 Can everyone hear me? OK, great. So welcome back for the second establishment of this lecture series. So let me recall briefly where we were before we started and tell you where we're going. So last time, we sketched the characteristic P version of this sort of motivational result of bot yanger and maw, where we showed that if you look at a netherian ring in positive characteristic, an analog of Koons' theorem is that it's regular if and only if it admits a faithfully flat map to a perfect ring. And so the goal now is to give a generalization of that or a twist on that in mixed characteristic, if you will, to show that if R is netherian and P is in the j, it's in radical. So either P is 0 or maybe you're in mixed characteristic, then R is regular again if and only if there exists a faithfully flat map to what's known as a perfectoid ring. Now, I haven't told you what a perfectoid ring is. And so in some sense, the point of this talk is to do that. But I want you to keep in mind, like I ended the last talk with, the easiest cases. So if your characteristic P, a perfectoid ring is the same thing, it's just a perfect ring. And then the interesting ones for us in some sense will be the sort of orthogonal direction. Let's say A is P torsion free, then the easiest case to say here is that A is perfectoid if and only if it is P complete and separated. And P has a P-th root up to a unit inside of the ring. And when I look at the Frobenius map, after getting rid of that P-th root of P, what I get is an isomorphism. So the Frobenius map on A mod P is surjective and the kernel is generated by pi. Is everyone happy with where we were? And so here's just a couple examples to keep in mind in both of these fall into this last framework down here. Just to give you a flavor with what these rings look like. So here you could look at, take the P-addicts, adjoin infinite P-th roots of P, and then P complete. Or for instance, you could take a nice mixed characteristic regular complete local ring. So the P-addicts series adjoin say x to up to x d. So d is the dimension of this ring because P is also a parameter in here. Now adjoin the infinite P-th roots of P and all of the x i's. And again, P complete what you've got. And both of those will be P-torsion free rings. So you can check what happens when you go mod P and compute the kernel Frobenius to check the perfectoid using this easy criterion down below. So but to give you the full generalization or the full statement here, a perfect means that the Frobenius map is an isomorphism. Isomorphism. So the traditional thing is to say that a ring with a surjective Frobenius map is semi-perfect. Any other questions? So let's get to the definition then if we can. And what I need are the Witt vectors. So to start here, let's say that R is any ring. So the Witt vectors, so the construction is maybe not so enlightening when you first see it. But I still think of it as the best way to start off. So is this out of all countable tuples of elements of the ring? Let's do. So if you will, just a countable product of things in there. I didn't write product for a reason, namely the addition and multiplication on these tuples is not going to be the usual addition and multiplication. So the twist is that addition and multiplication are determined by universal polynomials in the entries of the sequences so that the following maps are ring maps. I'll write these as capitals. So start off. For each fixed i, look at x0 to the p to the i, then drop the exponents by 1, put x1 to the p to the i minus 1, and add a p out front and make a pattern like that and write those out. And what you want is that such expressions give you a ring map. So these are called the ghost maps, so for all i. So knowing this, so if you believe that such a thing works, you can work out what the universal expressions have to be. So let's do just a couple of them. And I've written the x's here and tried to write capital x's because I want to think about this as just these xi's variables. And I want to force this to be a universal expression that forces addition and multiplication to be satisfied. So for instance, if I write down a tuple of the x's and a big sequence of the y's, I know I have to get some other tuple, say, with the s's. Well, if I start off here with i equals 0, that map is just the map which sends x0 to x0. And I want that to be a ring map. So the easiest thing to say is that I better have the map applied to each of these x or plus y0 equals the map applied to this guy. And similarly, I have to do the same thing, say, for i equals 1. So with i equals 1, I'll get an expression that looks like this. And the exercise is to go through and recursively solve these formulas to get out what the formulas for the s's have to be that force this to be an additive map. So for example here, so we already know that s0 is x0 plus y0. So I can plug that in over here. And now I have some formula that only involves s1 and polynomial in the entries of the x's and the y's. Now, if this is going to work, for any ring, in particular, it had better work in the case when my ring is, say, p torsion free. So when you solve for this and throw everything onto the other side, you'll find that you have some polynomial which has a p in front of all the coefficients. So let's just do that. I get p s1 here is p times this guy over here. And the whole point is, if it's going to work for a p torsion free ring, then I notice this expression has p's in all the entries after I do the expansion. I do that formally. And I just divide by the p's to get a formula that works for s1. Is everyone OK, or have an idea of how I got this formula over here? Similarly, I could do the same thing with multiplication and recursively solve, again, the first level. I know projection onto x0, projection into y0 has to equal projection over here, has to equal t0. And on the other hand, I have some product expression, which again, when I put in x0, y0 over here, that gets rid of the t0 and I can solve for t1. Just spit out an expression, which will tell me how to write down a polynomial for t. So again, the point of this is just to say that the traditional approach to exhibiting the vectors is just to say that, well, I have these vectors and I'm forcing each of the maps that look like this to be additive multiplicative. So after checking that this gives me a ring, I also have the first map that I forced, gives me a ring homomorphism from WR back to R, and just I see directly that this thing is surjective, because I can put any sequence on the left that I like, and these two things together are pretty easily seen to be functorial in R. So everything just involves these universal polynomials with coefficients in Z. And I would say the same thing, well, maybe I won't write it. I could also talk about the truncated vectors, where I only pick out certain numbers, so for instance, W2, where I only use the first two entries inside of this thing. That will give me a similar functor, W2, which is also functorial in the ring that I write down. So in particular, using the functoriality, if you know that the characteristic of R is p, well, so I have a ring map from the R given by R goes to R to the p, that lifts to a ring map upstairs, which on the entries just gives me the p powers. They just apply the ring map down below that commutes with the usual forbenius downstairs. In particular, note in the same vein that if forbenius is an isomorphism, IER is perfect, then the lift of forbenius up here also gives me an isomorphism for the same reason. So the inverse map will also give me the inverse. So with this, so the real magic happens is that when I input a perfect ring, what I get is really quite nice. WR is the unique p-complete and separated p-torsion free ring with a surjection down to R so that the kernel is generated by p. So it really has a strong uniqueness property here. So if you like, you can prove the uniqueness without going through the construction, just assuming that such a thing exists, in fact. And that will be implicit in much of my presentation later. But if you believe such a thing exists, you could also imagine trying to come up with formulas that define it. And of course, let me tell you a couple of other properties about this. So before we move on, the weight vectors come with a shift map, generally referred to as V, for Vershibang, German word for shift. So which just sends a tuple, sticks a 0 in front, and passes everything off to the right. Now, this is just an additive map, not a ring map, not a multiplicative map. But it has the nice property that when I do this shift, what I get gives me, in either order, f of v or v of f is multiplication by p on the weight vectors. So again, if you believe here that the kernel here was generated by p, obviously, the image of v over here is something so that if I project onto the first coordinate, I get 0. And this is somehow saying that that is related to how the formulas behave with respect to Frobenius. So the other thing I sort of need is the Teichmeler map. And I'm going to give a, we've done a little work already related to this in the first lecture, although you may have missed it. So you can define the Teichmeler map in the following way. So I'll denote it with just a bracket. Well, since R is perfect, it's isomorphic to R flat from the first lecture. OK? Now, our definition of R flat for an arbitrary ring was go mod p and then take the inverse limit over Frobenius. So this is just definitional. This is because R is perfect. All right? And we constructed the flat in the first part of the lectures. This musical notation came together with a sharp map. Whenever I had a ring that was, say, p complete, all right, got rid of the flat. OK? So the Teichmeler map here is exactly this string, right? So where I have a map from R back up to omega R. OK? And again, the key point of this thing was that this was multiplicative. OK? But you can explicitly write this down in coordinates as well. The image of little R gives me the tuple, which is just R in the first entry. OK? And in fact, you can show more generally, right? So that the image of sharp is always the set of, as we talked about last time, the set of elements that have compatible p power roots. So this precisely gives me the elements of WR that have compatible systems of p power roots. OK? Is everyone happy with this expression? OK? So let's just note a couple of things. So first off, well, if I stick R in the first entry and then I get rid of all the other entries, I get back R. I started off West. So in particular, pi is injective. As we said, the map here is multiplicative. More generally, you can show that it's easy to multiply times any sequence in the following way. So what I get gives me, right? So point being that while multiplication in general is complicated inside of the wet ring, if I give you a representative in terms of the dichmular map, it's easy to multiply times the tuple. OK? And in general, you can use it to show many other things. In particular, I give you an arbitrary tuple, say R0, R1, and so on and so forth, right? Well, you can realize this as an infinite sum where, in fact, you break this up into putting each one of the R's in the components and then shift, well, putting it in degree 0 and then shifting up the degree R. So this tuple can be realized as stick Rn in degree 0, then shift it up to degree n and sum all those up. And of course, I also know, since my ring is perfect, that I can always take roots inside of the dichmular representatives, since my ring R was perfect. So the Fn, or my ring with this dichmular map is multiplicative. It's another way of saying the same thing. And now, v to the n and f to the n, I know composed to give me a multiplication by p. So starting off with the tuple, R0, R1, yada, yada, yada, I can represent it over here, in fact, as a power series in p by looking at, well, I take the p to the nth root of the nth entry over there and then take the corresponding dichmular representative for that guy. So this gives you an alternate way to go back and forth or to take a tuple and represent it as a power series in terms of p, using that the ring is p complete. Is everyone happy with this formula here? But maybe in some sense, the right thing to say is this gives a unique way to represent things over here. So this is a separate set of coordinates in some sense on the width vectors. So these are called the dichmular coordinates. And in many ways, their manipulation algebraically is much easier to play with than the original addition and multiplication formulas. The other nice thing about the dichmular representatives is that it tells you a little bit about why you have uniqueness over here. So again, my representation of the dichmular map only involved the map sharp. And the map sharp only required from the first lecture some statement about the thing being p complete. So you can define an analog of the dichmular map for any ring satisfying these properties over here, leading to similarly unique expressions which tell you exactly how to do the isomorphism with WR. Okay, okay, great. Almost there, right? So one note in one definition. So note an element of WR written in these coordinates here is a unit if and only if the first entry here is a unit. So, well, again, easy to see. So in terms of sort of the functoriality of the representations. Okay, so with this, I make the following definition. So an element of the Witt ring here is distinguished if two things. So first R is complete with respect to the first entry D0 and the second entry D1 is a unit. Here's the representation in terms of the original coordinates on the Witt ring, right? Equivalently, right, so I could also write, right? So I can pick off the first entry as the dichmular representative of D0 in the expansion and everything left over, right, has a P in it, okay? And you can check that requiring D1 as a unit is the same thing as just requiring that this guy over here after I pick off all the other components of the unit. Okay, so here I've told you what it means to be distinguished in either of the two ways of writing this down. All right, so there's the first part of my definition is what a distinguished element is and the second part is perfectoid if it has the form WR mod D for some perfect ring R and the thing I kill over here is the distinguished element. So something of this form, right? Is everyone happy with this? So let me just make a couple of remarks. So this is the full definition. If you have a perfectoid ring over here, right, then in fact the ring R is uniquely determined, right? So let's just think about what happens when I look at A mod P, right? So if I kill P, what I'm left with is, well, WR, I've killed P D0, right? So of course I could have killed P first so this is mod D0, right? And if you go back to our construction of the flat map, all right? So what does this mean? Well, A flat then is R mod D0 flat, right? And in the first lecture, we argued again using some construction using the sharp map that whenever I take something which is perfect and I kill something principal, what I get back is the perfection of the first guy with respect to this guy. So this here gives me back R. So this uses that R is D0 complete, all right? So again, this was embedded in the first lecture. All right, so if I have such an expression, right, without seeing what the R is, I can tell you what the A, or without seeing what the R is, given A, I can tell you what R had to be, all right? And more than that, the presentation, right? So now we know that R had to be A flat, so what? This uses that my perfect ring R was D0 complete, right? I guess there's no brackets over there, I'm not. After I go mod P, there's no more brackets, right? So the presentation here also in the same way doesn't actually involve any choices, right? So I can tell you what that presentation has to be. It's determined by what's known as Fontaine's theta map, right? So what does that look like? Community square that looks something like this, right? So as we've said, everything in the wit ring has the form, has some type-mular expansion, right? So write it out as a power series in P, right? And what I'm gonna do is map that over to, right? So take the P to the nth roots of those representatives, again the ring's perfect, so I can take those at will, right? Times P to the n, right? And this gives me a map fitting inside of this square, which, all right, so here's my theta, which determines what this presentation is, right? So there's, the point of here of remarks one and two is to say that, well, the definition of perfectoid is that there is some ring R, right? So that it's WR mod a distinguished element. What the R is is completely determined by A and what the map, giving you that, is completely determined by the setup as well, and it's what's known as this Fontaine's theta map, okay? Let me think about that for a second. So I, this guy is, right? So, and I think this one is two, right, at the bottom, I think so, okay? Right, so I don't know that that's true for the other, well, yeah, right, so here is one definition of perfectoid, let me give you another equivalent definition, right? Which is maybe at least useful to note, in part because this is not how you'll see or recognize these things in the wild, right? So A is perfectoid, if four things, one, P is P complete and separated, the Frobenius map on a mod P is surjective, three, the kernel of Fontaine's theta map is principal, there exists a pi and a unit in A, so again, P is pi to the P times U, okay? So very similar to this other characterization in the easiest cases of what a perfectoid ring is, but slightly generalized, okay? All right, so, and I should say again that going through the equivalences of all these definitions is not so easy in the notion of what is perfectoid and whether it's relative or whatever has changed over the years, so maybe the equivalence of this definition with the others is due to Batmohr and Schulze. Okay, so, great. So again, the point of the theorem is, or my argument is supposed to say that you're supposed to replace in your mind the word perfect in characteristic P with perfectoid in either characteristic P or mixed characteristic zero P, all right? So you'd expect that perfectoid rings satisfy some of the analogous properties, all right? So here, maybe the first one, which I won't prove is not quite so straightforward to show as in the characteristic P case is that at least if you're perfectoid, then again you're reduced, just like you were for a perfect ring in characteristic P. Two, while it's not necessarily completely clear from our definition, okay? Maybe you'll see this in a second. If you're in characteristic P, then we would like to get again that perfecting perfectoid or the same thing, all right? So how do I see that down here? Well, I could take P as my distinguished element down here and I know that if I have a perfect ring, then W of R mod P gives me back R, the thing I started before and P is distinguished. This is the characterization I said before, right? So without writing it out for a third time here, so we have the characterization for P torsion free rings as well. So if you're P torsion free, then your perfectoid, if and only if you're complete and separated and you have a P-th root of P up to a unit, which gives you surjective or isomorphic Frobenius and maybe the fourth thing, and we'll end with this one. If you have a perfectoid ring, there exists a short exact sequence or a push out diagram, however you wanna phrase this thing, all right? So here this A bar is A modulo, it's P power torsion, all right? And one can show that, all right, here I've killed P or I've taken, killed a radical of P in particular, this is characteristic P, right? And the claim is that all of the rings, oh, I forgot the A. So A sits inside of two other rings mapping to a third, okay? These both of these rings are all three, these rings are perfectoid. Two of them are perfect in characteristic P and the third one is P torsion free perfectoid, all right? So in other words, A is the gluing of these two rings along the third, all right? So but in principle, this means that any perfectoid ring comes together by matching up cases two and three, all right? So in fact, to prove that the thing's reduced, you do it for torsion free and perfect perfectoids, all right? And then deduce it from such a sequence. To end maybe today's lecture, I wanna finish with my sketch of the proof of the Bat-Yinger Moth Theorem. Where here? So it's that guy, tensor that guy over this guy. So A bar radical P A times A bar, all right? Okay, so let's give some sketch here, all right? So without loss of generality, you can assume if I'm trying to show everything that the ring is regular local and maybe for simplicity, although this is pretty easy reduction, I'll also assume that the residue field is algebraically closed, all right? I'm trying to show that you're regular if and only if you admit a faithfully flat map to a perfectoid ring. So first start off by mapping R to R hat, all right? And there are of course three cases. Either it's a power series ring over K where K is characteristic P. So if this thing is regular local here or it's a power series ring over a whip ring, all right? So this is the unremifed case. It has the form, well, I have to add an extra variable and kill a relation, F is in the square of the remaining variables and let's say P does not divide F. Okay, so the ramified case. And in each of the three cases, the first direction of the theorem, right? So this is R regular implies you have a faithfully flat map for a perfectoid. You just exhibit it explicitly. So in particular, all right? So here we're in characteristic P, I can just take the perfection. Here you use my second example I used in the easiest cases from before, all right? So just to join all the P th roots of P and all the variables and then P complete, all right? So the only real interesting one to do is this one, all right, so the third case, you do something similar. I have all these variables, X one up to X D again, join all the infinite P th roots of all the variables, but not the P, right? Complete with respect to P and then kill the same relation, right? And of course you have to check that such a thing is perfectoid, right? So, but it's pretty easy to see in this case that this thing is P torsion free, so you have the easy check to see that this thing is perfectoid if you just play around with the elements, all right? So I'll leave that as an exercise. The other direction, right? So, we're gonna mimic the earlier argument we did when R is perfect and you'll find that you need to use the following sort of key lemma if A is perfectoid, right? And I have some ideal, which is the radical of the ideal generalized P, F two all the way up to F D, right? Again, when A was perfect, remember this was the ideal of all of the P power roots of all of those elements, right? Then the lemma says that the flat dimension is bounded and by the number of elements you had to take. So literally the same argument we gave goes through all you need to know is this finite flat dimension part to conclude that the ring A was regular, right? And the proof, I'll leave for you to figure out, right? But again, the point here is that, well, the radical of P, P has a piece root and in fact has a compatible system of, I can assume it has a compatible system of P power roots, right? So already has this form. So as we saw before, this is flat as an A module, right? So if I look at A bar, probably equal to one, right? But I know this thing has a two step resolution where both of the things are flat. So the flat dimension here is at most one. On the other hand, A bar is perfect so that implies that A mod I, which is, well, A bar mod I bar, right? So when I kill this radical ideal down here, I end up with A bar modulo. Again, something that looks like this, right? So once I know that A bar is perfect, right? So now I have a better expression for this thing and what we saw before gave us the flat dimension over A bar of A mod I or A bar mod I bar, however you wanna think about it was at most D minus one. So now if you use the standard, if you will, spectral sequence argument for how to compute or for torrent base change, right? So that'll give you, combining these two things, the flat dimension of A mod I over A, right? Is bounded above by D, okay? Great. So that's my sketch of the proof of this theorem, right? And we'll start there next time and move on to finally getting to big Kamakali things, the title of lecture series in the third talk.