 Can everyone in the back who wants to hear me hear me? It's the real question. OK, well, first off, thank you very much to the organizers for the invitation to be here. It's a pleasure to give you these lectures. So in some sense, the overall goal of my talks are to give an introduction to some of the new techniques that are allowing for advances in mixed characteristic, even though this conference is titled Things in Positive Characteristic. But hopefully, you'll see, as part of the theme of the lectures, often what that means is you're forced to come up with new proofs of known results in positive characteristic that fit into a framework that allow one to generalize them to mix characteristic. So I understand that a lot of things in this talk or in my talks may be a little farther away from things that people in the audience have looked at before. So I'm going to try and, at least at the beginning here, give as down to earth of an introduction as I can. But please feel free to ask questions. All right, so but throughout all the talks, I'll fix a prime number P. So that'll be true for the whole time. So P won't change. All my rings are commutative with one. And often, I'll talk about local ring, RMK, and the maximal ideal and KR mod M, the residue field. And most of the time, I'm going to assume that P is inside of the maximal ideal. So all that really means is that the characteristic of K is P. But of course, the characteristic of R need not be. So there are really two cases. So the equal characteristic case, so here where the characteristic R is also P. And it's important, I think, throughout the lectures to keep as many examples as mine as you can. So I might as well take a power series ring over Fp. And in contrast, I might take something where the characteristic of the ring from the residue field if it was local. So here, I want you to keep something in mind. I need the Piotics a join, or a power series a join, a bunch of variables. So note that in the second case, if you're in mixed characteristic and you are a domain, then in fact, just like in this example, your ring has characteristic 0. All right, so start off with just a definition you may have seen before. But so R is perfect if, well, this only makes sense in characteristic P. And the requirement here is that the Frobenius map is an isomorphism. In particular, part of the point of writing this is that I don't mean just that it is surjective. So and in some sense, the goal of my lecture today is going to be to tell you some things about perfect rings and tell you some new interpretations of theorems involving them. And then hopefully introduce perfectoid rings in mixed characteristic and try to mimic the same techniques. So let's start off with a couple of facts. So if R is perfect, easy thing first. R is perfect means R is reduced. If some power in element gives me 0, I might as well assume it's a large power of P, which means it goes to 0 under some big iterative Frobenius, which means it goes to 0 because Frobenius is an isomorphism. And two, well, if R is perfect, then R is an aetherian. If and only if, R is a finite product of fields. So a finite product of perfect fields. OK? So let me say a little bit about this one down here, just because it's going to come up with something that is a useful observation. So in order to show that if R is perfect in aetherian, it's a finite product of fields, it suffices to go modulo each of the finitely many minimal primes. So say Q is one of the minimal primes, and S is our mod Q. This ring here will again be a perfect ring. Certainly Frobenius is still surjective since it was surjective on R, but now S is reduced still. So it's Frobenius is still injective as well, right? So now my ring is domain. So it suffices to prove the theorem to check that this ring S is in fact a perfect field, right? So suppose I have some element, S inside of S. And what I want you to consider is the radical of the principal ideal generated by S, right? And the claim is that ideal, which I might write like this, S to the 1 over P to the infinity is the ideal generated by all of the p-th roots of this element S. So this S to the 1 over P to the infinity is not an element. This is just my shorthand notation for writing such an ideal down here, all right? And of course, again, my ring is perfect. So there's no real ambiguity about anything I wrote down here. Every element in R has a unique, or in S has a unique P or P to the E-th root for any E, all right? So but we also know our ring here is by assumption an Ethereum. So this ideal is equal to S to the 1 over P to the N for some large N, right? So these things all divide one another. So I might as well stop at some fixed value, all right? And as soon as I get to that, I have S to the 1 over P to the N plus 1 is going to be a multiple the previous guy, S to the 1 over P to the N, all right? But now I can factor this guy out and divide through. So let me write that out. So S to the 1 over P to the N plus 1, all right? So solve or moving everything to the left side and factoring out S to the 1 over P to the N plus 1. The point here is that 1 over P to the N is bigger than whatever P to the N plus 1, all right? So I have an equation that looks like this. My ring now, S is a domain. So that means there are two cases. Either, well, S to the 1 over P to the N plus 1 is 0, which is the same thing as saying S is 0. Or over here, this is 0, at which point, oh my goodness. I didn't do that. OK, so or this is 0, which is the same thing as saying it, well, some fractional power of S is a unit. So in particular, S is also a unit too, all right? So this shows that S is a field, all right? Is everyone happy with this? So maybe the next thing to say, I want to tell you two ways, one of which you've probably seen and one of which you may not have thought about before, to cook up perfect ring, all right? So starting at least from a ring of positive characteristic. So say you have a ring R of positive characteristic. So the first way I can cook up a perfect ring is the following. So I'll call this ring R-perf. So this is the, by definition, the co-limit over all of the different powers of Frobenius, OK? So in particular, so to show you that you've probably seen this before, so I claim this is the union. Well, take R reduced, and now the Frobenius map is injective. So I can view the Frobenius map as taking p to the east roots, all right? And the union of all of those rings gives me a perfect ring, namely our red perf, all right? So in particular, maybe the best case of this is when R is a domain, in which case I can view all of this as living inside, say, an absolute algebraic closure of the fraction field of R, all right? All right, so and this is the one you'll find in many of the other lectures, right? And it'll show up at least some in these as well. But in some sense, the new character I want you to focus on is, well, the natural thing to do here is instead of taking a limit over Frobenius this way, you might try and do it in the other direction, right? So this next ring, I'm going to call R-flat, and this, I might call this Fros, this is what's known as tilting later, at least in the mischaracteristic setting, all right? So here, by definition, this is now, instead of the co-limit, it's the inverse limit, again, over all the different powers of Frobenius, all right? So what does it mean to have an element of this thing? An element of this is some sequence, right? So I have elements of R with a property that Rn plus 1 to the p is equal to Rn, OK? All right, so just like when I limit over Frobenius this way, if I limit it in this direction, what I end up with is always a perfect ring. You can check that, OK? So just to have some example down here, let's say R is reduced. In that case, this is going to be the subring of all the elements of R that have arbitrary p-power roots, OK? So this is my attempt to have at least one example down here, but warning me that this is also not going to be particularly reflective of the cases where our flat is used, all right? So most of the time here, I'm going to use it on a ring R where Frobenius is surjective, but not injective, and so somehow very different things happen in that setting, OK? All right, great. So more generally, for any ring, I'm going to set a flat to be a mod p flat, all right? So given any ring, go mod p, all right? This gives me a ring in characteristic p, and now I'm going to flat that guy, all right? And again, so just to have it explicitly on the board, this is the inverse limit order for Frobenius on a mod p, all right? So I need to tell you a couple of more things about this tilting map, because this tilting procedure really is something which is at the heart of the theory, all right? So let's say that A is p-complete and separated, all right? So I'll add this whenever I say p-complete, even though it's maybe implicit just to make sure that you know that I mean it, all right? So the claim is in this case, all right? So well, what's one of the reasons to use the flat notation? So I have some kind of musical notation that goes on here, all right? In this case, the claim is that there exists a sharp map, A flat back to A, all right? And what is this, all right? So again, A flat here, I can think about as the set of sequences, all right? After I lift them is the set of sequences A and an A, so that A and plus one to the p is congruent to A and mod p, okay? At which point, what I want you to do is map this, well, all right? So I want you to map this just to A and the p to the n, all right? Or at least take the limit after you do that. So the claim is that if I have a sort of compatible system of roots in A flat, right? That the sequence A n to the p to the n, as I let n go to infinity, is a Cauchy sequence in A, right? And since this is p complete, this thing converges to something inside of A, all right? That's uniquely determined, all right? So I want you to check that this is well-defined. Say it again, p complete. So with respect to the pietic topology here, right? So, all right, so check this map is well-defined. I'm not claiming that it is a map of rings, right? But at least it's multiplicative, all right? If I have products over here, they go to the corresponding product inside of A. In particular, one can show that this implies that you get an isomorphism, at least of multiplicative monoids. I'm not saying anything about addition, all right? Between the following, well, in the definition of the flat map, all right? So I really was looking at the inverse limit over the Frobenii on Amod P, but I could look just at the multiplicative maps, A to A, given by sending elements to their pieth powers, not worry about the addition whatsoever. And that has a natural map to the A flat over here. So maybe more explicitly, an element to the left-hand side here is an honest sequence, An plus one to the p equals An and A, right? Without saying anything about addition, okay? And I can take such a sequence and just go mod p, right? To give me a similarly compatible sequence over here in A flat. And to see that this is an isomorphism, the sharp map tells me exactly how to write down an inverse, right? So simply take the sequence An mod p, right? And map that up to the sharp sequence over here. All right, so, and again, right? So the point here is to check that this is in fact, in fact inverse, okay? All right, so maybe we just want more remark. So similarly, if r is perfect, and I have some element d zero and r, right? Then you can check if r is d zero complete and separated, then I could take r mod d, right? So since r was perfect, r mod d at least still has surjective Frobenius, but probably not injective Frobenius. I can take this flat construction to make it a perfect ring. And the same arguments are gonna give you an isomorphism, r mod d flat with r flat, which is the same thing as r that started off with. All right, so maybe that's another thing that I didn't say earlier, right? In both of my two constructions here, so r lower perf or r per flat, if you start off with and take a perfect ring, then in both cases, what you end up with is exactly the same ring you started off with, right? So in some sense, right? So just follow the arc here. I've got r goes to r perf, gives me one way of producing perfect rings, r goes to r flat, gives me another way of producing perfect rings. And here I have one example of what r flat looks like, right? So but maybe not the most representative for these talks. And over here, I have one that was gonna show up quite a bit later, right? Namely, take a perfect ring, then r mod something is got surjective for binius, right? And this tells me that when I flat it, I end up getting rid of the quotient completely, okay? Yeah, right? So it has to have a compatible system of B power roots, okay? Okay, so like I said in these talks, in some sense, the theme is supposed to be when you work in mixed characteristic, one often needs to revisit things in positive characteristic. So I'm gonna come up with the new proof that generalizes to mixed characteristics. So with that in mind, I wanna tell you today about the following theorem, Bat-Iyengar and Ma, right? Which says the following, let's say r is an aetherian of characteristic p, right? Then r is regular if and only if there exists a faithfully flat map from r to a, where a is perfect, right? So, and if you were looking in the previous lecture series, the theorem Ilya had up on the board over here was Kunz's theorem, right? And so the nice thing about doing this is that what you're supposed to see is that this is a new way of revisiting Kunz's theorem that regularity is characterized by the flatness of for binius, okay? So let me sketch a proof, right? So let's do the easy direction first, right? So if r is regular, Kunz's theorem says that that's the same as asking that the for binius map here is flat and the co-limit of flat modules is flat, right? So, and so r-perf will be faithfully flat over a, okay? Is everyone happy with the easy direction here? So let's look at the other direction, which is the real meat of their proof, right? So, right? So, so any Noetherian ring of characteristic p, regular means regular at all of the local rings, so it's a product of regular domains, right? So, but feel free to focus on the local case throughout. You do lose nothing and all the proofs are gonna reduce to that anyways. In fact, for this direction to show, let's say that you have a faithfully flat map to some perfect ring, right? Without loss of generality, I can localize. So I might as well assume that r is local. I'm trying to show then that r localize that m is regular, right? So let's fix this system of parameters x1 up to xd, right? And as we saw before, right? So we know that the radical of the ideal generated by xi and a is the same thing as the ideal of all of the p, p th roots, right? And the observation here is that, well, I can build this guy up and view it as a co-limit over a bunch of multiplications by powers of xi, all right? So, and the point here then is that as soon as I can write that ideal as a co-limit, right? All of these ai's of course are free and so flat over a. So this thing will be a flat a module. So I leave it to you to check that you can exhibit it this way, but maybe I'll draw a little bit more to convince you that it's the case. So here, maybe the more obvious co-limit that would build up and give me this ideal of all the powers will be to take the ideal generated by xi in the first entry, xi to the one over p and just build it up power by power, right? And these exponents are exactly what one needs to do, right? So this is multiplication by xi, this one then is multiplication by xi one over p, right? These multiplications are exactly what you need to write down in order to make the corresponding squares all commute, right? And so the real core of this thing is that this, certainly you have a surjection from this co-limit onto the corresponding ideal, right? And the proof then is to say that that map is injective, right? Assuming just that r is reduced. So a similar statement will mean that all ideals of this form in any reduced ring are flat, right? Okay, and I should say, and based off of what these lectures and this conference is about, this is not a new observation, right? The fact that such ideals in a reduced ring are flat is due to originally Abrobach and Hochster. All right, so how does that help us? Well, in particular, look at the following complex, right? So stick a in degrees, degree zero, stick the ideal in degree one, right? Both of these things are flat. So each of these is a flat resolution of A modulo, the radical of XIA, right? So when I tensor those all together, what I get is a flat resolution was N or N is D, right? So this gives a flat resolution of A mod the radical of MA and again here, right? So this is a similar check. Namely, I wanna argue that this statement about the radical of finally generated ideals inside of perfect rings being the same as just looking at the ideal generated by the P to the ETH roots for all E. But since each of these has length one, this guy has length D. The flat dimension over A of A mod MA is at most D, okay? So what do I get from this? Well, that tells me that if I look at a torque of anything over A into A mod MA, this is zero for I bigger than D. So in particular, what I'm gonna stick in here is K tensor over R with A using that A is flat over R. This is the same as the torque over R from K into this guy. So I can put an extra tensor A in here using that that's flat, all right? So maybe I'll say it another way, right? What I can do is I can always take a projective resolution of K over R and tensor over tensor up to A. Since that's flat, that gives me a projective resolution of this guy over A and then I can use that to compute the torus. All right, great. So I have this torque vanishing statement, right? And the point here is on the other hand, this guy here is a K vector space. I don't know exactly how big it is, right? But it is some R module, which is killed by the maximum ideal, right? There's at least one copy there, right? From faithful flatness. So this tor breaks up as a bunch of direct sum copies of tor from K to itself, right? Over R. So the end result is that for I bigger than D, tor K into K, right? Is zero, which immediately gives that R is regular. Is everyone happy with the sketch of the proof here on the board? And so I'm gonna revisit this proof later. If I don't make it to it in this lecture, then I will at the beginning of the next, right? But I want you to take a look at it. And the key point here is once I have the faithful flatness to make the rest of the argument go through, all I really need is that the flat dimension of A mother radical of MA is bounded, is finite, right? So then I'll get some bound on the global projective dimension. Yeah, so let me tell you where we're headed. Sort of the first theorem then is I wanna take this and do the same thing in mixed characteristics. We've just done it in positive characteristic. And I'm gonna write the theorem without telling you exactly what it means. And then in some sense, that's our next goal, right? So same theorem, right? But let's say R is an aetherian and let's say P is inside of the Jacobson radical of the ring, all right? So in particular, as an example, take where R is local with P inside of the maximal ideal, all right? Then the statement is that R is regular if and only if there exists a faithful flat map to perfectoid ring. And so the first sort of goal in the lectures is to tell you, all right, what a perfectoid ring is, okay? And the argument is that they are the analog and mixed characteristic of what perfect rings are, all right? And so that is the foot in the door that allows us to make use of Frobenius techniques and things like that in mixed characteristic, all right? Is everyone happy with the statement of the revised statement of the theorem? I say revised because it's not clear that it is the same theorem yet until I make the following remarks, all right? So I'm gonna tell you what a perfectoid ring is in general, right? But in some sense, I don't want you to worry too much about the definition if you haven't seen it before. Instead, I'm gonna tell you the two easiest cases you should keep in mind that will give you a flavor for everything that's going on. All right, so what are the easiest perfectoids, all right? So, well, the first one we've already seen, if you're an equal characteristic P, then A is perfectoid if and only if it's perfect, right? IE, another interpretation of this is once you know this, this statement in equal characteristic P is exactly the previous statement, okay? All right, so that's not the interesting case, but maybe it was worth saying that maybe before this lecture, you might have had the impression that perfect rings were pretty easy to deal with and positive characteristic because you only work with R infinity maybe before and nothing else. And all right, so at least I'm wary enough at this point to know that the flat or the tilting construction leads to very complicated interesting examples of perfect rings and even in positive characteristic. All right, so second one, let's say that A is P-torsion free, all right? Then A is perfectoid if and only if, all right, A is periodically completely separated, right? And P is, has the following form. P can be written as a unit times pi to the P, right? IE, up to a unit, P has a P-th root inside of A, okay? And one more, right, so last thing, look at the Frobenius map on A mod P, right? What I wanna require is that that is surjective with kernel generated by pi, all right? Or is that in other words, the Frobenius map induces an isomorphism between A mod pi and A mod P, right? So P up to a unit has a P-th root and that generates the kernel of Frobenius on A mod P, right? Is everyone happy with this? All right, so and maybe I should also say that throughout all the lectures, certainly the definitions I'm gonna give are not the first definitions of perfectoids that showed up in nature, right? So I'll try and give some attributions, right? And so this one appeared first in the work of Batmora and Schultzah, right? So as I said, the idea of these boards were supposed to tell you some easy examples of perfectoids and maybe everything here with some theoretical mumbo jumbo that's while very easy to check and practice in many cases, not explicit, let's do one, right? So or at least do two. Take the Piatix that you started off with. So let's not think about this too much, but here's what I want you to do. Take the Piatix and enjoy all P-th power roots of P itself, okay? So each one of those extensions is in fact a free extension of degree P, right? So this is now an infinite extension here of ZP, okay? As such, while adjoining the roots at any finite step preserves the fact that what I have is P-complete after I go infinitely many times, now an infinite rank free module is no longer complete, so take the P-completion, okay? And so this is in some sense maybe the easiest perfectoid ring you can write down. You can do the same thing with some variables as well. Take ZP, right? So this is one that's gonna show up of importance later on. Take a nice beautiful regular ring and mix characteristic, right? And I wanna map this guy to a perfectoid in an obvious way. And what I'm gonna do is adjoin P-th roots of both P and all the variables again, all right? And then again, I have to worry that because I have an infinite rank extension, it's not P-complete, so the obvious thing to do is P-complete. And in both cases, right? All of these rings are Piatically complete and separated. They're P-tortion free, right? And it's not just that P has a root up to unit, but P has a P-th power root because we adjoin it in both cases, right? So it's very explicit, right? Of course, you should also check that when I go mod P, this thing is an isomorphism, right? So I leave it to you to go home and do the exercises of checking that what I get, right? Actually gives an isomorphism in these cases, okay? Okay, so let me end my talks with giving you what's to do tonight, right? So for exercises. So the first thing is to say that a lot of my, while I haven't followed it directly, one of my favorite sort of Clifnotes versions of a lot of the intro theory, right? Is in fact in this paper, the Botte-Ingermar paper, all right? So if you like, go ahead and take a look. There is a section on reviewing basic facts about perfectoid rings, right? And it's a good place to look at if you're trying to get some of this basic material down, right? And the second thing is, so I will give you the full definition of a perfectoid next time as we go through and give a sketch of the proof of the theorem, right? But in order to do so, I need to use the wit vectors, right? Namely the pedipical wit vectors, right? And I'm old school or just old, depending on your perspective. So my favorite source or source or where I learned the basic facts in fact is in Sarah's Local Fields book, right? So if you like, go look at the section on wit rings or wit vectors inside of Sarah's Local Fields and I'll start recalling some of that construction next time.