So it remains to prove that properties (1) and (2) are equivalent to (3), that is to say to the normality of the subgroup Galois of L over F. Well, let's take now let g be in G. And consider g of F which is in L. Then you see if H restricted if H restricted to F is identity, if H is an element of Galois group of L over F, right? Then obviously what we have is ghg^(-1) restricted to g of F is also identity. So this means ghg^(-1) is in the Galois group of L over g(F), right? g of F. So to say that F is equal to g of F is the same as to say that g times Galois group of L over F times g^(-1) is Galois group L over F. If we have this for any element g then this is our condition (2) and this is the definition of a normal subgroup. So apply this to all g in the Galois group of L over K and then this gives that the invariance of F is just equivalent to the normality. So finally, if all this is true, this, I mean, (1) which is (2) and (3), which is equivalent to (2) and (3), is true. Then, we can consider the map from Galois of L over K to Galois F over K because by (2) we have that F is stable by the action of G, so we can indeed restrict the gs to F, so this makes sense by (2) Okay, and this is a surjection by the theorem on extension of homomorphisms. So this is a surjection by extension of homomorphisms and the kernel is Galois of L over F. This is just by definition, because the kernel consists of things which are identity on F. So we have proved the Galois correspondence. So let me remark, that if the extension is not finite, then this Galois correspondence is not bijective. The maps which are in the theorem still make sense, but they will not be mutually inverse bijections and we shall see an example. But a bit later, now I just want to give you some very simple examples. So, simple examples. If we have extensions of degree 2. So, 1a, let's say. If I have an extension of degree 2, then obviously it is generated by any element not belonging to K, and this element is a root of a quadratic polynomial. The polynomial, minimal polynomial of x over K is quadratic. And in fact, if you look at the formula for the roots, you see that L is generated by the root of its discriminant, right? And what G does, of course, the Galois group of L over K, is a group of two elements, there is only one such group, it's cyclic group of order 2. So let's say Z by 2Z, cyclic group of order 2. So there is identity of course and there is an element which is not the identity. The non-trivial element just exchanges the root of the discriminant and the minus root of the discriminant. The non-trivial element permutes the root of Delta and minus root of Delta. Well, if I want to be very precise I would say, of course, I would not speak of the route of Delta, I would say that there are two elements which have Delta for square, and the non trivial element of Galois group permutes them, okay? So example 1b. Now if I have an extension of degree 3, L over K of degree 3. It does not have to be normal anymore. What happens is the following L is generated by a root x root of degree 3 polynomial. Of course, I have to make some assumptions on the characteristics here, in 1a, I assume that the characteristic of K is different from 2, because otherwise I willl have the root of Delta equal to minus root of Delta and there will be no non-trivial element of the Galois group, it will be a purely inseparable extension, so let's not deal with this case. Here, likewise, I assume that my thing is separable, the characteristic of K is different from 3, well, and from 2 just in case. We will see a later why. Okay, so L is generated by a root of degree 3 polynomial P. And it can be, so there are two cases: either P splits in L, then it means that L is Galois, and the Galois group is something of 3 elements it must be cyclic. So it's Z by 3Z cyclic group of order 3. Another case is when P does not split in L. Then we have M a splitting field. M is K of x1, x2, x3, these are roots of P. And L is just K of x1. So this M is of course Galois. And the Galois group of M over K is embedded into the group of permutations of those three elements, right? The elements of Galois group permute the roots. And then since M is strictly bigger than L, this means that the degree of M over K is not equal to 3, is greater than 3. So this is a subgroup of S_3 of cardinality greater than 3. And so they must coincide. It must be equal to S_3. So, in particular the degree of M over K is 6, okay? And if you see a polynomial degree of 3, How to decide whether it's Galois group is cyclic of order 3 or S_3? So if P... let P be a polynomial of degree 3 over K and irreducible. Polynomial of degree 3 over K then M be a splitting field of P, one usually calls the Galois group of P the Galois group of M over K. So how to decide when is the Galois group cyclic and when it is not cyclic? But it turns out that this is determined by the discriminant of the polynomial. So, the answer is in the discriminant. But we have to take a break now.