 This meeting is being recorded. Thank you. All right. So when young junk could continue these lectures for Monday. Go ahead. All right, so we're going to continue with our same. So that's a localization problem in touch closer. So I want to introduce the open problem to right away and my goal is to convince you that. So even though touch closer does not commute with localization in general, there are still many interesting open problems along this line. So. Okay, let me call open problem one. I'll try to write less and talk more maybe. Okay, so. So the content of problem is so does that closer commute with localization at one element. Okay, so we know that it doesn't commute to localization. In general, but what about I only localized one element. And this is still wide open. And so for the rest of the day, I'm trying to. In the first hour, I'll try to convince you this is still a in an interesting open problem. And to that, I probably want to. Should pay some homage. Amar to Craig's CBF knows I mentioned to result from that particular chapter in the book. Okay. So the first, I want to mention one sort of quick remark. And let's say R is not searing. And in character, the P and W is a multi. Pretty close upset. In our. So, if so W is disjoint from the union of all associated primes of all the four beneath power of this one I do I. So this should be for all. Then the conclusion is the title of. I can use with localization at W. Well, so how do you prove this. That's only one including requires a proof. So let's say. I have. An element. Z inside the title closer after the localization. Okay. So we may assume. Z is in our because a priority. This is a fraction with some element from a dubby in the denominator, but because that's invertible, we don't really affect the membership. So I can assume Z is in our. Now, like we did a Monday. So there exists. See, not the enemy of prime of our. Such as a C times some element from W. So W sub E this highly depends on E. Z to the P to the E belongs to. The forbidden power of I is for. Well, except W E depends on the. Okay. Now we want to get rid of W E. Now, since W E. Is not. In any. Assaulted a prime. Of the forbidden power. Of I. The C time P to the E is in. The forbidden power of I. And that means. Z is actually in the title of. Of I. So that means as long as your W. So the set you were localizing or inverting. Has nothing to do with. Assaulted primes of all the forbidden power, then. So that means. That means with localization at this particular. Set. Now, this actually tells us right away. So this implies or. implies. If. R is. So weekly. Regular. Meaning every ideal is totally closed. Then. R localize. At the maximum idea. So for any. Some mildew. Okay. All right. So. So any question on this remark. Now, if not, but this remark or the. The implication. Problems of questions or the question is the following. So let me. Write this as open problem number two. So. Assume. R is. Weekly. Regular. Is our local P also. Weekly. Regular. Of course. Now we probably have seen the definition of. Regular ring. So, which means. It is. Every localization is still weekly up regular. So. The definition of open problem two is that does weak. Regularity imply. Regularity. Okay. So the remark tells us if a local at any maximum ideal, it is. It's fine, but what about arbitrary prime ideas. So. So one thing I won't convince you is that. A positive answer to open problem one will imply a positive answer to open problem too. So at least as one way to justify. Problem one or localization at one element is still. An interesting problem. Now, I think I should mention that. If you have positive answer to problem one, then you can develop a theory. The principle of an open. The principle of an open. We are given by localization one element. And you can. Define time closer. On those principles about the opens. You will glue well. And so we can have a new theory. Where she. On the same schemes. Okay. Now, so very recently, like one week ago, I think. Which I haven't had a chance to read it. I believe. In his email, he mentioned that he had to give some sort of she fee. Criterion. For this. Open problem one, but I can't really comment on that because I haven't had a chance to read it yet. Yeah. All right. So let me mention. Sort of one result that's related to open problem. Two. Now, before that, let me mention one exercise. Because I'm going to need this. In the proof of the result. Okay. Sorry. I think there's a question that. Let me. The question is, can the union of. Sorry. I think there's a question that. Let me. The question is, can the union of all assorted prime of all the forbidden power be infinite and it is. Yes. I believe. One example is given by multi cast. Okay. So let me mention exercise. First now mention the result by motor. By motor. So. So. So let's say. Hey, so K. Is an uncountable. Field. Okay. And are. Is a funnily generated K. Don't man meaning is a K algebra. It is a domain. I'm going to assume the dimension. Of R is at least one. So the crew dimension is one. So W. Is a countable. Okay. So that would be the exercise. For you. So now let me mention the theorem. I believe in. Quick spoke. This was. Attributed to. For more days. Okay. So let's say K is uncountable. Okay. This is an uncountable field. Okay. So that would be the exercise for you. So now let me mention the theorem. I believe in. Quick spoke. This was. Attributed to. For more days. So this is an uncountable field. So again, the character is a P. No. R is a funnily generated K algebra. And assume. So R is weekly. After regular. And the concurrent is. Any localization of R is still weekly after regular. So this is. Weekly after regular. For any. Sorry. Maybe let me just. Believe it or efficient for a moment. So let's say then. My R. Is after regular. So again, after that means. Every localization is still weekly after regular. So in the localization, every idea still had to close. Okay. So that means we have a positive answer in this special case. So whenever your case uncountable. And you are either finally generated K algebra. Then we have regular implies F regularity. So. All right. So let's prove this. Now. Let's assume otherwise. So we're going to derive. Assume otherwise. So. So what does that mean? So that means. So that is. There exists a W. And an idea. I saw to that. If I invert W and look at the ideal and take a type closure. This is different from. The localization of that. So. I'm going to write W inverse than I start. Okay. So we know that there's one inclusion always. Because how to call it is persistent. So I can always. Pick. So pick. Say. Let's get you. In the type of code after localization. But not in a localization of that. We're looking for a contradiction. So. This tells us. By exactly same reasoning we've been doing. Twice already. So there exists a C. Not in any problem of our. Such that. So C times. W E. In W where W E highly depends on. This E again. And you to the P to the E. And this should be in the forbidden power of. Okay. And this should be true for all E at least one. Like. I want to emphasize again. So W E highly depends on. But in this. Inclusion. Only W E. Is involved. So I don't need any other elements in capital W. So I may replace W by. The multiplicative set generated by W E's. There are currently many W E's to the set. Generate by them is still countable. So replace. Capital W by. W E's. So by the one generated. By all the W E's. Okay. So how do you generate them? Well, you take a monomial in those W E's. So that's still count. That's countable because there are only kind of many W E's. Okay. So assume. W is actually accountable. You can see why. We have this exercise. It's not countable. And then you have a multi. That's countable. All right. So now we're going to pick. In maximal. Maximum ideal. Let's call them again. In this localization. Okay. So we know that there's a, there's some. Corresponding between the prime ideal in the localization. And the prime idea in the arena rain. Which do not meet with the multi predictive set. So that means the M corresponding to a prime idea. In R. So, let me write this away. So M corresponding to a prime ideal P. In R. Okay. So this exercise tells us P must be a. Maximum ideal in R. So. Exercise. P is. Also maximum. In R. This in R. So we can just write. And again. Okay. So the idea is that if this is not a maximum ideal. I can look at our modular P. Now the localization will be. W inverse R modular M, which is the field. So we're in a situation in the exercise. So you piece is not a maximum. Our model P will be a domain with positive dimension. Over an uncountable field is still finally generated. And we have a countable multi predictive set. And we have a field. Because M is a maximum. And that's not possible by the exercise. So. P must be maximum. I'm going to write P at M again. Okay. So now I'm going to focus on a special maximum ideal. So, so choose. Now what I said here about the maximum P and M. That's true for every maximum ideal. In the localization. So now I want to pick a special. A maximum ideal in a localization. So choose M. In the localization. And such is that you is in here. Of course. But not in I. Okay. So you is not in I to begin with in any localization. So I must. I'm not going to maximize you such as that. If I localize the maximum I do you is still not in I. All right. So what does that mean this means. This localization. So this tells me. To W. M. This is. Not. Weekly. Because the ideal. I is not tightly closed in this localization. Okay. So you is still in the target closer because. Todd. Persist. Persist. So it's in the type of order to begin with. Therefore, after localization is still in the type of color. But it's not in ideal. So. It's not how to close. Therefore the rings not weekly of regular. But. M. It's a maximum in both rings in our end in the localization of our. So this localization. It's actually the same as our localized. But we have seen that. If I localize the maximum ideal, it is still weekly. So that's a contradiction. So that will. That will be the proof. Okay. All right. So any question on this page. Okay. So like I said, so. Both questions are wide open in general. Maybe in the next hour. I'll mention more possible result to. Problem number two. All right. So if you don't have questions, I'm going to move to a new page. Wait. Just a second. Do I have to erase everything now? Wait, I thought I could turn page. You are in the zoom whiteboard or some other. I mean, zoom whiteboard. Does somebody have experience with zoom whiteboard. When I did it without. The videos and I could turn page. Now I realize. Not sure. Do I have to erase everything? Okay. I may have to erase something. It's okay. If I erase the left side. Yeah. Okay. Thank you. Sorry. Okay. So now let me sort of. Go back to the point I'm trying to make. Let me write this way. I think it should be clear. This implies. Open problem. I should say one implies two. Okay. So meaning a positive answer to question one. IE. If the cloud community will localization at one element. Then. A week of regulatory localize this. Okay. All right. So he is totally proved this. So let's assume. So are. Is weekly. Regular. Okay. So we're going to keep a localized localizing are at. Say first a prime. Maxim ideal. Then the prime ideal with dimension one and so on. So. So we can. Localize. At a. Maxim ideal. Of course it's still wicked of regular. And then. We can repeatedly localize. At the prime ideals. Of now dimension one. Right. So. Localize at the maximum ideal then localized at the prime idea of the mission one. If we still. We can. Regular. Good. Then we will. Go on. So we will local. We look at this localization. Then pick. A prime. Of dimension one. Which is now dimension two in the arena rank. But every time. Which is localized at the prime ideal dimension one. And then we. Do this repeatedly until. We hit. An example, which is not. Weekly. Regular anymore. Okay. So. The whole option is. So this reduce. The problem. To proving the following. So if. Are is weekly. Of regular. Local rain. Then. And. So he is. A prime. Of dimension one. Then. Are local. The P. Is still weekly. Regular. Okay. Because otherwise we can just keep on going. Do this by dimension. Okay. So we know that. Are is weekly. Regular. Even only if. All the primary ideas are totally. Close. So. No. Okay. This is. To show. Every. Okay. Let's just say. P. P primary. So P primary. Do I. Is totally closed. In our local. Okay. Okay. So. Open problem one is about one element. So now we're going to pick one element. Okay. And the whole object is now our model P had to mention one. So. We can be one primary. Because the mention is one. One. So we can pick. So any primary. So pick any. Primary. So it's a domain so we can pick any. Any non-unit. A non-zero element. You will suffice. Call the F. Okay. It just. F is not invertible and F is not in P. That's all. So then. This localization or locality. This is a localization. Of our localized at the F. At. A. Maximum ideal. P. Okay. Okay. So. That means we suffice this. To show. So the I. Is totally. Closed. In our local at the F. Because we know that. Todd closer behaves well. If you localized at a maximum. But now open. Problem. One. Tells us. So Todd close commutes with localization at one element. I. The same as I start to begin with. If we go all the way back to R. Because R is weekly at regular. To begin with. And so this is just. Okay. And that finish the proof. Okay. So any question on. This. Probation. Okay. So. If not, I'm going to introduce a more open problems. There's in something they are approached. To prove. Or to solving open problem one or two. And to do that, I need to erase the raw hand side. Of the board. All right. So let me. Call this open problem. Five. So in the literature, this has been referred to the LC problem. So he said the following. So let's assume we have a. No Syrian local ring. And I. Is primary. In something I'm just trying to extract the essence of the proof of problem for. Is primary to. A prime ideal. P. All the mention one. The question now is, is there. A constant. That's called the capital B. So we never say a constant. I mean, is independent. Such is that. The maximum ideal. To the B time P to the power. Kills. The zero's local knowledge of our modular. The forbidden power. Of I. And this should be true for all. At least one. Okay. So I think this is in something the same phenomenon as we have seen. We have a constant E. On Monday. So on Monday, we have this. Sort of instant where we have W. Partly depending on. E. And then we replace the W. By U to the L plus one, then to the P to its power. So even though it's still depends on the exponent, still depends on. But. So we have a constant. And this is the same phenomenon. If we look at. So we fix one E. The zero. Of our modular. The forbidden power. I. That has finance. So certainly a power of maximum ideal. We'll kill it. So you have it's bone. That depends on E. And the question is about whether you can find. A linear. Power linear with respect to P to E. And that can actually kill this module. Okay. So it's sort of the same phenomenon as before. But this. As stated, this is open. Of course, in the literature. The problem is sitting even more generally. Is stated without assumption on. On the height of P or. Or why the eye is from your own. Okay. Is general. General is just, you have a local ring. You have no ideal. Then. Can you kill the zeros? Local knowledge. Of our modular for being the power. By a linear power of the maximum. But. However. For us. This sort of special situation survives because in the proof of problem for we seen that. Only the height one prime. Sorry. To mention one. Prime ideal matters. Okay. And only those that's primary to P actually. Matters. Okay. So, um, Let me try to convince you. Why this problem is interesting. So let me call the proposition six. So let me write this away. So open. Problem. My five. Actually implies. Let me say open problem. Okay. And only those that's primary to P actually. Matters. Okay. So, um, Let me try to convince you why this problem is interesting. So let me call the proposition six. One. I need some qualifier now not in the food generality, but in the. In the situation that really, really matters. So. So in, so in the case. I is primary. To a. Dimension one prime capital P. Okay. All right. And. In the proof of proposition four, we seen that that's the only real case that matters here. Okay. If we want to solve problem number two, but it doesn't solve problem one in the food generality. Okay. So now let's prove this, but any questions on problem five or proposition. For. All right. Now let's prove this. Now, suppose I have a Z. In the top floor after I localized at one element. Okay. So. An F is a parameter in our model P meaning F is not in P and F is not invertible. Okay. So what do we want? We want. To show that Z. Is in the. Localization of that floor. I may assume Z and R by the same reasoning as before. And we may pick. So let's say we'll pick. C is not in any minimal problem or such that. C times. Z to the P to the E is contained in the forbidden power, but after the localization. Okay. Virtual I'm just repeating the definition of. So what does it mean? So mean this implies. Or equivalent to. So C time F to the N sub E. So this is the same phenomenon as the W E again. And just we can now focus on one element, but this phone and highly depend on E. Okay. All right. So. Since. Let me rewrite this. So F to the N E times C. Z to the P to the E. Is in I bracket. E. This tells me C times E to the P to the E. Actually belongs to the zeros local knowledge. Of our mod. The forbidden power. Because F is a parameter. So if we look at the ring our modular, the forbidden power. I. F is still parameter. So the radical of F is actually the entire maximum. So killing by the parameter is the same as killing by power of the maximum. So now we know this element C time. Z to the P to the E belong to a zero local knowledge. Okay. So now. So. So open problem. Five tells us there is a linear power. This actually implies. Exist B independent of E. So F to the B. Tom P to the E times C times. E. Is zero. But zero meaning this belongs to. I to the. P to the E. But that implies. If I rearrange this product of C. Times F. B. To the P to the E. That will be in. Okay. So that means. So I have to be times Z now belong to the top floor of I. And this. In Z is in. The localization of touch floor. Okay. So. And that's it. So that. We'll finish the proof. All right. So any question on probably position six. Again, like me, maybe back to my sort of takeaway. The takeaway is even though. A top floor does not mutable localization in general. And there are still many interesting problems. Okay. All right. So I have about six, seven minutes. So let me mention. Some partial result. Regarding problem number five. In our do that, I need to erase the left side of the board. Okay. So. Let me mention one sort of a. Partial result. That isn't great. The case. So the bottom line in the grid is the case. If instead of assuming. Are you the local? We're assuming R is an N graded. Now, say I mean. Where the degree zero piece is a field or positive characteristic. The answer is yes. So you do. You can find the linear power of M to kill this. So let me stay the serum. Then always say a few words about this. So let me. This will be serum seven. So if R. So this is the N graded. Okay. Now Syrian. And so the degree zero piece is a field. It's a field. So. A positive characteristic. Okay. So. We're going to set this to be. The party degree piece and. I'm going to state this in under some assumptions because this originally proved under the assumption. Then I'll tell you how to remove the assumptions. Dimensional and. So the mention. Is one. Okay. And of course I is your homogeneous. Then. There exists a constant. B. Such that. So M to the B. E. Are modular. This is indeed zero. Okay. So in the literature, there are at least three different proof to this result. Okay. I believe this was to say. I should. Attribute this to. He unique and to vorace. Oh, sorry. They prove it independently. Using different. Methods. Then much later, I gave a different proof. Without the assumption that our is actually like a dimension. Back in 2015, but they prove this in around 2000. Okay. All right. But otherwise this problem number five is wide open in general. Okay. Now I have three minutes. So let me mention one approach to this. Okay. So which has something to do with primary decommission. All right. So let's say I consider the primary decommission. Then and so be. So this is the primary decommission. All right. So every Q here depends on E and the number of children we need also depend on. Okay. Now. So here's a question I believe this was raised by arena and Karen. Okay. So let me attribute to. Irina can correct me if I'm wrong. Of course, I haven't stated the question yet. So does there. He exists. A constant. C such that for every such Q. So Q, I'm going to say E. I then I take this to the currency time. The power is actually contained back in. Hi. Okay. Now he is an exercise for you and you can see why this question. So say Q. Depends on E. Sorry. Let me say Q sub E. So Q sub E is. The M primary component. Of the forbidden power of eye. Okay. So I look at the same minimal prime decomposition. Meaning we got rid of all the redundancy. So I look at M primary component. And then. So this Q E will kill the zero slow homology. Is zero. So if you have a part of answer to the question raised by someone sent and Smith. Then you can see, well, okay, so that content C will be your content B. And then your power of maximum idea will actually kill the zero slow homology. Right. So I think my time is up. I'll stop here. Thank you. Are there any questions. Comments. Oh, okay. I think. Just tell me how to create a new page. Okay. I'll try that in the next hour. So you don't have to watch me erasing anything. Thank you so much for a nice talk and a question for everybody.