[MUSIC] So, construction of K1. Let S, be the set of all irreducible elements of k(x). So s is a very big set. Let A be the polynomial ring, generated overcame by all Xp where p is an element of S. So, it is a polynomial ring over huge number of variables. One variable Xp for every p in S, so this is polynomial algebra. Very big Polynomial ring. Let I in A be the ideal generated by P of Xp for any P in S. So it is an ideal generated by a huge number of elements. By all, right? Or take for any P in S, we take P(Xp), and this is one of the generators of our idea. So what I claim, Is that I is a proper ideal. Well, indeed if not then I can write 1 in A as a sum of Lambda i Pi(Xpi). So these are generators, some generators of i. These are elements of A, right? So also some crazy polynomials in crazy variables. But the main point is that this sum is finite. If my ideal contains one, then one is a finite, a linear combination over my generators. Okay. Well take L a splitting field of the product of all Pi. From i of 1 to n. So I take my Pi, and I take the splitting field of the product of the family of Pi's. If you want, I generate an extension of K. By all the roots of all my Pi which were irreducible over K, okay? So let alphai be a root of Pi in K. Then, My A is polynomial ring. And it's very easy to produce a homeomorphism of a polynomial algebra to some other algebra. One must just note where one sends their variables, right? So I can produce a homomorphism of K algebras from A to L. Just send in Xpi to alpha i's. These are just variables. My A is a polynomial algebra. Those are variables. And either Xp to 0 if P in is different from Pi. Okay. And if I look at phi(1) and use this identity. I obtain that it is 0 because each Xpi is sent to alpha i, so Pi. So, phi(Pi(Xpi)) is just. What is Pi of? Alpha i, so it is 0. But this cannot happen. Phi(1) must be equal to one, not to zero, okay? And so on, i is an idea. And then you know how to get a field. I hope you know. So a fact: any ideal, any proper ideal in any community for associative frequent unity is contained in a maximal ideal. Ideal M and the quotient. A/m is a field. So I just can take K1 = A/m. And then continue in the same way. To construct K2, K3, Kn and so forth. Okay. Well, maybe I should give you some on this fact, because some of you know it, some of you don't. And this is very important, and the technique is very important and will be used in these lectures. So maybe I make some comments on this fact. So digression, ideals in a ring. About commutative, associative with unity. I told you in the beginning that all of our rings will be like this. So, any ring, any proper ideal, is contained. In a maximal ideal. This is a consequence of which one cause Zorn's lemma, So let me give you the Zorn's lemma. Consider P not a polynomial anymore, but yeah partially ordered set. A subset c in P is called a chain. If it is totally ordered, that is for each alpha and beta in C, one has alpha less or equal than beta or beta less or equal than alpha. Where less or equal is our order relation. Okay. So Zorn's lemma says the following thing. If any chain, if any say well non-empty chain, any non-empty chain in a non-empty beam has an upper bound. So an element of p, which is greater or equal than any element of the chain, that is an M in P. So I show that m is greater or equal than x for any x in C, then P has maximal elements. Again I will not of course prove this Zorn's lemma because you will certainly have heard that this is the same thing basically as the axiom of choice or it's a melo theorem so this. Is relevant for Assessore foundations of mathematics, not to algebra and Galois Theory. But, we shall use it to prove that any ideal is contained in a maximal ideal. Now, let P be a set of ideals in A containing I. Well, proper ideals. Proper ideals are not equal to A itself, containing I. This is Of these of course contains I. And any chain has an upper bound. Any chain, I alpha where alpha is in some set of indices J that has an upper bound. This is just the union of all I alpha. Well you check that it is an idea or it is easy. I'll leave it as an exercise. So our says P has maximal elements. So P has maximal elements. So I is contained in M, M maximal ideal. And if we have a maximal ideal, if we take a quotient by a maximal ideal, then this is certainly a field. Well, otherwise it would have some proper ideals. Or some a in A/M. A/M would generate a proper ideal. And it's pretty much under the projection from A to A/M or would strictly contain M. [MUSIC]