[SOUND] The next example I would like to give, is that of Kummer extensions. So consider K, a field such that the characteristics of K is prime to a certain number n, and such that. X to the power n- 1 is split in K. So K contains all ns roots of unit and we suppose that n is prime to characteristics. So we consider a and k, so let alpha the nth root of a. Well an nth root of a of course, a root of x to the power n- a. The same in some algebraic equations, it does not have to lie in. And take d which is minimal i aside from that a to the power of i is k. Preposition d divides n, the minimal polynomial Of alpha is X to the power d minus alpha to the power of d, and K of alpha is Galois with cyclic Galois group of order d. Over k, of course. So let's prove this proposition. It's clear that K of alpha is Galois because all the roots of unity, all nth roots of unity are in K. So K of alpha contains. All roots of X to the power n minus a. So this is. A split in field of X to the power n minus a. So this is normal and also separable, because this is a separable polynomial. N is prime to the characteristics. So normal. And separable. Since n is. Prime to the characteristics. Okay, so this is Galois. Now, let's define a homomorphism from the Galois group to the group of nth roots of unity. We will send g to g of alpha over alpha. This is correct because g sends alpha to another root of the same polynomial. So g sends. Alpha to a root of X power n minus a, so the quotient is a route of unity. Okay, so this is well defined of course, this is injected. So let's give it a name, should we call it just f? But f is injective. Since g of alpha determines g. Alpha generates our field. So g of alpha determines g, and then what is it's image? In some sub group of un, un is cyclic a sub group of a cyclic group is cyclic. Both un cyclic, let delta be it's order. Of the image. So we want to show. This is the only thing which remains to show, that delta is d. Okay, so let's write g of alpha to the power of delta. This is f of g to the power delta times alpha to the power delta. F of g is a number, it's a root of unity. And it's exactly the root of unity by which the action of g multiplies alpha. And this must be equal to alpha to the power of delta, because delta is exactly the order of f of the Galois group. So f of g to the power of delta is that entity. Then it follows that alpha to the power of delta isn't K and. Alpha to the power i is not in K form for i less than delta. Since otherwise. [COUGH] The minimal polynomial of alpha will be of degree. So inferior to delta, but this is impossible, but it was. The order of the Galois group so the degree of the feud extension Is exactly delta. [COUGH] So this proves the proposition delta is equal to d and the minimal polynomial of alpha is x to the power d minus alpha to the power d. So proposition two and conversely. Any cyclic extension. Of degree n. Prime to the characteristics of k. Is obtained in this way. Generated by an nth root. Generated by. The nth root of a for sum a and k. Or let us prove this proposition. So take consider L an extension of K such that the Galois group is cyclic generated by sigma, then we have sigma to the power n is identity. Then by linear algebra. I shall not prove it, but you should know it from the linear algebra course, such a thing is diagonalisable. Sigma is diagonalisable. Now, let us show that all eigenspaces have dimension 1. All eigenspaces. Have dimension 1. Indeed if you have x and y in the same, eigenspace then sigma of x over y. Is x over y, because x and y are multiplied by the same number. So x over y is in K and this is exactly what means that the dimension of this eigenspace is equal to 1. So x and y proportional. Over K. So it follows that all roots of unity are eigenvalues. Of sigma. Now it suffices to take an eigenvector take alpha such that sigma alpha is zeta alpha for zeta some primitive root of unity. Then the orbit of alpha. Has. N elements. And also it follows. That the degree of k of alpha over k is n and alpha to the power n is nK. Since sigma of alpha to the power of n is zeta n times alpha to the power of n. So this is just alpha to the power of n, and we see that alpha is a root of X to the power of n minus alpha to the power of n. That well, it's better to write some here. And this is irreducible by degree reasons. So this is what we're going to prove. [SOUND]