[NOISE] [MUSIC] Let me give you a proposition which is a consequence of this lemma and summarizes properties of integral ring extensions. Proposition one. So first of all, if you have A and B and C, and you know that the B is integral, over A. Of course I forgot to say what it means. This means, of course, that all elements of B are integral over A, so the definition B is integral over A, with any alpha in B is. B is integral over A, and C is integral over B. Means for that C is integral over A. Secondly, if B is finitely generated over A as A-module, This means that B is A of some elements alpha 1 and so on alpha r where each alpha i is integral over A. And a third property says that, Elements of B, Integral over A, Form a subring of B. One calls this subring, the integral closure of A and B. Of A and B. So in the third part of the proposition, B is an arbitrary extension of A. So it's not necessarily integral over A. And we achieve all elements which are integral over A, and we affirm that this is a subring. I'll leave you as an exercise to deduce these propositions from the lemma. Knowledge you may see is a couple of words about the part. Three, for instance, how do you put three? You prove that if alpha and beta are integral over A, then A of alpha and beta is a finitely generated A-module. And it follows directly from the lemma and then you want to say that it contains, Alpha plus b and alpha plus beta and alpha beta and so on. [INAUDIBLE] Alpha plus beta, and alpha beta are integral over A. And this is exactly what we need to prove that this subset of integral elements is a subring. So maybe one more definition. A is said to be integrally closed, In B, if the integral closure of A and B equals to A. And if one says that A is integrally closed without mention B, So A is integrally closed if A is integrally closed in its fraction field. So for instance, Zed is integral closed. And more generally, Any unique factorization domain is integrally closed. So UFD is unique factorization domain is integrally closed. Now this is very easy to see, so let A be a UFD and x an element of the fraction field. So we write x as a quotient of two elements of A which are relatively prime. That is to say, they don't have any common prime divisor. X is, of course, nonzero. Pq = 1 that is this means no common prime divisor. Okay, and then if x is integral over A, We'll have P/ q to the power n = an- 1, P/ q to the power n- 1, and so on. A1 of P/ q + a0 = 0. And in denominators. And we obtain that something over q to the power n, that is to say p to the power n + qan- 1. P to the power n- 1+ and so on, + q2 to the power n x a0 = 0. But then it follows that q / P to the power n, which is a contradiction unless q is invertible. Unless our q is invertible. And this means exactly that x is in A. Okay, so let me now, Talk about the ring of integers, In a number field. So let K be a number field. A number field is a finite extension of Q. Now let's say the degree of K / Q is N. Then the ring of integers, Ok is the integral closure, Of Zed in K. So the integral closure of Zed in Q is Zed as we have seen, but we consider it in K. The basic properties are in the full elemental proposition. First of all, for any alpha in K, one can find a d, an integral number, nonzero, of course. Otherwise, it's not interesting such that d alpha is in Ok. Secondly, if alpha is an element of Ok, then the minimal polynomial of alpha over Q, Has integral coefficients. For the proof, we can write, The minimal polynomial of this alpha in Q, so it's x to the power m + am- 1, x to the power m- 1 + and so on, plus a1x + a0. So this is something with Q coefficients and remark that there is a d, The common denominator such that, Dai is integral for any i. Then, of course, d to the power m- i, ai, is also internal for any i. And this means, That (d alpha) m plus, well, let me denote these by bi and = b and- 1. (D alpha) m- 1 + and so on, + b0 =0, so this means that d alpha is in the ring of integers. And the second part is also easy. If we have such an alpha, it is a root of some monic polynomial Q with integral coefficients by definition, then the minimal polynomial divides this Q. So Q = P-minimal times R. If we pick P-minimal to be monic, then by an argument very similar to that of the Galois lemma, We conclude that both P-minimal and R in Zed(x). [SOUND] [MUSIC]