 All right, well, welcome back everybody. I hope you enjoyed the lectures last week. And this week's lectures will be by Dr. Kevin Tucker of the University of Illinois at Chicago. I first met him when he was a graduate student at University of Michigan and I was a postdoc there. And it was either during that time or shortly after that time that he solved a long wondered problem of whether the type, there was this notion of F-signature but nobody could even prove that it really existed. And he was able to show that it did. And then he's done lots of great, proved a lot of other nice things since then. But that was, so that's Kevin. So take it away. Okay, thanks so much for the introduction. Just checking, can everyone hear me? Okay, great. So first off, I would like to thank the organizers for the invitation here to speak in this. It's really a pleasure to tell you a story about one of my favorite objects in mathematics, honestly. So, yeah, the three lectures are sort of broken up and have very different flavors. And I've tried as much as I can to balance things and do things both for a somewhat more beginner audience as well as an advanced audience at the same time. So just a sort of blanket disclaimer. If you can't read something I write or you don't understand something I say, please do speak up in the chat or say something and I'll see what I can do to try and fix things. All right, so with that, let me get started at least a little bit. All right, so just a blanket assumption that I'm gonna use for pretty much all of the lectures. So throughout, I'm gonna let R denote a community gymnotheraean ring with a unit. And sort of for simplicity, although in many of the cases, you can generalize the results beyond that, I'm gonna assume throughout here that R is in fact a domain. All right, so sort of blanket assumption throughout R's domain. And of course, like in the previous week's lectures, basically all of the rings until maybe some side remarks at the very end, all the rings will be of characteristic P where P is a fixed prime throughout. Okay, so with that, let me say that the power of Frobenius, right? So the power of positive characteristic here is that you can use Frobenius to prove theorems, right? So again, so F throughout will denote the Frobenius anamorphism in particular the absolute Frobenius anamorphism, right? So our Pth power map and where does the power Frobenius really come from? It comes from the ability to use that it's an anamorphism and to iterate it as much as you would like, all right? So throughout, I'm generally gonna use the letter E to denote an iterate of Frobenius so that, and then F to the E is the P to the Eth power map, right? So, and just, although I've just been watching some pretty advanced exercises being solved, let me just recall for you the sort of main takeaway definition if you like from last week in our setting which is a little simpler than the general case, right? So if I take an ideal I inside of R, I can talk about its Frobenius or P to the Eth bracket power, right? And that's just the expansion of I under the Eth iterative Frobenius, IE it's the ideal generated by the P to the Eth powers of all of the elements of the original ideal, right? So, and with that, the tight closure of I, right? Is of course the set of all elements X and R, right? So, okay, so R is a domain so there is only one minimal prime and it's zero, right? So saying you're in the tight closure means that there exists some non-zero element C, right? With the property that C times X to the P to the E is in this P to the Eth bracket power, right? And while there's some ambiguity in exactly what you put here, right? Certainly it suffices to put down for all say E bigger than or equal to zero, right? So, and this is an equivalent definition. Okay, so I wanna sort of start off in a somewhat historically accurate manner here. So, let me give you the first sort of big definition here. What is the test ideal? And this came up at least implicitly in Neil's talks as well. So, and one of the themes of these lectures is that there are many, many, many different flavors of test ideal, right? And we're gonna see a bunch of them. So, but just again, maybe for the experts, this test ideal is the finiteistic tight closure test ideal. All right, so what is this? Well, in the definition of tight closure, right? So the point here is this element little C can depend on the X that you chose, right? And so at the most naive level, the test ideal of R is the set of all elements that are gonna work in all tight closure tests, all tight closure relations for all ideals and all elements and all tight closures, right? So, all right, so here just as a starting definition, if you will, right? So let's just define this test ideal to be the set of all elements C inside of R, right? So that C X to the P to the E is in eye bracket, P to the E, right? Okay, for all ideals, I inside of R, all elements X in I star, right? And for all E bigger than equal to zero, right? So the set of all Cs that work in all tight closure relations everywhere, okay? And as Neil mentioned, completely not obvious, right? Here that such, well, okay. So if you're being very careful here, you'll notice that there's always AC that works. I can put in SQL zero, right? So this ideal really consists of all test elements together with zero, right? But not obvious that this finiteistic test ideal contains any non-zero elements at all, right? That there exists such a guy. And in some sense, the most important argument I wanna sketch for you today and the main point of today's lecture is we're gonna go through and give a version of the proof that test elements exists, okay? So that's sort of what we're headed, okay? So let me immediately take this though and give you another way to write down and interpret this test ideal, right? So the following is really an exercise, right? So one can check that this finiteistic test ideal is the same thing as, well, look at an intersection over all of the ideals I inside of R and look at the elements that multiply the tight closures back into the original ideals I, all right? So intersection over all ideals I inside the ring, right? And I'm requiring that C times I star is inside of I for all of the possible ideals you can choose, right? Now one containment here sort of is obvious from my definition I have included E equals zero, right? In this definition here and that makes it sort of pretty clear just using equals zero we see that this guy is contained inside of this one, right? And sort of to do the easy exercise you just have to understand a little bit about how tight closure plays around with bracket powers. This is not a deep exercise, right? But maybe I should say in the notes which are up online I've put in way more exercises, right? Then I would expect you to finish and turn in and the range of levels of the exercises is pretty big, right? So I've designated seven of them I think as sort of star exercises and a star exercise in the notes is one of the official ones to turn in for the problem session. And otherwise some of them are designated as challenging and we'll run across them throughout, right? So but take care you might wanna go through and do the easier ones first as you go through and then come back perhaps even after this lecture series to try and complete the picture, okay? All right, so great. So here's our first main object of study this finiteistic test ideal, right? So the set of all elements C that work in all tight closure relations inside of your rank, okay? All right, so what's the advantage of introducing sort of this second perspective? Well, one of the reasons I like it is this makes it pretty clear just from face value, right? That you can in some sense see this test ideal as trying to get it away to use tight closure to measure the singularities of your rank, all right? So as was shown last week in a regular ring, all ideals are tightly closed, right? So in particular what that means is that all of these colon ideals are the trivial ideal, right? And so if your ring has the property that all ideals are tightly closed, then from this description it's very clear, right? That the finiteistic test ideal is the whole rank, okay? So I put that sort of as a separate thing here, right? So this finiteistic test ideal is equal to the ring if and only if R is so-called weekly F regular, right? So IE, all ideals are tightly closed, okay? And so great. So here's the first object of study. I've tried to say that this should be used in some way to think about, you know, give some ideal which is measuring the singularity of R in some way, but before really doing anything else, I wanna say that you sort of immediately see a bunch of massive difficulties with using this definition, okay? So let's talk about a couple of them. So the first, right? And in some sense this is to do with this word weekly right here. So this word weekly comes up from sort of one of the biggest problems that still remains open for in the theory of tight closure. I'm gonna move this up a little bit, right? So open question, does weak F regularity localize? IE, if I have a ring which is weekly F regular like a regular ring and I localize, does it stay weekly F regular? And we still don't know, right? So how do I see that from this sort of test and the ideal description? Well, when you look at this intersection, I want you to be worried about two things, right? The first is that this here is an infinite intersection and again, remember that infinite intersections do not commute with localization in general. You need them to be finite, all right? So the first problem you run into and trying to play around with localization is that you can't move your multiplicative set past this intersection. And then even if you do, all right, now you end up with the problem here that inside of this colon ideal as was referenced by Neil, right? Tight closure does not commute with localization either, right? So we have major problems showing up here and trying to deal with that, okay? All right, so I wanna make this question though a little bit more precise just to really hammer this home and again, there will be a later lecture series about this, right, but one could ask for something even stronger than asking that we have regularity passes to localizations. One could ask that again, when I'm talking about commuting localization beyond these two operations, I'd love to see that the finite statistic test ideal also is compatible with localization, all right? So what do I mean by that, right? You could ask does, all right? So if W is a multiplicative set inside of R, you would like to see that if I localize the test ideal, does that give me the test ideal of the localization for an arbitrary multiplicative set? Okay, well, we don't know that, but I'd be much happier in fact, or I'd be just as happy pretty much with some weaker versions of the same thing, right? So here I could also ask that just if you take a single element and invert it, does that commute with the formation of this test ideal, all right? And, okay, so, and similarly, I could also ask for what happens if I localize just a prime, right? So I could ask if I take a prime P inside of spec R and I localize the test ideal, I could ask do I get the test ideal of the localization of that prime, all right? So all of these remain open. And one of the reasons I say it this way is I'm gonna reference problems with localization throughout and feel free to think about any of these basically, but I'd be happy with some of the weaker versions. Even the statement that type closure does not localize, I believe the type closure is still not known whether it localizes well after inverting a single element. It's known that if you localize it a prime, it's bad, right? So, but these are very subtle questions, okay, great. Any questions so far? Okay, so as we start to sort of build up some more things here, right? So let me reintroduce some notation from last week, right? So the sort of Frobenius push forward notation now in the previous lecture series or in the previous set of notes, we have both push forward and pull back or a restriction and extension of scalars, if you will. So I've tried to minimize the amount of notation to be used. So, but I do need to use this restriction of scalars functor, right? So recall that, so I have the restriction of scalars of an R module M is for the iterative Frobenius, right? We just a note by F to the E lower star M, right? And I sort of wanna sort of tell you how I think about this. So on the one hand, right? So as an Abelian group, you've done nothing, okay? So here F to the E lower star M sort of as a set is still just in bijective correspondence with the original module. And moreover, how do I add two guys? Well, I add them just how I would have added them before, okay? What I've really done in doing this restriction of scalars is played around with the R multiplication structure and twisted it by the iterative Frobenius, right? So how do I multiply R times F to the E lower star of M? All right? Well, that gives me F to the E lower star of R to the P to the E times M, right? So R now acts by P to the E powers instead, okay? And one of the nice things about introducing this notation is before when I've talked about Frobenius, I've thought about really F to the E as being a ring endomorphism, right? But certainly when I write it in this way is not an R module endomorph or R module map at all, right? This map here is not R linear because the R module structure you should use on the right side is different. Namely, you should use this twisted structure, right? So here, another perspective on the iterative Frobenius is as the map from R to say F to the E lower star R, right? So again, this just sends say little R to F to the E lower star of R to the P to E, right? So, and you can sort of see this sort of in the back and forth, all I've really done is I've taken the right-hand side of this thing and added to the F to the E lower stars kind of as decoration, right? So it's like a painting, right? But in doing so, you see that we fixed this sort of R module problem, right? So here, this of course is R F to the E lower star of one, right? So this map here is determined by sending one to one, right? And it's now R linear, okay? You also saw last week another way to think about this here, R is a domain, right? So you can also think of F to the E lower star R as being the ring R to the one over P to the E, right? The ring of P to the E roots of elements of R. I use all of these notations in my daily life, right? So I've tried to minimize it here and to stick to just the F to the E lower star notation, but feel free just again to think of the F to the E lower star as P to the E roots everywhere and you really won't go too far afield, right? Great, okay, so we had at the beginning here some set of assumptions that R was the Neutrion domain of characteristic P and I wanna add one other one that I'm gonna use throughout all the lectures, right? Sort of for simplicity, right? So I'm gonna assume that R is so-called F finite. Now what is this, right? So this is just the assumption that F the E lower or F lower star of R, right? Is a finitely generated R module, okay? So, let's see, make sure that doesn't pop up. All right, so this assumption is pretty mild, right? It's satisfied by rings in most geometric and arithmetic settings. We'll see that in more in a second, all right? So, but while mild, it does have a bunch of important sort of implications that really make sure that you're working with pretty nice rings. Just as one example, Gabber has shown that an F finite ring is always a quotient of an excellent regular ring, right? So in particular, the theorem of Kunz follows as a result that R is automatically excellent. Moreover, since it's the quotient of a regular hence foreign-steen ring, it has a dualizing complex and works well with duality. So you get a lot of really nice things out of this, all right? Okay, so let me also say that, of course, if F lower star R is a finitely generated R module, then you can repeat this process with iterates as well. And so it's equivalent to ask that F to the E lower star R is a finitely generated R module for all E, all the iterates of Frobenius if you like, okay? All right, so let me tell you maybe one more case in which it's easy to show that something is F finite. We'll see throughout the lecture series that it's easy to see that either a power series or a polynomial ring over say a perfect field is automatically F finite and we'll do that explicitly a bit later on. But then this you can use to give a whole bunch of other examples in particular, all right? So using that sort of the push forward notation is compatible with localization, all right? It follows that if you take an F finite ring, then any localization of an F finite ring is also still F finite. And similarly, well, if R is local, right? Say with maximal ideal M and I look at the M at a completion, all right? Oh, I guess I should do this really in two steps. So I can think about the completion of that thing is an R module, all right? And that's the same as just tensoring with R hat because I've assumed F to the E lower star is finally generated here, right? Obviously, but you can also check that you don't need the F finite assumption for that first equality, right? But again, this push forward notation is compatible with completion. So again, you're allowed to take any finite, F finite local ring and complete it and you still stay F finite as well, okay? So very easy to come up with lots of examples and most of the things we'll run into throughout certain of the lecture series in that I run into naturally in life satisfy this assumption, all right? So, but... I just have a question real quick. Doesn't it suffice to check for a single E that F lower star, F lower star? That's right, so I could also take this for all and write for some, all right? And that also works totally fine, okay? All right, so, okay. So some of the implications of F finite and as I mentioned, at least verbally, a little more high brow, let me go back down to Earth here really, I really, really need throughout this lecture series one thing that we're gonna use over and over again, basically, namely that if I take an F finite ring, then for all E, the sort of home set from F to the E lower star, R back to R is compatible with localization and completion again, right? So here, what do I mean by that? So here, all right? So what this says is I can move this localization inside of the home, okay? And of course I've used again this property that I can move with the W inverse inside of F to the E lower star as well in this process, all right? And same thing in the completion case, all right? And again, why does this follow? Well, in general, whenever I take a home and the first entry here is finally presented, then it's compatible with flat base change, all right? So in our setting here, certainly localization and then all my local rings are in Ethereum. So completion of this local ring is also flat, all right? So I can just move these things inside very easily but I need a finite nest to do that, okay? Any questions? And where is this home set gonna come up? Well, we'll see that sort of immediately. I think about this home set here is the set of all potential splittings of the eth iterative Frobenius, all right? So let me recall here what F splitting is, all right? So an F splitting is simply our module retraction for Frobenius here, all right? So it's a map in home F lower star R back to R which is R linear and sends one to one. So put it another way. Here, if I look at the Frobenius map, R to F lower star R as we saw, then an F splitting is a map that heads back the other way so that the composition here is the identity map on R, okay? Why do you call it F splitting? Well, another consequence of this thing is of course that this then splits as an R module is R direct sum something else, okay? So it sort of splits this R factor off, okay? So and take this as a definition if you will, right? And sort of to follow up on the previous comment here, right? So you could also hear what this was a splitting of the first Frobenius map but I could also look for splittings of the Frobenius iterates, right? And it's easy to see that a composition of split maps is split and sort of to also go back the other way, so it's equivalent to ask here that R is F to E split for all E bigger than zero. And again, it suffices actually to check for a single E, right? So but I'm just gonna sort of use maybe the non-standard term F to the E split here, perhaps throughout, okay? Great, so the first sort of result I wanna sort of sketch for you is the following. So here we have a definition of what it means for R to be F split. And the result I wanna talk about is, well, if you're regular, then you're also F split, right? And in some sense, maybe the existence of test elements argument really comes down to saying something a bit stronger and we'll quantify that in a second. If you're regular, in fact, I want you to think in your mind that you are very F split, right? That there are tons of these things, okay? And in order to see that a regular ring is F split, I need to go back to the characterization of regularity in terms of Frobenius that was talked about last week, i.e. R regular or ring is regular in characteristic P if and only if Frobenius is flat or all of its iterates are flat, same thing, all right? And what does that mean, right? So well, I've added this special assumption here that R is a finite, right? So flatness is the same as saying that F to the E lower star R is a finitely generated projective R module, so because, okay, right? So here's another place where I've used the F finite assumption to do that. And of course, here, I could have instead of saying projective also said locally free, right? And that's really how I'm gonna use this characterization, okay? All right, great. So let's see here this statement that R is regular implies that it's also F split, okay? All right, so to do so, I want to look here at the evaluation at one map, and this is maybe the first time this is coming up, but it's gonna come up over and over again, right? So take this homset, which I wanna think of as the set of all potential splittings of the E iterative Frobenius, okay? And look at the map, which just takes a homomorphism and it takes the evaluation at one, okay? So that's an R module map here back from the homset. Now, when I do that, asking that R is F splits is asking if I can send one to one. So it's simply asking, is this map a surjective? Can I hit one, okay? So asking if this is surjective, of course, can be checked locally, all right? So without loss of generality, let's say that Rm is local and I have to just check that regular implies F split, okay? And again, again, for those keeping track at home, again, I've used that that homset there is compatible with localization to any maximal ideal, for example, okay? Now, in general, now that I'm in the local setting, F to the E lower star R is a free R module, all right? So I wanna think about the following sort of as an abstract question here. If I take any free R module, all right? I wanna know when can I take, given an element of that R module, when can I find a homomorphism that sends it to one that splits it off, okay? And it's pretty easy to see just thinking about the projection maps onto the different factors of this free R module that you can send something to one, if and only if one of the components of this free R module is a unit, right? So the projections onto the different factors generate the homset. And if all of those are inside the maximal ideal, then there's no hope. But if one of them is not, then you can just multiply by the inverse and send it to one. So easy to convince yourself, the things in a free R module that can be sent to one under some homomorphism back to R are exactly, whoops, the things that are in M times G, okay? What's the conclusion? Well, if I take an element, little R and R, then when can I find a potential F splitting that sends it to one? Well, this happens if and only if F to the E lower star R is not in M, F to the E lower star R, right? Which is the same thing as saying, okay, so of course, this is F to the E lower star of M bracket P to the E, right? So that's the same as asking that R is not in the eth for benius power of the maximal ideal, all right? So to conclude that my regular local rings F split or F to the E split, it simply suffices to observe that one is of course not in this guy. So it's F to the E split, okay? And there's our first sort of proof, all right? So sort of, but again, sort of the mantra of this first lecture is I want you to think of a regular ring here as being very F split. And we've seen here that it's much easier to send not just one, but lots of things. I can split off and send those to one. So that brings us to sort of our next definition, right? So R is strongly F regular. If the following condition holds, well, I wanna be able to take any non-zero element of the ring and I want that there exists some E bigger than zero and some potential F splitting with a property that I can send F to the E lower star D to one, all right? So in other words, R is said to be strongly F regular if given any non-zero element of the ring after taking enough for benius push forwards, I can send that element back to one, okay? So essentially it follows from the arguments we used here to show that R is regular that implies F split. We can see a much stronger statement. It's easy to get lots of things and send them to one. So that brings us to the sort of the next lemma here. R is regular, implies that it's also in fact strongly F regular. And where does this word really come from here, right? We saw weekly F regular before, strongly F regular is meant to be a stronger condition, right? So then this week F regularity, right? And it rose really from an attempt to show that F regularity localized, right? So in fact, strong ref regularity implies that all localizations of R are weekly F regular, i.e. this right condition was what historically was called F regularity directly, all right? Are there any questions about the lemma or the statement? All right, so let's go through a proof real quick, okay? So again, all right, so let's check the first one. So I wanna show that if R is regular, then it's also in fact strongly F regular using more or less the arguments we just looked at, right? So in the R regular implies F split, we looked at the evaluation at one map and I wanna do sort of a variant on that same trick. So if you take any non-zero element of the ring for large E, for instance, I could look at the map from F to the E lower star, the map from the hom set of potential F to the E splitings back to R, which is really instead of evaluation at one, it's the evaluation at F to the E lower star D instead, right? And again, saying that R is strongly F regular simply says that this ideal IE is trivial for some E, okay? So I'm trying to check that this ideal is trivial, okay? And, great. And in fact, it's a vices to check it. Well, once you've shown it for one E, we can check to see that it is true for all larger E as well, right? Yes, so just to answer a question in the chat, all the rings are at finite, right? Great. So using that R is F split, which we know since R was regular, right? In fact, these ideals are ascending, right? So let me pick an F splitting, let's call it chi, right? And let me argue to you that these ideals are getting bigger and bigger, right? So, well, let's say you take some element X in IE, then since IE is an ideal, certainly X to the P is in IE as well, right? So, and how is the ideal IE defined? Well, it's defined as the set of elements so that I can find a map from F to E, lower star R back to R, which sends F to the E, lower star of D to that element, right? And what I can do now is push this forward one more time under Frobenius, okay? All right, so hit this with F lower star, whoops. All right, so if the original map was Phi, right? I can find, I can push that forward and now I get a map from F to the E plus one, lower star R back to R, which sends D to X to the P. And now I can hit that with my F splitting chi, right? Now, if I think about that, I've rigged it, right? So this is X times F lower star at one, chi sends one to one, so this sends F lower star of X to the P here, just to X, all right? All right, great. Oh, doesn't finish the proof, finishes the first claim here, right? That these ideals are ascending, right? Okay, so if we're trying to show that some IE is trivial, right? I know they're ascending, checking that one of them is trivial is equivalent to checking that the here, here it suffices to check now, since they're getting bigger and bigger, to check that there's some, IE the union over all the IEs is R, right? So again, R here is an Ethereum, so a sum over these ideals, if they're getting bigger and bigger, is just equal to IE again itself for some large E, okay? Right, so again, we can check this locally, right? And now we saw before that you can split something off and send it to one exactly when it's not in the Frobenius power of the maximal ideal, all right? And of course, if zero, so D is non-zero inside of this assertion here, right? So we get that directly, okay? Any questions about that? Trying to move as fast as I can here, right? So for two, to see that if you're strong there for regular, then every all localizations of R are we there for regular, it suffices here since strong F regularity is compatible with localization, sort of from the definition, it suffices simply to prove that strongly F regular implies weakly F regular, IE that all ideals are tightly closed, right? So let's check that implication. Well, if X is in I star, then by definition, this means there exists some non-zero elements in that multiplies, whoops, that multiplies X to the P to the E back into I bracket, P to the E for all E, all right? And what I can do is I can find a map from F to the E lower star R back to R by assumption, call it phi, which sends F to the E lower star D to one, okay? And when I do that, it's pretty easy string of checks here. So, okay, so that means X is X times one, this is X times phi of F to the E lower star of D. I can bring that X in, right? And now I can use the tight closure relation to say that that's in phi of I bracket P to the E, right? And of course, I can then move that bracket power out so that thing is in I, okay? So easy to check that if you're strongly F regular, then you're also weakly F regular as well, okay? All right, so that finishes the proof of sort of this lemma here on the left. If R is regular, it's also strongly F regular and all the localizations are weakly F regular, okay? All right, great. So I wanna get as quick as we can here to sort of the next test ideal and to do that, let me introduce the following notion, right? So we say that an ideal I inside of R is uniformly F compatible if and only if, okay. So I've written it in some gross formulaic way here, right? But let me say it more in layman's terms, right? For all E and all potential P to the Eth splittings or splittings of the Eth iterative Frobenius, the ideals preserved under those splittings or potential splittings, okay? Or I said another way I could write this as some really kind of nasty sum, right? Look at a sum over all of the E's and all of the maps and I wanna look at all the evaluations on I and I require that you're still inside of I, okay? Great, so let me sort of reformulate this in a slightly different way, okay? And talk for just a second about the Cartier algebra of R, right? So well, this came up in our proof already, right? But one of the weird things that I ended up having to do was compose a potential splitting, push it forward and then compose with another actual F splitting in the proof, right? So the Cartier algebra here is just a sort of formalism that makes it a little easier to write that down, right? So and again, right? So I want you to think about a map inside of HOM F to the E Laura star R comma R, right? What does it mean to take such a map? Well, F to the E Laura star R is just R as a set. And so if I forget about the decoration, it's just an additive map here on R, right? And all I'm really doing is I'm rigging it so that P to the E powers pull out, right? So I've rigged it so that the map is so-called P to the minus E linear, okay? And the advantage here, if I forget about the decorations which I went into this effort to define earlier is really that as soon as I've rigged it this way, now the source and target here are the same. So if I take P to the minus E1 linear map and P to the minus E2 linear map, then it's easy to check that phi one of phi two, which I can just compose them directly. This thing is a E1 plus E2, P to the minus E1 minus E2 linear map, all right? So in other words, this is some formal way here of doing what came up inside of this previous proof, right? So what does this correspond to? Well, really this corresponds to if I write it in this sort of other notation with push forwards, I should think about taking the product phi one of phi two is really, well, how do I compose the maps? I first have to push forward, right? Under one of the iterates of Frobenius in order to do that, right? So the Cartier algebra is just a way of sort of forgetting about the Frobenius push forwards to make that easier, right? And so with that though, you can form what's called the total Cartier algebra, right? So just take a direct sum over all of these hams or over all of these P to the minus E linear maps for all E and under function composition, this becomes now non-commutative ring, okay? I use the word algebra here because it's traditional, all right? So, but let me just say real quick that there is some abuse of language here, right? So this ring that's total Cartier algebra is neither commutative nor is it in our algebra, right? So when you say something's an algebra, usually that means that R has to be in the center of the ring and that's not true. R is the zeroth graded piece of this ring. However, if I take an element R and R and some elements say phi and CRE, then by construction we have R phi is phi of R to the P to the E, right? So it picks up this skew linearity, what's going on here? Okay, okay, so what's the advantage of sort of putting this in? It makes it much easier for me to write down what it means to have uniformly F-compatible ideal, namely an R ideal is uniformly F-compatible if, well, I can think about R as being a module over this non-commutative ring on the left by evaluation and I is uniformly F-compatible just means that if I look at the whole Cartier algebra and evaluate that on I, then that has to be contained inside of I, right? So you can see here, I will use this notation later in some of the talks, right? This makes it sort of much easier and quicker for me to write down what it means for one of these ideals to be uniformly F-compatible, okay? All right, so here to sort of, as we build towards the conclusion of the lecture, I wanna sketch for you the proof, right? That test elements exists, right? So this is where we've sort of been building up to the whole time, right? And the first observation is that this finiteistic test ideal is uniformly F-compatible. And again, like earlier, I've sketched this out as an exercise in the notes, all right? So to show that test elements exist, what am I gonna do? I'm gonna argue for you using some of the machinery we've built up that there is a non-zero element C that's contained in every non-zero uniformly F-compatible ideal of the ring. And in particular, it'll be contained inside of tau finiteistic R, right? So IE, C is the test element, okay? Is everyone happy with the theorem as stated? Okay, let's see if we can get this done. Okay, so all right. Here's my proof, all right? So the first, what we're gonna use a bunch of the things we've already built up is you can find a non-zero element so that when you invert it, R is regular, R during one of our X is regular, all right? So that does require something, R has to be nice. So R is excellent here, so there's no problem with that, right? And in particular, we saw that regular rings are F-split, all right? And we also know that the humset of all the potential splittings is compatible with localization. So what that means is I can find a potential F-splitting. Okay, so it doesn't send one to one. I don't know that R itself is F-split, but it certainly has to send one to some power of X, okay? Let's call that element Y, all right? I'm gonna try to keep track of the powers and what goes on here, okay? And great, so once I have that, what I can do is I can take this sort of relation, phi of one is Y, and I'm gonna multiply the whole thing by Y, all right? So this implies now that, well, Y squared, right? Is equal to phi of F lower star of Y to the P, all right? And in particular, that is in phi of F lower star of Y squared R, all right? Since P is a prime, it's at least two, okay? And as soon as I've done that, I've rigged it so that I've got Y squared on both sides here. I can iterate that. So in fact, Y squared is then in phi to the E of Y squared for all E, where of course, phi to the E is this twisted iterate of the Frobenius splitting map, right? Like in the Carti algebra, okay? Right? And now I can multiply by Y one more time, right? And my claim is that that Y cube is gonna work, right? So that that is the C that I'm gonna take and show you that it's in every F-compatible ideal. All right, to see that, let's take any uniformly F-compatible ideal, right? Pick a non-zero element D in it, right? And let's use the arguments we had before. Well, again, we chose X so that R join one of our X was regular and we knew that it was not only F regular or not only F split, but that regular in fact implied strongly F regular, right? And if you're strongly F regular, you can send anything to one. So again, by this sort of same argument we had above, I know that there exists some E prime, oops, and some psi, let's say in HOM after the E prime lower star R back to R, right? So that, well, I can't send, I don't know that R is strongly F regular. I only know that after localizing it at Y, let's say, so I can send F to the E lower star of D to Y to the M for some large M, all right? So in particular, since J is uniformly F-compatible, D is in J so I can hit it with any map. I end up still inside of J so I get that Y to the M is in J for some large M. On the other hand, all right, if E is really large, that I still have P to the E plus two is bigger than or equal to M. So this means that Y cubed, which is phi E of F to the E lower star Y to the P to the E plus two, right? This here then is contained inside of J, which is contained in J again, not using the F-compatible, right? So there is the proof that test elements exist in this sort of rough form, right? Really building up from first principles, right? So I'm gonna finish here with one last definition and I'm not gonna get through everything I had planned here, right? So I will finish with the definition here and I'm gonna recall it here at the start of the next lecture as well, okay? So with that, let me just say that a corollary of this definition is, or corollary of the existence of test elements is that there is in fact a unique smallest non-zero uniformly F-compatible ideal, right? I'm gonna call that the non-finitistic test ideal, right? And that is sort of the second test ideal that's gonna show up here in this sort of lecture series, all right? So what is that, right? So how non-finitistic are is, well, look at the sum over all E, sum over all the maps, right? And take the image, right? Of whatever this element C was, this test element from the previous theorem, right? So when I've done this, right? So I knew that C was in every uniformly compatible ideal. So this sum is then automatically in every uniformly F-compatible ideal, but using that I can compose maps with one another, the right-hand side itself is a uniformly F-compatible ideal, right? All right, so let me write this maybe in an even shorter, more shorthand, all right? So this is also of course equal to take the whole Cartier algebra of R and just hit it with C, all right? And so the corollary of the existence of test elements, proof, right, is there is this ideal, the non-finitistic test ideal, which is the smallest uniformly F-compatible one that's non-zero, it's contained inside the finiteistic one, right? And it's gonna fix some of the problems we had with the finiteistic test-by-deal, right? So you can see almost from the description here that it's compatible with localization and completion, right? And I'll start with it next time and tell you a bit more about it, okay? Any questions? So it seems like C is dependent? C is, so what does C depend on, right? So let me reinterpret the question that way, right? So maybe from the definition here, right? So in terms of this definition, I want you to take out of this that the non-finitistic test-by-deal is just the smallest non-zero ideal that's uniformly F-compatible, right, inside of the ring. It's not clear a priori that such a thing exists and its existence follows from the existence of test element proof, right? So in that, I've explicitly exhibited this unique smallest guy, right? So from the test elements exist proof, what you basically see is if you take any element in the ring, let's call it little X, right? And you know that when you localize at it, what you get is regular and then some power of that will be a test element, right? In particular, find a big enough power so that it's in the image of some potential Frobenius splitting and then three times that power will be a test element for the ring. So it's even somewhat explicit, okay? But in particular, that says if I localize at any element and I get something regular, then a high power of that element is inside of the test ideal, right? But this really only, the X or the C here only depends on the ring, okay? Does that answer the question? Are there any other questions? Real quick before we, before it's, but maybe we should take a moment to thank Kevin for his testimony.