[SOUND] [MUSIC] So let me continue my example about the discernment. So let's say 1c. In general, if P is a polynomial, With roots in K bar, which are x1 and so on, xn, then the discriminant of the Polynomial P is of the product over i less than j, Xi- Xj squared and if you take G, which is the Galois group of P, you'll see that this data is preserved by all permutations, right? G is a subgroup of sn since it permutes the root. And any permutation, any permutation, Preserves delta, since, well It preserves each factor up to sine, so I have taken squares so this q sin has well because in a permutation so any element of Galois group preserves delta, it means that delta is an element of k actually. Now take the root of delta. This is the product of X I- X G. Something which depends of course. On the order in which I choose my roots. But let's fix some order. Then, this square root of delta is preserved only by even permutations, is preserved by even permutation. Even permutations. So proposition the Galois group is a sub group of even permutations group if and only if the square root of delta is an element of K. Since if the Galois group is even then, this will be preserved by an element of Galois group and so will be in K and conversely, if it is an element of K, then it must be preserved by the Galois group, but we know it is preserved only by even permutations. This is something which I forgot to write. It's preserved by even permutations and not by odd permutations. The odd permutations will change the sign of my square root of delta. Okay? So coming back to the case of degree three. We can now answer the question about the Galios group. Because we can compute the discriminate of the discriminate. Of x cube plus px plus q. Of course I can always q the coefficient of x square via variable change. So I may assume that my cubic polynomial is of this form, x cube plus px plus q the discriminant is, Easy to compute. This is minus 4p cubed minus 27q squared. All of this will be on exercise. Which you probably already know. So, what do we do? We look at the discriminant so e of delta is a square. In k. Then our [INAUDIBLE] group. Is a three, so sigma of order three. The group of even permutations on three elements is of course a cyclic group of order three generated by a cyclic permutation and if not then the Galois group of B is the full semetric group S 3 so it's a group of six elements. A non commutative group of six elements. And so you can also see in this case the subextention of the subextention of this. So one can see. The sub fields, the sub extensions of this written field. Of b over k. In the first case there is none, we don't have any nontrivial sub group of a flame. This is not in case one, in the first case. And sum in the second case. And sum, in the second case. We know that these subextensions are the same things as subgroups of Galois group. So there are three Of degree 3. Well, first by transpositions, this will be just K(x1), K(x2) and K(x3). These are fixed by non normal subgroups, subgroups of s 3 or order 2. You see if they are of degree three over k, then m is of degree two over them. So they must be fused by a subgroup of order two. So by a transposition. [SOUND] Of roots. And there is one extension of degree two, and one quadratic sub extension Fixed by A3 and S3, and this is of course K of the square root of delta. Galois correspondence tells us that there is no other sub extension. Because those sub extensions correspond objectively to subgroups of the Galois group. And in this case, it does not have so many subgroups. These are just three subgroups of Fourier 2 generated by transpositions, and one subgroup of Fourier 3 generated by a three cycle. Okay, example two, finite fields. We have seen that the Galois theory of finite fields is very very easy. In the Finite case, all Galois groups are cyclic. So Fq to the power n over Fq. So the Galois group Is cyclic generated by the Frobenius map which takes X to x to the power q. Okay? What is more interesting, in the case of infinite extensions over finite field. For instance, algebraic closure. Consider [NOISE] See fp bar is an extension. Of fp. Or if we take the invariance by Frobenues. In variance by Frobenius or by cyclic group generated by Frobenius. This is, of course, a theme. So if we had the exact analog of galore correspondence in the infinite keys we would conclude that the galore group of f p bar over f p was cyclic generated by Frobenius. But here are the Galois group of f p bar over f p is not cyclic generated by the Frobenius. So in this case there can be no bijective correspondence between sub fields and sub rule. In particular, If the Galois correspondence is not bijective. So how to see it, that the Galois group is not cyclic. What is the structure of this Galois group? Well, in fact, already a smaller group is not cyclic. Let us consider. Say, Fp in Fp squared in and so on. Fp to the power of two, to the power of n and so one. And we'll let school L The union of all those. I claim that already the Galios group of L over FP is not cyclic But I will explain this after a break. [NOISE] [MUSIC]