So let me continue my example about the discriminant. So let's say 1c. In general, if P is a polynomial, with roots in K bar which are x_1 and so on x_n then the discriminant of the polynomial P is the product over i less than j x_i minus x_j squared and if you take G, which is the Galois group of P, you'll see that this Delta is preserved by all permutations, right? G is a subgroup of S_n since it permutes the root. And any permutation, any permutation, preserves Delta, since, well, it preserves each factor up to sign, and I have taken squares, so this kills signs. well, because any 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. Up to a sign this is the product of x_i minus x_j. 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 subgroup 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 any element of Galois group and so will be in K, and conversely, if it is 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 discriminant, the discriminant of x^3 + px + q of course, I can always kill the coefficient of x^2 by a variable change. So I may assume that my cubic polynomial is of this form, x^3 + px + q the discriminant is easy to compute. This is -4p^3 - 27q^2 . Well, this will be an exercise which you probably already know. So, what do we do? We look at the discriminant, so if Delta is a square in K, then our Galois group is A_3, so cyclic 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 P is the full symmetric group S_3 so it's a group of six elements. A non-commutative group of six elements. So and you can also see in this case the subextentions of the splitting field so one can see the subfields, the sub-extensions of the splitting field of P over K. In the first case there is none, we don't have any nontrivial subgroup of A_3. so, none in case one, in the first case. And some 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. fixed 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 3 over K, then M is of degree 2 over them. So they must be fixed by a subgroup of order 2. by a transposition of roots. And there is one extension of degree 2, and one quadratic sub-extension fixed by A3 in 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 bijectively to subgroups of the Galois group. And in this case, it does not have so many subgroups. There are just three subgroups of order 2 generated by transpositions, and one subgroup of order 3 generated by a three cycle. Okay, example two is 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 F_(q^n) over F_q. So the Galois group is cyclic generated by the Frobenius map which takes x to x^q. Okay? What is more interesting, is the case of infinite extensions over finite field. For instance, algebraic closure. Consider F_p bar as an extension of F_p. If we take the invariants by Frobenius Invariants by Frobenius or by cyclic group generated by Frobenius. This is, of course, F_p. So, if we had the exact analog of Galois correspondence in the infinite case we would conclude that the galore group of F_p bar over F_p was cyclic generated by Frobenius. But here 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 subfields and subgroups. In particular the Galois correspondence is not bijective. So, how to see that the Galois group is not cyclic? What is the structure of this Galois group? In fact, already a smaller group is not cyclic. Let us consider Say, F_p in F_(p^2) in and so on. F p to the power of 2, to the power of n and so one. And let us call L the union of all those. I claim that already the Galios group of L over F_p is not cyclic But I will explain this after a break.