of order 8, but not in the field of order 2. So there are 6 elements whose irreducible polynomial is degree 3. And there are 2 elements whose irreducible polynomial has degree 2. And let's count up how many are left. Well, we have to take 64 minus 8 minus 4 plus 2, which is 56. So here we've got 56 elements. So 54, I can't do the addition. Whose irreducible polynomial has degree 6. And this gives the factorization of x to the 64 minus x, because it's equal to a product of 9 polynomials of degree 6 times, where we've got 1 polynomial of degree 2 coming from this field here, times, well here we've got 2 polynomials of degree 3. And we've got 2 polynomials of degree 1. And you see we're getting 9 polynomials of degree 6 because we've got 54 elements. And each polynomial of irreducible polynomial of degree 6 has 6 of these elements as its roots. So altogether we see there are 9 polynomials of degree 6 because you take all the elements that are in F64 but not in any other field and divide that by the degree of the polynomial. So you can write out the factorization of this polynomial explicitly. I'm not going to do it for 64 because that's a bit big. So instead I'll do it for the field of order 16. So you see that x to the 16 minus x is going to be the product of the irreducible polynomials of degree 1 coming from the finite field of order 2 times the irreducible polynomial of degree 2 times the irreducible polynomials of degree 4, which are these three. And again the picture will be something like this. Here we've got the field of order 16 and inside it is the field of order 4 and inside that is the field of order 2. And there are 2 elements here and 2 elements here and 12 elements in F16 but not in F4 coming from the 3 roots of this and the 3 roots of this and the 3 roots of that. Okay, so that's enough about finite fields for the moment, although they'll be turning up quite a bit later on in the course. The next couple of lectures we're going to talk about separable and normal extensions of fields in preparation for defining Galois extensions of fields.