[MUSIC] Let me summarize what we have done up til now. So we had a field K. And we said that alpha was algebraic, Over K, if it was a root of a polynomial. With coefficients from K. We said that L was algebraic over K if any alpha in L was algebraic over K. We said that F was finite, over K, if it was a finite dimensional K vector space. We have seen that finite implies algebraic. We have seen that in fact it is equivalent to algebraic, and generated by a finite number of elements, so I'm going to say finitely generated, and we have seen that if we generate our L by a single element alpha, we have seen that the degree of k of alpha, over K, was equal to the degree of the minimal binomial of alpha over K. So, given some algebraic element over K, a root of some polynomial. It is important to be able to decide whether this is the minimal polynomial of alpha over K. That is to say, it is important to have some irreducibility criteria, and this is something you probably already know. But since it's so important, I would like to remind a couple of things about this. So how to decide, That a polynomial is irreducible? Over K. Well in our example, we had a very simple polynomial. So x cubed minus 2 was irreducible over Q since it's a cubic polynomial, so if it was not irreducible it would have a root in Q. So this is easy since the degree is equal to three and there is no root. But, well, if you ask the question whether x to the power of 100 minus two. If reducible or not. This is already not so simple, right? Well, this is reducible. And here are a couple of facts which help to see this easily. So fact one. This is usually called the Gauss Lemma. If Pm decomposes non trivially Well by this I mean is a product of two factors of strictly smaller degree, that is P is Q times R, where the degree of Q, and the degree of R, are both less than the degree of P, or the same as to say that they are both strictly positive. So if this is the case over Q1, then it is also the case over Z. Well, of course I have to say that I am considering a polynomial with integral coefficients. So, if such a polynomial decomposes naturally over Q, then it also decomposes, over Z. Well let me give a proof. And so, Z over Q where P is equal to Q times R. They are not integral, but of course you can multiply by a common denominator, and they become integral. So lets say MQ is equal to Q1. In Z and NR is equal to R1 in Z. Then we have MNP is equal to Q1R1 over Z. Then take phi which divides mn. Then modulo P, we have 0 = Q1 bar R1bar, where bar denotes the reduction model of P, right. Well, but we are over that modulo PZ which is a field, so modular p means over fp, right this is a field. So we have Q1bar. Or R1-bar is 0. This means that P divides all coefficients either of Q1 or of R1. So P divides all coefficients C of q one. And then we can write unknown over p times our big P is Q two times R one in Z of x, and here, Q two use of course Q1 divided by P. Continuing in this way, We arrive at P is equal to, I don't know, Q LRS in zero effects. Okay, in fact 2, let me now prove it in full general team Let me just show this on an example, and then formulate maybe a general criteria for factor is Eisenstein criteria. Well how would you show is that X to the power hundred minus two is irreducible over Z. This is very easy. We will reduce modulo two, and so if X to the power of 100 minus 2, decomposes, [FOREIGN] yes two times R. Then modulo two, X to the power of 100, is Q bar time R bar in F two of x. Yes, but this means that Q bar and R bar are of the form X to the power K. Well, respectively x to the power L. So they have no constant coefficient modulo two. So the constant coefficient, constant coefficient of both Q bar and R bar is even and this means the constant coefficient. Of x to the power of 100 minus 2 must be divisible by 4. And this is not the case, so the general formulation of Isaac's standard criterion is as follows. In general, if you have a polynomial with integral coefficients, so P is an X to the n + a(n-1)X to n-1+, and so on. Plus a 0, and there exists P prime, such that P does not divide a n. P divides all ai except a n. And P square, does not divide, a zero, then this P is reducible. And the proof is exactly the same. And to conclude, I would like to say that both facts are valid replacing Z by some unique factorization domain. R and replacing two by its fraction field. And so on. I think this is a good point to finish the first lecture. [MUSIC]