 This meeting is being recorded. All right, welcome all to the next session of this school. The next speaker is Zolian Jiang from University of Illinois in Chicago. And he'll talk about the localization problem in tight closure. All right, thank you for the introduction and thank you all for coming, especially when we're competing with an excellent lecturer at ICM this morning. All right, so since the introduction of tight closure, there is one sort of persistent question. So let me just state that question first. And we will try to answer that right away. So does the tight closure commute with localization? Now, this question had been open for almost 20 years then Sohoga-Brenner and Paul Monsky, they answered in the next two. So it does not in general, okay? So perhaps let me just mention their example, but I won't go into any of the details because it's highly non-trivial calculation. So this belong to Brenner and the Monsky. So set the K to be the algebra closure of F2, so the field in two elements. And now we're gonna consider the following ring. So R is a hyper surface is K in four variables. And the polynomial is given by Z to the four plus X, Y, Z squared plus Z plus Y to Z plus T times X squared and Y squared. Now the ideal is generated by X to the four, Y to the four and Z to the four. Now the multiplicative subset W will be given by all the non-zero polynomial in T alone, okay? Then they prove for one particular element, Y cubed and Z cubed. Now this belongs to the type closure after localization. However, it is not in the type closure of I alone. So meaning after localization, the type closure may get bigger, okay? All right, so like I said, there's a highly non-trivial calculation. There's no way I can prove this in all details. There is an excellent expository paper written by Paul Monsky and he was self-contained and he's highly recommended. Now, instead of going through this example, I'd like to focus on open problems and the idea goes into some positive results. So today I want to focus on two things. So for today, I want to focus on regular sequence and two parameter ideas, okay? So the result will be, so even I do generate by regular sequence, then type closure commutes with localization. Now the proof is not hard, but there is sort of a phenomenon I'd like to point it out because it will be a sort of repeating same. We'll see that again on Wednesday. Now, in order to generalize the result on regular sequences, the natural candidate will be parameter ideas because the high condition, okay? Now, even just generalize regular sequence into parameter ideas, people have to invent new technologies. So there are two different approach to this. I want to discuss both. One is the phantom halal, homology and the other will be plus color approach, okay? All right, so that will be the plan for today. So let me start from the regular sequence. Now, so let me begin with the proposition. Let me call the proposition one. So let's say R is a North Syrian community brain with identity, okay? And W is a multi-colonial closed subset in R. Now part A says, now for each ideal I, there exists a U, a special U in the capital W, such that the following holds to the union of this colon ideal. So this union you're taking all small W in capital W and this turns out to be the I to the N colon with U. So there is one special U in the W which captures this all the colon ideal. So for all N leads to one, okay? So this is a uniform behave. One U captures all the colons. Now part B, by the way, in part A I'm not assuming I is generated by any regular sequence, but in part B I need that, okay? So now let's go to character P. So let's see now the characteristic of R is now P and I is generated by a regular sequence, okay? Now under this condition, let's say I'll say L element. Sorry, could you not immediately remove what you wrote about? Sure, what you want to do this is better? Yeah, yeah. And are you saying that one U works for all N? Right. So U is not just, U depend on I by U is independent of N. So that's sort of a crucial part. Okay, yeah. All right, so in part B, we are assuming now we are in characteristic of P and I is generated by a regular sequence of L elements. Okay, then now with U as in part A, the special U in W, now this U will capture all the colon with the forbidden power. So this is W, take the union of all W in capital W. Now this will be I to the bracket A to the E, colon ways now U to the L plus one, then to the P to the E, okay? So once again, a special element U to the L plus one, both U and L are independent of E, right? This captures all the colon between the forbidden power of I and all U, all W in capital W. This is for every E, I just want, okay? But that is the uniform phenomenon. So one special U in your entire multi-collective set will capture all the colon, okay? Now, so you might know that I had a proof but then Stashna told me that this was actually a sun as an exercise in Florence lecture. So I won't repeat the proof here, okay? Instead I'm gonna use, especially I'm gonna use part B to prove that if I have a regular sequence, then that color commutes with localization. The crucial part is this U and L which are independent of E. All right, now to do that, I also need easy proposition proposition two. So this is the true in general. I'm gonna just assume R is any no-zero. And W is same as before, W is just a multi-collective closed subset in R. All right, so the conclusion is each element say C0 in W inverse R-circ meaning C0 is in W inverse R and is not in any minimal prime in the localization. This can be written as C over W where C, the numerator can be chosen to be in R-circ. So C is not in any minimal prime. The difference here is in W inverse R, we may have fewer minimal prime. So C0 avoids fewer minimal prime potential. Now we want to say the numerator can be chosen to avoid all the minimal primes in R to begin with. And W in capital W. Now the proof will be an exercise. I believe Sunil Nash will present Monday, one week from now. So this is an exercise. Now, given the two propositions, now we can prove the theorem. So let me state it. It's not how it's felt. This was argued due to Hoxter and Hewney here. So again, so R is just a no-zero when the verb community with identity, so all prime characteristics P and I is generated by a regular sequence. So let's say of L element, then the conclusion is part-cloner commutes with localization. All right. So given the two propositions, now let's prove it. Because type-cloner is persistent, so only one important is needed because so the type-cloner after localization is potentially big surprises to show. Now let's pick on any element in the type-cloner after the localization. So say Z over say V is in the type-cloner is in the type-cloner after localization. So here Z is in R and V is in W. So that means just by definition, there exists a C zero in W inverse R serve such that, again, I'm just repeating the definition such that C zero times Z over V to the P to the E, this is containing I to the bracket P to the E in W inverse R. Now I may write, so by proposition two, so we may write the C zero as C over little W, I want to put everything back to R, okay? And C zero to begin with, only avoid the minimum prime that does not mean W. I want to see that in R serve, meaning C is not in any minimum prime of R, okay? All right, so now I can collect everything that's in W to call it W E, so then there exists a W E in W there exists a W E in W depend on E, okay? Such that C times W E then V to the P to the E is actually containing I to the bracket P and this all happened in R now, okay? Again, I look at this inclusion, then I collect everything that's in W, put them together, I call it W E. This highly depends on E. Now the first proposition tells me that I can replace this W E by something that's uniform or at least that's controllable. So now by proposition one, we can replace W E by U to the L plus one to the P to the E. Because W, so or predictively, I'm looking at a colonized U, that colonized U is I to the bracket P to the E colon with W E. That will be the U in the union over all W of the colonized U, I to the bracket P to the E colon with W. And the proposition said, okay, there's one very spectral U that captures all the colon and that is U to the L plus one to the P to the E. Now, so that means I E C times U to the L plus one to the P to the E times E to the P to the E is in I E. So this tells me, because C is not in any mean or prime, now tells me U to the L E times Z is in the time to the closure of I just by definition of that program. But U is in W. So that means my Z is actually in W inverse star. Of course, I still have a denominator V running around because V is invertible in the localization so it doesn't really play with all. So that shows, so this means the element will begin with Z over V must be in the localization of that program. So that proves the theorem. All right, so any question on this? So there's a point I'm trying to make here. So the proof is not hard, you can see. But there's a very crucial step. The crucial step is here. I replace something that is highly dependent on E, which is W E. W E come from fact, I need to look at this influence on the top and to collect everything that's in W to create the W E. So the W E is not really controllable. I don't know how it depends on it. However, the proposition one enables us to replace W E with something as more desirable or more controllable is a very special U to the L plus one. Which is not dependent on E at all, then to the P to the E. Meaning we're replacing something that's not highly dependent on E by something that's linear with respect to P to E. And that coefficient is independent of E. So on Wednesday we'll come back to a very similar phenomenon. So whenever you can replace something that's highly dependent on E with something that's linear with respect to E, then you have a nice zero lying around somewhere. That's the point I'm trying to make. So any question on the regular sequence part, if not, I want to move on to parameter ideals, which turn out to be much harder. But at least I like to mention the technologies people invented to extend this to parameter ideas. Now, there are at least two different approaches. So two to extend this to, okay? So one was done by Karen Smith, and which was used in classical, which I'll define in a moment. And two, this is, let me call it phantom homology, or I can say finite phantom projective dimension, but they phantom homology, okay? By Ian Abrahach, Miles Boxer, and Craig Hewitt. So let me mention the plus quote approach first. So this will be the plus quote. So maybe let me define it. All right, let's say the R is an integral domain. By the, so from now on, in order to somehow tone down technicality, if either for you to understand, you can always assume your R is a complete local domain, which is a finite, because the heart of the problem is actually in this situation. So I don't want to mention too much of a locally excellent and so on condition. If we focus too much on the leaf, we may actually lose sight on the point. So it is okay from now on to assume your R is a complete local domain, which is a finite. All right, so R is an integral domain. I is an ideal in R. So we say an element is in the plus closure. I'm to I plus of I if R belong to, if we go all the way to R plus. So by now I believe we all know what R plus is. Especially after Tomas and Gennady's lecture. At least Gennady proved that R plus is a big chemical algebra in character. This is the integral closure of your R in the algebra closure of the fashion field of R. Or an alternative way to see it is, R is in the plus closure of I, even only if there is a module finite extension, S such that R belong to I in S. If you go all the way to S, then that's fine. But of course, S is only module finite. For one element at a time. So S depend on all. All right, so what's the connection between plus closure and plus closure? So let me make maybe two remarks. One is, of course, I need now assume character P in order to talk about plus closure. So plus closure is always, always contained in the plus closure. So here is for one way to see it. So let's say R is in plus closure. So that tells me there exists a module finite extension. So R in S such that I R. So R even I once a way past S. Now, of course, because R is in I S, that tells me R to the P to the E will be contained in the forbidden power of I of course, this should happen in S. So now if I apply any R linear map from S to R, I will be contained in the forbidden power of I by any R linear map from S to R. So for any E this must belong to but this is contained in the forbidden power of I because my phi is R linear and the forbidden power of I is in R. Okay, that's the right hand side. Now what happened to the left hand side? This tell because it's R linear so this is phi of one times R to the P to the E. Now my phi is independent of E. So as long as I can convince you the E of phi such that phi one is not zero then this will be the equivalent to show us R is in the top closure. So now all I need to do is convince you that I can find the phi such that phi one is not zero. Now to do that so consider the evaluation map from this entire harm set back to R. So for every phi I evaluate phi at one that's R linear map. The goal is to show this map non-zero but if it's non-zero after localization so it cannot be zero to begin with. So all we need to do is to pass to K the field of fraction of R. This is localization and S is a module finite so harm commutable localization. That's crucial here. And that's why I'm not doing R plot directly I'm doing a module finite extension. Once we pass through a few fractions then this is just a vector space argument. Of course this is non-zero because I have a projection because localization as now turned into a K vector space of course I have projection and which cannot be zero. So that's our argument here. So any question on this part so the upshot is plus closure is actually contained in the top closure. Now the second part of remark is plus closure commutes with localization more like because R plots commutes with localization but I'll leave this as an exercise. All right. Now give this to let me mention Karen theorem. This will do Karen Smith's. Like I said I'm going to assume this is complete local domain if you look at the original statement it only require your ring to be locally excellent domain. Nevertheless. And the character is easy. Then for every parameter idea the plus closure agrees with top closure. Now because plus closure commutes with localization then this show for all parameter ideas top closure also commutes with localization. So this is one way to extend the result on regular sequence to parameter ideas using plus closure. Now this is a highly non-trivial result like won't be able to present the proof here but the idea was somehow already presented by Thomas. So let me mention that. So the one crucial ingredient is a following. You can see that the top closure was zero in the top local model. Then you show this is the same as the plus closure all zero in the top local collage. Now since this notation already mentioned I assume that plural module has been defined. So I won't repeat that here. But here is one exercise for you. Let's say take my R to be p x, y z modulo x q cos y q minus z q and p is not straight when p is not a domain and calculate so the top product was zero in H2 x comma y on R. Of course this is also a top local model module because the radical of x comma y even is a maximum ideal x, y, z and two having to be the dimension of the way. Now let me mention so I can't do the example here but let me just at least give you a reference. So we have seen this before in the same way. So this is again R modulo x q cos y q minus z q so p not three instance so z squared is actually contained in the part closure of x comma y now like I mentioned x comma y is a set of parameters so by Karen's result this should be the same as plus so by definition there should be a module finite extension such that z squared now belongs to the ideal genabytes and comma y in this module finite extension but that's not easy to find what that module finite extension is there was an easy example in I'll refer you to example 7.8 in quick CBM as well when p is 13 the module finite extension was more or less constructed here so all the necessary monic binomial are given in that example but it's highly non-trivial so it's not easy to find the module finite extension at all so there's just some indication why Karen's result is highly non-trivial so any questions on this Karen's approach use plus closure to generalize regular sequences to parameterize it so if not I'm going to move on to the other approach using phantom homology alright so so let me define a phantom or all the necessary notion here alright so here's the definition R is a non-trivial ring in characteristic P so we have this Paskin-Spierlfonser F of E okay can I use this notation this F E lower star notation is that okay so this is same as R as a bin and groove but the module structure is given by the E for banners so this module structure is is P to the E's power okay so this is the Paskin-Spierlfonser meaning if I want to move the little R across all the way to the left I have to turn into R to the P to the E that's all there is because the module structure on F E lower star R is given by E for banners power alright so given this function a complex say C of R modules this is called phantom a sacrilege if the kernel of i differential delta i from C i to C i minus 1 this is contained in the type folder of the image of the previous differential of delta i plus 1 okay this should be true for all I didn't say what C star looks like so C star so let me say this is C1, C0 because eventually I want to use the resolution so let me say I don't have anything in there to be great and I do want this to show that here so for every i at least one but it turns out there are complexes which are phantom or sacrilege but once you apply to the Paskin-Spierlfonser to it is no longer phantom sacrilege we need a notion called stable-late phantom sacrilege now if applied for being functor to C star this is phantom sacrilege for all E at least 0 or for F0 each is identically then this is called stable-late phantom sacrilege all right so given these notions now we can finally define a notion of a phantom projective resolution right so on our module M has finite phantom projective dimension if there is a stable-late phantom sacrilege complex of projective R modules let me call this C again such that one so H 0 of C is isomorphic to M so that is a resolution and two I don't want anything in that one I don't want to define it so there is an answer to that beyond that homologic index so this will be a finite complex so this is how we define the notion called finite phantom projective dimension all right so any question on the definition itself I know it's a lot to digest all right so let me do one sort of small example it's just a special case or much more general if not let's just say R is K you can say X to the 4 Q, Y, X, Y, Q then Y to the 4 so we know that there's another problem about calling ring X to the 4 now I can pick say two elements X to the 4 and Y to the 4 and the general M primary ideal now the clue complex of these two elements on R this is actually a phantom we know that X to the 4, Y to the 4 is not a regular sequence so the clue complex is not a sequence this state the man theorem then perhaps I'll do a hand waving clue using so here is the man theorem and this is what you do and if you need to I'm going to see them again my oversimplified assumption this is a complete local domain like I said the actual hypothesis is a much, much weaker the part one is if I is a parameter ideal or even your eye is generated by monomial in any parameters I would just say parameter ideal to fit the stem of the data then R mod I this has finite phantom project dimension so that's the first part which I will do a hand waving proof in maybe in a minute or two now the second part relates to the localization problem if R mod I has finite phantom project dimension then tag-clawder commutable localization for this particular eye so that means W inverse I star E the same as W inverse of I star for all possible multi-plated cross-set W so if you combine it to part together that shows tag-clawder commutable localization for parameter ideal okay so how do you improve this I still have three minutes so let me mention a criterion how you can check whether a complex is phantom acyclic okay this is called a phantom acyclicity criterion it's a modification of the well-known books from isombard criterion maybe I won't write it down I have only two minutes so in the books from isombard criterion you have two conditions the one is a standard conditional rank and the other is standard condition on depth if you change the depth to height then you have the phantom acyclicity criterion so the content is once you have a complex that's certified standard condition on rank the euro conditional rank and the condition on height then your complex is phantom acyclic okay now given this criterion now we can actually part one in this theorem so here is an idea I want to turn height into depth so I can apply the books from isombard criterion so how do I do that I use the fact r plus is actually a big chroma calling and in chroma calling rings height and depth are the same so I'm going to pass all the way to r plus because r plus chroma calling and I'm looking at a primate idea so the height condition now gives me the depth condition so now I apply isombard the books from isombard criterion because now I have the depth condition so that means once I pass all plus this complex is genuinely the kuzu complex for example the kuzu complex is genuinely a secret now now I put everything back to r that shows because plus chroma is contained in tiled chroma for example you can you can generalize the notion of plus chroma to the finite generality model so that means in your complex all the boundaries are contained in the tiled chroma on the full boundary and that's exactly how we define phantom acyclic that's just a very quick hand-waving tool but like I said this is another way to generalize the result on regular sequence to primate ideas thinking my time is up so I'll stop thank you for your attention thank you very much Wenyang are there any questions