 Okay, welcome back for lecture three, right? So a couple of things. So first off, for those of you being a digit at home, you may have noticed that my speed in lecture two was probably slightly greater than my speed in lecture one as I'm trying to cover more and more advanced material, right? And unfortunately that will also continue in lecture three. So I've pre-written a lot more and that makes it even more important. Please, if you don't understand something I say or have a question about something I've written, please do speak up. So, but also let me also just say that in the previous segment, I was working with duality and comicali rings, right? And if you'd like, take a step back and we're now going to go back to just saying that R is an f finite domain. I'm not assuming in comicali unless explicitly stated later on, okay? Okay, so, but in the previous segment, what we saw is some special case where there was really one map on the canonical module, right? Or P to the minus c linear map that was controlling everything. And we use this to develop some version of a test ideal theory, right? And I wanna do that formally in general, right? So, all right? So let me talk a little bit about test ideals of P to the minus e linear maps here, right? So, okay, so again, recall, we talked about what a P to the minus c linear map was is just a map from f to the e lower star R back to R or I could think about it as a map from R to R which is additive. And so that P to the ETH powers pull out. It has this Q linearity, right? So, and again, I wanna define a test ideal that depends on the map, right? Instead of on the ring, okay? And so what we're gonna do is sort of the most naive thing you could do here which is the test ideal of a map is gonna be the smallest non-zero five compatible ideal, right? In order to make that definition once again, I mean, you could just make it but then you, the first thing you have to do is show that it's well-posed, i.e. that there is some non-zero element in all non-zero five compatible ideals, right? To show that that is a well-defined test ideal, right? And so in a sense, I'm starting this lecture with revisiting the first lecture once more, okay? So in a sense, I wanna say that the argument for the existence of test elements, right? Immediately carries over to this situation. And so I'm gonna take this as an excuse to sort of revisit that again, right? Cause it is a super important argument, right? So the claim is that given any non-zero p to the E linear map, right? There is a non-zero element C inside of the ring, right? Which is inside of all of the non-zero five compatible ideals, okay? We'll use that to define the test ideal of five, right? So, well, again, remember from the first lecture how all of this went, right? So in particular, what we did is we said, okay, take a non-zero element Y, so that R join one over Y is regular. And the key from the first lecture was this means that it's also strongly of regular, okay? And now I wanna sort of take something from the second lecture, right? So R join one over Y now is regular. So in particular, it is also comacoli. And if I'm willing to localize further, it is sufficiently local comacoli. So that means here that the home set for the omega, which is just R join one over Y again, right? Has a generating map, right? You can localize further to assume that your given map, all right, so maybe I'll fix that, cause it looks weird. Your given map phi, when I localize it, in fact forms a generator for that home set, right? So again, put it another way, if I've localized enough so that I have a generating map to begin with, any given phi is a pre-multiple of that thing. And then I can localize it, whatever that pre-multiple was, and that pre-multiple is now a unit. So you might as well assume that your given map is the generating map, okay? All right, and then sort of the last condition here, right? Was that phi, let's see if I have this right at my notes, right, was in the image of phi of f of the alloster R. Now I've done something a little that I didn't do in the previous version of the proof, right? Namely, we know that after I localize it Y, R is F-split, right, so that means that there exists a P to the minus c linear map on R, so that some power of Y is in the image, okay? And what I'm doing here is I'm replacing that power, Y with its power, so that it's in the image, right? And I'm also using that I could have taken as my left the phi itself, because all of the maps after I've localized are just given by multiples of the localization of phi, right? So using all the machinery built up so far, you can assume that after replacing Y by a power, right? Y is in the image of phi of F to the E lower star, okay, and now let's just revisit the proof once more, right? So with this, we had that Y squared is then gonna be in phi of F to the E lower star, well, I should put Y to the P to the E there, but P to the E is at least two, so certainly Y squared then is F to the E lower star, Y squared, so certainly Y squared then is inside of phi of F to the E lower star, Y squared, so I just multiplied by my equation by Y, but now I can iterate this, so in fact Y is in the image of phi to the N, F to the E, N E lower star, Y squared for all N, right? And so then I get that Y cubed is in phi N, F to the N E lower star, Y to the two plus P to the N E R for all N, right? And this element Y cubed is the C that we're gonna show works in all the things, so this is mimicking exactly the existence of test element proofs as best as I could, right? Okay, how does it proceed? Well, again, if you take a nonzero phi compatible ideal, call it J, pick a nonzero element D inside of that ideal, right? The fact that R join one over Y is strongly irregular and that phi is a generating map implies that you can, after taking sufficiently many iterates, send D to one after localizing it Y, so before localizing, I can send D to a power of Y, i.e. there exists some large iterative Frobenius and there exists some large power of Y, right? So that's phi of N prime after the N prime E lower star of D gets sent to Y to some power, okay? In particular then, since D is inside of J and J is phi compatible, that means that Y to the M is inside of J for some large M, but now we already know that I can send all large powers of Y under phi back to Y cubed i.e. to C, right? So this gives that C equals Y cubed is inside of J, i.e. this is something which is inside of every non-zero phi compatible ideal, okay? So again, this is probably very fast, right? But again, I'm mimicking exactly the same proof and it's written up fully in the notes here, right? So good excuse to revisit that proof once more, okay? Same consequences before i.e. We can define tau of phi to be the smallest non-zero phi compatible ideal. I have some element C that's in any of those, right? And I can then use sort of the obvious iteration recipe. So take C and hit it with all the iterates of phi and this produces a phi compatible ideal as you're sort of by observation, which is gonna be contained in all of the others, right? So tau of phi is given by the expression here in this sum, okay? In particular, it makes sense, okay? And what I wanna do now is, okay, so why have I introduced this? One, it's gonna make some of the later statements a little easier, right? But my claim is that looking at test-ideals maps again tells us some non-trivial statements about the big test ideal that I'm really interested in. It's one of my fundamental objects of study, okay? So to do that, let me introduce some other sort of very convenient formalism. We talked about the Cartier algebra on R, right? In the first lecture. So this was the graded ring where the ETH graded piece was the set of all potential F-splittings of the ETH iterative Frobenius, right? So that I called CR, okay? And what I'm gonna do now is look at certain sub-algebras of that thing, okay? So what is the Cartier sub-algebra? Well, I wanna look sort of at taking only subsets of maps and do the same constructions, right? So a Cartier sub-algebra is a subring, a graded subring of the total Cartier algebra. I still wanna require that the zero-th graded piece is R, right? So this is just the set of all hams from R to R, so multiplication by R, okay? And I wanna require for all of my guys that they have at least one non-zero map in them. And you'll see why that I wanna require that in a second, right? So there is at least some DE, right? So that DE is non-zero for some E, okay? Is everyone happy with the definition here? So again, let me just say this a little more informally. D is determined by taking some collection of potential splitings of Frobenius and all of their compositions and iterates with one another, right? And make a ring out of that, okay? But I'm not requiring you look at all of the potential splitings anymore, okay? And again, same... I start to feel like a one-trick pony here, right? So tau of D is just the smallest non-zero ideal that's compatible with all of the maps in this sub-algebra that you chose, okay? All right, so, and this is my definition. Sorry, real quick, in general, for the full Cartier algebra, is it true that it's sort of generated in degree one position? Absolutely not. So one, again, be careful because this thing is not punitive and even the worst use the word algebras bad, so this is a little bad to think about, right? So the answer is, in fact, it's worse than that. Let me just say that in general, it is not funnily generated. It is not funnily generated even in nice situations, right? I.e., for me, in the anti-canonical ring is funnily generated. It's not funnily generated, okay? So that was a question for a while. An analogous local question was asked by Smith and Lubesnik and Stanley Reasoner rings were shown to give the first counter example and a four-authored paper that I don't know have the names out at the top of my tongue, right? But this is studied and how not funnily generated is also something that people have studied, all right? All right, so it's really a pretty kind of gross thing in many ways. On the other hand, if R is Gorenstein, it is, right? And in fact, it is sort of generated in Degree 1 by the trace map that we talked about before, right? And sort of the opposite ring of R joined Frobenius, right? And it is a very concrete and understandable ring, okay? All right, but great question. Okay, so I want to show you, we've defined now tau of D, right? It's not so clear that the thing exists again, right? I have to prove for you that such a thing exists, except it's going to be pretty easy to see that it exists. We know that by definition, let's just assume for a second that tau of D exists, right? By definition, it's compatible with any map inside of the Cartier subalgebra, right? But tau of phi is the smallest thing that is phi compatible. So it's easy to see, right? Sort of just from the definition that the test ideal of D, right, contains tau of phi for any map inside of there, right? Simply because, right, this ideal is phi compatible. And so the smallest one, tau of phi, has to then be inside of that, all right? So in some sense, it's easy, right? So I've already given you a proof that tau of D exists. If you take something which is decompatible, right? Then it contains the element C from before with respect to the sum phi inside of the thing, right? So the existence of this thing follows again directly from the previous theorem, right? And moreover, the same sort of formulas all work, right? So IE, then this is going to be the same. Is the sum over E look at DE and evaluate, say, at... Whoops, right? Evaluate at C, right? CR, if you like, for C as in previous theorem, all right? So again, these things exist in general and it follows from, once more, the existence of test elements argument from the first lecture in many ways, okay? In fact, I want to say something a little bit stronger, right? Right away, and this is a theorem of Schwede, right? Not obvious at first glance, right? But the test ideal of an arbitrary Cartier subalgebra here, in fact, is completely determined by and is the sum of the test ideal of all the maps that you started off with and took inside of it, right? So one of these inclusions here is obvious, right? So the right side is contained in the left, right? But the left... The other conclusion requires some actual serious work, right? And again, my point here is to say that this actually tells us something which is not so immediate, right? From the original definition for the non-finitistic test ideal again. So the non-finitistic test ideal, which is the test ideal of the total Cartier algebra by definition, then is going to be the sum of the test ideals for all of the maps, for all of the potential Frobenius splittings of all iterates of Frobenius, okay? Is everyone... Any questions about the statement of the theorem here? And if they did it, right? So again, the point here is some sense that this is really a non-trivial statement, right? Giving us some more structure about the test ideal that we didn't have before, right? So I want to say something a little bit more, right? So in order to do that, let me just say that I've sort of aimed these lectures as an algebraic audience as I could, right? So given what my expectations were, right? And we'll see this more at the end of the lecture as well, right? But ultimately, part of the importance of test ideals is that they are very closely related to a bunch of geometric constructions as well, right? I don't want to go into a whole lot of this, but I can't not reference them because to me they are fundamentally important, right? So when R is normal, there is a so-called divisor correspondence arising from duality that associates to every potential Frobenius splitting a certain effective Q divisor on spec R. And if you don't know what that is, just forget and stop listening here for two minutes, okay? But if you do, right? So this gives me a so-called boundary divisor on spec R. It's not an arbitrary one. So it's one so that p to the e minus one times kx or kr plus delta is linearly equivalent to zero, okay? I am in this correspondence using this efficient local to write this thing down as I had before. So in the Coma-Caulay segment, but this is a pretty minor assertion, okay? So when you do this, this gives you a way, really the best way to do it is again, using Cartier subalgebra to define a test ideal for an arbitrary Q divisor, even without these conditions here, right? And an interpretation of this statement here that's how non-finitistic R is the sum of the test ideals of all the maps then, gives you a very non-trivial geometric statement, namely that the non-finitistic test ideal is the sum of the test ideals for all the boundary divisors, right, that are sort of, so that k plus delta is Q-Kartier of index prime to p, right? So a sort of very important application in geometry as well, okay? So this formalism of Cartier subalgebras and looking at the test ideals of maps has a long story here and really is, even for the usual non-finitistic test ideal of R, gives you a whole bunch of important theorems, some of which are important because they connect up to a statement in geometry that are not so obvious at first glance, okay? Okay, so with that, that's sort of all I want to say about test ideals of maps and test ideals of Cartier subalgebras, other than in the next segment, I want to give you another example of where I can define for you another Cartier subalgebra and use that to produce a whole bunch more test ideals with super cool properties and tell you some of my favorite applications thereof, okay? So if you space that for a second, come back, right? So here is an important, again, starting definition here for much of the stuff we're going to do, okay? So the following really are based on a set of test ideals that were first introduced by Harun Yoshida and so I want to talk a little bit about test ideals of ideal pairs, all right? So again, same standing assumption, so R is an f finite domain here, right? But the idea is that now I'm going to, in addition to taking R, I'm also going to consider and write down a test ideal using the input data of a non-zero ideal A inside of R and a non-negative real number T, so an AT test ideal, all right? And to do this sort of in the cleanest way possible, right? I'm just going to use Cartier subalgebra to do it, okay? All right, so I'm going to define for you a Cartier subalgebra where the ETH graded piece of this thing, C E AT, is given by, well, I take the, oops, ETH graded piece of the total Cartier algebra and I just look at all pre-multiples by some specified power, A to the roundup, T P to the E minus one, right? Of my ideal A, okay? So said another way, if I don't use the language of P to the minus C linear maps, it's, well, look at the total, the homestead of all potential splitings of the ETH graded Frobenius and then just multiply that by this specified power, right? Of the ideal A, right? And I want to really say this a lot more concretely, right? What does it mean for an element in here? So A potential, right? What does it mean for a potential splitting of the E iterated Frobenius to be inside of this Cartier subalgebra? Well, it means it can be written in the following form. So I will have phi of something is equal to some finite sum of things of the following form, right? So it's a sum of a bunch of other maps, phi j pre-multiplied by elements, A j, right? So that each of these A j, well, okay? So phi j is another potential iterated splitting, right? And A j is in this ideal A t p to the E minus 1 all rounded up, okay? Right? So it's a sum of maps where you take a potential Frobenius splitting and you multiply by this power A to the roundup t p to the E minus 1, right? And in fact, so it's pretty easy to check this gives you a Cartier subalgebra, right? So what do I mean by that? Again, so the Cartier subalgebra is just a graded subalgebra of the total Cartier algebra. So the total Cartier algebra, the multiplication structure on that thing is just function composition. So to check that this is a Cartier subalgebra, basically what you have to show is that I can take two things in here and compose them, right? And I will end up still inside of the same subring, right? So in fact, this is not so hard, right? So I don't want to go into these details too much, right? But the roundings are specifically done and the sort of exponents are done exactly, right? So that's the, this composition works, right? So IE, if you look at CAT E1, one of these graded pieces and multiply times CAT E2, right? That function composition is automatically inside of CAT E1 plus E2, okay? And that's just really a rounding trick that you have to use to check that, okay? Great. So again, this is at this point already been defined, right? But just to put it down here explicitly, we've defined then tau of CAT, right? And I'm just going to let tau of AT be that for simplicity of notation, okay? That is the smallest nonzero ideal, which is compatible with everything inside of this fixed Cartier subalgebra, right? So in particular, it's equivalent to say it's the smallest nonzero ideal, which is compatible with all things of the form phi F to the E lower star A blank, right? Where phi is a potential iterated Frobenius and little A is in this ideal ATP to the E minus one, okay? Is everyone happy with the definition here? Are there any questions? Okay. So as I said, the roundings are specifically chosen here so that you get to Cartier subalgebra, right? However, one of the features of the theory is that exactly what roundings you use don't matter, right? So there are a bunch of alternative roundings, right? That often make it easier to prove and write statements down, right? And so I want to give you sort of the alternate one, which makes proving a bunch of things very easy right off the bat, and is itself also a non-trivial alternate characterization of the AT test ideal, right? So said another way, right? You could take this lemma as a definition of the AT test ideal again, based solely on the non-finitistic test ideal of R, right? So the input on the right hand side is this non-finitistic guy that we've already defined. And the other thing that you're supposed to observe here is that I've switched my roundings instead of having a P to the E minus one everywhere, I've switched that for a P to the E instead, right? So otherwise, it really looks very similar to the previous descriptions we've got, okay? But it's a very useful alternate description that you have. So let me say it another way, right? So sort of in two other ways. So another way to write this down is, I could write it in terms of the total cartel algebra, right? So instead of using the F to the E lower star notation, which gets pretty cumbersome to write down, I can write it in terms of P to the minus E linear maps, right? So now I can write it here like this, right? Now I've written tau of R where I did not write tau non-finitistic or tau finiteistic anywhere, right? And for the very careful listener and reader, you will notice that at this point, I have made a definition of tau of R, which didn't have those decorations by taking A equal R and T equal one, right? And my definitions mean that tau of R here is non-finitistic test ideal. I would like to tell you that that's just an accident here, but unfortunately, that is certainly a personal bias. I tend to write tau of R for the objects that I know has the best properties, but historically, one should be a little careful here if you refer to test ideal, the right thing to do or historically is to mean the non-finitistic, or is to mean the finiteistic test ideal, which I'm kind of throwing by the wayside in much of the theory I'm talking about now, okay? All right, so, and let me write one more quality down. Another point here is that we also know that you can build up the test ideal from lots of other things. Given a single test element, you can use it to build up tau again. So I would say that you can do more. In fact, the same description works if instead of putting tau of R here, you put any single test element down, anything in tau of R works, right? And that's part of the statement of the lemma. In particular, this is extremely useful, all right? When R is strongly F-regular, in which case one is a big test element or the non-finitistic test ideal is R, right? So you can remove the test ideal or see completely from the description and then you get that the test ideal tau AT is just given by taking images under all of the Cartier algebra with the appropriate powers of A, all right? And you don't have to worry about the extra test elements at all, okay? Is everyone happy with the statement of the lemma? Okay. So one of the other reasons I wanted to do this lemma in some sense and say something about it is really the following. We have two ways to sort of play around with and take things from test ideals and it's very important and essentially all arguments about them come from using the two directions. I can use that the test ideal is uniformly compatible with respect to some set of maps and that gives me a way to bound the test ideal from above, right? So we'll do that first and then I also know that I can produce things in the test ideal once I've found any single thing in it, right? By taking images of maps over and over again, right? And so sort of much of the power of the theory comes from being able to go from two directions. All right, so let's just see that. All right, to start off with looking at the lemma let's just let J be the right-hand side of this thing. So I've defined J to be this sum where I've switched the exponent to be P to the E and I've put in the non-phonicistic test ideal and I'm trying to show that that is equal to tau AT, right? So the first thing I'm going to do is check that it's compatible with all of the maps in CAT, right? And this is just an explicit check. You just write down and again just uses some rounding checks, not hard. Immediately, since you know that tau AT is the smallest non-zero ideal which is compatible with CAT this gives you that J contains tau AT. All right, so one of the conclusion one of the inclusion comes for free, right? Just from some simple check. On the other hand, right? We also know that if you take any elements of well, okay, so I know that, right? The test ideal of all of the maps, right, is tau of R, right? So I can pick some test element in there, right? So pick some element C which is in tau AT which is in particular in tau of R, right? Any test element I can use to then build up the test ideal again, right? So I use sort of the standard description here, right? So take iterates under all the maps for a fixed test element and that produces the non-financial test ideal of R, okay? And from that, it's pretty easy to check just by plugging in to the definition of J here. So I'm gonna plug in for this guy, right? And expand. If you do that, the naive formula spits out something which is manifestly in tau AT again, right? So I've written it all out here. I don't think it's worth going through here right now, right? But let me just say that this is just, again, an explicit computation. You just write it down and just falls out completely, okay? I've been a little careful about how I picked the C and that I picked it in tau AT to make this all work. But once you know that it works for such a C, you could replace it with a C which is in the slightly bigger ideal tau of R and the same string of equalities will work to give the full theorem as well, okay? So again, what's the conclusion, right? The conclusion is that it's not too hard to give this slightly alternate description of tau AT and in fact, let me say that some people, if you were just worrying about trying to do this in the simplest way possible and only concerned with tau ATs, you could start off with this as your definition and try to develop a theory just from here instead of how I've done it, okay? All right, so at this point, right? So I want to produce for you a sort of standard list of properties, right? I've got a list of all these test ideals and part of the magic comes that they satisfy some pretty amazing algebraic properties. Let me just say the four I've listed here, none of them are particularly hard, right? These are all going to follow from relatively elementary arguments, but let me go over them sort of one by one. The first one says, well, if I take a bigger ideal A inside of B, right? Then for any T, the test ideal of A to the T is contained in the test ideal of B to the T. All right, so it gives me a bigger ideal, right? Moreover, I also get a quality, oops, what in the world's going on? I'm on the wrong thing. Moreover, I get a quality. If A is a reduction of B, i.e. if A agrees with B up to integral closure, right? So I'm not going to say a whole lot about integral closure here, right? And I think, again, one of the later lecture series will probably say a bit more about integral closure than I've got time to do here. All right, so but the point is, is that there's a natural algebraic condition where quality also holds as well, okay? Second property, right? If I take a test ideal, tau of B to the S, right? And I multiply times A, then I can bring that A inside, right? So A times tau B to the S is contained in tau of A times B to the S, okay? So, and moreover, you also get a quality again if the ideal A, in fact, is a principal ideal, right? In general, you just get containment, okay? Next property, so the so-called SCOTA theorem, right? Let's say that A is generated by R elements, right? And I take an exponent T that's at least R, right? Then, in fact, tau of A to the T, oops, let's fix the statement, okay? So then what you can do is you can pull out one power of A at the expense of reducing the exponent of A by one. Now, I've done something in at least B and C here, and also in D, that is a little shady, right? I.e., I've defined tau of A to the T, and now I'm looking at things that look at, like, tau of A to the T times B to the S, right? So this requires some justification, i.e., you need to write down and define a Cartier subalgebra for not just an A to the T, but an A to the T, B to the S as well, right? So Cartier subalgebra defined for two ideals and two exponents, right? Of course, you can easily do this for finitely many ideals and finitely many exponents. There are some compatibilities to check here, right? So if A is equal to B, this is gonna give you two different definitions, right? Of things to check, right? So they're not so straightforward. Well, they're pretty straightforward. They're not so hard, right? So this is one of the last exercises that I put down in the notes to be turned in. So I'm asking you to formulate these sort of extended definitions and prove some analogous statements to the earlier description up above, all right? And the last statement here, the last property, right? So the first two properties hold in complete generality, but the last one is really special and is a bit more different, right? It requires that the ring is regular and it says that if I took, look at a product or I looked at tau of A to the T, B to the S defined in the way I just described, then that in fact is contained in tau of A to the T times tau of B to the S, okay? Okay, so I want to get to some of the main applications that I'm going through here, but again, you could have a whole lecture series just on these ideals, tau of A to the T and their applications, okay? So I definitely don't have time to go through a lot of the details here. I've written all of them down in a way that hopefully if you're really interested, you can follow even at this stage, but I think it's instructive to come back after the later lecture series and take a look at much of what goes on, okay? But let me describe some of the statements here real quick and do some sort of highlights of the proofs of some of these properties, all right? So again, just for A, so again, the property A was that if A is inside of B, then tau of A should be inside of tau of B, okay? Okay, so do I have my list up here? No, I have to keep going, all right? So as we go through and look at this, let me just say that if A is inside of B, then the same contaminants hold for all the powers of the ideals. So in fact, whoops, you get a containment of the corresponding Cartier subalgebras just directly. So then it's very easy using the same argument we saw before, all right? So the test ideal for B to the T is automatically compatible with the Cartier subalgebra for A to the T. And so the defining property of the test ideal for A to the T means that it must be inside of that. So you immediately get the appropriate containment, right? So no thinking involved, okay? Moreover, if the ideals agree up to integral closure, then it's pretty easy to show that you can find an element X, right? Which is gonna send all the powers of B back into all the powers of A for all powers, right? You're just using that A is a reduction of B, right? And again, from our formulaic descriptions of the test ideals, right? I can use that in our lemma, right? To just throw that in and with the appropriately chosen test element X, right? You get the other containment as well, okay? So here I'm using that X times all the powers is contained in X times all the powers, right? But that basically requires some understanding of the theory of integral closure of ideals to get this. Okay, all right, so that's all I wanna say about A. We're just gonna run through these one by one, right? So for the second property, right? So I would just say that this is an exercise in roundings, right, or in bracket powers, however you wanna say it. If I multiply by on the outside and I bring it into the description of the test ideal from the previous lemma, right? A comes inside of the Cartier subalgebra by picking up a P to the ETH bracket power, right? The P to the ETH bracket power is contained in the regular power and that exactly gives me the description of the test ideal where it's inside, all right? Moreover, if A is principal, then the bracket power and the usual power are the same, so a quality holds throughout, all right? All right, so last one or the next one, right? So the so-called SCOTA theorem, if A is generated by our elements, right? Then I wanna show that I can, oh, I did it again, right? Show that I can pull out one power of A and reduce the exponent by one, okay? And roughly speaking, this involves a very useful fact, right? So if I have a big enough power of an ideal A but the ideal A is generated by our elements, right? Then I can gather the monomials and collect enough powers of a single element, right? And I split off a bracket power from that power, right? So if N is at least R times P to the E, right? Then I can pull out a bracket power from this ideal, right? This is just an explicit monomial check and again, this I'm sure will come up in later lectures as well, right? So, but again, once you've got that, if you write down the obvious, or this description of the test ideal we had using the appropriate power of A, I can pull out one of the bracket powers from that guy, right? The bracket power now moves outside of the whole thing, right? And I exactly get that the A pulls out of the description, right? So the point here is that this is a purely elementary argument, right? So great, I approve an A, B, and C. Let me postpone saying anything about D just for a second to give you sort of the main application of A, B, and C, if you will. This is the so-called Brinson-Skoda theorem or a version of it. And again, I'm not assuming that this is something that you, if you haven't seen it before, great. See, this is a highlight for a later lecture series. But again, there's a whole lecture series, right? On Brinson-Skoda that's coming up later that you'll return to this, all right? And I'm not giving the strongest statement by far, right? So if R is strongly a regular, right? One can then use test ideals to show, right? That if A has a reduction by most R elements, then A to the R plus N bar is contained in A to the N plus one for all N, right? So in particular, the sort of non-trivial one, the first one, you get that the inner enclosure of the arth power of the ideals contained inside of the ideal A, right? And I've traced this through, right? So, but you can go through in the notes, right? So this one I believe uses, let's see if we do this right, right? So, all right. So, given A bar R plus N, I can multiply by the test ideal, which is trivial because I've assumed R to be strongly a regular. Now I can pull that into the test ideal using property B, which says I can pull anything inside of the test ideal, right? Property A says that I can remove the inner enclosure and I don't change the test ideal. And then I can apply Skoda as many times as I want, or as many times as I can, right? And that pulls out a bunch of powers of A. And at the end, I just forgets the last test ideal to get the containment in question. So all I've used here is properties A, B, and C, okay? Another very famous application of test ideals, right? So, let me run out of time here. So let me not say anything about the proof of property D, but use this as an advertisement again, right? For Skoda, right? So, using essentially just elementary properties of the test ideal again and that F lower star R is free locally, right? That's what allows you to prove sub-additivity. And so, let me just stop there. Am I pulled on on the left? Right? So sub-additivity can be used to show another feature of another one of our later lecture series, i.e., if you look at the R nth symbolic power of an ideal inside of a regular ring, right? Which is defined in terms of all of the associated primes of the ideal if you've not seen this before, right? So here is a nondrival containment in a regular ring that the R nth symbolic power of an ideal is contained in the nth symbolic power of the ideal in a regular ring, right? And once more, right? The key point here is you can easily see that the only thing you need are properties ABC and this last property D, sub-additivity, showing up on what's there, right? And these really slick proofs are really due to Takagi, right? So I probably should give, I'll do some more work in putting, or do some work in putting references down before they're all said and done, okay? And sort of to end the lecture series, I want to say just a couple of applications or a couple of things which have some more geometric flavor, right? So the first one is, okay, so I've defined a sequence of ideals here. We've talked a bit in the first four properties about what happens as you vary the A, right? But what you could also do is you could also think about varying T, right? And sort of by construction again, you can check that if you take larger exponents, the test ideals for a fixed ideal A are only getting smaller and smaller. However, they satisfy some sort of semi-continuity property in some sense, right? So at any point, you're allowed to increase T just by a little bit, right? Without changing the corresponding test ideal, okay? And again, this doesn't change, these are pretty elementary to show. So maybe it's best to describe this by a graphic. This is the so-called F jumping numbers, right? So if I look at the T number line and all of the test ideals tau of A to the T, right? Well, I start off with T equals zero when I get the big test ideal of R, right? But then as I increase at discrete jumps, right? Certain jumps call the CIs. What will happen is the test ideal is constant, right? And then at some point, it jumps and gets specifically smaller at some value, call it C1. And then I go a little bit further and it jumps. It jumps in size, goes down and gets something strictly smaller, right? And these numbers, right here, are called the F jumping numbers of A, right? And so there are some numerical invariance just defined in terms of the ideal A. There's a particularly famous one. If R is regular or say strongly of regular, then the first one is called the F pure threshold of the ideal A, right? And again, it is an object of intense study, right? And so I want to mention another theorem here, right? When R is Q-gorenstein and again, say it's normal, right? Take any ideal and completely not obvious, you get that the F jumping numbers, right, are a discrete set of in particular rational numbers, right? Neither of which are apparent at first glance, right? This, again, probably should have attributions by a large number of people, but at least in the way I've stated it here, right? This theorem is written, is a theorem of myself and Carl Schwede, right? In this setting. So there are many other cases, particularly the principal ideal case that were done earlier by a large number of authors, right? And the proof of such a theorem again really uses a lot more geometric techniques than we've really referenced in these lectures. But I did want to mention that the sort of best proof for the polynomial ring involves only degree, right? And is quite elementary, right? So I've gone through and I've sketched the proof in the notes, which is a proof of Blicola, Blicola, Mestazza and Smith, right? For just the polynomial ring case directly, right? So, and I'll say more about that in a second, right? But the proof only uses degree, okay? And again, I've referenced this a couple of times now, but I'm going to end sort of with the following sort of two theorems, right? Okay, so I've made vague references to the geometric applications of test ideals, but I really, for me, think that they are at the heart of why they have been so useful and continue to be so useful, right? So the connection is to something in characteristic zero called the multiplier ideal, right? And so if you've seen that before, those before, great. If not, I'm not going to go through the details here of this theorem other than to say that there is a description. You can provide, again, an alternate definition of the test ideal, right? Which if you take the same definition and you work over the complex numbers, right? Produces something called the multiplier ideal, right? So there is a uniform definition that will allow you to do both at the same time. And the definition involves so-called regular alterations, right? So an analog, due to the young of resolutions of singularities that work in arbitrary characteristic, okay? But really references much more geometric things than we're really talking about here, okay? Similarly, trying to come up with definitions of tight closure and test ideals that work in characteristic zero has been something that was studied from the very beginning in some sense of the theory of test ideals and tight closure, right? And I would be remiss without stating that there is a lot of very important work that under reduction mod P, right? You can say even stronger statements relating the test ideal and the multiplier ideal. And if you're particularly interested, Kent, the TA for the lecture series, I'm sure is someone who would be someone who would be happy to talk to you more about very strong theorem here, which says something about reduction for a fixed characteristic P instead of just for all large P, right? So again, take these theorems, if you will, as just an advertisement that there is a lot about test ideals that I haven't had a chance to cover. Two things, I'll put references in the notes to one of or two or three other survey articles out there, all of which approach test ideals from sort of a different perspective in particular. There is a very low level one by Smith and Benito where they really try to approach these things just when the ambient ring is regular and they do the discreteness and rationality statement using very elementary techniques while simultaneously connecting up to the theory of multiplier ideals and things in characteristic zero, right? There is an older one by Carl Schwede and myself where we sort of try and do things from a map perspective as well as an alternate by Schwede and Blickler about P inverse linear maps. There is a very recent one from a very algebraic perspective by Ma, Linchua Ma, who will be the next lecture series speaker in fact, right? And Thomas Polstra, who is a later lecture series speaker, right? And I expect there are some books being written as we speak by some distinguished authors which will have a lot of overlap with this material. So there's no lack of sources from this material from lots of different perspectives, right? So to go over and look at the theorems I've tried to highlight here. And with that, let me stop. All right, so there is a question in the chat that says, I mentioned reduction generated by less than our elements. So a reduction of an ideal is a subideal that has the same integral closure as that ideal, right? And so when I say it has a reduction by that, I mean that there is a reduction that has, at most, our generators, right? One of the super useful facts that I'm sure will come up in the later lecture series is that if I'm working in, say, local ring of dimension D, then every ideal has a reduction by at most, okay. Let's say the ring has algebraic flow as residue field, so it's relatively nice. Every ideal then has a reduction by most dimension number of elements where dimension is the dimension of the rings. You can get some very nice uniform bounds this way. Such uniformity, again, goes back to much of the applications of that closure from the beginning. So a later lecture series will also talk about so-called uniform art and reese, right? And that uniformity is really also wrote some very amazing papers in the 90s focused exactly on such uniformity, right? And some even more recent survey articles with Raikou, right? Where they really hammer this point home, right? But this uniformity is really at the heart of a lot of the techniques and theory here that's really cool. Yeah, I have a question real quick. Back to the, where you're discussing, like, test study all of the map to an algebra, right? So it's nice when the algebra is solid. I'm kind of personally interested in using the perfect closure there instead of all p-thruits, you know, p-to-d-thruits for every key. Is that typically known, is that known to be solid, or is that... Great question, right? So this is closely related to recent work of Dada and Moriyama. So in particular, so it is not true, even for an excellent local regular ring that our perf, the perfect closure is solid. Wow, right? So crazy. Their example is beautiful and involves sort of the so-called Tate algebras, which are the analogs of the polynomial ring and analytic geometry, right? So, but let me also give you a, so that's a negative answer to your question. Let me give you a positive answer too. Let's say that you're complete. In that case, you can define the whole theory of test that deals basically over a complete local ring using maps from the perfection instead. Okay. And in that case, the perfection will be solid. What's the paper to come in? Something about Frappini's non-splitting or something, right? If you look up my most recent paper with Rankaia with sort of the Dada Tucker paper, I'm sure it will have every paper I've written with Rankaia has the reference to the paper to come in as well. Awesome. Thank you. I guess the other thing I could also... Yeah, go ahead. Hello, can you hear me? Yes. So in this last theorem, a multiplier ideal versus test ideal, this p large enough, does it depend on p or... It does depend on... So the answer is sort of yes, right? As written there with enough discreteness statements, right? So essentially using the analog of discreteness and rationality, you can find one that will work for all t, right? But you need to use some sort of non-trivial statements in that process, okay? Okay, thank you. That's a great question. And this problem of not knowing how big of t or p you have to take to show anything, right? The non-constructivity of how big p is is at the heart of some cool recent work by a number of authors, particularly Linkramal and Karl Schwede and Trinckit Kagi and Kandisato as well. Okay. I guess the other... Just going back again to the other question about whether our perf is solid, the other thing I would be remiss to say is much of my presentation here is motivated by thinking about analogous statements. When you're trying to show things for a fixed characteristic p and by a reduction, you're naturally led to wondering what happens in mixed characteristic, which is where a lot of really cool and exciting recent work is going on, right? The questions about solidity only get worse in the mixed characteristic setting, right? So in particular, the right thing to say is whether our plus is solid and I don't think that's... I think it's expected that it's no for a zp adjoined two variables. So even in super nice rings, right? But the theory of test-to-deals has grown now to have a number of different versions in mixed characteristic. In particular, the R plus description I mentioned earlier, one of the reasons I stated is that it's related to recent work of Ma and Schwede, where they essentially use that as a definition of the plus test ideal in mixed characteristic, right? And these questions of solidity all show up there as well, right? So put another way, sort of the reason I sort of thought of that from the question is my advisor has told me there are really two kinds of people that work in characteristic p. There are the kinds of people who work in characteristic large p. And in theory, they really are people who are almost complex geometers. They really are working geometrically for most things. If something bad happens for a fixed p, you just switch to a bigger p, right? And then there's the second kind who work in mixed characteristic where, or in a fixed characteristic p, where you refuse to move, right? So p is fixed. And if something bad happens, you must just accept it and try and figure out how to work your way around it, right? And there's sort of a, definitely that tension in many of the theorems and much of the theory here, right? You're approaching it from different perspectives. Any other questions for Kevin? All right. Thank you. I'm going to thank Kevin one more time for excellent, excellent hints at finite letter. Sorry. I find it's excellent. Talk.