Our next goal is a theorem you probably already know. The theorem saying that the multiplicative group of a finite field is cyclic. To make it instructive I will prove it in a slightly more general version. So let K be a field. And G, a finite subgroup of the multiplicative group of K, then G cyclic. And let us prove this. The idea is to compare G and the cyclic group of order N where N is the cardinality of G. So, let me denote psi(d) the number of elements of order d in G. So we need to prove that psi(N) is not 0, and we know that N is equal to the sum of those psi(d), and let me denote phi(d) the number of elements of order d in Z/NZ. But Z/NZ is a cyclic group, and it contains a single subgroup of order d, which is itself cyclic. So, as subgroup of order d, which is itself cyclic. So, as Z/NZ contains a single (cyclic) subgroup of order d for each d dividing N. Namely this is the subgroup generated by N over d, so Namely this is the subgroup generated by N over d, so Namely this is the subgroup generated by N over d, so the one generated by N over d. phi(d) is also equal to the number of generators of Z/dZ. This is well known to be the number of numbers This is well known to be the number of numbers between 1 and d - 1, which are prime to d. So we know, of course that phi(n) is nonzero, so I claim that either psi(d) is 0, or psi(d) is equal to phi(d). This is of course sufficient since the sum of the phis and of the psis must be equal. So, proof of the claim. If there is no element of order d, then of course psi(d) is 0. If there is no element of order d in G, then psi(d) is zero. If there is one element of order d in G then this x is a root of the polynomial x^d -1. of the polynomial x^d -1. And if you look at all the rules of such a polynomial you see that they form a group of this polynomial form a cyclic subgroup of G. So G as well as Z/NZ has So G as well as Z/NZ has a single subgroup of order d, or cyclic (which is cyclic) Or no such subgroup at all. Or no such subgroup at all. So this is a comparison between G and Z/NZ, but the difference is of course that Z/NZ has such a subgroup for any d dividing N. And G not necessarily a priori. Okay, so this implies that if, this in fact implies the claim. If phi(d) is nonzero, then there is such a subgroup. And the number of elements of order d, phi(d), is the number of generators of that subgroup. Well, obviously this is the same thing as phi(d). So whenever psi is nonzero it is equal to phi, okay? In particular psi(d) is always less or equal then phi(d), but in fact there must be equality, because the sum of those are equal. So the sum of psis is equal to the sum of phis, and then the psis all must be equal to phis. In particular, psi(n) is nonzero. Well, and we have proved the theorem. There is an element of order N in G, so G is cyclic. Okay, this has a couple of corollaries about field extensions. Corollary 1 is that if K is an extension of F_p of degree N, so K is F_(p^N), then is generated by one element. There exists an alpha such that K is equal to F_p(alpha). Well, you might object and say this has already been shown. We have seen that F_(p^N) is a stem field of any irreducible polynomial of degree N, but this corollary is actually stronger because when we've been discussing those stem fields we did not say that such polynomials existed. So, in particular, there exists an irreducible polynomial of degree an irreducible polynomial of degree N over F_p, okay? This is the thing which we don't know yet, but the proof is of course very easy, since we know that the multiplicative group is cyclic it suffices to take a generator of this group. So for the proof it suffices to take for alpha a generator of K*, okay. Corollary 2, the group of automorphisms Corollary 2, the group of automorphisms of K, well, F_(p^n) over F_p of K, well, F_(p^n) over F_p is cyclic generated by the Frobenius map. where Frobenius map F takes x to x^p. As we have already remarked this is a field automorphism. Well, let me prove this. Of course, X^(p^n) is X for any x in Of course, X^(p^n) is X for any x in F_(p^n), this we have already seen several times. So, Frobenius to the power n is the identity. On the other hand the order of this Frobenius is exactly n. Since if m is strictly less than n, Since if m is strictly less than n, then F^m is not the identity because x^(p^m) - x = 0 has only p^m roots. And p^m is strictly less than p^n. So it cannot take all elements of our field to itself. Okay. Well, and finally F_(p^n) is F_p of alpha finally F_(p^n) is F_p of alpha where alpha is a root of irreducible polynomial of degree n, so it cannot have more than n automorphisms. Well, this alpha goes to another root of P Well, this alpha goes to another root of P Well, this alpha goes to another root of P under an automorphism. So there can not be more than n of them. So the cardinality of this group is at most n. So the cardinality of this group is at most n. Well, then it is n and the group is cyclic generated by F.