 Let me just pick up with a couple examples on what we were, so we finished last time proving the, finishing with the sketch of the proof of the body anger on Ma, about characterizing regularity in terms of 50 flat maps to perfectoid rings. And now we have gone through all the definitions and really started to build up the theory of what perfectoid ring is, so just to have some examples fresh in your mind. Most of these we've, at least the first couple we've talked about already before again, right? So we have these easy examples, right? The pedics to join the p-th roots of p, hat p, I can do the same thing where I join with some variables as well. And my exercise for you is now that you have all the definitions to go through, compute the flats of all of these things and actually try to check that they satisfy their various equivalences, right? This one will be useful for us later on in a second, so or came up in our proof, right? So here take a regular ring, a complete regular local ring and mixed characteristics, say where the coefficient field is the width vectors of our perfect field, right? I join all the roots of p and all the variables again and then hat p, right? So that's something very similar to the example up above here, right? And so let me give you two other examples, right? To think about to sort of jumpstart what we're going to do today. So the first one, right? So take any complete local domain of either characteristic p or mixed characteristics, so p inside the maximal ideal, right? And let's assume that the residue field is perfect, k equals k to the p. Let's show it up and how the earlier talks as well, but let r plus be the absolute integral closure of r. So the integral closure of r inside of a fraction, an algebraic closure of its fraction field, okay? And one can check that r plus hat p, right, is perfect. So of course, if r is characteristic p here, then I don't have to do this hat p at all, and I just get r plus, but if you're in mixed characteristics, you have to add this down, okay? And the exercise for you to check is, you know, in the definition here, you want to check that the kernel of Frobenius is generated by the pth root of p, right, in mixed characteristic, okay? So, and then sort of just as a last example, right? So let's say that a is perfectoid and I happen to have a whole bunch of elements that admit a system of p power roots for all powers, okay? Then what I want you to do is take a and kill, we've seen these ideals earlier in the perfect case, but I want you to think about this also in mixed characteristics, kill that ideal, right? When you do so, what you'll get may not be pietically separated, so you will still have to do a p completion to satisfy that, right? But then after you do what you get is perfectoid again, okay? All right, so more generally, all right, so as you've seen sort of already in the theorems, part of the game that you want to play here is to find as many ways to construct these perfectoid rings as you can, all right? And when you can do so in some nice way, often you get theorems coming out of those. So with that, part of what I want to do today is essentially sketch for you a bunch of packages of how to produce perfectoid rings with certain properties, all right, that sort of show how they're used in the study of singularities, right? So to that effect, let me start off by quoting a result for you, right? So of Barton Schultz, let's say you've found a perfectoid ring and I give you a map to, so give you an R algebra S, where this is p complete. And one of two things holds, either S here is finally presented or let's say S is integral over R, well at least up to p completion, right? So just like in this example over here, right? So you want S to be the p completion of something which is either a finally presented algebra or an integral map, okay? In this case, there exists the conclusion of the theorem, right, is that there exists a unique, call it S-perfect or the perfectoidization of S, right, which is the smallest perfectoid ring over S, i.e., it's universal for maps from S to perfectoids, so maybe draw it in a diagram, right? So it comes for free with a map from S and if I give you any other S algebra that happens to be perfectoid, there exists a unique arrow making the triangle commute, right? So there's a unique smallest perfectoid, right? Is everyone happy with the statement of this? Really unique. It's really over S, I think it's, well, unique up to the map, yeah, yeah. So, okay, it's really unique, all right? So now it's unique, again, assuming you've started off and you've found a perfectoid to begin with, okay? Okay, so I couldn't have started with some local ring, I really had to do a bunch of modifications before I got this, okay? All right, so just a couple of remarks, all right? So, well, the first is just like in that last example, somehow over there, if R to S is surjective, well, if R is perfectoid, then its perfectoidization is itself and the conclusion of the remark is R perft surjects onto S perft, so what you end up with is a quotient of R and, all right, two, all right? So, if S happens to be petortion-free, all right, then so is S perft, right? It's not so easy to show or get your hands on what these things are, all right? So, and maybe it's not so easy to give attributions also for where all the various properties are going to show up in the rest of my talks, so please ping me if you would like more references, but this one shows up in an appendix of a paper of mine with a bunch of co-authors, so Lang Schoenma, Karl Schwede, Joe Waldron, Waldron and Jakob Budaczek, all right? So, as the notation would suggest, right, so here you might say, well, in my theorem I assumed I took things that were over S over here or over R, right? So, but then my notation S perft doesn't make any reference to R. So, Bonnie Schultz also proved that what you get this S perft is independent of R, meaning, right, so, so long as you know that S admits a map from some perfectoid ring, the perfts you get are all the same, all right? So, really doesn't really depend on the base ring, and maybe the last thing to say is that more generally, without either one or two, right, Bonnie Schultz, right, so formulate, right, so you use a derived prismatic homology to formulate a notion of S perft as a derived ring, right, so some complex, right, over S. So, in some sense, the point of the setting over here without getting into a lot of technical details is that this is a case where you can check that that complex is discrete, is really concentrated in degree zero, and so it gives me an honest ring, right, and not some more complicated homological object, okay? Okay, so, so in some sense, part of this is to, we'll use this, this perfunctor later on, all right, so, but it gives me a way to construct a bunch of perfectoids, and we'll see some more, right, so I want to sketch through today, right, so tell you a little bit about, right, so, and a lot of my theorems that I'll write going forward, I'm trying to combine as many results from people all into one nice formulation, right, so it says the following, let's say I have a local map of complete local domains, all right, so then I can map R and S to their absolute, you know, reclosures, R plus and S plus, and then I can map each of those further to rings B and C so that all of these diagrams commute, all right, in such a way that B and C are, in such a way that B and C, respectively, are BCM R algebras and S algebras, all right, so maybe just a shorthand, whenever I want to say that B is a BCM R algebras that dominates R plus, I'll just write BCM R plus algebra, meaning that it's BCM over R, okay, and the contribution of some of the other people are, well, right, so Andre, Gebra and Bott, right, can even take B to be the piatic completion of R plus and C to be the piatic completion of S plus, right, so, and some, right, so this, this is a big theorem here, so this is saying that given a complete local domain, RMK, there exists a big, call the R plus algebra and that construction can be done functor, or weekly functorially over, over the map R to S, okay, all right, so, great, so in particular, the first thing I want to sketch for you over here today, right, right, so is Andre's result really is, well, Hoxer and Hinokie proved in characteristic P, right, and if you were at the summer school and looked at the lectures last summer, right, that certainly R plus is big comacole over R in positive characteristic, right, so that's their contribution to this theorem, right, Andre used these perfect writing methods to show the existence of big comacole algebras, right, in, in mixed characteristic zero P, all right, and that's the, what I want to tell you a little bit more about today, okay, but before I do that, are there any questions or any, I'd like to ask about the statement of the theorem here, okay, so, so I want to sketch for you the essential ingredients of, of our current perspective on Andre's proof, right, and really, but before I do that, let me say maybe a little word about what big comacole means, right, so, let's say you have a big R module M, right, so, big meaning possibly not finally generated, and let's assume that M is not equal to little m times M, right, then, what I mean by BCM is balanced big comacole algebra, so, all system of parameters are regular sequences, of course, if all system of parameters are regular sequences, then that implies that some system of parameters is a regular sequence, so, you might also talk about being comacole or, comacole, big comacole with respect to another fixed system of parameters, and that implies something which is a, a very even weaker, namely that if I look at local comology, right, so, then the local comology modules vanish for i less than than r because I can use any fixed system of parameters to compute the local comology modules, all right, in most cases for a lot of the modern proofs, the weakest notion of these will work to make things go forward, but historically this is the one maybe that is most useful to keep in mind, right, and sort of to relate all of them, there is an implication back the other way if, if the module is complete, right, so, if the module you have is complete, right, so, meaning here ematically complete, then all of these notions in big comacole are the same thing, all right, so, in practice what often happens is you can produce something that, say, satisfies the local comology criterion here, right, and you then need to ematically complete to get a big comacole algebra, all right, so, all right, so, as I said I want to give some kind of a sketch of, of the proof at least of the existence of big comacole r algebras, right, so at least sort of sketch on raised proof, and to do that the key ingredients what Andre really figured out how to do here and that no one else had figured out before is the following, so, he says the following, say you have a perfectoid ring and you have some element, his lemma says that there exists a map to another ring A prime where this A prime is still perfectoid, the corresponding map is faithfully flat, well, modulo any power of P, right, so, and in such a way, all right, I'll write it like this, in such a way that your fixed element G admits a system of compatible P power roots inside of A prime, right, so of course in positive characteristic perfectoid is the same thing is perfect, in which case all elements have unique P roots and so these systems already always exist, so, but in order to mimic a lot of the proofs and the techniques from positive characteristic, all right, you often then need to figure out how to take P roots of particular elements, but, but in mixed characteristic those things are unique, there are choices all the way across, right, and it's not clear that, right, so, and all the perfectoids I've written down before, I've given you a particular often finite list of elements that have compatible P power roots, right, but, for instance, in the earlier examples, if I look at x1 plus x2, it's not clear that that thing has compatible P power roots in mixed characteristic, okay, so, all right, this is the lemma that allows you to do that, all right, is everyone happy with the statement of the flatness lemma, all right, so, let me try to get through this, all right, so let's try and give a sketch of the proof of the existence of big comb and call the algebra, all right, so, but for simplicity, right, in this case is just as pretty much the heart of the whole thing, so, let's say that, okay, as algebra closed, and we're in, right, so, we're in mixed characteristic zero P, all right, all right, so, since I've assumed that my rings are all complete, all right, so, let's write my ring R as a quotient, say, S mod Q, right, where S is a power series ring over, over a, over a weight ring, so, right, so whatever it is, and here, of course, P is not in Q, okay, all right, so, let's let C be the height of Q, all right, and standard prime avoidance arguments allow you to choose elements, say, F1 up to FC inside of Q, so that P, F1, F2, up to FC are an S regular sequence, okay, and I can extend to a full regular sequence of S by picking, extending P to a full system of parameters of S mod Q, all right, so, extend, right, to a full system of parameters on S, all right, so here, by, by numbering, right, so C was the co-dimension, D should be the dimension of S mod Q, namely R, right, so C plus D is the dimension of S, okay, okay, so, in particular, all right, so P, X2, up to XD now are a system of parameters, right, for both S mod the F's and my original ring S mod Q, all right, so, and I can pick, and you'll see this is where we're going to use Andres' flatness lemma, right, so, I've done it in such a way that Q is a minimal prime of the ideal F1 up to FC, in particular, it's an associated prime, so, there is an element G which is not in, in Q, or G not in, so, there is an element here of this quotient which is annihilated by Q, all right, so there's an element G in here, so that the G times Q is contained in the F1 up to FC, okay, so, roughly speaking, the proof is going to proceed in the following way, all right, so, the theory we built up is going to make it pretty easy to come up with something which is big comacole for this quotient here, where I've killed this fixed F1 up to FC, all right, the co-dimension number of elements, and then we have to use some tricks, right, in order to bootstrap off the existence here to go to S mod Q, so, set S to infinity, all right, to be, all right, so, remember again, S was this nice regular ring, right, so, we have our standard trick to create a perfect ring out of that, namely, adjoin the infinite piece roots of P in all of the variables, right, instead of this thing, and then P complete it, right, and as you've seen, this is perfectoid and a fifthly flat over S, okay, but in this ring, right, for instance, if I were to take any of these elements, F1 up to FC, or the element G, for example, it's not clear that those elements inside of S have infinite piece roots inside of here, all right, so, by the flatness lemma, right, can find some fifthly flat mod P to the N map, all right, to some other perfectoid, right, so that G, F1 up to FC have a compatible system of P power roots inside of S infinity prime, so, apply the flatness, say, C plus one times, okay, you can check, all right, I'll leave this to you to do that, well, since S infinity is P torsion free, S infinity prime is, be separated, this gives that S infinity prime is still P torsion free, all right, so this uses the flatness mod P to the N for all N, so if you go through and try and do the proof for yourself, all right, so more generally there are statements you can make about P complete flatness, right, which is a similar statement here, great, and so in particular, I know that, well, P F1 up to FC and X2 up to XD were a system of parameters for S, so in particular there are an S regular sequence, all right, but now S goes to S infinity is faithfully flat, S infinity to S infinity prime is faithfully flat mod P, so it's easy to see then that, okay, when I, I don't have to worry about the non-zero divisor P to start off with, I've already checked that separately, but once I kill P, right, faithfully flatness mod P means that these guys still remain a regular sequence in S infinity prime, right, so, okay, so that shows that these guys when I pass to S infinity prime still say a regular sequence, but of course when I have a regular sequence, right, or some elements, then asking that it's a regular sequence is the same as asking that any power of those elements remains a regular sequence, so I could take P to the E th roots for each of these F's and it would still be a regular sequence, that works for any E and that's what I mean here by putting the infinity down on each of the steps, and so last point here, right, so this means that P X2 up to XD are in fact a regular sequence on, all right, so I'll, I'll write this off from another spot, all right, so P X2 up to XD are going to be a regular sequence on, well, S infinity prime modulo, right, this example we had before where you look at this ring and kill the infinite P th roots and then take the P completion, now I've used something a little slick here, right, if you have two elements AB that are regular sequence on a module, in general I can't permute those, right, however it is always true that the first element is a regular element modulo the second, right, so if AB is regular then A is a regular element modulo B, right, so that allows me to take that P which I know is regular on S infinity prime and deduce that it still remains a regular element on the quotient over here, so as we've said before this guy here is, is perfectoid and as we've set it up, if I were worried about S mod F1 up to Fc, right, here I produced a big comacole algebra for S mod F1 up to Fc that is perfectoid, right, so or at least after I amatically complete, right, I have some regular sequence on S mod F1 up to Fc that remains a regular sequence when I pass on this guy over here, okay, so I have a system of parameters for S mod F's, right, that becomes a regular sequence on T, so the question is how then can I transport this guy to give me something for Q and we're gonna use the following trick, so consider the map T to T prime, right, which has the following form, look at the set of homomorphisms over T from G1 over P to infinity, this ideal back to T, okay, this is a little crazy, but check that this is a commutative ring, right, in general, right, you'd expect this is just some T module, right, if you will, right, not at all clear when I write homs down that I can compose them, the little loan that they become, I can switch the order of how this thing all works, right, and the point here is this is really the first place in everything where I can't, I think I have to use almost mathematics in order to actually make the statements, the point here is that, well, this is a very special kind of ideal, right, it's an ideal whose square is equal to itself, right, and when you have such a guy, this is the framework for doing what's known as almost mathematics, right, so in particular in this case, this map from T to T prime is, is, well, G1 over P to the infinity almost an isomorphism, right, so, meaning if I multiply by G1 over P to the E for any E, right, on either the kernel or the co-kernel, what I get vanishes, okay, so this ring you can think about in some sense as multiplying, right, T times G to the minus 1 over P to the infinity, right, so sort of extracting some extra roots from this thing, all right, so once you've checked this, and this is fun to check, it's not, I would check this directly, what this means then is, well, P x to up to xd, there were a regular sequence on T, this is almost an isomorphism, so the end result is that these elements are almost still a regular sequence on T prime, right, so they're, right, so meaning, right, what does it mean to be almost a regular sequence, well, an almost non-zero divisor is one so that the kernel of multiplication by that element is almost zero, i.e., is killed by all powers G1 over P to the E, right, and once I have that, now I require that when I go modulo P, x2 is almost a non-zero divisor, so on, so forth, and also it has to be checked separately, right, so that quotient, right, is not almost zero, right, I didn't just multiply and kill everything, okay, well, our choice of G was such that GQ was contained in F1 up to FC, but now I'm in a situation where both G and all the F's have roots, right, so you can check that this, again, this is in some perspective multiplying by G to the minus 1 over P to the infinity, so the end result is that, well, if I have this relation and I pass up further, this is somewhere which has roots here, right, so, well, up to P completion, right, this will give me that G to the infinity is contained in this after taking roots, right, so the punchline is that I have this string of maps here and this Q, which really lives inside of R, right, or really lives inside of S, right, after I map it all the way up to T prime, goes to zero, so T prime is an R algebra, you can see that we're almost done, all right, so now I've constructed you something, all right, so, so this T prime here is perfectoid, it's an R algebra and I have a system of parameters, it doesn't become a regular sequence on T, but it almost is a regular sequence on T, right, and so the last trick is the following, right, so we call this Gabber's trick, because he showed us all this in a note after a hot topic MSRI SL math workshop on the homological conjectures, the handwritten note that's still posted on the site, okay, all right, so now, right, I have to, I want to take this almost regular sequence and force it to be a regular sequence, right, so here's the trick, let B be a localization of a countable infinite product of copies of T, okay, and the localization is the multiplicative set generated by this, the roots of G where this is giving us our almost-mathematics system, and at this point, you can check, right, so one, Px2 up to xD were a G almost regular sequence before and the point of this trick is that it takes things that are almost true, right, and forces them to be true on the nose, right, so in fact these guys are an honest regular sequence now on B, and two, right, it's still proper, right, so I didn't kill, kill everything, all right, all right, so this is now, right, well, it's a BCM R algebra with respect to this fixed system of parameters, so I can get an honest BCM or balanced BCM, BCM R algebra by P completing or by M completing, right, so what, T prime, thank you, all right, so B hat M will be a BCM R algebra, right, and that completes the sketch, okay, so let me do a couple of remarks, right, so, so at this point we know BCM R algebras exist, right, so if I give you some collection of them, right, then Mon Schwede showed that you can find perfectoid BCM R algebra B, all right, dominating all of them, more over, this can be done compatibly with the direct limit of the B lambdas, assuming that was a directed system to begin with, right, it was a consequence, it's exactly this compatibility with directed systems that allows you to, to create a BCM R plus algebra in a, in a systemic way, right, so say for each finite extension of R, right, I can find a BCM algebra, right, and the finite extensions of R give me a directed system and I can now find something dominating all of those, right, so these are the kinds of extensions you have to do once you even know existence in order to keep going further, right, all right, so again these lectures build off of in some way some of the talks from last summer, right, so in particular there was a lecture series by Polstra on some of the applications of the existence of big comacol algebra, and I'd be remiss if I didn't say at least one of them, namely the direct sum-end theorem, right, so here's the statement that if RMK is regular local then R is a splinter ring, i.e. whenever I have a module finite domain extension, there exists a splitting, so splinters are a little enigmatic in general, right, so in equal characteristic zero, i.e. if you have a ring that contains Q, splinters are the same thing as being normal because I can use the trace map to split off of a finite extensions, right, in positive characteristic or mixed characteristic that's no longer true, right, so and in general often some very easy properties with planters are not so easy to check, right, so for instance it was shown by myself on RKData that if you take a local ring, right, that it's a splinter if and only if its completion is a splinter assuming the ring has a geometric irregular formal fibers, right, so but that in general the statement fails, okay, so let's finish with the proof of the corollary, so it's pretty easy to reduce to the case, right, so while it's not true that R is a splinter if and only if its ametic completion is, if the ametic completion is a splinter then the original ring must be, right, so it's pretty easy to reduce to the case where R is complete and say I have a module finite extension which is, my ring R was complete local so since R is a, since S is a domain extension, right, it also remains local, right, so if I take a system of parameters, it's a system of parameters for R but also for S, all right, since the map is local, take S to B BCMS algebra, right, so in here we've got R mapping to S, maps all the way to B, all right, so this map here is automatically faithfully flat, right, so again remember R here was, was regular so let's say that this was a regular system of parameters for R, that system of parameters is a system of parameters for S and then also is, remains an regular sequence on B since B was big comma collie, so as a result if I look at say the Ith causal homology of the X's on B, right, this vanishes for, for I less than the dimension, right, because the X's are a regular sequence on B but of course that's the same thing since the, I can take the causal resolution of the residue fields K to get a resolution for R, right, right, so now I use the fact that whenever I have a faithfully flat map or even pure, right, map from something which is a complete local ring, the mat thing splits, so say take the splitting map phi and then of course what I can do is take that splitting and restrict it to S to get a splitting for S, all right, so there's my sketch of the proof of, of the theorem, all right, so maybe we'll stop the lecture there and let me tell you, so at this point now I have access to the big comma collie algebra, so in the last lecture I will tell you as much as I can, as quickly as I can about how to use the existence of these big comma collie algebra to define some classes of singularities that mimic many of the classes that we've been talking about in the other lecture series, all right, so, hi Craig. Hey Kevin, hi. It's good to see you.