 This meeting is being recorded. We're going to start with the introduction. Should I go ahead? Yes, so I'm supposed to introduce you formally. Okay. It'll be very brief. Don't worry. No problem. Thank you. It's my, it's my pleasure to introduce Professor Ian Arbarback. His PhD is from Michigan and Arbor. He has been working, I think, first of his work. I saw the phantom homology, I think. And the tight closure of signature. Of course, HK multiplicity and Brian's con is called a type of theorem. So today is going to give a lecture series today and on Wednesday on Brian's con is called a type theorem. So I welcome Professor Robert. Oh, thank you, Professor Triveni. Let me start by thanking the organizers for inviting me to be a part of this. It's a great honor. And I want to thank Suprajo Doss for helping me in preparation of the manuscript. For, for the lectures. And I certainly want to thank professors, Hoxster and Uniki for their. Contributions to community of algebra and their contributions to, to me in particular, I was fortunate enough to be Mel Hoxster student and have Craig as my postdoc advisor. And certainly almost everything I've done since then has stemmed from, from their work. All right. So let me go over a few conventions. Well, first of all, we want to talk about brands and SCOTA theorems. I am going to stick mostly to characteristic P. Of course, these theorems exist in all characteristics. So let me just start with a few conventions. You need to share your screen. Oh, yes, I need to share my screen. Let me do that. Thank you. All right. Sharing my screen. And there we go. That's probably going to work better, isn't it? All right. Good. So let's talk about brands and SCOTA theorems and characteristic P. By RMK, I mean a local, which includes Noetherian ring. The M is the maximal ideal and K is the residue field. As we've been doing R naught is R with the union of the minimal prime, primes removed. We're mostly going to be thinking about domain. So this is usually just R minus zero. I'm going to use E when care, sorry, when characteristic of R is prime, I'll use E for a natural number. And Q is the corresponding power of P. And then we've got the I to the bracket P to the E powers or I to the Q powers, bracket Q powers as usual. And mu of M will be number of minimal generators. So let me start with some history. So in this case, let's let O of D be, we'll start with the complex numbers, early brackets, Z1 through ZD. And this is the ring of convergent power series. Okay. So in this case, if we have a nonunit, so if F is a nonunit, then it was known that F is in the integral closure of its partials. So if the ideal generated by, by F partial with respect to Z1 through the partial with respect to ZD times OD. And so therefore we've got some integer K such that F to the K is in this ideal. So FZ1 through FZD OD are the integral closure. And so Mather asked the following question. He asked, so is there a bound on K? That's independent of F. So let's move that up a little bit. And so that question was answered in a paper that appeared in 1974 by Branson and Skoda. And so their theorem says this. It says the following. So suppose that we have an ideal contained in this ring of convergent power series. And let's say the number of minimal generators of I is L. Then for all W greater than or equal to zero, we have that. If we look at the I to the L plus W power and take its integral closure, that is contained in I to the W plus one. So if you take W to be equal to zero here, we can apply an L to be D, then the bound above, the bound in this question is that K can be taken to be D. So therefore K equals D works. And that was the initial Branson-Skoda theorem. So there was a conference in, I think in 1979, this was before my day, about analytic techniques and commutative algebra and algebraic geometry. And Mel Hoxter had noted this theorem and he said he basically saw that this is a theorem about regular local rings. Okay, so I'll use RLR going forward for that. So prove it with algebra. So I guess I should have said in talking about the Branson-Skoda theorem here, this was proved using analytic techniques. Proved with analysis. So Joe Lipman, who began working on this and along with several co-authors, succeeded in getting the following theorems. So the first theorem, both of these papers appeared in 1981. So we have the following theorem of Joe Lipman and Avinash Satya. And it is the same theorem stated for regular local rings. So it says the following, let RMK be a regular local ring. And let me get the exact statement correct. So if we have an ideal in R and J contained in R is a reduction, I'm going to review the terms here in a couple of minutes. So don't worry if you've forgotten what a reduction is or don't know. And with the number of generators of J being L, then again for all W greater than or equal to zero, we get I to the L plus W integral closure is contained in J to the W plus one. So that is a direct generalization of the Branson-Skoda theorem. And in slightly more general context, what if your ring isn't regular? Then there are still, there was still a result for what are called pseudo-rational rings, which I believe Florian Inescu talked about. So I'll refer you back to his lectures for any definitions. This next theorem again appeared in 1981. This was Joe Lipman and Bernard Tessier. And that theorem says let RMK have pseudo-rational singularities. And well, let's give it say D is the dimension of R, then for all ideals I contained in R and W greater than or equal to zero, I to the L plus W is contained in I to the D plus one. Sorry, D, I to the W plus one. And I made a mistake here. So let me correct that. This L is not an L. There is no L in that comes up immediately in this theorem. That is a D, I to the D plus W. So notice D is the dimension here. So this is a bit weaker than the version in the regular ring. So I'll just write, this is weaker when we have a smaller analytic spread. Again, I'll say what that is in a minute. Okay, so this was the situation for a while. And let me, like I said, let me review some information about integral closure and the reductions. And then we'll get to the characteristic P version. I have a question. Does the theorem of Liebman's state assumes that the residue field is infinite? No, no, it doesn't. I'll mention later, there, there's the kind of the situation here with reductions is you usually you pass to a ring with infinite residue field. And yeah. So I'll mention a little bit later. Okay, so let me do a quick review of integral closure and reductions in no theory in rings. So this actually, I guess, if you've all watched Craig Hewneke's introductory lecture. He spoke some about this I started to get worried as I was watching it that I wasn't going to have anything to lecture on, but fortunately he held back. But let me go over some of this. Okay, so integral closure. Let's let, let's let R be a no theory in ring. And I contained in R. Then X is in the integral closure of I if equivalently we have X to the K is in. Let's see in what I times I X to the K minus one K. The method that we're going to actually use is that there exists a sequence of integers, zero less than or equal to n one less than n two less than three. So in a sequence of integers going off to infinity. And a C in R not such that C X to the NJ is in I to the NJ. So it's this ladder one that we're going to use to show that things are in the integral closure work. Not to show that they're in the integral closure, but that we have this and use that. All right, so that's the idea of integral closure. It's a, you know, gone goes back a long ways it has a lot of information in it. So what about reductions. So let's see. So, let's label that I guess reductions. So again, R is no theory and I contained in R is an ideal. So, so J contained in I is a reduction of I if there exists are greater than or equal to zero such that I to the R plus one is equal to J to the R. That's what it means to be a reduction. So then we've got that the reduction number of J with respect to R to I so with respect to so I don't have to write that whole thing out is the smallest such are all right. So that is the reduction number. So now suppose we have a local ring if our M is local then we can define the so called fiber cone. So the fiber cone of I is I'm just going to use a regular f of I a lot of times that's a script F. So what that is is it's a graded ring. It's the sum of I to the N mod M I to the N. So this will be a graded K algebra. And then the analytics spread of I is I'm going to use L of I and script L and by definition that's the crawl dimension of f of I. So let me give you a few quick facts. I'll call it that are known about reductions and about analytics spread. So all of this was there's a very nice paper of Northcott and Reese. It's very readable that it's the that's in the references in my write up and and it's certainly a great paper for anybody to go through and read to make sense of this stuff. So here is some information. I suppose we have a local ring. Okay, is a local. I'm going to assume basically it's a domain just for simplicity. It says that I and J should be stretched into definition of reduction. It says, oh, Oh, yeah, yes. Okay. So we have a reduction number of I with respect to J. Thank you, Jugo. That's there we go. Okay. Thank you. All right. So and suppose over here we have J contained in I is a reduction. Then what can we say well first of all, that the height of I is a lower bound for the analytics spread and the dimension of our is an upper bound. So there's some sense in which the further away L of I is from the height, the less nice the situation is. So what else can we say. In this case, we've got that the integral closure of I and J are the same. And J is still a reduction of the, the integral closure of I. And lastly, going back to the question I think about Alessandra is if the in the residue field is infinite, then there exist reductions J contained in I which are minimal with respect to containment such that the number of generators of J is this analytics spread. Okay, so these, these are kind of the nicest possible reductions that you can get in in the sense that they're the smallest with respect to the number of generators. So, yeah, let me get a little bit more information here so among the exercises that I've proposed is the show the following. So suppose that we have a reduction J. Let's say it's a one through a, I'll use T instead of L, just because I don't assume it's minimal and I is a reduction. Then here are some things we can do. So first of all, J to the N is a reduction of I to the N for any N. Also, a one to the N through a T to the N is a reduction of I to the N. And another thing that will be useful for us is if the reduction number of I with respect to J, I don't think I put that down here. So let's go back in the notes to the previous page. This is denoted R of I with respect to J. So if this reduction number is R, then move this up a little bit, then we have that I to the N is J to the N minus R, I to the R. So you see when you have a reduction, what's happening is that asymptotically large powers of I are behaving like large powers of J. Or maybe we should think of it the other way when you have a reduction to large powers of J are like the large powers of I. But your original ideal I may have many, many, many generators. Okay, there's no bound on the number of generators for rings of dimension greater than one in general. But the number of generators of J, at least when you have an infinite residue field, is quite small. So that's all the background I believe we're ready to actually get to the theorem. So let's finally talk about the Branson-Skoda theorem in characteristic P. So this is one of the things that we're going to talk about. Branson-Skoda theorem in characteristic P. So this is one of the first theorems to appear here. This is my copy, my well-battered copy of the paper of Hoxton-Hunike. I'm out in 1990, but this is essentially the very first paper other than the announcement in the bulletin that Craig and Mel put out. And I was privileged enough to be able to actually see the manuscript for this as a PhD student in 1988. It really helped to through-freed it. And I learned a huge amount from reading this paper. There's all kinds of things I'd never really known anything about. It's a beautiful paper. And so one of the earliest theorems in this paper is their proof that the Branson-Skoda theorem has a very straightforward proof in characteristic P. Okay, so that's what we want to prove to start with. So here's the theorem. The 1.8, that's just what's in the manuscript. So this was, again, this is in 1990 from Hoxter and Hunike. So let R be Noetherian of characteristic P. Let I be an ideal of R. Let J contained in I be a reduction with mu, the number of generators of J being L. Again, I'm using L as if it's analytic spread, but this doesn't have to be the analytic spread here. Then for all W greater than or equal to zero, we can look at I to the L plus W. We can take the integral closure. And we are contained in J to the W plus one. Now, what's written so far is actually wrong. Because you notice the statement here is about any Noetherian ring, and one certainly shouldn't expect something of this strength. But what it says is that the tight closure captures the stuff that's not actually in J to the W plus one. So the integral closure here of this power is in the tight closure of where one would have it in a regular ring, or a pseudo rational ring, if one were fortunate. And this is this is a theme that you've seen continually probably already and is that that tight closure kind of takes care of the stuff that you would that that doesn't work correctly. In your ring, because you don't have Cohen Macaulayness or regularity or something like that. So, so there is the statement of the Branson, let's go to Sarah. Okay, well, one of the, we're going to prove it in a minute, but let me put one of the key exercises here this is this comes out as exercise to in the notes. Okay, so let's let R be a Noetherian domain. And J contained in I a reduction. So what let me, I want to use different letters and J and I so I forgot I didn't the notes, I didn't change it but for this I want to change it so let's let's call this be contained in a reduction. And then there exists a C in our not which remember here is just our minus zero such that for all and see maps. If we take the integral closure, I did it again. I take the integral closure of a, then that C times a bar to the end is contained in B to that. Okay, so we will use that repeatedly. And that to the student says an exercise. And so now we can go ahead and do the proof. I am going to do the proof when W is equal to zero, because I like to do the easy things. And I will leave the proof for larger W to you guys as part of the exercises. Okay, so how does this proof go. Right. So, let's start where so this is what we want to show. We want to show that I to the L bar is contained in J to the zero plus one so that's just J to the one star. So, let's let X be an I to the L bar. So, then J to the L is a reduction of I to the L. So, choose C as in the exercise J to the L contained in I to the L. And then what we have is that C X to the N is in, whoops, J to the L and for all and okay. So, I guess, maybe we better let's talk about generators so let's let J be a one through a L. That's our hypothesis that we have this L generated ideal that's a reduction. All right, and take N equal Q, which is our P to the E. So, what do we have we have C X to the X to the Q is in a one through a L to the L Q. So, this ideal, this ideal is, we can just think of the a one through a L as they're not variables but when we're trying to compute this we can think about computing with variables and then just replacing by the a is. So, this ideal is generated by manomials in the a is of degree L Q. Okay, so this is generated by manomials in the a eyes of degree L Q. And so, so, if the degree of each. I is too small, so is less than Q for each of these so for one less than or equal to I less than or equal to L, then the total degree is too small. So then it's it's less than or equal to Q minus one L, which is strictly less than L Q that's a contradiction. And good I've given myself enough room here. And so, so that says that C X to the Q is actually in a one to the Q a two to the Q through a L to the Q. And that is exactly the Q bracket power of Jay. And therefore, we have that X is in the tight closure of Jay and that's that's the proof. You know, I encourage you to read the paper of Lippman and Satya, but it's, it's a challenge that that paper is quite difficult. Whereas, this is an argument that you can explain to somebody in high school, if they kind of have the concept of thinking about degrees a little bit. And it's pretty amazing. I think. And so there you have the, the proof, we just did it in a few minutes. Okay, and the proof to expand it to the positive W's is not not difficult either. Okay, so let's see was there anything else I want to say about that before we go on. Okay, well, I guess one application is that we now have a full grand since go to theorem for any weekly F regular ring. So, so corollary. Let's let R M be a weekly F regular ring. Remember that means every ideal is tightly closed. So, for instance, one example where that would happen is, for example, a direct some and of a regular local ring. Then we get the same statement. We get that I to the L plus W bar is contained in J to the W plus one. And the tight closure is the ideal. Okay, so I didn't write it all down but it's the hypotheses are the same as theorem 1.8 I didn't write up all down. So let me address that earlier question actually here. So what if the residue field is finite. So, so in this case, I'll put it in quotes there's still you can still talk about minimal reductions. They're they're not necessarily ie not analytic spread of I generated. Okay, but you can pass to a faithfully flat extension with infinite residue field field and get most of what you want. And this is in the exercises. Okay, so I won't, I won't say much more about that I'll often just assume we have an infinite residue field for that. Okay. So, so that was where things stood in 1990 we now have a very easy proof in characteristic P of the Branson scota theorem, and that then by the methods as as he talked about in his opening talk that that allows one to use reduction to characteristic P for rings containing a field of characteristic zero so one one gets, excuse me one gets the, the tight closure version of the Branson scota zero, as long as one knows how to define tight closure by reduction to characteristic P which is an entirely non trivial type thing so I don't, I don't. The question then becomes, what can we do with with the stuff we didn't use. So what what what do I mean by that what's the stuff we didn't use. So, let's look at the following situation. Say that Jay is a one through a L. Right, so at the in the proof. Right, that's, that's the same Jay here that I want to think about and we computed this. So, I then when we look at Jay to the L Q. This is contained in a one through a L. So let me put cues on those to the queue. Okay, but if let's say M is a monomial in the a eyes of degree. L Q. For some I, what we said, is that for some I, a I to the Q divides. So, say that we write M is a I to the Q. M prime. Right. Well, then this M prime is J to the L Q minus Q. So that's J to the L minus one Q. So really, what we have is that J to the L Q. Well, it's not just contained in but it's actually equal to the Qth bracket power. And then these then times J to the L minus one Q. And the question becomes, can we make use of this extra information and how much time, let's see, I don't am I supposed to go for for 50 minutes. So I'm about at the end of my time. Yeah, it's your talk is for 50 minutes. Okay, so, so, so I guess I'm right at the right place then for getting to lecture to. So let me just say include closing today. What we want to talk about on Wednesday is, how can we use the extra coefficients, things like these guys to make more precise statements about the Brienne since go to zero. So that'll be, that'll be coming up on Wednesday. Thank you. So are there any questions or comments. I don't see any question but I, I'm not sure how do you prove if your ring is regular than every ideal is tightly closed. Do you require a lot of work or it's essentially the flatness of Frobenius in the regular case. So that that C. Right, if you have C. X to the queue in I to the queue. It tells me that C is in I to the queue colon X to the queue. Now, in a regular root ring. That's I colon X to the bracket queue. And, and then this is contained in and into the queue. That'd be for all queue, and that would tell you that C is equal to zero. That's, that's the proof in the regular ring that every ideal is tightly closed. Okay, so are there any further questions. So let's thank Ian for the