 Okay, this is part three of lecture two and will be mostly on Hilbert's Nor-Stellen-Satz. So the Nor-Stellen-Satz is a word meaning zero's position theorem and describes the relation between ideals of a polynomial ring and algebraic subsets of the corresponding affine space. So what we have is we have subsets of affine space a to the n and we have ideals of the coordinate ring kx1 to kxn and we want to know what is the relation between these. Well if we've got a subset y, we can map it to the ideal i of y. So this ideal is just functions vanishing on y, polynomials vanishing on y. Conversely, if we've got an ideal, we can go back to a subset, so suppose a is the ideal, it goes to the subset z of a, which does just the set of zeros of a, meaning the set of zeros of all polynomials in the ideal a. So this map from subsets of a to ideals of a and from ideals of a back to subsets of i are certainly not inverses of each other. So the question is what is the relation between them? So first of all there's one easy one. Suppose you start with the subset, take the ideal of all elements vanishing on it and then take the common zeros of those. So what we're doing is we're taking a subset, we're taking the ideal things vanishing on it and then taking the common zeros of this ideal. Well this is just the closure of y in the Zariski topology. This is fairly easy to prove, it's more or less the definition of the Zariski topology, so it just consists of unwinding the definitions a bit and we won't worry about it anymore. So what about the other directions? Suppose you start with an ideal, take the subset of zeros and then go back to the ideal. If you get back to the ideal you originally thought of, so we can ask is i Z of a equal to a, the answer is no in general. And there are some obvious examples, we can just take the ideal to be the ideal generated by x squared in k of x. Then the zeros of x squared is just the point zero and the functions vanishing on this zero is this ideal i Z of a. But it's obviously just all multiples of x which is certainly not the same as all multiples of x squared. More generally if f to the n is in some ideal a, then f certainly vanishes at all points where all elements of a vanish because f to the n does. So then f is in i of z of a. So the radical of a is certainly contained in i of z of a. And I will just remind you what the radical of an ideal is. So the radical consists of all elements f so that f to the n is in a for some n equals one, two, three and so on. By the way, the radical of an ideal also forms an ideal. The radical is obviously closed under multiplication by elements of the ring is not quite so obvious that the radical is closed under addition. So we know we point out that if f to the n is in a and g to the m is in a, then f plus g to the m plus n is in a because all coefficients of this are either divisible by f to the n or g to the m. So we see the radical is also closed under addition. Anyway, so we've seen the radical of a is certainly contained in this. So the first question to ask is, is the radical of a equal to i z of a? And the answer is again, no. And it's fairly easy to find a counter example to this. All we do is we take the ideal a to be the set of multiples of x squared plus one in the ring polynomials over x. Then you notice that the set of zeros of a is empty as there are no real numbers whose square is equal to one. So i of z of a, insisting of all polynomials vanishing on this empty set is just the whole ring R of x and does not equal to the ideal a. And this is problem is obviously something to do with the fact the real numbers are not algebraically closed, so we can't find a square root of minus one. And it turns out this is true if the field k is algebraically closed. This will be Hilbert's nullstellensatz. Anyway, before doing that, we're going to look at a slightly easier theorem, which is the problem of finding what are the maximal ideals of a polynomial ring k x1 up to xn? Well, there are some obvious maximal ideals. Let's take any point a1 up to an in a to the n and then look at the ideal x1 minus a1 x2 minus a2 and so on. And you can see this ideal is, of course, just the functions vanishing on this point a1 up to an. And it's obviously a maximal ideal because the quotient by this ideal is just the field k. So since the quotient is a field, it's a maximal ideal. And the question is, are there any other maximal ideals? So we can ask, are these all maximal ideals? And the answer is no, again. And it's no for exactly the same reason as above that this ideal here is maximal. And it doesn't come from a point of the real numbers sort of vaguely related to the complex number i. Well, again, this is true if the field k is algebraically closed. So this is the weak null Stelenzatz and this is the strong null Stelenzatz. I'll just write as n because I'm getting a bit lazy about writing out this long word. So what we're going to do is we're first going to prove the weak null Stelenzatz, which as the name suggests is actually weaker than the strong null Stelenzatz. And then we will use an amazing trick found by Rabinovich, which shows that you can actually reduce the stronger strong null Stelenzatz from the weak null Stelenzatz. So now let's prove the weak null Stelenzatz, weak null Stelenzatz. Oops, let me adjust that so it's on the screen. And we first want to show that if k is a field, which is finitely generated as an algebra over a little field k, then k is a finitely generated module over k. So saying it's finitely generated as a module is just the same as saying it's a finitely generated vector space. So you've got to be a bit careful because the term finitely generated can mean at least three different things. You can ask if is big k finitely generated as a field, is it finitely generated as an algebra, or is it finitely generated as a module? And in general, these three concepts can be quite different, but if k happens to be a field, then these two turn out to be the same. And we're going to cheat. So let's say we cheat at this point, and we get the cheaters we're going to assume a little k is uncountable. The result is still true if little k isn't uncountable, but the proof becomes a little bit harder. So out of laziness, we're just going to do this special case, which will more or less show the main idea. And if little k is uncountable, this is very easy because k is of at most countable dimension as a module or vector space. And this is very easy because k is a finitely generated algebra, and you can easily check that any finitely generated algebra is at most countable, countably generated as a module because you can take a basis to be all monomials in the finite number of generators. On the other hand, if k is not algebraic, so if x is in k is transcendental over little k, then the elements 1 over x minus a for a in k form an uncountable linearly independent set. And this is a contradiction. So all x in k are algebraic over little k. So since big k is finitely generated as an algebra and all its elements are algebraic, this implies by some easy field theory that k is finite dimensional as a module or equivalently as a vector space. So this proves this lemma that if finitely generated algebra, it's also a field as a finitely generated module. So here this collection of elements is uncountable because we assumed the field little k was uncountable because there are an uncountable number of elements here. If we don't assume little k is uncountable, you've got to work a little bit harder. The idea is roughly speaking that the finite number of generators of big k only have poles on a finite number of irreducible subsets of the affine space over k, but they're an infinite number of such irreducible subsets. But anyway, we won't worry about that too much. So this proves this little lemma. Now we're going to use that to deduce the weak nullstellens. So this is very easy. So suppose i is a maximal ideal of little k x1 to xn. So we want to show that i is equal to x1 minus a1 x2 minus a2 and so on for some a1 a2 and so on. Well, all we do is we put big k is equal to k x1 up to xn modulo i. So k is a field and the reason it's a field is it's the quotient of a maximal ideal. We assumed i was a maximal ideal and any quotient by a maximal ideal is a field. It's also finitely generated as an algebra and it's finitely generated as an algebra because, well, here's a finite set of generators of it x1 up to xn. So by the previous sheet, k is finitely generated as a module. In other words, big k is algebraic over little k. Well, now we're done because we assumed k is algebraically closed. So little k must be equal to big k because if little k is algebraically closed and big k is algebraic over it, then they must be the same field. Well, if big k is equal to little k, this implies that each xi is actually in little k in this field k. So xi equals ai for some ai in little k. So this is equal in in the field k, which is quotient by i. And this implies that i is actually contained in the ideal x1 minus a1, x2 minus a2, and so on. So as little i, sorry, as i was considered to be maximal and it's contained in this ideal, so i must be equal to this ideal. So this is the end of the proof of the weak nullstellensatz. It shows that all maximal ideals of this ring for k algebraically closed are just the obvious ones corresponding to points of the affine space. So in the next part of the lecture, we will show how to deduce the strong nullstellensatz from the weak nullstellensatz and Rabinovich's trick.