 You couldn't tell me when I'm supposed to start. OK. Welcome to the second session of the day, right? It's a pleasure to introduce Ling Xuan Ma from Purdue University. Ling Xuan got his PhD under the direction of Mel Hoxter in 2014 from the University of Michigan and has taken off. And while he was talented as a student, has only become even more talented since. Now at Purdue University, he was awarded a prestigious Sloan Fellowship. And on his website, you can see, has already over 40 papers and preprints, all right? It is a real pleasure to hear him speak about Direct Summands, and take it away, Ling Xuan. OK, thank you. Thank you for the very nice introduction. It's my great pleasure to speak. And sorry, my other device have some issues, so you cannot see me. But I'm just using my iPad right now. But after the lecture, I'm happy to handle any questions. So all right. So the title of, OK, so this lecture, I'm going to, the lecture that was assigned to me was about Direct Summands of Regular Rings. Actually, I feel a large part of the materials, a large part of the lecture, was already contained in this Kevin's Notes and Neil's Notes. So I'm just going to, perhaps, I'm going to do this question from a different point of view. So let me change my title slightly. So the title of my talk is going to be this lecture. We will focus on strongly F Regular Rings and Direct Summands of Regular Rings. All right? So, OK, so to start it off, let me just quickly remind you some basic concepts. So first of all, throughout my talk, all rings are commutative noetherian with multiplicative identity 1. So I'm not going to move beyond this noetherian setup. And so we will focus on rings of characteristic p. And so if r has positive characteristic p, so then there is a natural Frobenius anamorphism, then we have a natural Frobenius anamorphism f, which goes from r to r. And of course, it sends little r to r to the p. And sometimes we will iterate this map f to the e. And then the map is going to send little r to the p to the e. And at least in my lecture, I'm going to use a notation you've probably already seen in Kevin's lecture. That is important that we need to distinguish the source and target of the Frobenius map. And so the notation I'm going to use is OK. So here's a notation. So I'm going to use fe lower star of r to denote the target of the Frobenius. So in other words, instead of writing down this fe, that map goes from r to r, so we're going to view the Frobenius map fe that goes from r to fe lower star of r. OK. And so let me just quickly remind you. So in this notation, this one is esomorphic. This is the same as r as in the bidding group, but the r module, if I view this copy as a module over the left of the source, maybe let me just even write it down. So elements in fe lower star r are denoted fe lower star little r, where little r inside r. And the r module structure on fe lower star is defined using the following. So if you have r1 times fe lower star r2, it's going to be fe lower star r1p to the e times r2. So this is the Frobenius. But on the other hand, let me also point it out. So this one is also esomorphic to r as a ring, obviously, because it's just, in some sense, it's just r, but I just used this awkward, a little bit awkward, so this is the same as r as a ring. So it sends fe lower star r back to r. So that's the ring isomorphism. OK, so now here's the definition. I believe you've also seen this in Kevin's talk. So a ring r of characteristic p is called f finite. If for some or equivalently, or e bigger than 0, fe lower star r is a finitely generated r module. So in other words, a finite means the Frobenius map is a finite morphism. OK, so I will leave this as an exercise to show that in this definition, you can require this for some e, or you require this for all e bigger than 0. It's the same thing. It's not hard. So let me just give you, OK, let me just mention some examples of a finite, and so for example, if you just have a field k, a field k of characteristic p is a finite, right? So k to the 1 over k is a finite dimensional vector space over k, so it's a finite extension. So this is because, well, I didn't mention that, but hopefully you already see this in the earlier lectures. While this one r is reduced, you can view this Frobenius map as the natural inclusion from r to r to the 1 over p to the e. And so for field, the 1 over p to the e is just the field extension. So for example, a perfect field or algebraic closed field are always a finite. And I leave an exercise that hopefully you have seen this before, that if r is a finite, then all rings. Maybe let me just write it down here. So this will be an exercise that if r is a finite, so then the localization of r are a finite, the quotient of r are a finite, adjoining a power series or polynomial variables are a finite. So in summary, so for all rings, essentially a finite type over r will be a finite. So if r is a finite, then all rings, essentially a finite type over r are also a finite, because you can join a bunch of variables, and then you modulate some ideal, and then you localize, invert some multiplicative system. All these procedures are preserved at finite property. And let me also, for example, so then I mean, you can start with a perfect field, and then you'll get lots of examples. All the essentially finite type algebras over perfect field. Just a comment. You say that a field is a finite if and only if, if you consider the extension of k by k over 1 over p, it's a finite dimensional k vector space. But then you can also say that k is a finite if and only if k is a finite dimensional k to the p vector space. Both things are equivalent, and maybe this is a more direct point of view. Yeah. Or if and only if this is a more straight point of view, if you want to avoid p roots at this point. Yeah. OK. Good. Yeah. You can also say k is a finite dimensional kp vector space. OK. So are there any other questions, comments? And so the other thing I want to point out, I just want to give you plenty of examples of finite rings. So complete local rings of characteristic p, of course, with a finite residue fields are a finite. So this also follows from one of the exercises. And basically you use the co-instruction theorem. So you write this complete ring as a quotient of the power series ring over a coefficient field. And so as I said, adjoining a power series variable will preserve the finite assumption. So this is going to be fine. But anyway, this is an exercise. OK. So and throughout my lecture, when we talk about strongly irregular, just go back to this title, when we talk about strongly irregular rings, I'm going to assume the rings are always a finite, just for I just want to avoid this technical issue. And there are some results saying at finite rings are nice. For example, I'm going to just state it in words. So Coons have proved that at finite rings are excellent in the sense of words indeed. And Gabber has shows that at finite rings are homomorphic image of at finite regular rings. So these rings are, so in particular, for example, they admit canonical modules, et cetera. These rings are not pathological. OK. Now let me recall you. I think this already shows up in Kevin's lecture. But because this is a key concept that I will study. Did I label this? Or I didn't label this. Let me just remove this. OK. So here's the definition. And at finite ring R of characteristic P, of course, is called strongly irregular. If for every c inside R, not in any minimum prime of R, there exists e bigger than 0 such that the map R to f e lower star R sending 1, 2, f e lower star C splits as a map of the homomorphic. So this is going to be the definition. And so clearly, let me just put it out. So OK, let me let's perhaps stick with the definition a little bit. So the definition requires that for any c not in any minimum prime, the map R to f e lower star R sending 1 to this map, sending 1 to f e lower star C, this map splits as a map of our modules. So in particular, let's just try some simple case. If you take c equals to 1, let's see what happens. If you take c equals to 1, then the map is going to be 1 goes to 1 f e lower star 1. So this is just if you take c equals to 1, this is a natural Frobenius map. So in particular, let me just say remark, so strongly irregular rings are always f split. So f splits just means the Frobenius map splits. R to f e lower star R sending 1 to 1 splits. So in particular, so whenever the Frobenius map is split or even much weaker, as long as the Frobenius map is injected, it's an easy exercise that you can show the R is reduced. So in particular, strongly irregular rings are always reduced. They are always reduced. OK, so and remember, well, I mentioned this earlier, but I didn't write it down, but you can view the Frobenius map as the natural inclusion from R to R to R to 1 over p to the e when R is reduced. So and so since here, all the rings we consider, the ring we define, there will be automatically reduced. So this condition, if you don't, this is in particular, if you first see this or if you don't like the f e lower star notation. So this condition is really saying what? It's saying if you identify this one with R to the 1 over p to the e. So then this condition is really saying strongly as regular, this is really saying what? It's saying like the map, the condition you want to put is that for every c, not the n minimal prime, not the minimal prime, not the n. So the map, the natural inclusion R to R to the 1 over p to the e. Well, but I'm not going to use the natural inclusion. I'm going to send in 1 to c to the 1 over p to the e splits as a map of R module. So this is completely equivalent to this. It's just a different notation, but same thing. So f e lower star R is identified with R to the 1 over p to the e. And under this identification, this f e lower star c is identified with c to the 1 over p to the e. So saying this map splits is the same as saying this map splits. But of course, the only small point I want to make is that if you want to use R to the 1 over p to the e notation, your ring better be reduced. But it tends. So strongly as regular rings are always reduced. So there's no ambiguity using this 1 over p to the e notation. But throughout my lecture, I prefer to use this f e lower star R notation because this, so first of all, it's compatible with Kevin's lecture. And secondly, it's going to be more compatible with the modern language that in the character p coming around to it. OK, so here's the definition. So I gave you the definition of strongly F regular rings. And the first remark is that they are always f split. In particular, they are always reduced. And so the first lemma is that, so for local rings, you can actually say more. So they are not only reduced, they are actually domain. So let me just state this. So I let R mk be an f finite and strongly regular local ring character p. Then R is an integral domain. OK, so let's prove why is this. So since R is reduced, to show is a domain, you only need to show there is only one minimal prime. It is enough to show R has only one minimal prime. OK, so now suppose you have more than one. So let p1, so we're going to prove a contradiction. We assume you have more than one minimal prime. When p1, pn, you have minimal primes of R. And let's say, at least two, let's just try to arrive at a contradiction. So the idea is that you want to, OK, so the only condition you have is the splitting condition. So you want to split off some elements not inside all the minimal primes. So you have to pick this element sort of cleverly, OK? So now we pick. So now you pick. So for each minimal prime I, you pick fi inside all these other minimal primes. But it's not in the PI. It's not in the I minimal prime, OK? So then if you look at the sum of all these fi's, so this is not, so then this is not, I claim this is not contained. This is not in any minimal prime, right? Because I hope this is easy to see because this is basically by the way you choose all these fi's. So for example, let's just look at p1. So f1 is not inside p1. But all the other, like f2, f3, fn, so all the others will be inside p1. So there's only one element not inside p1. So the sum is going to be not inside p1. And you do this for other i's too, OK? So you choose this element not in minimal prime of r. And so now by the assumption, you can pick some e such that split this element, OK? So let's just write it down. Since r is strongly, I will use sfr to be a short hand for strongly F regular. So since r is from F regular, you, so there exists e bigger than 0 and an r linear map from f1 lower star r to r such that this, let me call this map phi, such that phi sends this element to 1. So phi of fv lower star is 1, OK? So this is just the concrete meaning that the map r to fv lower star r sending 1 to fv lower star of this element splits. So I'm just writing down, so split means there's a map back sending this element to 1. OK, so now let's just try to arrive at the contradiction. So far, we haven't used any of this. So far, we haven't used the local condition r. And now I'm going to use the fact that r is a local way, OK? So well, so OK, so first of all, this map is additive. So this is certainly equals to phi fv lower star of f1, OK? And so now, of course, this is equals to 1. Remember, this is by our assumption. And so you have a sum of element, which is a unit. And so now I'm going to use the fact that I'm working in a local way. And so at least one of these elements must be a unit, because if otherwise all of these will be contained in a unique maximum ideal, so this can be a unit. And so since RMK is local, at least one of this phi fv lower star fi is a unit, OK? And so therefore, well, without loss of generality, let's assume, we may assume just phi fv lower star f1. It's u, which is a unit, OK? And now where is the contradiction coming from? So remember, we have at least two minimal primes. So this little n is like this 2. So what I'm going to do, I'm going to pick an element, OK. So let me just write it down. So but since f1 times f2, so this is equal to 0. So this is, again, by the way we choose this element. So I choose this f1, f2, fn such that their sum is not inside any minimal prime. But if you take the product of any two of them, you got 0. And so the reason is that if you let's just look at f1, f2, f1 is in all the minimal primes except p1, and f2 is in all the other minimal primes except p2. So f1, f2 is inside all the minimal primes, but r is reduced. So the intersection of the minimal primes is 0. So if you take f1 times f2, or more generally, you'll take the product of any two of the fis, you got 0. So f1, f2 is 0, OK? So then, well, you almost see the contradiction. So we have, OK, so now I'm going to, you look at u. So this u is this phi of f e lower star f1. So you just think about what is u times f2. So let me just write it down. So u is, by our definition, u is phi of f e lower star f1 times f2. But phi is an R linear map. So phi is an R linear map that goes from f e lower star R to R. And so I'm multiplication by f2 outside. So you can pull this f2 inside. So this is phi f2 times f e lower star f1. And so now, what is f2 times f e lower star f1? Well, this is, remember, there's a Frobenius f e lower star here. So when you have f2 times f e lower star of something, if you want to pull the f2 inside f e lower star, you have to raise it to p to the e. So this is phi of f e lower star f2 to the p to the e times f1. OK. But now this is 0, because f1, f2 is 0. So f1 times f2 to the anything is 0. So this is phi of f e lower star 0, which is 0. So this is a contradiction, OK? Because u is a unit and f2, we suppose f2, I mean, we begin with f2 not 0. So this is a contradiction. OK. So I do this little lemma here basically just to, well, first of all, show you like for local rings, strong F regular rings are better. They are domains. And we'll see later, actually, like they are always normal. But let's just stick with domain for the moment. And also, I want to get you familiar with this f e lower star language, because it's useful. OK. Any questions? Sorry, I cannot see the chat. If you have questions on the chat, maybe the organizers can remind me. But feel free to just unmute yourself and ask questions. OK, so let's just, we define strongly F regular rings. The maps like r to f e lower star r for any c, OK, so for any c there is just e such that the map sending 1 to f e lower star c splits. Excuse me, there is a question in the chat asking why strongly F regular implies reduced. Right, so let me just repeat that part. So if r is strongly F regular, so the condition is that for every c, for every c not any minimal prime of r, this map splits. So in particular, the map from r to f e lower star r sending 1 to f e lower star 1. So this map better be splits for some e, OK? And so, well, so this means the natural Frobenius map is split. So in particular, I mean, to make sure that this map splits, it better be injected. And it's fairly easy to see. Actually, I think I leave it as an exercise. If the Frobenius map is injected, then the ring is reduced. This is, well, you will see that it's not a hard exercise. But anyway, OK, any other question? OK, so we have this definition. There's an easy remark that the F regular implies the strong F regular implies F splits, and they are always reduced. And now for local rings, you can do a little bit better. These rings are always, the strong F regular local rings are always domains. OK, so the next thing, because we want to study strong F. I mean, we want to use, so basically, the goal of my lecture is that I want to use this theory of strongly F regular rings to study this, well, to prove that the X amount of regular rings are comacodyl or whatever. So this is like the final goal of this lecture. And so we want to study some basic properties of strongly F regular rings, first of all. OK, so here's a lemma, the second lemma. So it says strongly F regular ring is a local property. So let R be an affinite ring of keras of p. Then R is strongly F regular if and only if Rp is strongly F regular for every p inside spectrum, for every parameter. OK, so by the way, let me just make a comment. So perhaps you have learned in this Niels or maybe Kevin's lecture that there is also a notion of weakly F regular. So obviously, because of the name that we define them, so strongly F regular always implies a weakly F regular. So weakly F regular is defined in terms of tight closure. And so actually, in my lecture, I leave it as an exercise to I will ask you to show that strongly F regular rings are always weakly F regular. So this is indeed a stronger notion. But you can see that. So for example, this lemma is like one reason, at least, like why strongly F regular rings behaves better. Oh, of course, it is conjectured that the weakly F regular strongly F regular, they are ultimately they should be all equivalent. But we still don't know that. And so this lemma indicates the strongly F regular rings are some sort of easier to study because it is indeed a local property. Well, for weakly F regular rings, we still don't know whether they localize it well. So I hope you have seen that open question part or some of the mentioned that before. But anyway, to check strong F regular, to check something, some ring is strongly F regular is really check it locally for any prime. OK, let's prove this. So there are two directions you have to show. So first, let's suppose R is strongly F regular. So suppose R is strongly F regular. OK, I want to show all the localizations are strongly F regular. So let P1, P2, Pn be the minimum primes. So OK, so now I fix the prime P and I want to show, let's say I want to show Rp is strongly F regular. And so now we want to use the definition. It is enough. OK, so let me just say, so what's happened? OK, so to show our local as a P is strongly F regular. Let's just think about the definition, what we need to show. Well, we need to show for any little C, I mean, maybe I should say, for any C, not inside any minimal prime of R local as a P. So any C whose image in R local as a P is not in any minimal prime of Rp. So you need to show for any C like this, there exists e bigger than 0 and an Rp linear map. From f e lower star Rp back to Rp, sending f e lower star C back to 1. OK, so this is definition. So to show something strongly F regular, I basically just copy the definition for it. You need to show exactly this thing. OK, all right, so now what's the, there's only like one, OK, this is the easy direction, by the way. So the only small subtlety that you have to think about is that, OK, so if you can, if this little C is not inside any minimal prime of R, it's not inside this P1 to Pn. If this is not inside any minimal prime of R, you can do this. Because we assume R is from a regular, you can pick a splitting from f e lower star R back to R, sending f e lower star C to 1. And you just localize that map. It induces a map from f e lower star R local as a P to R local as a P. OK, so if this C is actually not inside any minimal prime of R, this is obvious. You have a map on the level of R, and they will localize that map. That's pretty straightforward. But the only, as I said, there is a small subtlety here is that we are only assuming that C is not in any minimal prime of R local as a P. So what are the minimal primes of R local as a P? Those correspond to the minimal primes of R that are contained in P. So there could be some minimal primes that are not inside P. So this is the only small subtlety. So the claim is that you want to avoid those cases. Because if your C is inside some minimal prime of R, so you can't apply your assumption to split C off under R level. OK, but this is not a problem because the idea is that you can always replace C by some other. You can replace C. So the idea is that you can replace C by something else whose image in R local as P are the same. But the other thing is not inside any minimal prime of R. OK, so let me just give the details. So the claim is that we may assume C is not in any minimal prime of R. OK, so let's explain why is this true. So suppose C is contained in, let's say, P1, PI, but not in, say, the other, not in the other minimal primes. So I mean, remember, I'm assuming P1 to Pn are all the minimal primes. And so, for example, if you pick your C whose, I mean, the image in RP is not inside any minimal prime of RP. But unfortunately, C is inside some minimal prime of R. That could happen in prior. Let's say C is inside P1 to PI, but not in PI plus 1 and Pn, since it's not that. And so now the trick is that, so now we pick C prime inside the intersection of, OK, so Pj, j equals to i plus 1 to n minus the union of PI. Let's say Pj, j equals to 1 to i. OK, so this is by standard prime avoidance. So you pick a C prime inside all the other minimal primes, but not inside P1 to PI. And you replace C by C plus C prime. And replace C by C plus C prime. And the advantage of doing this is that, so there are two things we have to track. So the first thing is that C plus C prime is not inside any minimal prime of all. Because, well, C is not inside these Pj's. And you add, so C is not inside minimal prime from i plus 1 to n. And the thing you add is inside the other minimal primes. So C plus Ci is not in the other minimal primes. But this is also not inside any P1 to PI, because C prime is not in these minimal primes. But C is inside these P1 to PI. So this is the first thing I have to check. And the second thing I have to check is that, so C plus C prime have the same image at C inside our local asset P. OK? So maybe I leave this as an easy exercise. So check. You have to check this. Well, maybe let me just say C prime, the image of C prime in our local asset P is 0. OK? So this is just basically by the way we choose C and C prime. You notice that none of this P1 to PI, none of these PI's could be contained in P because we assume the image of C in our local asset P is not inside any minimal prime of our P. So P1 to PI cannot be contained in P. You have to use this fact to show this. OK? All right. But anyway, so the idea is that you can modify your C. You replace C by C plus C prime to assume that the C is not inside any minimal prime of our. And the image of C plus C prime and the image of C are the same in our local asset P. And so then once you do that, so now you use your assumption, as I said, then it's very easy. But since R is strongly irregular, you can split off this C. So there exists E bigger than 0, such that R to Fv lower star R sending 1 to Fv lower star C splits as a map of R modules. And then we'll just localize that. So and localize this splitting. We get RP linear map sending it was our C to 1. OK. So this direction is the easy direction. So as I said, the only subtle is that you have to modify your C a little bit. OK. So that's all right. So now we prove the converse. This is a really interesting direction. So now I'm going to assume all the localizations of R are strongly irregular. And I want to show R itself is strongly irregular. So let's see why is that. OK. So now we prove the converse. So now we fix a C inside R not in any minimal prime. OK, so now this time our assumption is that so we know for every P inside spec R, there exists E bigger than 0. Well, let me just emphasize this little E may depend on P, OK, such that Rp to Fv lower star Rp sending 1 to Fv lower star C splits. OK. So we want to show OK, so now our assumption is that all the localizations of R are strongly irregular. We want to show R is strongly fixed C not in any minimal prime. So of course, this time since C is not in any minimal prime of R, the image of C inside Rp is not in any minimal prime of Rp. So you can use our assumption. So for every P, you can find E, but E may depend on P, such that this map sending 1 to Fv lower star C splits. So this is our assumption that our local P is strongly frag. OK, so now you want to sort of spread out this splitting. And what do I mean? So since R is at finite, OK, so if you look at the harm set, so harm Rp, Fv lower star Rp, Rp. So this is equal to Rp times R, harm Fv lower star RR. OK, so this is the easy fact that harm commutes with flat base change if the first module is finally presented. But our rings are no theorem. I mean, finally generated is the same as finally presented. And our assumption is that this is the finally generated R module. So we are using the affine list here, OK? So once you have that, this is a flat base change. So you can just say it's the harm R, Fv lower star R, tensor with Rp is the same as you just tensor each thing inside with Rp. So you get this, OK? So I'm sorry. All right, so what was the upshot? The upshot is that, so OK, so we know that there is a map. OK, so we know like the map Rp to Fv lower star Rp is sending one to, we know this map splits, OK? So that means there is a map here sending Fv lower star C to 1, OK? And so now every map here is essentially, well, what I'm telling you here is every map here is essentially coming from a localization of a map of this level, OK? So the conclusion you get is that there exists a map V inside harm R, Fv lower star R, R sending. So you know like you have a map here such that after you localize, it is a split. It sends Fv lower star C to 1. And what you know, you know that before you localize, it must send Fv lower star C to something outside P to F, which is outside P. Because after you localize, you want this map to be subjected. Fv lower star C have to send to a unit. So before you localize, you know like it must send Fv lower star C to something outside P, because otherwise it's after you localize, it's not going to be a splitting anymore, OK? So this is what you can do, right? But then what you know is that our localization app to Fv lower star R localization app sending 1 to Fv lower star C splits, OK? So while this is, sorry, I'm already so I will stop after I finish the proof of this lemma. So the point is that so you have a map from Fv lower star to R sending Fv lower star C to some F not inside P. So as long as you localize this map at F, so the map is going to send Fv lower star C to a unit. So in particular, this map sending 1 to Fv lower star C splits, OK? So now for every P inside spec R, we can find F. So for every P inside spec R, we can find such an F. In particular, well, maybe I should really say an Fp here because F depends on P. F is not in P, not inside P. And hence, if you take the union of all the basic open sets, so that's equal to spec R. So to take the union of all these basic open sets, it really contains all the prime ideas of R. So this is equal to spec R. And so now since spec R is quasi compact, this is the very first thing we learn in community of algebra about the spectrum of community of rings. So hence, there exists like F1, F2, to Fn, such that the union of the basic open sets is spec R. And for each Fi, there exists an EI bigger than 0, such that the map R looks at the Fi to FeI, lower star Fi, sending 1 to FeI, lower star C splits. Because this Fi is coming from some P, so we have it, OK? And so now, so this is another exercise question. So now it's easy to see that you just take the E that are larger than all these EIs, so then the map will be split. So it is then easy to check. This is an exercise that for E0, which is the maximum of this E1, En, so the map R2 of E0, lower star R, sending 1 to Fe0, lower star C splits. So this has finished the proof. So let me just explain a little bit about the last step. So the point is that if you pick this E0, OK, so this exercise that I assigned you is that, so if this map sending, if you have some EI sending like 1 to FeI, lower star, there's a typo, RFI. So if you have some EI, such that R2, FeI, lower star, sending 1 to whatever FeI, lower star C splits, so then you can always enlarge the EI. You can pick for every E bigger than this EI, the map R2, Fe, lower star R, sending 1 to Fe, lower star C also splits, OK? And so now basically, if you pick, so now the upshot is that if you pick this E0 to be the maximum of all these, so then the map from R to Fe0, lower star R, sending 1 to Fe0, lower star C, it splits after localized at each of the FI, OK? And then again, it's an easy exercise that if you know it splits after localized at each FI, so then it has to be split because the obstruction, just look at the harm set. So if it does not split, then the image have to contain in some maximum ideal, so you just localize it at that maximum ideal. But our FIs, they cover the spectrum, so you can't find any maximum ideal, OK? So anyway, so I'm a little bit in rush, so maybe this last step requires some exercise, but anyway. So we prove that the strongly afregal ring is a, packing the strongly afregal is a local property. OK, sorry, I will stop here today. I'm not entirely sure who the chair of the session is, so let's clap for Ling Xuan. Are there any questions? There's one in the chat. In the statement, can we replace spec R by max spec R? Yeah, you can replace, that's a good point, because if you know it's true for maximum ideals, you also know it's true for prime ideals, but for any prime because it localizes. It's the same information. Another question. So this shows the strongly afregal localizes, but we say that weekly afregal doesn't, and we know the call me call it does. No, no, no, I mean, weekly afregal, we don't know. I mean, you don't know, right? That's what I'm sorry. But what about afrational? Afrational localizes too. OK. Do you need to assume that the ring is local? Yeah, well, OK, so first of all, I don't know whether people have already defined afrational, but for me, your first defined afrational for local rings. Are you defining local rings? Yeah, and then you just define it. OK, for non-local rings, I would just define it to be all the localizations are afrational, because I don't know how to define afrational. Well, maybe you can, you can use the canonical. But yeah, but in either case, I think afrational is also a local property, at least under some excellent type assumption. But it will be afrational. Yeah, it's a local property. And Florian will of course go over that in a lot more detail. Right, I think the next lecture will be focusing on afrational rings. Thank you. OK, if there are no more questions, then let's thank Klingchon one more time, and we'll resume again on Wednesday.