 So in today's session, we will have two talks by Irina on uniform art in this lemma continuing from the last time. So, you know, you may begin now. Thank you. All right, so our goal is to prove the weak version of the uniform art in this lemma. So I'm restating a few things that we did last time. We don't know all the conditions under which the weaker the strong version hold. In my notes I wrote it wrote up a few strong version uniform strong versions that necessarily true for all ideals but when you restrict ideals to maximal ideals or the principle ideals then that has been proved. But what we did for a week version. For the week version we already made some reductions. We don't have to worry about all ideals. We it suffices to prove it for the primary for ideals that are primary to a maximal ideal. That would not be good enough for a strong version that the inclusions just don't go the same way. So strong version is very different kind of animal. So anyway, our goal is to prove. I'm trying to show one screen. No. Okay, so these are right here. If why isn't this working. There. So if we want to prove, we cannot possibly prove all three parts in all the detail because that would be much too much but I will make some indications. Another thing that we reduced to last time is we had this wish definition. We related this uniform Martin we slam our we made a digression that we could just prove that we have this analytic on ramifiers that are not zero. And there exists elements that multiply the end power of indoor closure, when I deal into just the ordinary and minus case bar if we can, if there exists one K that works for all ideals are all and that would be key of our. And so the first goal today is to prove so this is. The first one is T of R, not zero. So no we haven't yet proved the week uniform Artemis lemma, but we're making a distraction which is actually an important ingredient for the proof as well as, I think it's of interest of its own kind. So first of all, for T of R to be non zero. R needs to be reduced. And the reason is that. It's not reduced. All internal closures always contain the, the nil radical of the ring, but ordinary powers do not do not contain the nil radical in fact high powers do not contain it at all. So we really need the ring to be reduced. So the theorem that proves so let R be reduced be reduced. No theory and ring. And of characteristic P. Then T of R is not a zero. If any of these conditions happening. So, the first one is our has a test element, and the cruel dimension of our is finite. So, I do not know if there exists. There might be a theory and rings of characteristic key that have a test element but are infinite dimensional. There might be a theorem of that form but I'm not sure. So I am imposing this finite dimensionality right now. In the next cases are is finitely generated over an excellent. We know that such a ring is finite dimensional. Or the third condition, if R is a finite. And it is. If it is a finite then there's a theorem of points that it has finite dimension. And so, both of these two imply the dimension of our is finite. And then the other thing that's also true conditions. Part two and three imply that our has a test element. The proof of of for part three was already given by Kevin Tucker this was theorem seven. I also wrote it up in the notes and the proof of that, in case to our is finitely generated over an excellent local ring but then have a test element. That is, a lot of pages, many of the 14 page section six in humickel hockster paper the first one, the first big one. So I cannot possibly justify this so take two and three was now somebody would have to check the proof but I'm not doing it today. So, so what essentially I have to do is. This is all I have to prove. I've reduced to the proof of only part one, and the rest is by previous work by various people. So we have a test element and we have a finite dimensional ring. So, So, the first thing is and one thing I in my notes, I just 15 minutes before my lecture realized I need a little bit more work. The first thing to do is step one. We can assume, suppose that there exists a K impact actually. We will not prove that K is to be where D is dimension. So D. That is the dimension of the ring. Suppose there exists case such that for all zero dimensional ideal zero dimensional primary ideals. We have this inclusion, I, and for all. And David milk of the pain, we take the inter closure by to the end that's containing it to the end minus K. Then it's, we already proved last time the uniform and the weak uniform martini slama. It suffices to prove it for and primary ideals or and there is over maximum ideals. It's the same kind of proof. So what we actually have that the same holds for all ideals so that means that T sub K of R is actually the whole ring. And one multiplies the inter closure and arbitrary of the inter closure of the end power of an arbitrary ideal into n minus K power of that ideal. Okay, so that is, I'm leaving that. Step two step. Step two is, we need to find some kind of reduction of eye. So that we will have a uniform D. So the first thing is, is by prime avoidance. So, so now we only have to prove it for that I be M primary, arbitrary and primary, where M is a maximum ideal. And now we want to find some kind of reduction of our eye. So first of all by prime avoidance. For prime avoidance, I don't need M primary at all, and this would work for any ideal. There exists a D plus one generated ideal J containing I such that the radical of I equals the radical of J. So this is just prime avoidance and I will assume that you have seen something like this. Or, I will leave it as an exercise. By the way, the exercises there perhaps too many exercises I just added a few which haven't been uploaded. But I'm also saying don't do all of them. And I write what I think of exercises. Good for the soul. So then. So for this we don't need the M primary ideal. The what we need for M primary ideal is that now we can localize after localization at M, then I is M primary after localization. So if we know that the ideal is, if we know that the ideal is, sorry, if we know that the residue field is infinite, then we can find the degenerated reduction, but I didn't assume that the, the residue field is infinite and there's a little bit more work. And so here I will also leave it as an exercise that we can we have that the integral closure. Sorry, I don't understand why this happens sometimes. I must be touching something here. This is in this side. What did I do with it. I started you know, okay. All right. By an exercise. What we have is that the, if we take the integral closure right to the end so this is locally. That's contained in the tight closure. I guess I should be locally. Tight closure after localization, and mine is deep. And even though the dimension of our localize it and might be smaller, but it's at most D so this D works. Okay, so this is the usual grievances code of theorem. So we know this. We know, Ian talked about the grievances code of theorem, and it works whether they have an infinite residue field or not because it can first pass to the infinite residue field so it's again one of those tricks. Okay, so now what when we left to our Then, oh, I didn't need this at all. When you lived to our, then the inter closure of Oh, and this is to fall in the inter closure right to the end. Well, we know that the closure does not commute with localization, but it does commute with localization. For all and primary ideals, there's only primary ideals primary to maximum ideas. Okay, so actually, either made a mistake right now I discovered this only after a gap. So maybe I don't need this at all. Anyway, so we have the inter closure right to the end is contained in the title right to the end minus D. And then what we did assume is that our has a test element. So we have a test element. This is true for all and. And so let's. Okay, we're now. Now this is final step, let, let C be a test element. And then C times. I to the the inter closure C times inter closure right to the end is containing C times the inter closure and minus D to the title closure of this power, but C is a test element. So this is just in I to the end minus D power. All right, so now, if we go back here. What I said, in this side I have to correct. So we didn't prove if we, we didn't quite prove this in step one. We just prove that there's C times, I to the, the inter closure is in I to the end minus K. And so then we can conclude that for some non zero C. And then we know that this year is also not zero. So a little bit of back and forth. If we know that this T sub K of ours, not zero when we only restrict to and primary deals then it's not, it's not zero for otherwise. So, so C is in this T sub D sub D sub D of R, and if I go back here under here, I, I was also wrong on this side here. Where did I say it was just D. Okay. I should be minus one. So that proves this. And then I want to state another theorem. And this one, again, we cannot prove this is due to Lippman and satay. And it's quite restrictive I write the statement in 4.2 in the notes. And so the previous theorem the previous pages we started with the reduced string of characteristic P, Lippman and satay as handle. As handles characteristic zero. So let R be a regular domain with infinite residue fields, and whose field of fractions field of fractions as characteristic zero. It's not saying that our contains a field of characteristic zero. Just, it has field of fractions as characteristic zero. T of s is not zero for any finitely generated our algebra as that is that contains. And as finite crew dimension. So a proof of this. Lippman satay theorem, the T of s is not zero. This would take us a while as well. So, sorry, Lippman satay proved some other things but then he unique approved. The T of s is not zero. In this case. So this is this ingredient that we came stumbled upon last time. And now we need another ingredient that will help us prove the uniform art and re-slamma. Okay, so these are now do new section co and the qualifiers. It's not an official name but obviously it was meant. So, this was. First, we need the definition. So we start with the complex that has F zero F one. And it stops. So we have F sub T, F sub T, and then zero. So this is a complex of finitely generated free modules. So each of the capital F, I should call these G sub I so so the F's are finitely generated free modules and the G's are maps in the complex. So we know what the rank of a free module is. I guess ours over the ring are. So we know what a rank of a free module is. How about the rank, rank of a map. So you have a free module and you have a map here. This is the maximum of R such that the, we can think of G as a matrix. So if we take the ideal of R by R minors of this matrix that is not zero. So this is rank of a map. So let's say that two things. The complex F satisfies a standard conditions on rank. So first of all, over here at the steep spot, the rank of F sub T is the rank of G sub T. And everywhere else, the rank of a rank of G sub I plus the rank of G sub I plus one is exactly the rank of F sub I. So this, if we had vector spaces here, this is exactly the exactness of a complex. But we don't have vector space of fields. There's general not a field. And then we say that F satisfies standard condition on height. If the height of the ideal. So the ideal of maximum possible minors of G sub I that don't give us the zero ideal so this by definition is a non zero ideal. If height of this is at least I, you might have seen F satisfies standard condition on depth in the depth of these ideals is bigger than equal to I. R is going to call it then height is the same as rank. So we have this would be an exactness, a criterion, but then the second part of the different or the next definition is under these conditions. So let's say that the set C M of R the column home coli piers of R. This is the set of all elements X in our such that. So it's the set of all elements X in our such that for all complexes at satisfying standard. And height conditions. C kills this X kills all the homologies. This is zero for all I bigger than equal to one of course. So, in the column of coloring, one would be an element of this column of color. This ideal versus. This is actually an ideal. It's not hard to see. So in a column of coloring, this ideal would contain one, but in general, it may not be so big. Okay, and the, I will start this. The next theorem now, but I won't I will be able to finish the proof only after the break. So here's our lemma is the level. Let R be an Ethereum ring. Let P be a prime ideal in our. And let C be an element of the ring not in P such that its image in our mod P. Lies is in the coma qualifier of our mod P. So let J be any ideal generated by a one so the elements. And what we want is the height of this ideal. When we go mod P is at least D. And suppose that there exists an integer K such that for all N. So the N intersected with P plus CR is containing J to the N minus K times P plus CR. So in other words what we were saying that at least for this ideal J would have uniform art and reads for the module M is the whole ring and M is the ideal P plus CR. And the conclusion is that then we have at least for this J. And we have the uniform art and reads also for P with the same. With the same K. Oh. Oh, so my iPad is showing is it 931 or is it 941. How much 31. Okay. Oh, so my iPad is showing the wrong time. Okay. All right, so I have plenty of time to read this. All right. All right. So, proof. So, what do we have to show? Well, this is certainly contained in P plus CR. Actually, let's, let's make it equal. So this if I also intersect with P. And then by assumption this is containing J to the N minus K P plus CR intersect with P. Now, by the same trick we've done several times this is J to the N minus K P. So this is J to the N minus K C intersected with P. But, but this year, this ideal here, I don't hurt anything if I write intersected with C. So this year is J to the N minus K P plus. Oh, C times J to the N minus K. Oh, this here is just C times P right because P is a prime ideal and C is not contained in it. So we can factor C out so we're left with this. So we have to show to show C. And this, I have to erase this bar so you don't see into it's not into a closure. I had it underlined. So we have to show is that C multiplies J to the N minus K intersect with P into the J and minus K P. So this is where the column of qualifiers coming so C is a column of qualifier mod P. So we can just show that now we have some math or some complex of finitely generated free arm modules. Free arm of P modules. So we need arm of P modules that satisfy the standard rank and height conditions. And that maybe. So here is J and minus K intersect with P J and minus AP is this equal to H sub I of some F of satisfying standard rank and height conditions. So this is the question. And that's the goal for when we do that, then we'll be done. So, let's see. So proof of this proof of this that it actually true. So, so now let's start with. We started with complex and I'll call the maps, maybe I should call them G. F's get map G's and G's get map ages. This be a minimal homogenous free resolution of the ideal generated by D the nth power of the ideal generated by D variables in the polynomial ring. So T, T is generated over the integers by excellent through XD. So, there are many ways of producing such a resolution and take for example, the Egon Northcott resolution. I'm, that's another thing that you might have seen it if not. If anybody knows more recent recent reference like Bruce Feather have a pretty nice proof but there might be another one. I'm not in print some other nicer proof. So if you know a nicer proof and want to type it up and the great, we can include it, or we put it in your next expository book. And so what we know is, at least for this Egon Northcott resolution. We know that if we take the ideal. So, for each, for each of the maps. So if we take the ideal of maximal minors so I omitted writing that this is the ideal of maximal minors just implicit that it is the ideal of maximal minors. So we know that this ideal, at least up to radical, it's equal to X1 through XD. So that's what makes this Egon Northcott resolution very powerful. And. And so now let F be the advantage of this ring T is that it maps to every ring in particular maps to our R. So not arm of P but R well it could also map to arm of P but we're tensing with our, and the map is that X sub i of course goes to our a sub i sub i sub i is J is this ideal generated by base. So this is F. And so what do we do is now we compare. We compare. So on top we have G the bottom we have F. So all I have to really write is there's are the ego Northcott resolution. resolution, I should have said of r mod. So here is r to the n plus, wait, we're doing f here. This is f. So I already answered and d minus one choose d minus one. And here whatever it is, let's just call it f two. We will need more details than this. And then, so this is f. We don't know if f is an exact complex. We know that because the a's are unlikely a regular sequence. So this is not necessarily an exact complex. But down on the bottom, we will have a similar thing. But h here in this row is a free resolution of r mod j to the n. So and here we have h sub two. Again, it doesn't matter what the rest are. Now these two are equal and whatever the map here is, we have the same map here. We call it g one. It's the same map. And this bottom thing h is an exact complex. And f is a free complex. And whenever you have a free complex going, well, at least the start, we have here a start of the map from one complex to the other. We know that there exists a map going down here that makes everything convenient. So at least we have a map from f two to h two. So now we really need something mod p because c is a column a column by a mod p. So tensor with r mod p. And what do we get? So f tensor r mod p. And at least the start of the map r mod p. What we have is we have a surjective map. This map is a surjection at least in the second spot. In the second spot. Not second. So this is here. This is in the first spot. First spot. So we have a surjection that should be a different color. No, this is color. We have a surjection after tensing with r mod p. I should just do it here. So we have its identity here, its surjection on the homologous. So we have h one of f tensor r mod p surjects onto h one, g, sorry, h tensor r mod p. So we have a surjection. But what is happening with this thing here? This thing here is still is a complex of finitely generated free modules, free r mod p modules and the height condition holds. So the radicals of all the tensored maps. Well, before tensing the radical contained all of the contained the all the x is so now after passing to r the radical contains all the a's. And so now it still contains all the a's. All the a's. But these are the radicals in r mod p. So we have the standard height conditions are satisfied. Right. Condition is satisfied. But also the rank condition and rank conditions are also satisfied because they were satisfied by the original Eagle North by complex. And the maps didn't go to zero the and the ranks stayed the same. Okay, so what that says. So what this says, thus, c kills this homology of the tensored f. So this is zero. But we have this surjective map here. And by the surjection see also kills this. This is the red complex. That's also zero. So what we have to figure out what is this homology here, but this is exactly j to the n intersected with p over j to the n. So that requires a little bit of work. All right, so let's go back to where did we start. So what we have on this side, we have to show this. Well, we showed it for all powers of n. So we conclude that j to the n intersected with p, we follow through over here. This is contained in j to the n minus k p plus c times j to the n minus k intersected with p. We just now prove that this is in j to the n minus k p and that does it. So let's go back to what we proved. So remember, on Monday, we had some cases where if we had some kind of reductions mod p and I call them j there too, we had some kind of reductions mod p and if we can get a good handle on them, then then we can we get some leverage on ordinary other ideals, or at least in primary ideals. So here this lemma really gives us that leverage. And by the previous section on trying to find t of r is nonzero. So if if we know the t of r mod p is not zero and the these column of qualifiers of r mod p are nonzero and we can do this for all primes p, then we then we are in pretty good shape for being able to prove the uniform art in the Islam. Okay, so when is the next section would be when is this column of qualifiers not zero? When is c of m of r not zero? So that's the other big question. And for the t of r to not be zero, remember we needed a reduced strain for the column of qualifiers we don't need reduced strings. And the next section section six in the notes has quite a few results where it is not zero. I give fairly complete proofs where the coma qualifiers are not zero. But I will probably skip that part. It's kind of technical to get to the juicy part in the next section is when is the when can you prove the uniform art movies lemma? Okay, are there any questions? Okay, well, then we can take a break now. Yes, yes. Okay, we'll meet after 10 minutes. Yeah. Okay. All right. See you.