 Our next speaker, so Arina Swanson from Purdue, tell us about the Uniform Martin Reese. Okay. Thank you so much for having me here and I especially thank Juga Verma for organizing everything and putting so much effort into organizing this and just having the big vision for everything that we're doing and this is not the only course that we're doing in our tight closure. So I want to, what about, how do I get back to the, somehow it scrolls way down. So I'm talking about Uniform Martin Reese Lama. And in blue there is something that wouldn't that be nice to have. Let R be an Ethereum ring and N a subset, a sub module of M, they're both finally generated our modules. And is it true that there exists a positive integer K such as for all ideals I and all integers and bigger than equal to K. We either have this strong version of the Arina Reese Lama I to the end in times M intersect with N is equal to I to the N minus K times the sum the module I to the K M intersect with N. So maybe we have this, or maybe we only have the the weaker version. And so we only have this inclusion here. Oh, that's why it's not right. Maybe we only have this inclusion. Well, the way I wrote this right now, neither of these two versions is true. And what happened here. Okay, so what is true is this uniform Martin Reese Lama is true if we impose certain conditions. And so Craig Hewnecker wrote a paper and published a paper in 1992. And he proved that this is true if I was essentially a finite type over in the theory local ring. If I was a finite of characteristic P or if I was essentially a finite type over Z. And I cannot possibly go through all three of these in the three lectures that I have, especially I cannot go through part three because that requires reduction of characteristic P. But these three cases are not the only places where we have some version of uniform Martin Reese Lama. And there are some open parts you can possibly prove another one. But it is not true in the big generality. And so let's first talk about our first talk about that but before I get to that. Let's talk about the classical. Art and Reese I think we have to start with that. This was published by David Reese in 1956. And it was a lemma inside a paper that he had for the proof of the crop. And it was lemma one and ever since then this has been called a lemma, but it's a very important result so we are usually told that if something is a big result we should call it a theorem. Yet here we have a lemma, and it's staying a lemma I think maybe that's for affectionate not just historical reasons, and you really like something then you call it cute and little even if it is giant and big importance. So it has the name art in there, apparently according to David Eisenbach's book, commutative algebra with a view towards algebraic geometry, he has a footnote that Emile Arton lectured on this in the greater generality so art David Reese proved it for rings. And so there exists he proved that there exists K, depending on N and M. Well, this is the general statement that will have such that for all ideals I and for all integers and bigger than equal to K, I to the end. No, no, no, no. This also depends on I. I to the end M intersecting M is I to the end minus K, I to the K and just say the end. So David Reese only proved this in case M is the whole ring and N is an ideal, and then Arton lectured in this in generality and Matsumura writes in the book that this was proved independently by Arton and Reese. Okay, and we have all seen many uses of Arton Reese Lemma so when you're doing completions, Hilbert functions are for the while you could can be used to prove the current crew principle ideal theorem co intersection we compare iodic topologies with this. It has also been used for existence of superficial elements and sequences the prove Reese is there for comparing valuations to determine final generation of ideal transform rings this was in some lecture notes by Peter Schenzel. So there are many notes and exercise zero which will not appear in the final version of this notes but if you can give me other examples of uses of the classical Arton Reese lemma then I'll add them here and move exercise zero from the back. Well, so this classical Arton Reese lemma is already powerful. But why would we want to have this uniform version where K doesn't depend on I. The first time this appeared in print but only as a question was in a paper of Eisenbad and hawk hawkster and hawkster. It does the question, but they only want said when I varies over maximal ideal. And they in their paper, they were proving the. One of the risk is main lemma so the risky has several main lemmas and in fact, there was a main lemma he had them homomorphic functions. We don't need to worry about what that is. But the risky had a main lemma inside there and then, again, another lemma perhaps it should be a theorem and hawk Eisenbad and hawkster then gave another proof in a greater generality but it said well. Another possible proof would be if we had a uniform Arton Reese of this former the ideals could vary over maximal ideals. And then that version was proved. The weak form, which is all that was needed week form was given by O'Carroll him O'Carroll. And that was in 1987. And, and in that paper, he used arbitrary Eminem, but he used to prime filtration and we'll get to infiltrations in the proof of the uniform version of Unicus as well. And then Duncan and O'Carroll had a paper. Two years later. They proved it in a strong form. And then there was I want to mention one more thing. O'Carroll proved another strong version. And that was later I can't remember the year, and he proved it when I varies over principal idea. And for this he needed some primary compositions of what it was doing. So, so this is what I'll say about the background and now we will go towards proving the this uniform Arton Reese lemma well, we will only be proving the week version. And, and, but I don't want to start with these restrictions here, I want to start with an arbiter in the theory and ring and do as many reductions as I can, until I'm forced to make some extra assumptions. So, that's how we'll proceed. Okay, so what we will do in the step one step one, we will do a reduction to the case where M is the whole ring, and N is a prime ideal. Perhaps it's not clear why we would want to do that but in general working with some the whole ring and prime ideals or some ideals in general is much easier than working with arbitrary modules. Okay, so. All right. So the first thing to do is one part a is we do and is a sub module of M and let's call this L zero and now form a prime filtration. And maybe we finished it else of us. Okay. And so we have a prime filtration that means that for all I, the relevant li mod li minus one is isomorphic to our mod PI for some prime ideal piece of our. So, I trust that you have seen this before, so I will not go through the proof. And then one part so still reduction to M the whole ring. I suppose that if there exists K sub by such that for all ideals, and I, and for all integers and bigger than equal to K sub by L sub I plus one, I to the. Maybe I should call them because ideals are I so indices should be J ideal I to the N times L sub J plus one intersect with Elsa by. So if we know that this is containing I to the N minus case of I times Elsa by. And this is for all of we're doing J's here. And this is true for all J for all J. Then it's an easy exercise that for all and bigger than or equal to the sum of these case of J's I to the end. And for all ideals I, I to the end and intersected with and intersected with and is containing I to the N minus the sum of case of J's when he didn't know that's it. So we have a prime filtration. And then if we can prove the result for each one of the adjacent modules, then you know it also for the big chain. Right. And then the final reduction is. M mod n. So we reduce to the case with M mod n is isomorphic to armot P and P is a prime ideal. And, and if there exists K such as for all integers and bigger than equal to K. And for all ideals I, I to the end. And then intersect with P is containing I to the end minus K P. Then we conclude that I to the end and intersected with N is containing I to the end minus K P. That is supposed to be. So, this is, I'll actually prove this. So, go up here. This is our reductions. Okay, so proof of one C. One C. Okay. So, what we have is, since M mod n is isomorphic to armot P. We can write M as N plus R times some element X for some X in N. You know, X in M, such that the set of elements, the ideal of elements that multiplies X into N equals P. So that, that, thus we have I to the end and intersect with N. Well, I to the end and M is just N plus R X intersect with N. And now we use the following fact, we have a plus B intersected with C. And if a is a subset of C, then this is just a plus B intersected with C. So we will use that up here. So this is just I to the end N plus I to the end X intersect with M. And now we have to work with this part. So, let Y, okay. Let Y be in I to the end X intersect with N. So we can write Y is ZX for some Z in I to the end. Okay. And then, but Z also multiplies X, Z is an element of R, multiplies X into N. So, Z is therefore in I to the end intersected with P. So we have an assumption. We're assumption here this is in I to the end minus KP. So that's where Z is. So why, which is the X is therefore in. This is in I to the end minus KP and then we still have X. But P times X, that's in N. Okay. So we proved this with one part C. So what we just accomplished, what we just do. Why are you deleting things? We just reduced, I changed the red color didn't work. It only works in one screen. All right. So we may manage to reduce to this from now and we don't have to deal with arbitrary modules. We only have to deal with the whole ring and a primary deal inside. Okay. Any questions about this? Okay. Then I guess I should scroll. Now, so now let's go step two. Red color. Okay. Step two. What are we doing step two. Oh, I skipped what is in my notes. Step one was a reduction to faithfully flat extensions. Okay, so I'll write here. Okay. I'll write step zero or is we may, we may always pass to a faithfully flat extension. So if we can prove the uniform married and we slow mine faithfully flat extensions, then it's true wherever we started. So now I'll call the new step to agree with the notes on Moodle. I'll call this not step two, but step three. Step three is a reduction to only using I being just a primary ideals, ideals that are primary ideals that are primary to maximum ideas. And maybe I should say something more about step zero step zero is trivial. I said a lot more. Go so let's go back to step three. So, if we can work with M primary ideals then you have a little bit of a power about understanding of primary decompositions and when we lift things from various parts. Things will be, we can force things to be contained in those high dimensional high height ideals. So this is a pretty an important step as well. So, first, let I be arbitrary, I'd be an arbitrary radio. Then, if we need to understand this. Well, take an arbitrary maximal ideal. Actually, let's take an arbitrary maximal ideal, raise it to the elf power L could be large, and then raise this to the end power. We don't need m's and m's anymore. Well, I could just work with RSNPs, but it's fine if you have modules. Okay, and this is true for all positive integers L and for all maximal ideals of the ring. And what is true. If we know the weak uniform art and re SLEMA for M primary ideals for zero dimensional primary ideals, then there exists a K. There exists K that makes this intersection here. It's contained still over all L and all M that's containing I to them L and minus K times M. But then we can be generous. This again L M. This power here is containing I to the M minus K plus M to the L. We're very generous, but it's okay. Because now, when we vary over all L and all N, all maximum ideals M and all exponents L, this is just I to the M minus K. Right, so that, that at least we have only N primary maximum ideals so accomplished. We accomplished this part here. Any questions about this. And so we're here. And so in particular, in the class of generality. We only take I is primary to a maximum ideal. And we need to show there exists. So, without last reality, we have to show that for all prime ideals P. It's okay, such that for all primary ideals that are primary to maximum ideals, I to the N, such that for all and this is containing I to the M minus K P. Now, we may be tempted to say well why don't we just localize what is true that if I is M primary. Then, I to the N intersect with P is I to the N minus K P, if and only if it is true locally. After localization at M. Well, globally, then it's true locally. If it's true locally at this M and I is M primary. Well, globally, when we take this thing and this thing here. We don't need to localize it any other prime ideal, any other maximal ideal, if they don't contain I, then the left side is just P localized that maximum ideal and the right and so it's the right side so of course, this inclusion here words that works after localization anywhere else. So, well, this looks pretty promising. So is it true that it's suffices to prove this. It's suffices to prove this reduction to using a weak art and reslam uniform art and reslam only after all the localizations. And that's not quite true, because if you first localize it all these maximal ideals, then you get K that depends on this localization. And we want a global K that works before localization. So what is really happening. So, what we want is global K, but inclusion can be proved locally. And sometimes locally we have quite a bit more leverage for proving this. So we need it. Yes. Professor I guess there is a correction here pointed out by somebody in the chat in the if and only if statement when you're localizing. Yeah, is it. Oh, right. Thank you. Yeah. So the intersection should be with prime ideal. Yes, this should be localization that and was that the question. It says intersect with our localization. No, this we need. Oh, here. Yes. Yes. Thank you. Yes. Thank you. Yes. Thank you very much. Thank you, Alessandra. Okay. Now it might be true. Yeah, it's true. Okay. So the important thing is, we need a global K, we can just first localize find a K, we need a global K, and then we can localize to prove this inclusion here. And then, and the consequence of this. Oh, you can just prove things. Other consequence. Okay, here I can show more. Well, it's no good spot. And here I can go up. Another consequence is, if you have to prove inclusion locally, then we may in addition assume that locally. We find a global K, then we localize at a maximum ideal we need to prove the inclusion there locally. And then now we use the, that's step zero that I forgot to say, first, that that locally we can pass to a faithfully flat extension. Because to get infinite residue field. So, I won't say much about this but if you start with an Ethereum local ring, then you, you join a variable X. That's not a local ring anymore. But if you invert all elements that are not in the ideal extended, the prime ideal extended from M, then now localize. This is a faithfully flat extension. And this ring here has infinite residue field. So, maybe we want to prove something for all ideals in here, if we could prove that there exists that this K, this global K key here works for the uniform art and reslam after localization first in our localize the M and then after passing to this, then we will know it also in the original. Okay. So what's the advantage of infinite residue field. Advantage reductions exist. Minimal reductions exist. Well, with minimal reductions. That's why do we need minimal reductions, we're trying to prove the art and reslamma. There are, there's nothing about integral closure at all. Minimal reductions have to do with integral closure. However, at least in Unicus proof. And actually in his paper, he shows that this uniform art and reslamma and uniform, integral closure inclusions are very tightly related. And we'll talk more about that. So, for now, let's just keep it at this and we will use it eventually. Any questions about this first. Okay, so let's go for step four. Well, step four says reduction to minimal reductions. Okay, so this is two different meanings of the word reduction. Except this is not quite true. Not completely we won't just replace. Said, all right, we only need maximal ideals that are primary to maximum ideals. We won't say now we only need ideals that are minimal reductions of primary deals to maximum ideal so no, it won't be so it won't be a complete reduction. However, it will be. It will be good enough for whatever we want to do. So, here's a lemma in my notes it's lemma 3.8. And in Unicus notes it's lemma 3.1 in Unicus papers lemma 3.1. So, let I J. PB ideals in the ring R. So this is global ring. I is contained in J, and P is the prime ideal. And let H and K be positive integers. Or at least non negative such that for all integers and I will stop writing and bigger than equal to K or and bigger than equal to H because I to a negative far as a whole ring so it all works. So for all integers and we have two parts one is J to then intersect with P is in J to the n minus K T. So here you may want to think about the J is probably I read. Oops, I said it the wrong way. J is J. J is a sub ideal of I. And J is, to some extent, a reduction of I is not quite. But here I'm not stating that it's a reduction but in the future maybe we want to think of J as a reduction. So here, supposedly now we may have a uniform martin Reese for this reduction ideal maybe for all reductions. So suppose that we know uniform martin Reese for J. And now we need to know something about I suppose that I to the end intersected with key is containing J to the n minus H plus key. Okay, so here, maybe we're getting a hint that J is not quite a reduction of I but maybe J is a reduction of I mod P. So this is true for all. This part two is true for all integers, large integers and that means that mod P. Well, no, it doesn't quite say that. But at least the mod P something good is happening. All right. And so, if we have these assumptions, so this is and then what's the conclusion. I to the end intersected with key so now we get the uniform martin Reese for ideal I which might be arbitrary. So this is in I to the end minus K minus H minus one P for M. This is true for all me. Okay. So if we know it for these, maybe today's a reduction of some sort, then, and we know some good relation between I and J so these are assumptions one and two. And we also know the Artemis, a uniform argument is for arbitrary ideals. Okay, so proof. First to claim that for all and I to the end is containing J to the end minus H plus I to the end minus K minus H minus one P. And this, the first end that we have to worry about is when any escape plus H plus one. So then this is just I to the zero. And I to the zero is just the whole ring. So this is just. This is the assumption to. Right, so my assumption to this whole so now we have a base case. If any slower than that then it's still base case. So, if we know the claim for some end. It's bigger than or equal to K plus H plus one. Then, well, what about the next and I to the end plus one. So since we know it. So, so we if we know it for some end, then we have this J to the end minus H plus I to the end minus K minus H minus one P. I didn't mean to you have extra intersection. You wrote an extra intersection in two intersection says no so I'm a suit. I'm assuming this. Where's intersection. This is lost here in hypothesis to you don't have intersection. You don't need into six right there. This is not an intersection. But you wrote I to the end intersect with me. I mean the left hand son. Yeah, which part are you talking about. Yes. Condition to this and. Yeah, there's no intersection here. Here there's. Oh, thank you thank you. There's no intersection here. Thank you. Yeah, there's no intersection there. Thank you. Who's that. Thank you. Yeah. Okay, thank you. Yeah, because otherwise this wouldn't be to over here. Right. No, this wouldn't be to here. All right. All right. So if now by assumption we do have the base case because of the correction that we just saw. And now suppose that we, we have induction step here. And then let's multiply through by I. And then we get this. And we also know by, by part two. This is contained in J to the N minus H plus B. So, oh, I see if I can. I have more space here. So we have a now this again. This part here is a sub ideal of the left side so I can put this in here. So I'm going to pass the intersection of the rest. Interceptual J to the N minus H plus B. Oh, no, no, no, no. This, because I have N plus one here I also have to have N plus one. Plus one. Except that I don't need that N plus one now I will be a little generous. This is contained. I to the end some, I keep copying this. And then we're here, this is generously containing J to the N minus H intersected with J to the N minus H. I guess plus one plus B. Now, this here is a sub ideal of that so we get I to the N minus. This is a sub ideal of the left side. And this is a sub ideal of the right side. Oh, right. I'm still just proving the claim. Okay. N minus K. Plus H. P. And this is J. N plus one. Minus H. Plus the intersection of the rest. Intercept. And this here is by our assumption one. This is now contained in I to the N minus K plus HP plus J to the N plus one minus H plus J to the N minus H minus K. I approve the claim. I to the N plus some going back here. I to the N plus one should be contained in J to the N plus one minus H plus I to the N plus one minus K minus HP. And that's, well, it's, I need just one more line here. So today is a sub ideal of I. So this is a sub ideal of I to the N minus K plus H. So that proves the claim. Let me just go back to blue color. So now let's go back to we really want to prove this, this inclusion then. Right. So then what do we have. I to the N intercept was P. Well, we use the assumption to this is containing J to the N minus H plus P intersected with P. Actually, no, we will use the claim. What does the claims say. I to the N is containing J to the N minus H plus I to the N minus K minus H minus one P and we intersect that with P. And this again, this is a sub ideal of peace so we can pull it out. Plus the intersection of the rest. And now we use assumption one. This is contained in. So first a couple of this. J to the end minus H minus K. And J is sub ideal of I. And anyway, this is contained in there. So we crude what we want it. So this proves the lemma. So what this lemma will enable us to do is that well, there are sometimes when so called reductions or almost reductions will be the workhorse and if you can prove it for those almost reductions which we have to make more precise then things will be pretty good. Now, this is, I only have two more minutes so let me finish. So let's look at this assumption. I to the N is containing J to the N minus H plus P. So if we knew that so this means that J, which is a subset of I. This is a reduction mod P. So, if this is true for all in sufficiently big. This means that we have a reduction mod P. So we do have a reduction. Well, so well, why don't we always use reduction. The problem is that here age, perhaps depends on the reduction, or at least on the ideal I. And then that would not satisfy uniformity. So what I will make here is like a wish and definition. And I think this really shows the power of how unique it thinks. And sometimes I met he has these definitions, and maybe they're true. And, and then he works with them and well when are those wish definitions to, and that's why the, the uniform mind we see Lema doesn't work for all rings because these wish definitions are not always true. So, for every positive integer for every ring R and every positive integer K, define T sub K of R to be the intersection. You vary over all the ideals. So for all integers, and you take the integral closure of I to the end, and see what elements in our multiplied into the ordinary. And then you take lower power of the ideal I so that's T sub K, and then let's define T of R to be the union of all of these T sub case. So, one big thing that can unique at the proven we'll get to it next on Wednesday is these T of ours are not always zero. In fact, they're not always continuing minimal prime there, they can be pretty big. So this gives us like a uniform. So, analytically unremifed. So, these rings have to be analytically unremifed for these to exist. So, uniform analytically unremifed, unremifed elements, elements or something. I don't know they he he did not name them. So, so this is where integral closure does come in well that's still not convincing that we will need it but we'll get to it next time that it is indeed needed. Okay, I'll stop here. Are there any questions. There are no questions. Thank you so much arena. Thank you. Thank you. See you later. See you all in sick.