 This lecture is part of an online course on commutative algebra and will be about the art in re-slammer. So we will be stating and proving the art in re-slammer and then using it to prove that if M is a maximal ideal of a notary in local ring and you take the intersection of all powers of M then this is equal to zero. So before stating the art in re-slammer, we first need to define filtrations of modules. So a decreasing filtration on a module is a series of sub-modules of M decreasing like this. A typical example is where you take module MN to be I to the N times M where I is an ideal of R. So this filtration is called the I-addict filtration. The name I-addict comes from the special case when you take the module M and the ring R to be the integer Z and you take the I to be the set of all multiples of a prime P. Then the I-addict filtration is given by putting MN equals P to the NZ. And this is the usual P-addict filtration on the integers used to construct the P-addict numbers. If you know about completions, then the P-addict numbers are just the completion of the integers with respect to this filtration. We'll be covering that a bit later in the course. So the name I-addict just comes because it's a generalization of the P-addict filtration on the integers to an arbitrary ring. Any filtration gives rise to a topology on the ideal. Here, the topology is just given by taking the filtration and saying that's equal to a basis of the neighborhoods of nought in N. So, for example, a basis of the open sets in M will be the set of all sets of the form X plus MN where X is some element of M. And there's a basic question. Suppose M is a sub-module of N. Then we can put the I-addict topology on N N, which is the set of things of the form I to the N times M. And we can put the I-addict topology on M, which is the set of things of the form IN times M. And if we restrict the I-addict topology on N to M, then we get the sets M into section I to the N times N. So we seem to have two different topologies on M and we can ask, do these give the same topology? And the answer is, in general, no. To see an example where these topologies are different, let's just take R to be the integers, M to be the integers and N to be the rational numbers and I to be the ideal of all multiples of two. So M contains N is just Z contained in Q. And now we can look at the two-addict topology on Q and we notice that two Q is equal to Q. So in fact, the only open sets are the empty set in the whole of Q and the topology here is indiscreet. So the open sets are zero in the whole of Q. So if we restrict this to Z, we will get the indiscreet topology on Z. On the other hand, the two-addict topology on Z is Hausdorff, as you can easily check. So it's certainly not the indiscreet topology. So here, the two-addict topology on the rational is restricted to Z is not the two-addict topology on Z. In fact, even when the topologies are the same, the filtrations may be different. For an example of this, let's just take R to be the integers and N to be the integers and N to be four Z and I to be the ideal two. So M is contained in N. And now let's look at the two-addict topology on N. So we have N contains I, N contains I squared, N and so on. And these are just Z, two Z, four Z, eight Z and so on. If we take M intersection with these sets here, we now get Z, sorry, four Z, four Z, four Z, eight Z, 16 Z and so on. On the other hand, if we take I to the N, M, we get four Z, eight Z, 16 Z, 32 Z and so on. So these two filtrations are obviously not the same. They're different filtrations. On the other hand, you can see that they're the same topology because every set in one filtration actually does occur in the other filtration. It just might be shifted a bit. So these are different filtrations, but the same topology. And this is the sort of phenomenon that the Arten-Ries theorem, allows us to control. So what we need to do for that is to find stable filtration. So a filtration M, N of M is called stable. If I times M to the N is contained in I in M, N plus one N plus one and equality holds for all large integers N, or sufficiently large integers N. So for example, if we take M, N equals I to the N, M, this is stable. And it's also obviously the smallest stable filtration because any stable filtration must be at least as big as this by this condition here. And furthermore, any two stable filtrations give the same topology. For example, we have M naught contained in M one contained in M two. Suppose this is stable. We can compare it with the filtration M contains I M, which contains I squared M and so on. And now we know that I to the K M is contained in M K for all K by what we just said up there. On the other hand, if N is sufficiently large so that M N plus one equals I M N. So suppose we choose some big number N such that this holds then M N plus K is equal to I to the K M N, which is contained in I to the K M. So while these submodules may be bigger than these submodules, they're all bounded by these modules shifted by a finite amount. So in particular, these two, every module here is contained in one of the modules here and every module here is contained in one of the modules here. So they give us the same topology. Well, having discussed to find stable filtrations, we can now state the art and Reese lemma. This says suppose that M contained in N are two finitely generated modules over a notarian ring are with some ideal I. Then there are two forms of this. We can have a strong form, which says that any stable filtration on N restricts to a stable filtration on M. There's a weak form of the art and Reese theorem, which says that the I addict topology on N restricts to the I addict topology on M. So the weak form obviously follows from the strong form because as we said, any stable filtration gives you the I addict topology. In order to prove the art and Reese lemma, we need the following key step. So the key point is that a filtration M0 contained M1 and so on with I times MN contained in MN plus one is stable if and only if the module M0 plus M1 plus M2 so on is finitely generated over the ring R plus I plus I squared and so on. So these are graded rings and modules so that the degree is zero, one, two, three, and so on. And it's very easy to check that this is a ring and this is a module over this ring because of this condition here. And to see that stability is equivalent to this thing being finitely generated is quite easy. So for example, if this module is stable, then at some point you have MN plus I, MN plus one plus I squared MN plus two and so on. And you can see that if the filtration is stable from this point onwards like this, then this module is finitely generated because you just take a finite set of generators of all the modules M0 up to M1 which exists because the ring is notary and M is finitely generated. And you can see these former finite set of generators of this entire module. So stable implies finitely generated. On the other hand, if this module is finitely generated, then you can take some N here such that all generators are in M0 up to MN and it's then easy to check that these generators form M0 up to MN. A set of generators for the entire module. So finite generation of this module here is equivalent to the filtration being stable. Okay, with that, it's now really easy to prove the art and race theorem. So we just say suppose the filtration N0 contained in N1 and so on is stable. Well, this implies that the module N0 plus N1 plus N2 and so on is finitely generated over the ring R plus I, plus I squared and so on. And now we notice that this ring is actually notarian because it's a finitely generated algebra over the notarian ring R. And it's a finitely generated algebra because you can easily see the set of generators for this ideal I form a set of generators for this algebra over R. So since this ring is notarian, any submodule of a finitely generated module is finitely generated. So M0 plus M1 plus M2 and so on is a finitely generated module. And finally, this immediately implies that the filtration M0 contained in N1 and so on is stable where I should have mentioned that MN is of course just equal to M intersection NN. So that proves the art and race lemma. The proof looks kind of simple, but it's actually really tricky to find. You've got to have this idea that stability of a filtration is equivalent to this module being generated over this algebra. This algebra, by the way, is sometimes called the blow-up algebra. And I'll probably say a bit more about it in the next lecture. So to finish the lecture, we will now show the theorem I mentioned earlier so we're going to have this application. Suppose R is a notarian local ring with maximal ideal M. Then the intersection of all powers of M is equal to zero. So in the previous lecture, I gave some examples of non-notarian local rings where this equality failed. And this follows really easily from the art and race lemma together with Nakayama's lemma. All we do is we put J to be the intersection of all powers of M. And we notice that M to the N intersection J is equal to J for all N, completely obviously. So the M addict topology on R restricted to J is indiscreet. However, J is a finitely generated module because R is notarian. So by the art and race lemma, we see that the M addict topology on J is indiscreet because it's the restriction of the M addict topology on R. So this means that M J is equal to J because that's what the M addict topology being indiscreet means. And now we notice that this output from the art and race lemma is exactly the input required by Nakayama's lemma. So J is finitely generated and M J is equal to J. So we just apply Nakayama and we find J is equal to zero. So the intersection of all powers of M is equal to zero because that's what J was defined as. Okay, next lecture will probably be about the blow-up algebra and various other algebras you can construct from filtrations.