 This lecture is part of an online course on commutative algebra, and we will be discussing the geometry of the set of associated primes of a module M. So I'll just start with a quick review of the background. So first of all, M is going to be a finitely generated module over a notarian ring for the whole of this lecture. Then we just recall that the set of associated primes of M is the set of primes P such that our modulo P is isomorphic to a subset of M, which will sloppily denote like this. And we recall this set as finite. And we were looking at the problem of decomposing M into a series of modules like this with each quotient M i plus one over M i isomorphic to something of the form of modulo of prime. And what we showed is that if P is in the associated prime of M, then R over P has to occur in every chain like this. There's a sort of converse to this one can ask about. Suppose we've got a module, can we find a chain like this such that every quotient is of the form R over P for P in associated prime? And the answer is no. In fact, we had an example of this last lecture. You remember if we take M to be the module or ideal X, Y in the ring R, which was polynomials in two variables, then we saw the set of associated primes of M. It's just zero. On the other hand, as we saw last lecture, we can't really find a sequence like this such that all the quotients are isomorphic to R. So the exact relation between the set of associated primes of M and the quotients in the sequences like this is a little bit messier than in the case of modules of finite lengths. In fact, counter examples of this are quite common. If M is a torsion-free module over an integral domain R, then again, the set of associated primes of M is equal to zero, just the zero prime. And torsion-free modules are fairly common. Over principle ideal domains, they're just, they have a very simple structure, but in general, they're a bit messy. For example, even if you take R to be, well, another standard counter example to everything is the ring generated by the square root of minus five. So this is the standard example of a dedicated domain that is not a unique factorization domain. And in particular, it's got plenty of non-principle ideals. For instance, you can take M to be the non-principle ideal generated by two and one plus root minus five. And then again, M, the associated primes of M is just zero, but you can't build up M by taking extensions of R. M is just non-principle and this just doesn't work. So what we want to discuss for most of this lecture is the geometry of the set of associated primes of M. So let's just recall that the spectrum of R is the set of prime ideals. And we want to relate the associated primes of M to the spectrum. So the set of associated primes is in fact a subset of the spectrum of R. As well as the set of associated primes, we're also going to talk about the support of a module, which is denoted by the support of M. And it's just the set of primes of P with M, P, none empty. So you can think of this as being the stalk of M at the point P of the spectrum. So this is just a module over the localization of P. And it's easier to check the support of M. It's just the set of primes containing the annihilator of M which is the set of elements of the ring that just kill M. So let's just look at a few examples. First of all, let's take, let's start with R as the simplest case. And here, let's start by taking M equals Z over 12 Z, say. This isn't very difficult. If we draw the spectrum of R, we've got the ideal zero, which is this one-dimensional generic point. And we've got the point two, three, five and so on. And it's pretty obvious that both the support of M and the associated set of associate primes of M just consist of these two points here. So we have a sort of M equals support of M equals two, three. So for modules of finite length, there's not a lot of difference between them. So let's have a slightly more complicated example. Let's take M to be Z, module Z itself. And now if we draw the spectrum, we find the support of M is just the whole of the spectrum of Z. So everything is in the support. Let's draw the support like this. On the other hand, the set of associated primes is much smaller. So the set of associated primes of M is just zero. So here, the set of associated primes is just this generic point, but doesn't include these closed points here. So the support of M is much bigger than the set of associated primes. And if you look at a third example, suppose you take M equals Z plus Z over two Z. Here again, the support of M is obviously just the whole of the spectrum of Z. So if we draw the spectrum of Z like this, the support of M is going to consist of everything here, and it's also going to consist of all these points here. On the other hand, the set of associated primes of M consists of zero and the prime two. So we've got zero, which looks like this, but we've also got this extra point here. So what you can see from this is the set of associated primes is sort of giving you more information than the support of M. Here, the support of M is a very crude and variant that once your module contains M, the support is already everything and no matter what else you add to M, the support can't detect that. However, the set of associated primes can sort of notice that M has a little bit extra apart from Z here. If you look at this, you notice the support of M always seems to be the closure of the set of associated primes. And in fact, this is true in general. So let's just check this. So this is for, as usual, for finitely generated modules over notarian rings. The support of M is just the closure of the finite set of associated primes of M. And this is easy to show. First of all, if P is an associated prime of M, then P is equal to the annihilator of A, some A in M, it's a prime ideal. And this implies P contains the annihilator of the whole of M, which implies P is in the support of M. So the associated primes of M is contained in the support of M and since the support of M is closed, it must be, this is contained in that. On the other hand, if P is in the support of M, then this implies by definition of MP, the localization is none zero. And now what we do is we look at at the set of associated primes of MP, which is not empty because MP is none zero. So let's pick some prime Q in the set of associated primes of MP. So Q is an element of the spectrum of localization of R at P. And then you can check the, we've got a map from R to RP. So this gives us a map from the spectrum of RP to the spectrum of R. So if this map is F, this map is sort of taking the inverse of a prime. And if we look at the inverse of Q, we see that this is an associated prime of M contained in P. So P is in the closure of this prime. So anything in the support of M is in the closure of some associated prime of M. So let's have a look at another example. Let's take R to be the ring of polynomials and two variables. And then you remember the spectrum sort of looks like a two-dimensional plane. So it's got some closed points, which correspond to the points of the plane. And it's also got some one-dimensional points which correspond to primes generated by irreducible polynomials. And finally, it's got some sort of huge generic two-dimensional points corresponding to the ideal zero. And it doesn't matter if you can't read this very faint yellow because I don't really need it, but I'm drawing it very faint to avoid distracting you too much. And then if we take M to be, say, the ideal KXY, so not the ideal D, the module KXY over Y, then the support, sorry, the set of associated primes is just the ideal Y. So you can think of the support as just being this... Sorry, the set of associated primes is just going to be the X-axis where Y is equal to zero. And the support, of course, would be the closure of this. So you should then add on all the points on this, but I don't really want to worry about the support too much. And what you can do is you can think of this module as being somehow living on its support. So the module sort of sits here and is zero outside the support. Now let's look at a slight modification of this. This time I'm going to take the module to be KXY, modulo the ideal generated by Y squared and XY. And if we draw this, then the support is again going to be the support this time is just the same as before. So this is just the support. Support is actually going to be the closure of this, but the set of associated primes contains not just the support, but also the ideal zero zero. So here the associated primes of M is Y and the ideal XY. So you can see if you write down a basis for M, you can see it consists of the M that's one X, X squared and so on together with the element Y. And this ideal is the annihilator of the element Y and this ideal here is the annihilator of the element X. Now, what's going on here is this is an example of something called an embedded prime. So let's note to that. So this is called an embedded prime and this is called an isolated or minimal prime of M. So we're going to divide the associated primes and the two sorts of primes, embedded ones and minimal ones. And the minimal ones are the minimal elements of the set of associated primes and the embedded ones are the ones that aren't minimal. So here you can see that this prime is contained in that one, so this one is embedded. And if you're wondering about the terminology for being embedded, it's pretty obvious from this diagram because you can see that this prime is kind of, it's in the closure of this prime here. So this is the prime Y and this is the prime X, Y and you can think of it as being sort of embedded in it if you view it geometrically. And what happens is the most problems about modules are caused by embedded primes. If a module doesn't have embedded primes, it's a lot easier to deal with. In fact, if we go back to the previous example, you can see we actually had some example of embedded primes and minimal primes. For instance, you can see that this is an embedded prime and this is a minimal prime. And again, it's sort of the embedded prime is sort of contained in the minimal prime. And here there are no embedded primes and here there are no embedded primes. So what we want to do now is, well, we sort of shown that a module M can be visualized as a, it can be rather weakly visualized as a subset of the spectrum of R. So you can think of the module in this case as sort of, it's sort of sitting on this point here, but then there's more of it sitting at this point here because there's sort of more of it coming from the embedded prime in some sense. However, this is, although the set of associated primes of a module is more refined than the support, it's still a very crude invariant. For example, if you take this module and you take two copies, some of two copies of this module, it's still going to have exactly the same set of associated primes. So what we want to do is to find a more refined way of looking at the module. And this is going to be something called the co-primary decomposition of a module, as usual finitely generated over a notarian ring. And the idea is to kind of break up M into modules, each associated to one element of the set of associated primes of M. So back in this case here, we somehow want to break up this module here into a module associated with this element of the set of associated primes and a module associated with this element. So let's just see what happens over Z. Well, if we've got a finitely generated Abelian group, then by the structure theorem of finitely generated Abelian groups, we can write A as A naught plus A two plus A three. And so on, where this is going to be free Abelian, Z to the N, this is going to be a two group. This is going to be a three group and so on. And here the set of associated primes of Z to the N is just zero, assuming it's non-empty. And here the set of associated primes is going to be two if it's non-empty. And here the set of associated primes is just going to be three. So we can write any Abelian finitely generated Abelian group as a product of Abelian groups, each of which is associated with just one prime in the set of associated primes of A. And before we go on, we should notice that these groups here are unique. So A two and A three just has to be the set of elements of order of power of two and so on. However, Z to the N is not unique as a sub-module of A. I mean, it's unique up to isomorphism, but what I mean is that there's not a unique sub-module of A corresponding to that. For instance, if you take A to be Z plus Z over two Z, you could take the element Z to be generated by one zero or you could take a different element Z to be generated by one one. So this decomposition is, well, some bits of it are unique and other bits of it aren't unique. And we will see that this non-uniqueness tends to be caused by embedded primes, which as I said, cause all problems. So the next question is, can we do this for finitely generated modules M? So can we write M to be a sum of modules M i where each M i has exactly one associated prime. So we can do this for a billion groups. And the answer is for general rings, no, we can't. And we've seen an example of this before. It's the one we had earlier where you take M to be the module R, modulo Y squared X, Y. And if we draw, if we write down a basis for M, we can write a basis as one X, X squared, X cubed and Y. And now we have the associated primes of M. So one associated prime was R modulo Y. And we can take the sum of this with the module R over X, Y. And we have a map from this module to the sum of these. Well, this is a basis one X, X squared and so on. And this has a basis just of the point one, so that should be a Y squared, point one and Y. And you see this isn't actually an isomorphism. This module doesn't map onto these two. The best we can do is to say that M is a sub module of sum of modules M i. So this will be the co-primary decomposition of a module M. The reason for this, it being called this, is that these are co-primary modules, which we'll define later. For finitely generated modules over no two in rings, a co-primary module is one that has exactly one associated prime, although for general modules there's actually a slight difference between them. So you can see that this isn't really quite a decomposition of M because it's only a sub module of this, but it still gives a more refined picture of M. So each of these modules is going to be associated to one of the associated primes of M. So here this is a sort of sum over associated primes of M. So this is about as close as we can get to the usual structure theorem for finitely generation of billion groups for modules over a notarian ring. So historically, people didn't think of this as a theorem about modules. They thought of it as being a theorem about ideals. And for ideals, what we get is essentially the famous Laska-Nurter theorem. What this says is that if I is an ideal of R, where R is notarian, then I is the intersection of primary ideals where this is a finite intersection. So what's a primary ideal? Well, a primary ideal J is an ideal such that if X, Y is in J, then X is in J or Y to the N is in J for some N greater than or equal to one. So this does like the definition of a prime ideal except we allow some power of Y here. So this rather weird looking definition was actually invented by Laska. So another way of putting this is that if we take the ring R modulo J and it says that every zero divisor is nilpotent, whereas in a prime ideal every zero divisor would be zero. So for example, if we take the ring of integer Z, the prime ideals are zero and P for P prime in the usual sense, whereas the primary ideals are just the ideal zero and the ideals generated by powers of a prime. You might think from this that primary ideals are just powers of prime ideals. Well, they're sort of related to powers of prime ideals but they turn out not to be the same for more complicated ideals. I just finished by saying a little bit about Laska who was one of the more interesting mathematicians. He was actually at one point, the world chess champion. So here's a picture of him and here's his book on chess. In fact, he's in some sense the most successful world champion of all. He held the world chess championship for longer than anybody else has ever held it. And in fact, he proved the Laska-Nurter theorem while he was world chess champion. He sort of took time off to do a PhD in mathematics. So this is definitely the best mathematics theorem ever proved by a world chess champion. And I guess he's also the best world chess champion who was ever a mathematician. He also did various other things. He attempted to write books on philosophy. In fact, if you're interested in seeing his books on philosophy, he mentions them here, comprehension of the world and the philosophy of the unattainable. These were apparently completely ignored by 20th century philosophers although given what 20th century philosophers are like that probably doesn't mean very much. I have an interesting challenge question. Can you think of an idea produced by 20th century philosopher that is both comprehensible and true and not obvious? Anyway, Laska also wrote a play which has been lost without trace and he tried breeding pigeons and this was completely unsuccessful and he tried various other things as well. So he did quite a lot of things apart from the Laska-Nerta theorem. Nerta is of course the famous Emmy Nerta. Laska originally proved the Laska-Nerta theorem for polynomial rings over either fields or the integers and his proof was nightmarishly complicated. His original paper on this was 100 pages long and it was one of the longest papers ever written in mathematics. We're not going to prove Laska's version of it. We're going to prove Nerta's version of it. So Nerta proved a stronger theorem because she proved it for all Nertaian rings and her proof is just a couple of paragraphs long. It sort of condensed this 100 pages into just a very short almost trivial result. So anyway, next lecture we're going to explain what a co-primary module is and what it has to do with a primary ideal.