So the Galois correspondens. Let's take a Galois extension. Let L over K be a Galois extension. The group of automorphisms of L over K is called by definition the Galois group. And is denoted by Galois of L over K. Theorem by Galois So, if L is finite over K, then there is a bijection between the subfields of L, sub-extensions of L and the subgroups in the Galois group. sub-extensions correspond bijectively to the subgroups. H in Galois. So, how do they correspond? If we have a subfield F, then we just send it to the Galois group of L over F. L over F is Galois, because it's separable and normal. It's F, which is not necessarily Galois over K, okay? And then, in the other direction, if we have H, then we send it to its fixed field L^H. The second part. As I just said, F is not necessarily Galois, but one can say when it's Galois. So the second part, F is Galois over K if and only if for any g in Galois of L over K g preserves F globally g of F is F, so if g fixes F pointwise than this means of course that F is equal to K, but here we ask for something weaker, we ask that g fixes F as a set globally. Okay, and this is also equivalent to saying that the Galois group of L over F is a normal subgroup in Galois of L over K. In this case g goes to g restricted to F is a surjection from Galois of L over K to Galois of F over K and the kernel is just Galois of L over F. Okay, so let's prove it. In fact the first part has been more or less done, right? So, L fixed field by Galois of L over F is F, this has been done. for any separable extension. For separable extensions. Then also H is a part of the Galois group of L over L^H. This is obvious by definition. But also by Artin's theorem. They have the same cardinality. The degree of L over L^H is equal to the cardinality of H. But this degree, as we have seen, For normal extensions they have as many automorphisms as to the degree of the extension is, right? The cardinality of Galois group of L over L^H so by normality and separability, of course. We have seen that normal and separable extensions have many automorphisms. That in general the cardinality of the group of automorphisms is less or equal than the degree of the extension and in the Galois case they are equal. So, we have a subgroup which has the same cardinality as the group. And so they must be equal. So one must have H is equal to the Galois group of L over L^H. So this means that the maps which we had in the theorem, the map which sends F to the Galois group of L over F, and which sends H to the fixed field L^H. Well, this means these are mutually inverse. Inverse. And if a map is invertible, it's a bijection. So, in particular, these are bijections. So, the first part was more or less already observed before when we were discussing Artin's theorem and other things. Now the second part, something has to be done here. So we had three equivalences. First statement was that F was Galois. The second statement was that any g g of F was F for any g in Galois, and the third statement was, of course, that the Galois group of L over F was a normal sub group. This is a notation for a normal subgroup in Galois of L over K. So we must prove equivalence. Well, let's show that 1 implies 2. Let's take an x in F. We have the minimal polynomial of x over K. It splits in L, but it has a root in F. And has a root in F. So it must have all rutes in F by normality. So, F is normal since it's Galois. This means that this P minimal splits in F. This means, of course, that any map from Galois group preserves F since it premutes the roots. Any g in the Galois group, permutes the roots of P minimal of x over K, right? Since this is true for any x in F, this means that g of F must be contained in F. Right? Since this F is generated by all such roots. F is generated, consists even by such roots. Now it's also very easy to remark that 2 implies 1. If g(f) is in F, then all roots of P minimal x K for x in F are in F. since again, g permutes those roots. Well, since the Galois group acts transitively on roots of an irreducible polynomial. This is something which is very very important. We have seen this last time, it results from this theorem about extension of homomorphisms and it is very important to memorize it, that the Galois group is transitively on roots of any irreducible polynomial. Okay. So, this means that any polynomial which has a root in F is split over F and this is the definition of normality. This is the definition. I will continue after break.