 This meeting is being recorded. Okay, so Professor Abirbakh will resume his talks on Brionskan Skoda Theorem. All right, well, thank you very much, Professor Trivedi. Hello, everybody. I said good morning earlier, but I think I'm one of the few people for whom it's morning. All right, so last time we discussed the Brionskan Skoda Theorem in character from holds up to type closure. And we ended the talk by noting that this reduction J if it's L generated, then a key observation was that J to the LQ is contained in the Q's bracket power. And then we said, well, what about the kind of remaining powers of J can we get any more information? So, let's suppose that we don't have the type closure here so that we're in say a regular ring, then if we look at I to the L in a real closure that's contained in J and say that's generated by A1 through Al. So, then an element X in I to the L bar implies that X is equal to R1 A1 plus R2 A2 plus RL Al. And the question then becomes what can we say about the R? So what can we say about R1 through RL? So, for instance, can we say something really nice like are the R's in I to the L minus 1? Well, that's probably unlikely. But it turns out we can make some statements. And one has to be careful. It's difficult to do the computations here because of course the any information you get is up to say causal relations on the R's with respect to the A's. And it's not necessarily that easy a computation. So that's the beginning of today's lecture is to talk about Branson-Skoda theorems with coefficients. In other words, we're looking to say ask to get some information on those R's. Let's start with some, a couple of definitions. If I is contained in R, and here I just want to assume that we're in a netherian ring, then let's let I un-mixed be the intersection of the minimal component of the ideal I. So this just removes the embedded components. Okay. And the second definition we need is something related to the height, what we call the big height of I. So this is equal to the maximum of the height of P such that P is a minimal prime of I. So this is exactly the height if this ideal has minimal primes of all of the same height, but in general, we can have heights, different heights. And so the big height might be strictly larger than the height. So here is the first theorem. I'm using again, I'm using the numbering from the write-up. So this was proved in 1993 by myself and Craig Hewneke. And what it says is the following. Let's let our M be a regular local ring that contains a field containing a field. Say we have I is an ideal of R. Let's let L equal the analytic spread of I. And let's let H be the big height. I've usually used H for the height, but here it is the big height. So and J contained in I a reduction. So then for all W greater than or equal to zero, if we look at I to L plus W integral closure, this is contained in J to the W plus one. This is a regular ring. So that, that by itself is the Branson-Skoda theorem. But what we can say then is that this is in J to the L minus H unmixed. So in other words, we can take the Rs on the left-hand page to be in this ideal that is related to I itself. And we were interested in this in part because it. Hewneke that he had proved about the connection between Cohen Macaulay graded rings and re-sranks over Cohen Macaulay rings. I think maybe there are some people who could mute who that might be helpful. So I want to do the proof of this. Let me make some observations. And so what are my observations? I'm already forgetting. Okay, so in, in this setting. So if, if we have J contained in I a reduction, then the minimal primes these two are these two ideals are the same up to radicals so the minimum of ideals, primes of J are the same as the minimum primes of I. We have that the big heights are the same. So that's one thing we'll need. And the second is that in a regular local ring, well in a regular ring of characteristic P. So for any ideal a contained in are the associated primes of our mod a to the bracket Q are these are equal to the associated primes of our mod a that again that's the flatness of Frobenius. So, most of the work is in the following lemma. Let's prove this lemma. So we're really working on this serum so let's let R I J L. Maybe H should show up here too. As in the statement of the serum. And what is our lemma say, it says, yeah, let's throw in H. It says. This is an F. Not equal to zero such that for all Q sufficiently large. If we multiply J to the L minus one cube by F. Then we are in the following ideal J to the L minus H. Unmixed all taken to the Q bracket power. Okay, so what this is saying here is, you look back at our question. Over here, this extra information. If we have, we'd like to get something from it. And so this is saying there's a uniform multiplier. That takes that quantity and puts it in a Q's bracket power and sure enough, right. It's the Q's bracket power of the coefficients and so that's how we're going to prove our theorem. Okay, so. Let's look at the proof of this. All right, so let's, let's let P be a minimal prime of J, which is also a minimal prime of I. So, then, if we localize, so J localized to P. This has a minimal reduction. That is generated by the, the analytics spread of J RP. But that's less than or equal to the dimension of RP. And that is less than or equal to the big height because we took a minimal prime of J here. So, Okay, so let L sub P be such a minimal reduction. And right. And of course, the number of generators of L sub P is less. So, we can choose. I'm going to call it F sub P. This is not a localization now this is an index to P. As in exercise to, and that's this LP contained in J RP. Okay, and without loss of generality, we can assume that F P is actually an R we can clear denominators. So, then, what do we get. So we do the computation. As in the proof of theorem 1.8. Okay, and now using the work that you guys are going to do so in exercise three. Here is the what the computation gives us, we have that F sub P times J RP to the L minus one Q. Where does that sit. Well, you see the L in when to localize the analytics spread is dropped so what's what the now L has become a W that's bigger than or bigger than zero. Okay, so this is contained in. We're putting things in a power of the reduction. And so where, what power do we get. Well, once we finish the computation we're in a cute bracket power, but the power is, we've got L minus one minus H plus one. We're taking that whole thing to the cute power. So again this is just a right this is this is a W in the in the argument. And then this is where subtracting off something that's at least as big as the analytics spread, and then the plus one. Okay. So where is this. Well now we can make our. So this is, once we simplify. I guess it's we're just doing a simplification. This is LP to the L minus H to the bracket Q. And that's contained in J are P to the L minus H to the bracket Q. So, the thing is now that we apply observation to. By observation to, and take the product of all the FPs to get our result. Okay, so perhaps I haven't been as clear as I could be but I'll let you think about it or go back to the notes where it's maybe written up a little bit more can clearly. And so this is the key to the whole thing. So I think I gave myself an extra page, meaning I was going to say more. Now we can go to the proof of the actual theorem. And again I'll do the W equals zero case. So, let's prove that I to the L integral closure is contained in J times I to the L minus H quantity and mix so we again remember when we when we take powers of an ideal even if we started with an ideal without any embedded primes were likely to get embedded prime so we have to remove them in in this argument. Or in this theorem. Okay, so let's let J contained in I be our minimal reduction I guess that's actually stated in the theorem. And what do we want to do take C. Not equal to zero as in exercise to. And that's for the ideal I to the L are F not equal to zero as in the lemma. So we start with some x in I to the L bar in the integral closure, and then for all q sufficiently large. We have the following we look at C times F x to the q. And so where is that well, using the C that's in J to the L Q, which is now contained in J to the bracket q J to the L minus one q times let's put back that F. Okay, and now by the lemma. This is contained in J to the q bracket q times J to the L minus H on mixed to the bracket q. Bracket cues are nice in the sense of multiplication so that's J times J L minus H on mixed all to the q. That's exactly the criteria for being in tight closure. So, X is in J times J to the L minus H on mixed closure, and that is equal to J times J L minus H on next, because we're in a regular local ring and so all ideals are tightly close. And one of the early kind of revelations to me was the idea that you know in a regular ring or in weekly F regular rings in general ideals are tightly closed. And so you might think well, you can't really use tight closure to prove theorems in such a ring all that easily. But it turns out that using the tight closure is incredibly powerful even in rings, or especially in rings, where every ideal is tightly closed. So, I did include in the notes a some information about how we use this stated graded rings and column calling us but I decided that in this talk, I wouldn't be able to squeeze that all in so I'll refer you to the notes for that. Okay. And so I want to turn my attention to F rational rings and go back to trying to understand the Branson-Skoda theorem in F rational rings and if you remember, even in the regular case, sorry, not in the regular, even in the pseudo rational case or rational singularities case, the full Branson-Skoda theorem was not known. And so let's talk about that. And so we want to switch our attention now to the case of F rational rings. And so as I said, this result of Lippmann and Tessier is only partial. There wasn't quite the full Branson-Skoda theorem. So that leaves the question, is this full Branson-Skoda theorem true? And the answer at least to some extent is yes. So, Uniki and I were able to prove the following theorem in 2001. And so let me state the theorem. Let's let RM be an F rational local ring. And in this case, by F rational, that means it's characteristic P. Suppose I contained in R is an ideal. J contained in I is a reduction and say L equals the analytic spread of I. Then as usual for all W greater than or equal to zero, we do get the full Branson-Skoda theorem. So the L plus W bar, the interval closure, is contained in J to the W plus one. That then in characteristic zero, this gives the same result for algebras of finite type over, finite type over a field of characteristic zero rational singularities. So again, that's that's a reduction to characteristic P result, which I'm not going to discuss in these lectures. I want to try to go over the proof of this. There's some technical information that we'll have to deal with. So we'll get to that in a minute. So remember that F rational says that parameter ideals are tightly closed. Since we're only dealing, we're only going to deal here with local rings. It's a little nicer. So it's basically kind of any, any ideal generated by part of a system of parameters is probably the best way to think about that. So, but not all ideals. Okay. And so the thing is that if the analytics spread of I is strictly bigger than the height of I, and J is a minimal reduction. J is not contained in a parameter ideal. Okay, so there's, there's, you can't get at this directly. That's, that's the problem. So here is the idea. So given our J contained in I, that's a reduction. And find a parameter ideal. I'll call V sub n such that. Well, you can't get J to be inside V sub n, but you can get them to be ematically close so that J is congruent to be and at least once we mod out the maximum ideal to the end power. Okay. And then, then if our is F rational. That implies B sub n is tightly closed. Right. And now, let N go to infinity and use the parole intersection theorem to conclude that we get our result. I'll say, I guess I'll say that again as when we do the proof. Okay, so in order to do this, we'll need to technical results are both rather long. And so what I've done is I've already written them out, because I figured I would just mess up writing them. But let me try to go over them and tell you what's going on. And I'm going to suggest for those of you if you're taking notes, you may not want, you may want to not try to write all of this down. But it is all in the notes that are online. So, let me start with lemma 2.8. So suppose we have our local ring. And I will be assuming here that we have infinite residue field. And as usual, we have our minimal reduction J contained in I and L is the analytics spread. So it turns out that the key is to pick a very nice generating set for the ideal J. And so what qualities does that have. That's these one, two and three, which I'll go over. So suppose, whoops. So it turns out, there exists a generating set a one through a L for J. And in the next few arguments, let me use J sub I to mean the partial ideal generated by a one up to a I. So the properties that this has is suppose we take a prime containing I. Okay, and that prime has height up to the analytics spread. Okay, so we've got this height, H ideal, the analytics spread is larger. And we've got a prime kind of somewhere in that range containing I, then, then J sub I, let's, even though I wrote it ahead of time, I still manage to mess up. So suppose I is the height of that prime. So then if we take the first I elements, we do get a reduction of the ideal locally at that prime. Okay, now it's not necessarily minimal, but it is a reduction. Okay, so that turns out to be very powerful. All right, so second for if we take n large enough. If we look at the following ideal we look at J sub I, I to the end colon I to the n plus one. If we look at that ideal, and then we also add I, then the height of that is at least I plus one so that says that there's no primes too small containing both I, and this colon ideal. Okay, and notice, this is very much related to thinking about reduction numbers that colon is related to reduction numbers. And the third property is that if we take the ai's and we change them by elements of the square of the ideal. So to get ci's, then both of these properties still hold. Okay, so once we've got this nice generating set, you can use elements of I squared to modify them at no cost. So that is the limit 2.8. And then proposition 2.9. Right, so we need a little bit more information on our so suppose our local ring are is equidimensional and catenary. So just, I think, complete domain. That's as nice as we can get. And again with infinite residue field. So suppose again we have J is our minimal reduction of an height, each ideal analytics brand L suppose we've written J as in limit 2.8 but I'll use primes on to start with. But this is a little bit more technical. So let's fix an n greater than zero that n is like in our B and where I'm going to want to think about m to the end and a w that comes from the brands and scota theorem. So we can change the a primes to to a to a is in J. So, again, just modifying by elements of I squared, that's the first condition, we can also pick teas. So th plus one through TL in the maximal ideal. And for the lower ones, th through T one, I'm just going to set those equal to zero. Those T is we can pick an M to the end. Alright, so we're going to do this for a sequence of ends going to infinity. Then, if we take the A is and we add the teas. So be we get B one through B one BL these will be part of the system of parameters. If we take the ideal of bees and we're in an F rational ring, that ideal will be tightly closed. I don't know that we're going to use it very much in our argument but these teas are actually part of the system of parameters in our mod I. The last one that is a little hard to gauge what it really means but we will use this is that we for for some M, some exponent large enough in the range of eyes between H and L minus one. So what happens to T I plus one so this is, in other words, this is th plus one through TL. It is in this nice colon that multiplies high powers of I back into powers of J sub I to notice, in other words, so here, we are using the index here, right from I plus one to I, right. And this happens in a certain range that that we need. Okay, so again, reminding you J sub I is the ideal generated by the first I elements. I'm going to. So now what I want to do is I want to go through the argument in the simplest case, which is when L is h plus one. And show you how this works. Let's keep 2.9 available. Let's do the proof of theorem 2.6 when L is h plus one and W is equal to zero. Okay. Let me catch up on my notes to where I am. All right, so first off, let's fix fix and bigger than zero and let's let's choose. A equals a one through a h a h plus one. All right, and T h plus one in M to the end, as in proposition 2.9. So, I think I'm running maybe a little sift. I think I have time. So I won't try to condense the argument too much. So, then, let's make the following observation. Well, let's let's also choose M large enough, such that we have T h plus one. T h to the M plus K. This is a bad notation. My, my little K here is an index. It's not the residue field. So where is this this is in J h to the K. I to the M. Okay. So where J h is just the first each of the days. All right. So, then, we can check. I'll let you guys do the work that for all, let me call it N prime greater than zero. I to the H plus one to the N prime. I to the M plus K and prime. Well, we just repeatedly use the inclusion above. And this over and over again and where does that land that lands in J h to the K and prime. I to the M. Okay, so that's the key observation. All right. So now, let's pick X. I to the L bar. And C as in exercise two. Okay. And C prime in I to the M. Not in our not in this case in our F rational ring or not is just everything but zero. Sorry, not minus or not. That should be. Let me just say the, this just needs to not be zero needs to be in our not. Sorry, not, not out of work. Okay. So, now we want to compute. So, so what do we get, we get that, let's look at C times C prime X to the Q. Okay. So, where is this well, we're not going to use the C prime yet. C times X to the Q is in J to the QL. Right. And so, where is this. So this is in C prime. This is the bracket Q times J to the L minus one. Q. So let's say that the CC prime X to the Q is equal to, we still got the C prime. We've got our one. A one to the Q. Plus R H. A H to the Q plus R H plus one. A H plus one to the Q. And these, the R's, the R survives are in J to the L minus one. That's what we have. Okay. All right. So, let's think about what happens when we multiply this last coefficient. We really only care about that coefficient. So this coefficient, we can multiply by T H plus one. So, so then let's look at T H plus one to the Q. Times C prime. R H plus one. That's what we want to look at. And where does that land. Well, this is in T H plus one. The C prime R H plus one is in I to the H Q plus M. That's where the C the C prime gets us that plus M. And that is contained in J sub H. To the H Q. Over here by that observation. All right. But this ideal is H generated. And we've got the H Q power. And so this is contained in J H to the bracket Q. Right. So what we're going to do is in the sum above. In this sum. We're going to add and subtract that value. So that's, that's adding zero. And we're going to use the fact that when we. That the result lives back in the earlier part. So. So. Add and subtract. T H plus one Q. C prime. R H plus one. Okay. To get that. C C prime X to the Q. So that's what we're going to do. Is in a little short on time. So, so we're going to get in a one to the Q. A H to the Q. Plus. Well, we've added in subtracted. So we have. A coefficient of a H plus one Q and a coefficient. So this is going to be in that. This ideal plus the ideal generated by. A H plus one to the Q. Plus T H plus one to the Q. And so this. Is a one through a H. B H plus one. This guy is a parameter ideal. And so what we've now shown is that we have. I to the L plus one bar. Is contained in. A one. Through a H. B H plus one. Star. Which is equal to the ideal. A one. Through a H. Oops. B H plus one. Because we're in a national ring. And so. So now we let. And go to infinity. And we've got. I to the L bar. Is contained in. J plus M to the N. So maybe over here. This is contained in. J plus. M to the N. For all N. And so. What we get is that. I to the L bar. Is contained in J. B H plus one bar. And that's the proof. So. Looks like I'm out of time. So let me, let me take one more minute if I may. Just to say the, what is the general strategy? You know, what if, what if the analytics spread were H plus two. What we'd have is. We'd have a T plus one bar. And we'd have a T plus one bar. And we'd have a T plus one bar. On the, over here in this sum, there'd be one more piece of the sum. Right. And we'd also have a second T. And so we just work backwards. We'd start with T H plus two. And we would multiply by the Q's power of it. And that would push. We would be able to push. With some information about the coefficients on a H plus one. That's good enough to then. Multiply by T H plus one and push stuff back into here. And that's the general argument. It works. Nicely, but it's, it's a little technical. So. So there you have. The Reansons go to theorem now in F rational. And it's full. Beauty. I don't know in mixed characteristic that anybody has actually proved this. I'm not sure about that. No, thank you. So. Are there any questions or comments?