So, construction of K_1. Let S, be the set of all irreducible elements of K[x]. of all irreducible elements of K[x]. So S is a very big set. Let A be a polynomial ring, generated over K by all X_P, where P is an element of S. So, it is a polynomial ring over huge number of variables. One variable X_P for every P in S, so this is a polynomial algebra. Very big polynomial ring. Let I in A be the ideal Let I in A be the ideal generated by P of X_P generated by P of X_P 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(X_P), and this is one of the generators of our ideal. 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 of X_Pi. a sum of lambda_i Pi of X_Pi. 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 1, then 1 is a finite, A-linear combination of 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 alpha_i be a root of Pi in K. So let alpha_i be a root of Pi in K. Then, my A is polynomial ring. And it's very easy to produce a homomorphism of a polynomial algebra to some other algebra. One must just note where one sends the variables, right? So, I can produce a homomorphism of K-algebras from A to L. Just sending X_Pi to alpha_i's. These are just variables. My A is a polynomial algebra. Those are variables. And other X_P to 0 if P is different from Pi. Okay. And if I look at phi(1) and use this identity I obtain that it is 0, because each X_Pi is sent to alpha_i, so Pi, so, phi(Pi(X_Pi)) is just Pi of so, phi(Pi(X_Pi)) is just Pi of alpha_i, so it is 0. But this cannot happen. Phi(1) must be equal to 1, not to 0, okay? So, I is an ideal. Then you know how to get a field. I hope you know. So a fact: any ideal, any proper ideal in any commutative associative ring with unity is contained in a maximal ideal ideal m and the quotient A/m is a field. So I just can take K_1 = A/m. And then continue in the same way to construct K_2, ... K_3, K_n and so forth. Okay. Well, maybe I should give you some comment 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. Digression: ideals in a ring commutative, associative, with unity. I told you in the beginning that all our rings will be like this. So, any ring, any proper ideal is contained in a maximal ideal. This is a consequence of what one calls Zorn's lemma, So let me give you the Zorn's lemma. Consider P, not a polynomial anymore, but a 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 non-empty chain, any non-empty chain in a non-empty P has an upper bound, an element of P, which is greater or equal than any element of the chain, that is an M in P such that M is greater or equal than x for any x in C, then P has maximal elements. I will not of course prove this Zorn's lemma because you certainly have heard that this is the same thing basically as the axiom of choice or Zermelo's theorem, so this is relevant for set theory, 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. a set of ideals in A containing I. Well, proper ideals. Proper ideals are not equal to A itself, containing I. This is non empty because contains I. And any chain has an upper bound. Any chain, I_alpha where alpha is in some set of indices J, has an upper bound. This is just the union of all I_alpha. Well, you check that it is an ideal, it is easy. I'll leave it as an exercise. So, our set 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. Some a in A/m. would generate a proper ideal. And it's preimage under the projection from A to A/m would strictly contain m.