Let me summarize what we have done up till 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, 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(alpha) over K was equal to the degree of the minimal polynomial 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 3 and there is no root. But, well, if you ask the question whether x to the power of 100 minus two is irreducible or not, this is already not so simple, right? Well, this is irreducible. And here are a couple of facts which help to see this easily. So Fact 1. This is usually called the Gauss Lemma. If P decomposes nontrivially 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 is to say that they are both strictly positive. So if this is the case over Q, 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 nontrivially over Q, then it also decomposes, over Z. Well, let me give a proof. So, over Q write 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 Q_1, in Z and nR is equal to R_1 in Z. Then we have mnP is equal to Q_1 R_1 over Z. Then take p which divides mn. Then modulo p, we have 0 = Q1 bar R1 bar, where bar denotes the reduction modulo p, right? But we are over Z/pZ, which is a field, so "modulo p" means over F_p, right? This is a field. So we have Q1 bar or R1 bar is 0. This means that p divides all coefficients either of Q1 or of R1. So p divides all coefficients, say, of Q_1. And then we can write mn over p times our big P is Q_2 times R_1 in Z[x], and here Q_2 is of course Q_1 divided by p. Continuing in this way, We arrive at P is equal to, I don't know, Q_l R_s in Z[x]. Okay, and Fact 2. Let me not prove it in full generality, let me just show this on an example, and then formulate maybe a general criteria. Fact 2 is Eisenstein criteria. Well how to show that X to the power of hundred minus two is irreducible over Z? This is very easy. We reduce modulo 2, and so if x^100 - 2 decomposes nontrivially as Q times R, then modulo 2 X to the power of 100 is Q bar time R bar F_2 of x. Yes, but this means that Q bar and R bar are of the form x to the power of k, 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^100 - 2 must be divisible by 4. And this is not the case. So the general formulation of Eisenstein's criterion criterion is as follows: In general, if you have a polynomial with integral coefficients, so P is a_n x^n + a_(n-1) x^(n-1) + ... + a_0 so P is a_n x^n + a_(n-1) x^(n-1) + ... + a_0 so P is a_n x^n + a_(n-1) x^(n-1) + ... + a_0 and there exists p prime, such that p does not divide a_n, p divides all a_i except a_n. And p^2 does not divide a_0, 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 Q by it's fraction field. So, I think this is a good point to finish the first lecture.