 Okay, this is lecture eight of algebraic geometry on the strong null Stellensatz. We should say I've changed the numbering system for lectures, so under the old system, this would have been part two, part four of lecture two. So I'll stop by just recalling that we can have strong null Stellensatz. So we're taking K to be an algebraically closed field and we have the weak null Stellensatz, which states that the maximal ideals of the polynomial ring K X1 to Xn just correspond to points of affine space. And we want to reduce the strong null Stellensatz, which says that IZ of A is the radical of A for A and ideal in the polynomial ring. So you remember Z of A was the set of zeros of the ideal A and I of A was the ideal corresponding to this algebraic set. Well, in order to prove this, first of all, it's very easy to check to check that the radical of A is contained in IZ of A and the hard part is to prove, is to show that IZ of A is contained in the radical of A. And we're going to use a very cunning idea due to Rabinovich called the Rabinovich trick as follows. What we do is we suppose A, the ideal A is A is generated by elements F1 up to FM. And we suppose that F is an element of IZ of A. So we want to show that F is in the radical of A. Well, this means that F vanishes if F1 up to FM vanish, F1 up to FM and 1 minus X naught F and F1 up to F1 minus X naught F have no zeros in A to the N plus one. Well, what's this X naught? Well, this X naught is Rabinovich's cunning extra idea. So Rabinovich's trick is this, we add an extra variable. It's kind of funny because you're proving something in N dimensions by jumping up to N plus one dimensions. That's kind of obvious because if any common zero would have to be a zero of F1 up to FM. So it would also have to be a zero of F in which case this wouldn't vanish. They have no common zeros. Now we apply the weak nullstellensatz. We are applying the weak nullstellensatz to A N plus one. So when I said the weak nullstellensatz implies the strong nullstellensatz, it's really the weak nullstellensatz in one extra dimension implies the strong nullstellensatz in N dimensions. Well, the weak nullstellensatz says that if they have no common zeros, this implies they generate the unit ideal. They means these elements here because if they have no zeros, they're not contained in any of the maximal ideals. Because by the weak nullstellensatz, the maximal ideals just correspond to points of A N plus one. So we know that F1 up to FM one minus X naught F. The ideal generated by these is just the unit ideal one in K X naught up to XN because we've added this extra variable. So one is equal to G naught one minus X naught F plus G1 F plus GN, sorry, GM FM for some GI. Now put X naught equals one over F. So we're now working in a ring of rational functions and F1. So we find one is equal to G1 F1 plus G2 F2 plus GM FM. G1 F1 plus G2 F2 plus GM FM where now all the elements GI are, they're not quite in the ring of polynomials from X1 up to XN because they might also involve one over F. So they're actually rational functions with denominators that can be written as powers of F. Now, since each of these is rational functions with denominators of power of F, we can just clear denominators by multiplying by a high power of F. So we find F to the power of N is equal to H1 F1 plus H2 F2 and so on, where each HI is equal to GI times F to the N and the HI polynomials. Well, this just says that F is in the radical of the ideal generated by F1 up to FN, which was what we wanted to prove. So this is the end of the proof. So the weak and the strong nullstellensatz give us the sort of correspondence between affine space A N and the ring KX1 up to XN. So this is the coordinate ring of this affine space. And remember we said earlier that anything you could do for affine space had an analog for this ring and so on. And we can now fill in some of these points, A1 to A N of A N correspond to maximal ideals of this, the maximal ideal would of course be X1 minus A1 to XN minus A N. So this is the weak nullstellensatz and we say that algebraic sets correspond to radical ideals. These are ideals A that are equal to their own radical. So this correspondence here is the weak nullstellensatz and this correspondence here is just the strong nullstellensatz. Well, you may wonder why do we, what happens for non-radical ideals? Well, until about 1950, most people didn't pay too much attention to the non-radical ideals unless they were doing commutative algebra. So all ideals turn out to correspond to sub-schemes or closed sub-schemes. So this is something we'll see later on in the course when we do schemes that radical ideals correspond to classical algebraic sets. All ideals correspond to schemes and this is partly why you introduce schemes rather than just working with algebraic varieties. It's because you want to include all ideals. So let's see a few examples of this. So it's the first example. Let's take a line. So this corresponds to the, this is just the line Y equals naught. So it corresponds to the ideal Y and we can take a parabola which corresponds to the ideal Y minus X squared. So here we have two perfectly good radical ideals. And now let's look at the intersection of these two varieties. Well, you can try forming the intersection by taking the unit, by taking the thing generated by these two ideals. So we take the ideal Y minus X squared and Y. So this corresponds to the intersection. So it corresponds to this point here. However, there's a bit of a problem because this is not radical. So it's the ideal Y X squared and it's not a radical ideal. Well, obviously if we take the root of this ideal Y X squared it's just the point X Y and this corresponds to the point naught, naught in a fine space. I'm afraid the notation is a bit confusing. So here I don't mean the point X Y. This means the ideal generated by X and Y. And this is the point with coordinates zero, zero. I'm afraid left and right parentheses are a little bit overused. So they're sometimes used for ideals and sometimes used for points. And you have to figure out which is which. So here we have a non-radical ideal turning up very naturally as the intersection of two curves. And the fact that this ideal is not radical kind of corresponds to the fact that if you intersect the parabola and the line well it sort of looks as if there's only one point there but it's really a double point. So it ought to be counted as two points. And somehow the fact that this ideal is non-radical is trying to account for the fact there are really two points there in exactly the same place, whatever that means. For the next example, let's look at nil potent matrices in let's take all n by n matrices over sum of field K. So matrix A is nil potent if A to the n equals zero where this n is the same as that end. If matrix is nil potent of some power of it equals zero and you can see that if some power is zero then h to the n is zero. So the matrices are in m and K which we can think of as being affine space of dimension n squared. So we want to write down some natural equations for this matrix to be nil potent. Well, if A is equal to A11, A12 and so on. So here we have some matrix and A to the n will be something complicated. These will be some homogenous polynomials of degree n in all these coefficients here which I'm not going to write out because they're such a horrible mess. And now we can say these generate an ideal I in the coordinate ring of A n squared. So we're just going to take all these coefficients of A to the n and say that's going to be the ideal. And in some sense this ideal describes the set of nil potent matrices. So now we can ask is the ideal I equal to the radical of I? And the answer is no. So it's a very natural ideal. It's just the most obvious way of defining nil potent matrices. And yet it turns out to be not radical. In order to see this, if A is nil potent this implies all eigen values zero which implies the trace is zero. Well, the trace is just A11 plus A22 plus A33 and so on. Well, this is not in I because I is generated by homogenous polynomials of degree n. So you're not going to get a homogenous polynomial of degree one over that. And not only the trace is, sorry, what am I saying? So this element here is in the radical of I by Hilbert's nullstellens. That's because it vanishes on all nil potent matrices. And not only the trace vanishes but some of products of pairs of eigenvalues also vanishes and so on. So there are quite a lot of things in the radical of I that are not in I. So let's take n equals two and just see what happens. So here A, the matrix A is going to have four coefficients, A, B, C, D. And A squared equals zero. Well, A squared is just the matrix. A squared plus BC, B times A plus D, C times A plus D and D squared plus BC. D squared plus BC. So the ideal I is equal to A squared plus BC, B times A plus D, C times A plus D, D squared plus BC. And we have seen that some power of A plus B is in I. So the question is what is the, sorry, that should be A plus D, not A plus B. So we can ask what is the smallest power of A plus D in I? And again, it's not obvious. You might guess the answer is two, because the number two is turning up very often, but this fails. A plus D squared is not in this ideal I. In fact, we find A plus D squared is not in I, but A plus D cubed is in I. And I'll leave this as a very short exercise. So, you know, Hilbert's Null Stellensatz is actually telling us something that isn't at all obvious. You know, I mean, it's certainly not obvious to me that some power of A plus D lies in, in this ideal. And it's, I think it's not at all obvious what the smallest power of it is, for example. I mean, so it's, it's, quite a deep theorem. The next example is the following. Well, when we looked at Null Potent Matrices, what about commuting matrices? So let's look at the condition AB equals BA for AB, both N by N matrices. And we want to find the space of matrices that commute with each other. Well, what we do is we set A equals A11, A12, et cetera, B equals B11 and so on. And then AB minus BA will be equal to some big matrix here. So here we have A11, B11 plus A12, B21 plus something minus something. I mean, the coefficients are rather complicated messes. We're gonna take I to be the ideal generated by these. So I is in the polynomial ring of A to the two N squared because a matrix A is given by something in N squared dimensional affine space and so is B. So here we've got affine space of dimension two N squared and we've got some ideal defining the subset of commuting matrices. And now we have the question, is I equal to the radical of I? And the answer is, well, I don't actually know what the answer is. This seems to be a very hard open problem. And the point of this is that it can be really difficult to find out what the radical of an ideal is. I find this a bit surprising. I mean, what you're asking for is you're looking at the ring R over. If you've got an ideal I and you want to know whether it's equal to its radical, you're asking whether the ring R over I have nilpotent elements. And I would have guessed that nilpotent elements in a ring are actually rather easy to find. They have this rather striking property that some power of them is zero, which ought to make them obvious. But even in a natural example like this, it's really, really difficult to see whether the corresponding ring R over I has nilpotent elements. Incidentally, the space of pairs of commuting matrices is a notoriously difficult spaced study. Okay, that will be all for this lecture. Figure out how to...