[SOUND] [MUSIC] We still have to prove that Di surjects to the Galois group of k1 and ki over Fp. So we may of course assume that i is equal to 1. By the primitive element theorem, we can find a z such that Fp(z) is k1, such that z generates k1 over Fp. Now, by Chinese remainder, We can find y in A such that y is in Ji for i different from 1 and y is equal to z modulo j1. Let's consider polynomial Q, Which is the product of our g in the group. G in the Galois group of X- g(y). This is a polynomial with integral coefficients. We know that the coefficients are G invariants. This means that these are rationals and then they are also integral over z so they are contained in z since z is integrally closed. Now, let's study Q bar which is Q modulo J1. If g is not in D1, then there exists an i such that g(Ji) = J1. And we have y which is in Ji, so in particular, g(y) is contained in J1. So for g in D1, the correspondent factor of Q bar will be equal to x. For such g, x- g(y) reduction modulo g1 is equal to x. So we have Q bar. This is an element of Fp(x). The product over the complement to D1 of x's times the product over g1 of x- g(y) modulo J1. So this thing. Has z as a root. And D1 acts transitively. On its roots. But now, recall that z generates k1. This means that an element of the Galois group of k1 over Fp, Is determined by the image of z. And we have an element of D1 which sends z to any possible image of it. Of D1 sending z to any given possible image of it. So this means, of course, that D1 surjects to the Galois group. I have to prove the second part of the theorem now. In this part, we assume that D bar has no multiple roots. So alpha one and so on, alpha n were roots of P, denoted by alpha one tilde and so on, alpha n tilde, roots of P bar. It's not alpha i tilde, but rather alpha i bar. Of course which I wrote here, so denote like this the reductions of those roots modulo J1. And let G in D1 act as identity on k1. Then of course, g of alpha i bar are equal to alpha i bar. But go now back to the roots of P. g of alpha i is also a root of P. So it belongs to the set of alpha j's. Alpha one and so on, alpha n. And it cannot be alpha j different from alpha i, since these two would have different reduction modulo J1. Reduction modulo J1. So g(alpha i) = alpha i for any i, which means that g is identity. So, conclusion, D1 to the Galois group of k1 over Fp is an isomorphism. And finally, the same argument, Gives that the Galois group of k1 over Fp with alpha i bar adjoined. Is trivial, then by Galois theory, of course, we have k1 = Fp(alpha 1 bar and so on, alpha n bar). So the theorem is completely proved. Now, how does one apply these to the study of Galois groups? One uses this theorem to construct elements of a certain type in the Galois group to show that the Galois group is large. So let P be an irreducible polynomial and suppose that there is a prime, Such that P is also irreducible modulo this prime. Irreducible, then, The Galois group of p contains a subgroup, Which is isomorphic to the Galois group of P bar. So this is irreducible of degree n. This is irreducible of degree n. But we know the Galois groups of finite fields, and we conclude that this Galois group contains an n cycle. Because such a group, the Galois group of P bar, is a cyclic group generated by an n cycle. Sometimes, there is no such prime, but of course, a variant of this argument exists also in other cases. Suppose, for instance that P is irreducible of degree 5 and that P bar = R2R3 where Ri is irreducible of degree i. Then the same argument, Gives that Galois group of P contains, The permutation 12, and then 345, up to a numbering of rows. For some number in a fluid, there will be such an element in the Galois group. And in this way one can construct elements of particular type in the Galois group and use this to show that those groups are very large. At this point, I finish my lectures. [SOUND] [MUSIC]