[MUSIC] So it remains to prove that properties one and two are equivalent to three, that is to say through the normality of the subgroup Galois of L over F. Well, let's take, and now let G being G. And consider G of F, which is in L. Then you see if H restricted to F is that entity is H, if H is an element of Galois group of L over f, right? Then obviously what we have is G, H, G minus one restricted to G of F is also identity. So this means G H, G minus one 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 L over F times G minus one plus Galois group L over F. If you have these for any element G then this our condition too and this is the definition of a normal sub group. So apply this to all G in the Galois group of L over K, and then this gives us that the N variance of F is just equivalent to the normality. So finally, if all this is true, of this, I'm in. One which is two, and three which is equivalent to two and three is true. Then, we can consider. The map from Galois L over K to Galois F over K because by we have that F is stable by the action of G so we can indeed restrict the G to F, so this makes sense by Okay, and this is a surjection by the theorem, or an extension of homomorphisms. So this is surjection, by extension of homomorphisms, and the kernel is Galois of L over F. This is just by definition, that was the kernel consists of things which are identity on F. So we have proved, then, Galois correspondence. So let me remark, that if the extension is not finite. Then, this Galois correspondence is not by objective. 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 will do it later, now I just want to give you some very simple examples. Some simple examples, it will have extensions of degree two. Now 1a, let's say if I have an extension of degree two. Then obviously it is generated by any element not belonging to K, and this element is a root of a quadratic polynomial. Of 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, as the Galois group of L over K, a group of two elements, there is only one such group, it's simple group of order two. So let's see, set that to that secret group of [INAUDIBLE]. So there is identity of cause and the reason element which is not of 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 variant precise I would say, of course I would not speak of the route of delta, I would say there were two elements which have delta four square, and the non trivial element of Galois group permutes them, okay? So example one b, now if I have an extension of degree three, L over K of degree three. It does not have to be normal anymore. What happens is the following L is generated by a root X, a root of degree 3 polynomial. Of course, I have to make some assumptions on the characteristics here, in the 1A, I assume that the characteristic if K is different from two, because then otherwise I'll have the root of delta equal to one. Minus root of delta and there will be no non trivial element of the Galois group, it will be and a purely unseparable extension, so let's not deal with this case. Here, likewise, I assume that my thing is separable, and the characteristic of K is different from three, well and from two just in case. We will see a later, right? Okay, so L is generated by a root of degree three polynomial L. [INAUDIBLE]. And it can be, so there are two cases other P splits in L. Then, it means that L is Galois, and the Galois group is something of three elements it must be cyclic. So it's Z by 3Z cyclic group of order three. Another case is when P does not split in L. Then we have M as split in feud, M is K x1, x2, x3, these are roots. Okay, and L is just K of X one. So this N is of course Galois. And the Galois group of M over k is embedded into the group of [INAUDIBLE] over those three elements, right? The elements of Galois groups, Galois group permit the roots. And then, since M, is strictly bigger than L. This means that the degree of M over K is not equal to three, is greater than three. So this is a subgroup of three of cardinality greater than three. And so, they must coincide. It must be equal to S3. So, in particular another degree of M over K is six, okay? Under, if you see a polynomial degree of three. How to decide whether it's a Galois group is three or S3. So if P mim, let B be a polynomial of degree three of our key and irreducible. Polynomial of degree three over K then M the split and field of P, one usually calls to 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. [MUSIC].