 Okay, all right. Thank you. Go ahead. Okay, so this will be the locks lecture I believe. So in the in the second lecture basically were introduced to the problem, whether we have regularity localize this or so put it differently so to two, one of the focuses was whether we have regularity implies F regularity. So, so now we're going to sort of look at a stronger conclusion. So let me put this open problem. Here. Right. Well, you don't let me write. Sorry. Let me start over. Apologize for that. So let me. Okay. So let me mention this open problem. Okay, so in literature this was having referred to the weak imply strong problem. Okay, so I believe we have seen the notion of strong of regularity. Right. So, or let me probably recall that. We call. Now, I believe when Lincoln define this, there is a qualified so the ring assume to be a finite I'll continue to do so. Okay. So, and a finite, let's say it's reduced. Okay. R is strongly of regular. If for all see not in any minimal primes. There exists the E such that this map. So R goes to because R is reduced. I can take peace route and then repeat. So I take P to the East so one goes to see to the one over P to the E this split. Okay, that's how we define. Okay. All right. So, so I like to mention some approach to this problem to this and I want to maybe characterize both properties. We're using modules. Okay, so the first theorem. I want this, of course, was due to Hawks and uniquely. So the following. So R is weekly of regular. Even only if for every pair. So, and is contending M with M funnily generated. So N is actually tightly closed in M. So any the same as the type of closure of an email. Okay. Now the keyword here is finally generated. So because in a moment we will see a very similar characterization of strong F regularity but we're going to drop this qualifier funded generated so the final statement for strong F regularity will be R is strongly ever even only if for every pair or modules. Find a general not every some of you will be tightly closed. Okay. All right, so let's prove this. Some direction requires a proof because if I assume the statement for modules I can simply replace M by R and M by I. So assume. So R is weekly of regular. So let's assume there exists such a pair. So N is in M, but finally generated. That's crucial here. And so N is not tightly closed. So that means I can pick. So there exists a Z, which is in the target closure but not in M. So we want to reduce to the, to the ideal case. Here's how. So consider all some modules of M that doesn't contain Z. The set is not empty because N is in there, right, our assumption. So you can buy sales one llama you can see there is a maximum element. So there is a maximum such module. We're going to call it L. Okay, so certainly and is contained in L. So Z is not in L and is contained in L because he's containing L. I know that the type closure of and it will be contained in the type of over L. So that means he's doing the type of L but he's not in L. So I can replace M by L now, because the condition is preserved. All right, so we may replace M by L. Okay. So we're looking at those triple. So I have a Z, I have L, I have M, right. I can. So by, by modding out L, we now consider Z, because he's not in L. So Z is still non zero. And, and, okay, so I'm replacing L by zero and M by M model L. Okay, so now we're looking at the condition so Z is not zero and Z is contained in the type of zero in L. Okay. So what's the benefit of doing this benefit is, well, because L is the maximum sub module, which does not contain Z. So once I kill L, I know Z is contained in every single non zero sub module by the construction of L. But put it differently. If I consider a cyclic module generated by Z, so R dot Z. So M is an essential extension of the cyclic module generated by Z, so R dot Z. So now let's consider any associative prime of M. So say P is an associative prime of M. So my claim is P must be a maximite. Okay. So because P is an associative prime, so I know there is a natural embedding from R model P to M. So I may consider R model P as a sub module of M. And Z must be contained in R model P, not only that, Z is contained in every single non zero sub module of P. So if P is not a maximite, so the dimension of R model P is at least one, then R model P cannot contain one element that is in every single ideal. Okay. So now if dimension R model P is at least one, then it does not contain an element that is in every non zero ideal. Okay. Because every ideal of R model P now can be considered as a sub module of M. So that implies P is a maximite ideal. So now from now on I'm going to write P as M. Okay. By the same reasoning you can show there's only one associative prime, a maximite ideal. Okay. All right. So now I, you can, because M has only one associative, so a similar reasoning, by the same reasoning, or you can say, you will want local R to M as totally fine as well. So, you know, let me just do that. Now consider R local at M. Because M is a maximite ideal, so R local at M is still weakly irregular. Still weakly irregular. Okay. All right. So M will still find a generated and has only one associative prime with the maximite. This means M has finite lines. So this implies M has finite lines. So that means a power of the maximum ideal will kill M. Okay. All right. So the next step is probably the only step I won't prove for you. But I do need the property of weakly of regular ring. So R M. This is weakly of regular. So we know that this implies it's normal. Okay. I believe we have seen this before. It must be called a normal ring. Okay, it's normal. And now there's a theorem by Milhochster back in 1977. This tells us so every power of M is an irreducible primary ideal. Okay. And this is where he introduced notion called a prosimistic or instinctive. All right. So normal rings are normal domains are approximate or instinctive. So every power of the maximum ideal must contain an irreducible and primary ideal. Okay. So that means there exist an irreducible and primary ideal. So Q that can tend in M to the N. Okay. So now so M is an R mod Q module. Okay. So because Q is irreducible and R mod Q is a tinier. So R mod Q is a Gorenstein and it's self-injective. So this implies one and two. So R mod Q is self-injective. So E is a Gorenstein. But this is where the name approximate Gorenstein come from. Okay. So let me do a quick proof. So one way to characterize Gorenstein-Renyi is that it has a primary ideal. Well, primary means a general by a full set on parameter that is irreducible. Now, because Q is irreducible. So the zero ideal in R mod Q must be irreducible as well. But zero ideal is a primary ideal of you because we're looking at our team. So the ring must be Gorenstein and our team in Gorenstein are self-injective. So meaning R mod Q is actually injecting module. Okay. So now the assumption is Z is containing every single non-zero cell module and my M is finalized. So Z must be contained in this non-zero cell module. Okay. This is called the soccer. Now the soccer must be one dimension of our assumption because if the circles are sort of K vector spaces and the one dimensional K vector space if it are different, the intersection must be zero. So Z can only be containing one copy of the K vector space. But Z is containing every non-zero cell module that will force the soccer to be one dimensional and only one copy of a K. Okay. So now let's combine these two things together. R modulo, so Z general soccer. So the secret cell module generated by RZ, so sorry, I mean one more step. So this tells us, so the cyclic module generated by Z is actually one copy of the relative field. So M is an essential extension of the relative field. Now R mod Q is actually injective that contains K. That tells us M must be containing R mod Q because the injury, for example, injury of how all relative field is the maximum essential attention of K. So given any other essential attention, you better have injection from that essential extension into the individual. So that tells us. So M can be embedded into R modulo Q. Okay, so now the zero sum module in M now corresponding to zero in R mod Q, therefore corresponding to Q in R. But R is weakly irregular. So Q is tightly closed. That tells us zero must be tightly closed in M. Now since Q is tightly closed. So zero must be tightly closed in M. But that's a contradiction. Because we look the presence of element Z. So Z is in the toddler zero in M, but Z is not zero. So that would be contradiction. Okay, so any question on this page? I do a question. Sure. Can you explain again why M is finite length? I mean it's finally generated. Right. So, so if you have a final general module, there's always a prime filtration. Oh, I see. Right. But the prime module can only be maximized. Because the maximum idea is only associated prime. I see. Okay. That means that prime filtration actually gives you the competition series and it's finite. Okay, thank you. You're welcome. Any other questions? Well, at the end of the day, am is just the objective, how the residue field of arm modern. Yes. Because of course, if this is an essentialization of blah, blah, and then you do the whole argument and at the end you are saying okay, I'm injecting site these and but what I am understanding if I am not wrong, and it's injected. M is the injective hall of arm of M. Capital M, capital M, I mean. Oh, capital M is just an essential extension. So it may not be. Okay, so you don't know whether M, capital M is the injective hall. I'm pretty sure you don't know. We don't know that. Let's just go here. So M is actually a sum module. We find the general sum module of the injective hall. So R module Q, that's the injective hall. Ah, okay, okay. Okay. Like I said, so R module is self-injective. So R module is actually injective in the local. So it doesn't split. So that forces it to be the injective hall of. Thank you. Okay. Any other questions? Okay, so if not, I'm gonna try to add a new page. Okay. Yay, it works. Okay. All right. So now I'm gonna categorize as a strong f-reglarity, but there's a point to make and just an approach to prove whether or to prove that weak implies strong. So here will be the, so theorem three. Again, this will do to Hoxter in a unique way. Okay. Okay. So, let's say R is, all right. So I mentioned the principle. I'm gonna plot the principle here. It's a complete local domain and these are finite. So since I mentioned finite, so must be in character B. Okay. Then the following are equivalent. Okay. So part A is, so R is strongly, probably efficient for a moment, is strongly of regular. Okay. And part B is, so N is tied to the closed in M, right? So this is for every module. Okay. Instead of finite, general, every module M and every sum module N in M. Okay. And C is, well, we don't have to look at every R module. There's one module that captures this property, which is the ingenious power of the rat you feel. So this is just EEs. Sorry, let me just say this is R and K. Okay. Okay. So, okay. I won't prove, I need to prove three implications here. So A implies B, B implies C, then C implies A. B implies C is trivial. So I'm gonna skip the implication A implies B. Now the proof is quite similar to our proof on pre-the-page and plus something about this, the forbenism map. Okay. The idea is quite similar to the pre-the-page plus the idea in the implication C implies A. So I'm gonna just prove C implies A, right? So proof of C implies A. So say, so U is a generator of the circle. E is a generator of the circle of E. Because this is in general, how we know that is the circle of the one-dimensional, okay? Now, since, now E is an, is, so that means, so the secret sum module generated by U is just one copy of K, that's what means. Now, the indiate of how of K is certainly either an essential, it's a maximum essential extension. Right, okay. So that tells us this condition holds if and only if U is not in the total closure of zero because U is contained in every non-zero sum module of E. Okay, so if the total closure of zero is not zero, it must contain U, okay? So that means it's the final part of the check, U is not in the total closure of zero in E. All right, so to do that, we're gonna go back to the definition of total closure for modules. Now, to somehow ease the notation, I'm gonna use R to the one over P to the E notation, okay? So one may view the Frobenius or rather the E for Frobenius as this inclusion, okay? Right, so once you do that, because in the definition of total modules, we need to apply the Paschian Spiro function to the module. So this F E, now this becomes the R to the one over P to the E now turns there over R is E, okay? And so the actions, so here's R, I need to tell you the module structure. So this R acts where this is Frobenius, okay? All right, so okay. So just by the definition of total closure of zero or total closure of the modules in general, what is this sign now? I'm just changing the notation from the Paschian Spiro function F E to this fraction power of R, okay? So U is not, so U, sorry, U. So U is not in the total closure of zero in E amongst to, for all C in R third, well, meaning C now there. For every non-zero element C, there is an E such that to basically I'm just doing the negation of the definition of total closure of C to the one over P to the E turns there with U. This is not in the total closure of E, okay? So P to the E turns there with U, this is not zero in this tensor product, okay? Now, which is equivalent to the following. So we're gonna tensor this R linear map where one goes to, so one go to C over W. So one to the P to the E. So tensoring this with E, the image of U is not zero. Okay, I'm just translating this condition on the bottom on the left, okay? Now, because E is an essential extension, so since E is an essential extension of the circle which is generated by U, this condition here to the non, the image being non-zero, so the image of U being non-zero, now E is equivalent to the injectivity of this map. So tensor over E goes to R over E, okay? So it's identity on E, but on the other hand, the map from R to R to the one over P to the E is sending one to C to the one over P to the E, okay? Now, I want to check this is actually injective. So why is it equivalent? Because if there's, assume there's a kernel, and this is isomorphic to E by the way. So if there's a non-zero kernel, so if there's a non-zero kernel, then U must be in the kernel because U is containing every single non-zero sum module, right? So the image of U being non-zero means this map must be injective. So now, so harm this into E, so we turn injectivity into search activity now. So it's equivalent to the search injectivity of, you know, I mean not skip any stuff. Into E goes to harm P and E, okay? All right, this is isomorphic. Now here I'm using the fact my ring is complete. So harm E into E just R. So for the harm module on the left, I'm using a junction between harm and tensor. So this will be isomorphic to harm. R, P to the E into R, and now this is isomorphic just to R, okay? So that means this is surjective. Well, surjecting means that I can hit one. So that means, so E, there exists, phi in, let me write this way. Back to R, such that, okay. So let me give this map a name, maybe I call it state C comma E, okay? Such that if I can pull this to C, this is identity on R, okay? So that's exactly the condition we're looking for. So that means this map from R to R to one over P to E where send, we send one to C to one over P to E. This map actually, because I find the inverse, the left the inverse for it. So that shows R is strongly of regular, okay? So that will be the proof of C in point, okay? And you're watching on this, okay? So if not, I'm gonna move to a new page, okay? Now, if we combine these two characterization together, they tell that one approach to proving that we can apply strong, okay? So these two characterizations tell us, we may consider the following. We may consider the union of the particular of zero, the m where, so m is a finitely generated sum module of E, okay? Because individually, if we consider the particular of zero in m, m is finite generated. So individually, this should be captured by the weak of regularity, okay? Then we take the union, the whole bit, this union will capture strong of regularity. So let me state that. Certainly, this is open problem and this is actually equivalent to the weak implies strong condition. So I believe this is four now. I'm gonna write this away. So, so type of zero in E, whether this is in the same as the big union because this union certainly is contained in the type of zero. Okay, so I should put the question mark here, okay? Now, this problem is equivalent to the weak implies strong problem. So here the one we do see it, so quick proof. So if R is weakly F regular, then by the characterization, every individual one now will be zero because m is finite generated. So m is finite generated, okay? But that means the union is zero. If the union is zero, then based on the open problem, that says zero is totally closed. And so by the characterization of strong of regularity that shows the ring must be strongly irregular. So this will provide at least one approach to attack the problem whether weak implies strong, okay? Now, but can we actually achieve anything using this approach? The answer is yes. So this brings us to the theorem by Bernoulli-Lubasnik and Nicaraguan Smiths. So, but this has a name. This is called, I should say, this is called the finite test take type color of zero, so finite test take on type color. So meaning you take the type color inside the finite generated sum module and take the union over that. So let me mention the result, five. So the content is un-engraided. So the reason for sort of for us to look at the graded ring is the following. In the previous lecture, we've seen that even for the LC problem, you can actually solve it for the, in the graded setting. So that's at least one indication. The graded setting should be somewhat easier, okay? So F finite is weakly F regular, even only if it is strongly F regular. Now, I won't be able to prove this in like 10 minutes. Initially in my notes that I send to Stashna, I had to completely prove, but I don't think it can fit in 10 minutes. So let me just mention sort of the crucial idea here. So was the actually prove the following? So at the technical heart, here is their serum. So the technical core is the following. Again, this is due to the basnik and Karen Smith. Well, the proof is that the following under the same assumption. So R is still un-engraided and finite, okay? And say any Athenian graded ring. Sorry, Athenian graded R module. Then the following holds. If you take the time to close out with zero in M, this turns out to be the following union, okay? So delta or delta is ranging all delta. So means we truncate M because M is Athenian. So we know that there is upper bound on the degree. So after a certain degree or beyond certain degree, M is entirely zero. So if we truncate at a degree, now we have only finite many degree pieces and that is at a finite generated. So this is finite generated. So that's their idea. So instead of considering all the arbitrary finite generated some module, they consider some special one, okay? The truncation by degree. And they succeeded improving that. The tauticulory is actually the union of tauticulory in such truncations. And that's of the problem in the graded case, okay? All right, so like I said, upshot is once we have the accreditation, you do have this approach. But how do we actually generalize their approach in the graded setting back to say the local setting, which is real, the harder problem for weak implies strong. So here, let me mention sort of an open problem. So the analog of this truncation. So here we take the truncation, right? So what's the analog in the local case? The analog of this in local case is the following. I can take the pieces that's secured by a power or maximum idea, okay? So we may call this M for example, okay? That'll be a local analog. And so sort of the open problem or just if it's equivalent to before, so I'm gonna write this as four prime, maybe. So whether the tauticulory of zero in E is the same as the following union, I take the tauticulory of E sub N, okay? So it's four over N. Again, E sub N means the sum module that is queued by N's power of M. And that's finally generated because our module M to the N has sine N lines. And so you harm into E that you have sine N lines. All right, but this is wide open. So this is wide open, certainly, okay? So that means this weak implies strong problem is wide open in the local case. So now I probably move on to say some partial results. But before that, any questions on this page? If not, let me move on to the partial result. So now that's the partial result. I won't be able to prove any of this in that five minutes. So let me call this theorem. Okay, this five, I believe. Oh, sorry, the five was the theorem by this was six. So I'm just gonna mention a whole bunch of partial results now. So weak implies strong if, so one, if the dimension of R is three, this will do to Williams. So the protein is also somehow to extract a uniform bond on the kernel. Oh, sorry, let me step one, take one step back. So local mileage. So the indivisible E can be viewed as a local homology of the canonical module. Now, local mileage is the drug's limit of kazoo homology. So we need to understand the transition maps in the drug system consisting of the kazoo homologies. So the idea is that to analyze the kernel of those transition maps and to extract some uniform behavior, okay? And then he's to succeed in dimension three. Now, in dimension four, but it's not quite a local case, I need to put in some qualifiers. So R, so this will do to Arbobar and Postro, very, very recent, okay? So he's published in early this year. So assume that this is a finite generated KL in dimension four and K has infinite transcendent degree over the finite field LP. Over the finite field LP, okay? So under the assumptions in dimension four, we imply strong. Now, their approach is a finer analyzed analysis of this transition map of the kazoo, between the kazoo complexes. It's too technical for me to state that, okay? But now maybe lastly, why did I say four? This too, because dimension four, okay? Now, if R is comacoli with unisolated, oh, just sorry, let me take that back. Let's say if R is Gorenstein, okay? And this will be an exercise for you, okay? Now, four, this is due to, again, due to Dubasnik and Smith in a different paper, okay? So if, so R is comacoli with unisolated, non-Gorenstein point. So meaning, so R local at the P is Gorenstein. If P is not a maximum ideal, okay? And lastly, let me just mention this result by any rock sink. So if R is Q Gorenstein, but of course, he put something even stronger, he showed that if you are Q Gorenstein, then splinters are strongly of regular. Okay, I think I need to stop here. Thank you for your attention. Thank you, Wenliang. Wonderful talks. Are there any questions? Just a little curiosity, who proved the result for R Gorenstein? R is Gorenstein? The one that you left as an exercise. Oh, sorry, I should, yes, let me write. This was already proven originally by Paul Hawks doing Hewney, okay? Are there other questions? If not, well, thank you very much, Wenliang, for wonderful talks. All right, thank you for the invitation. And we will see everybody on Monday. All right, see you on Monday.