 This lecture is part of an online Galois theory course and will be about the Frobenius endomorphism, or possibly automorphism. So first of all, we just look at the case of finite fields. So suppose r is a ring of characteristic p. In other words, p is equal to zero, where p is some prime 2, 3, 5, whatever. Then we know that in this ring, if we define phi of a to be a to the p, this is actually an endomorphism of the ring because a plus b to the p is a to the p plus b to the p, and a b to the p is of course a to the p times b to the p. So for finite fields, so if you look at the finite field fp to the n, then phi is an automorphism. It's obviously injective and the field is finite, so it must be a bijection. And we saw earlier that it generates the Galois group. For general rings or even fields, the Frobenius map need not be an automorphism. For instance, if you look at the field k of x for some field k of characteristic p, then the image is contained in k of x to the p, which is definitely smaller than the field k of x. So this map is not in general an automorphism of a field. Well, it seems as if the Frobenius automorphism or endomorphism is only used for characteristic p. I mean, if you're in characteristic zero, then this construction just doesn't make sense. However, there is a way of getting a Frobenius automorphism in characteristic zero. And this gives one of the very few easy ways of constructing explicit elements of Galois groups. I guess the other easy way is to use complex conjugation, which is a sort of analog of the Frobenius automorphism for infinite primes. So let's do the easiest case. Suppose we have a Galois extension of q, so q contained in m. And we're going to assume that m is a finite extension. And the problem is to find elements of this Galois group. Well, we suppose that m is given by polynomials. So we take polynomials over q and quotient out by some polynomial f of x. And we're going to assume that f of x has integer coefficients and has leading coefficient one. So it's equal to x to the n plus a n minus one x to the n minus one. So one plus a naught, where the a i's are integers. And the idea is we want to reduce m mod p. Well, this makes no sense because p is invertible in m. So there's no central way to reduce m mod p, where m is a field of characteristic zero. However, instead of defining m to be the rationals module of this, we can define an integral form of m to be z of x modulo f of x. And you can see that this is now a free z module as a z module of rank n. And now we can reduce this mod p. So, for example, we might take f of x equal x squared plus one here. And then m would be the Gaussian, the field of Gaussian numbers. But here, z of x modulo x squared plus one is just the Gaussian integers z of i, which we can reduce modulo primes. We're also going to assume for simplicity that all roots of f are in the ring r. And the second assumption we're going to make is that we assume f is separable modulo p for some prime p. So what does this mean? It means it's got no repeated roots modulo p. So we recall from that a polynomial f has a discriminant, which is a product of the differences of the roots possibly squared. And you see, this is none zero if f has no repeated roots. Now, in the case of our polynomial f, the discriminant is some element of the integers because we can write it as some polynomial in the coefficients of the polynomial f with integer coefficients. So its discriminant is in z and all we're asking is that p does not divide the discriminant. So in number theory, we say that p is un-ramified. So if you've seen the phrase un-ramified in a number theory course, it's essentially this condition that it doesn't describe the discriminant. There are some slight differences. We're not bothered to assume r is integrally closed and so on, but never mind that. And now we can form the ring r modulo p, which is going to be fp of x modulo f of x. Now, we notice that f is irreducible over z by assumption, but need not be irreducible mod p. For instance, if we take f of x equals x squared plus one and we take p equals five, then f factors as x plus two, x plus three mod five. So if we reduce the ring r modulo p, we can ask what it looks like. Well, f factors in fp of x has f1x, f2 of x and so on. And notice these polynomials are distinct and in particular they're co-prime because we assumed that f is irreducible modulo p. So by the Chinese remainder theorem, fp of x modulo f splits as a product of fp of x modulo fi of x. So this is just a product of finite fields. If f has multiple roots mod p, then more complicated things happen. We don't want to worry about that now. Moreover, if alpha one up to alpha n are the roots of f in r, then we can look at the images in fp of x over fi of x. And they must be distinct because we assume that f is separable modulo p. So in particular, any automorphism of the ring r, which was z of x over f, must act, so it permutes the fields fp over fi. And if it fixes, so if it maps say fp over f1 to itself, it must act non-trivially on fp over f1 unless it is the identity on r. And this is because if it's not the identity, it's a non-trivial permutation of the roots of f in r and must therefore be a non-trivial permutation of the roots of f in fp because we can identify the roots of f in r with the roots in here. Similarly, any two automorphisms of r mapping say fp over f1 to fp over fi, so that should be fp of x, that should have been an fp of x, it should all have been fp of x. So if it maps that to fi, if these are distinct automorphisms, must be distinct homomorphisms from fpx over f1 to fpx over fi. Well, what's the point of all this? Well, now we're going to count the automorphisms of r. Well, the number is going to be at most the number of maps from f1 to fi, where we sum over all i, which is going to be at most the sum of the degrees of the fi over the constant field fp, which is equal to n, which is the degree of f, because the product of the fi is just r modulo p, which has dimension n. On the other hand, the number of these automorphisms is equal to the order of the Galois group of qx over f, which is equal to the degree of f, because we assume this field was a Galois field. So all of these must actually be equalities, because we're assuming the extension was Galois. So what can we deduce from the fact that these are equalities? Well, first of all, all the fields fp of x over fi of x are isomorphic, because there must be the maximum number of maps from f1 to fi must actually be the degree of fi, which is positive. And secondly, we see that the subgroup of the Galois group mapping, say, fp of x over f1 of x to itself maps onto the full automorphism group of fp of x over f1 of x. In particular, we now get to the key result we really want. The Frobenius automorphism, which maps a to a to the p of, say, fp of x over f1 of x is the image of some element of the Galois group of q of x over f of x. Because we said that any automorphism of this field has to lift to some automorphism of r, and these are all induced by automorphisms of the Galois group. To summary, under the assumptions we put on f, we get an automorphism fi of r, which was z of x over f of x, or for that matter, q of x over f of x, for each factor of f modulo p. Here, as usual, we have to assume that p doesn't divide the discriminant of r and all the other assumptions we put on it. So this automorphism fi is also called a Frobenius automorphism. Notice that it doesn't raise every element to its p's power because that's only an automorphism mod p. However, it does raise elements to a p's power modulo p. So let's see an example of it. So here's an example. Let's take f of x equal x squared plus 1, so roots are plus or minus i. And let's find primes p with f separable. And that's quite easy. Any p not equal to 2 will do. For p equals 2, this polynomial becomes inseparable. For any odd prime, you can see its derivative is a multiple of x. So what does the automorphism do? Let's call fi p the Frobenius automorphism corresponding to p. Well, it must map i to plus or minus i because these are the only two roots of x squared plus 1. Now the elements plus or minus i map to the finite field f p x modulo some factor of x squared plus 1 mod p. They sort of map injectively to this. And on this finite field, the Frobenius automorphism takes a to a to the p. So in particular, these fourth roots of unity are going to map to fourth roots of unity in this field. And in this field, it must map i to i to the p. So it must have the same action here. So fi p of i is equal to i to the p. And let's see what consequences this gives. So we've got this Frobenius automorphism fi p of q of i maps i to i to the p. So let's see what it does. Well, if p is congruent to 1 mod 4, the Frobenius automorphism p of i is equal to i because i to the 5 or 9 or whatever is just i. If p is congruent to 3 mod 4, then the Frobenius automorphism of i is equal to minus i. So it's complex conjugation. Well, if the Frobenius automorphism fixes i, then it acts trivially on the field where we've reduced c of i modulo something. So this implies that x squared plus 1 splits mod p because the field corresponding to any factor of this must have its Frobenius automorphism acting trivially. So it must just be the finite field of order p. On the other hand, if 5p is minus i, the Frobenius automorphism is non-trivial on this field. So this implies that x squared plus 1 does not split. Well, x squared plus 1 splitting mod p means minus 1 is a square. And x squared plus 1 not splitting means minus 1 is not a square. So from properties of this Frobenius automorphism, we've got the result that minus 1 is a square mod p for p odd is equivalent to saying p is congruent to 1 modulo 4, which is a well-known result you come across in a number theory course. So here we see it's a consequence of looking at the Frobenius automorphism of the polynomial x squared plus 1. Okay, next lecture we're going to be looking at cyclotomic polynomials, which have roots that are roots of unity. And we're going to be using the Frobenius automorphism in order to prove that the cyclotomic polynomials are irreducible over the rationals.