 This lecture is part of an online commutative algebra course and will be about the famous Lasca-Nurter theorem. So just before stating this, I just recall some notation. So M is going to be a finitely generated module over a notarian ring R. And we recall that co-primary for a module M means M has at most one associated prime. And the theorem we're about to prove says that if we've got any finitely generated module, then M is contained in a direct sum of modules MP where the sum is over associated primes of M. Of M and each module MP is going to be co-primary with only one associated prime, which is of course the prime P. Well, before giving this theorem, I want to explain what a primary module is. So we can ask, you know, we've got these co-primary modules. So obviously there should be something called a primary module. And the answer is very simple. There is no such thing as a primary module. And the terminology in this area has got completely messed up for various historical reasons. And I will now explain why the terminology has been messed up. Again, I think a little bit of the history of the subject. So I'm going to go through three different versions of the Lasca-Nurter theorem. And by looking at these, you'll see why we've ended up with this funny term co-primary for something that really ought to be called primary. So the original version says that if I is an ideal of a ring R, then I is a finite intersection of primary ideals. So although there isn't a primary module, there's a perfectly good notion of a primary ideal. So you remember primary means that if X, Y is in the ideal J, then X is in J, or Y to the N is in J for some integer N. So this version of the theorem was proved by Lasca for polynomial rings over fields or the integers. And then ME-Nurter generalized it to all notarian rings. Well, another version of the Lasca-Nurter theorem generalizes it from ideals to modules. So version two says the following. Suppose we've got modules N contained in M. So these are both R modules. As usual, we assume they're finitely generated. Then version two says that N is a finite intersection of primary sub-modules. Okay, well, you're thinking to yourself that I announced a couple of minutes ago that there is no such thing as a primary module, and now all of a sudden I'm talking about primary sub-modules. So what's going on? Well, the answer is I put this word sub in and there is a notion of primary sub-module, but there's no notion of a primary module. So in order to understand this, I'd better actually give you the definition of a primary sub-module. So X is called a primary sub-module of N. M, if R M in X implies M is in X or R to the N, big M is contained in X with some N greater than zero. Here R is in R and M is in M, of course. And the first thing you notice is this is not a terribly memorable definition. Next thing you notice is that if M is the ring R, then the sub-module is just an ideal and this just becomes Lasker's definition. So this is essentially Lasker's definition of the primary sub-module, except I don't think Lasker ever actually talked about modules. The next thing you notice is that this is not actually a property of the module X. It's the property of the way X is embedded into the module M. In fact, it's really a property of the quotient M over X and let's call this quotient Y. So what this says is that if R is a zero divisor of Y, this means RY equals naught for some Y not equal to zero, then R to the N Y is equal to naught for some N greater than zero. So this property of a module Y is, if Y has this property, we say Y is co-primary. Well, if you have been paying attention, you will have alertly noticed that I've now given two completely different definitions of what is meant by co-primary modules. So earlier I said that Y was co-primary from the one associated prime and now I'm seeing it's co-primary if it's got this weird somewhat technical condition. Well, we're soon going to sort this out by showing that for finitely generated modules, the two definitions are equivalent. But before doing that, let's just give the third version of the Lasker-Nerter theorem. So the third version says that M is contained in a product of co-primary ideals. So let's see why this is more or less the same as the second version. So to show the second version implies the third version, what we do is we just take N equals zero. And by the second version, we can put naught is the intersection of JP where this is primary sub-module. Well, what this means is the natural map from M to the product of M over JP is injective because the kernel is just the intersection of all these ideals here. And now, although this is a primary sub-module, this whole thing here is a co-primary module. So you see the second version involving sub-modules and modules is unnecessarily complicated. We're talking about two modules and really you don't need to talk about two modules at all. All you need to do is to talk about one module, which is the quotient module. And similarly, the definition of things being primary is a bit silly because you're talking about two modules and all you really need is that quotient, which is just one module. So the clean version of the Lasca-Nurta theorem ignores primary sub-modules completely and just uses co-primary modules. Well, now we should get back to a problem we mentioned earlier, which is that I've managed to come up with two apparently totally different definitions of a co-primary module. So here are the two definitions of co-primary. The first definition says there is at most one associated prime. And the second definition says that Rm equals naught implies m equals naught for R to the nm equals naught for n greater than zero, some n greater than zero. So we should check that these two conditions are the same and as always this lecture, m is finite in generation over a notarian ring. So let's show that two implies one. Let's suppose m is none zero and m is some element of the module m. And let's put p to be the annihilator of m. And we're not assuming p is prime yet. So this might not be an associated prime, but if p is in the annihilator of little m then by condition two, if you decipher it, this implies that p is contained in the radical of the annihilator of m. And that's more or less just a rephrasing of condition two. So let's note that in green, so that we can see it later. Now let's assume p is prime. If p is prime, then we notice that the annihilator of m is contained in the annihilator of little m, which is the element, which is equal to p by assumption. So the radical of the annihilator of m is contained in p as p is prime. So let's mark this condition in green. And if you compare these two conditions, you'll see that if p is prime and is the annihilator of some element of m, in other words, if p is an associated prime of m then p is equal to the radical of the annihilator of m. So there is at most one associated prime and as a sort of extra bonus, we found out what the associated prime is. It can be given as the radical of the annihilator of the whole module m. So next we want to show that condition one implies condition two. Well, we can assume that the annihilator of m is equal to zero because if the annihilator of m isn't zero, we just quotient out the ring by it and carry on as before. So let's put p is the only associated prime of m and this contains the annihilator of m whenever m is not zero. And the reason for this is that all maximal elements of this form are associated primes of m. And since there's only one associated prime of m it must contain all of these. So we just need to show that p is nilpotent because condition two just says that if something annihilates some element of m, then some power of it has to be in the annihilator of m which is zero. So showing that p is nilpotent is more or less equivalent to showing condition two holds. So suppose some a in p is not nilpotent. If every a in p is nilpotent then the idea of p is nilpotent because it's finitely generated of course. Then the localization m8 to the watt minus one is nonzero. And this follows because if x in m8 minus one equals zero, sorry if, and sorry I should have said if x in m is zero in m8 minus one, this means x8 to the n is equal to zero for some n. So m8 minus one equals not implies the n is in the annihilator of m because for some n because it kills some power of a kills any particular element of m and m is finitely generated and the annihilator of m is not by assumption. So a would have to be nilpotent which it isn't. So this module here is nonzero. Next we see that since m8 minus one is not zero, it's set of associated primes is also not empty. So these are primes of the localized ring are a to the minus one of course. So we pick some prime t that's an associated prime of m8 minus one. So this is an ideal of r8 minus one. And now we let q be the inverse image of t in r. So we just recall we've got a map from r to r8 minus one and we've got an ideal t here and we're just taking its inverse image q here. And these are both primes. This t is prime because it's an associated prime and q is the inverse image of t. And q you can check is the union of all the ideals. We can take the annihilator of m which is contained in the annihilator of mA squared and so on. And you can see that from the definition that q is everything which annihilates m times some power of A. This is an increasing sequence of a notarian ring. So the union must be equal to the annihilator of mA to the n for some n. Because eventually the sequence becomes constant. So q is an associated prime of mA because it's the annihilator of some element of m and it's also a prime ideal. But we notice that A is in p and A is not in q. I mean, we assumed A was in p and it's not in q because A is a unit of mA to the minus one. And if A were in q, then this would be zero. So p is not equal to q because of this. So m has at least two associated primes. Well, this contradicts our assumption that m has only one associated prime. So we see that if there's only one associated prime, then Laska's condition for being co-primary holds. Okay, now we get to the proof of the Laska-Nerta theorem. We get the proof of the Laska-Nerta theorem. So this says that every finitely generated module m over notary ring is contained in a finite product of co-primary ideals. And the proof of this is actually embarrassingly trivial. As I mentioned, Laska's original proof of this was a hundred pages of really hard calculations in elimination theory and so on. But by using the correct definition, so we use that the definition, we do this thing for notarian rings rather than polynomial rings. And we use a definition of co-primary, which is much easier than Laska's. So our definition says that something is co-primary if there's only one associated prime. The proof becomes amazingly easier. So let's have a proof. If not true, let's pick a maximal sub-module n so that m over n is not contained in a product of co-primary ideals. I guess that should be a finite product. And we may as well assume that n equals zero just by quotient out by n. And we're going to get a contradiction. First of all, we notice that m is not co-primary because if it were co-primary, then it would obviously be contained in a product of co-primary ideals in other words itself. So it has two sub-modules m one isomorphic to r over p one and m two isomorphic to r over p two, where p one is not equal to p two, and these are both primes. And now we notice the annihilator of x for x not equal zero in this module is equal to p one because r over p was an integral domain and the annihilator of any element of an integral domain is just zero. And similarly, we see the annihilator of x, any element x in here that's none zero is p two. And p one is not equal to p two. And from these two facts, and the fact that p one is not equal to p two, we see that m one intersection m two is just zero because anything in m one intersection m two, its annihilator has to be p one and has to be p two, unless it's zero, but p one isn't equal to p two. So now we look at the following exact sequence. We take m one intercept m two, maps to m goes to m over m one plus m over m two. So this is an exact sequence which holds for any two submodules. We can map m to the sum of two modules and the kernel will be the intersection. Well, we've just said the intersection here is just zero. So this map is injective. Moreover, we're assuming that zero was the maximal submodule that's not a product of co-primary ideal. So each of these two modules is contained in a product, find that product of co-primaries. So now we've found that m is a submodule of this module, these two modules, and each of these two modules is contained in a product of co-primaries. So m must be contained in a product of co-primaries. So we've reached the contradiction by assuming that it wasn't. So that's the end of the proof of the Alaska Nert theorem. As you see, it's one, two, three, four, five, six, seven, eight. It's about 10 lines long, give or take a line or two, which is a reduction of the complexity of the proof of a factor of maybe a thousand or something. And I sort of look at things like that and you wonder what's gonna happen in 100 years? Are the proofs we have now that take hundreds of pages going to look equally trivial in another century? Or are they really that hard? I mean, what's gonna happen to the classification of finite simple groups, which is 20,000 pages long? Are we gonna have a 10-page proof of it in another century? Probably not, but given what happened to the Alaska Nert theorem, who knows? So let's just finish by showing that the Alaska Nert theorem in the form we've given it really does imply Alaska Nert as original form. So here we're just gonna take m to be the quotient of r by an ideal i. So the form we have says that r over i, which is equal to m must be contained in a product of co-primary ideals. And these co-primary ideals, sorry, co-primary modules and these co-primary modules are all gonna be quotients of r. I mean, we can just replace them by the image of r otherwise. So they're all of the form r over ip for some ideal. And we know these modules must be co-primary. Well, if the whole module is co-primary, then it means the ideal here must be primary in Alaska's sense. So we've got a map from r over i to as a sub-module of a product of r over ip's. And you can immediately see that this implies that i must be the intersection of these ideals ip. So the original version of the Alaska-Nurter theorem really is a special case of this theorem for modules.