 Let me start by defining what a perfectoid field is. So we're working in this set up sort of non-archimedian geometry. So we are working with non-archimedian fields throughout. So we call it a non-archimedian field is a topological field whose topology is induced by a non-trivial valuation of rank 1. So in particular, you have sort of a norm map from k to the positive fields, which is essentially unique. And I don't require that it's complete. But now for a perfectoid field, I will make this requirement. So a perfectoid field is a complete non-archimedian field k. With the following requirements, its residue characteristic should be positive. So it's a fixed prime number of p greater than 0. One has to require that the rank 1 valuation that is sort of corresponds to it is non-discrete. And the crucial condition is that the Frobenius on the power bounded elements modulo p, so x match to x is a p, is subjective. So their sort of examples are k is the completion of one of the following types of fields. You take qp, but then the Frobenius would be surjective here because it's just fp. But this sort of, we don't want this. So we have to join lots of p power roots, for example. Or one could also join the p power roots of unity. Or you could just go all the way up to cp. Or you could also do a similar construction in characteristic p, so take a long series field and join all the roots of g. So somehow this condition here should exclude unremifed extensions of qp. We don't want them. So one thing that implies is, for example, the following. So the yu group of k cross inside the positive fields, this is a pd versus a b group. So just in the naive sense that any element has a pth root. And it's very easy. So it proves as follows. So let gamma be this yu group. Then gamma is not just the powers of the integers just because we exclude the unremifed extensions of qp, or we assume that it's not discrete here. And so it follows that gamma is generated by the absolute value of x for x elements, which satisfies that the absolute value is strictly larger than p, strictly between 0 and p. And so we only have to find pth roots of such elements. And then there exists some y such that the absolute value of x minus y to the p is less than the absolute value of p. By this requirement, it's convenient, it's a surjective module p. And so then the absolute value of y to the p is exactly the absolute value of x by the strong triangle inequality. So some of the perfected fields separate into two cases. So there's the characteristic of case p. In this case, this requirement simply amounts to saying that k is a complete perfect non-acumenian field. So it's something pretty easy to write down. And so in characteristic 0, it's sort of in some more subtle notion. But now we want to define this tilting functor, which takes any perfected field of characteristic p. And then we will also sort of, in this whole series of talks, see that the whole theory over this perfected field of characteristic 0 is parallel to the theory of a characteristic p field, where everything can be expressed in more or less elementary terms. So we have the tilting functor from all perfected fields to those of characteristic 0, of characteristic p. So it will be in a noted k maps too, k flat. So let's construct this. And so let us choose some sort of uniformizer. But as a valuation is non-discreet, we see that it doesn't quite make sense to talk about uniformizers. But let's use some element that is not too large. So it's not too small and less than 1. And we consider the ring just from tense construction. It takes the inverse limit over Frobenius of the power bound and elements modulo pi. So this condition precisely ensures that this ring is of characteristic p, meaning that p is 0 in this ring. So it makes sense to take this inverse limit here. And so this is a perfect ring of characteristic p. And it also has a topology. So it has the inverse limit topology, noting that these things naturally have the discrete topology. And so now we cook up the tilt out of it. So we'll state some lemmas about how this behaves. So there is a multiplicative homomorphism from you can take the inverse limit of the power bound and elements now under the piece power map, which now in general is not a ring homomorphism anymore. And you can project this to this ring. And this then turns out to be a nice homomorphism. So in particular, we get a map from this inverse limit of k0 mod pi to k0, which I denote by x maps to x sharp. So somehow by taking the inverse of this isomorphism and then projecting to the first coordinate. So the second part is that there is a somehow we want a corresponding sort of uniformizer on the other side. And so we choose some, there's an element which is then called pi flat in this inverse limit, such that it's corresponding element in k0. This is a sharp element. It's not necessarily equal to pi, but it has at least the same absolute value. It will be enough. And then we let now this tilt be we take this inverse limit over Frobenius. And then we invert this element pi flat. And then the third part is that k flat is indeed a perfectoid field of characteristic p. We still have this isomorphism that k flat can be, one can get it by taking the inverse limit over the piece power map on k. And so in there we have, again, the power bounded elements. They are similarly this inverse limit of kth Turk, and which by this isomorphism is then also this ring here. And if we want to, we can also sort of consider the maximal ideals of these local rings. So m flat, which is the set of elements of norm less than 1. And this also is then the inverse limit, or corresponds somehow in here. And what else can we say? So this provides us with a map k flat to k x maps to x sharp. Some are extending this map that we have here on the power bounded elements. And if one takes this other algebra and takes the stilted algebra, and I would use it again when you lose this uniformizer, then this becomes canonically isomorphic to the small part, some of our characteristic 0 local, of our characteristic 0 perfectoid field. And finally, it's pretty clear if the characteristic of k was already p, then the whole process didn't change anything. If that is just k, well, maybe I give a short proof of this. So for one, so first we construct the map from this inverse limit over Frobenius to k0. So this x maps to x sharp map. And how do we do it? So we have some element of this inverse limit. And we lift each xi bar just arbitrarily to some xi in the power bounded elements. And we define x sharp as the limit as n goes to infinity of xn to the p to the n. And so we have to see that this makes sense. And so we have to check that, or it's enough to check that if xn is congruent to xn prime modulo pi, then xn to the pi to the n is congruent to xn prime to the pi to the n modulo pi to the n plus 1, for example. And so somehow each term of this sort of approximates this x sharp to a larger extent. And so this can be sort of proven by induction on n. That's OK. And so then we have this map. And then we get now a map also to the inverse limit over the p-spaw map of this thing. Just sending some element x here to x sharp, x1 over p sharp, and so on. So it's clear somehow that this map, sharp is multiplicative and continuous. So let's say this. And so this is a map inverse to the projection. And one easily deduces part one. And so for part two, we have to find this element. And so we first take some pi1 such that the norm of pi and k0, such that the norm of pi1 to the p is equal to the norm of pi. So this exists because our value group is p-divisible. And then we take any pi flat, which is also form 0, pi1, and then something in this inverse limit over Frobenius in this thing. And then the proof of part one shows that if you want to calculate this pi flat sharp, then as a first approximation, we can take the piece power of this pi1. And the error term will be pi squared. And so this already implies that this s desired. And then we wanted to see that all of these things are satisfied, but because this map is multiplicative, it's easy to see that it extends to this map on k flat, giving this isomorphism. And one can also see that the norm on k flat is given by the norm, which I denote by dot flat, is given by evaluating the norm at the sharp element. And then it's easy to see for the other things, somehow. Other claims follow. So I mean, to see that it's a perfectoid field, we only have to see that it's perfect. That's clear. And we have to see it's complete, but it's also pretty clear. And then we know that it's perfectoid field of characteristic p. And the last part is that if it's already of characteristic p, then it's the same as k, then clear since somehow this is the inverse limit over the piece power map, which now is k is perfect. In this case, it's just k. So usually, we now redefine our uniformizer pi to be of this form pi flat sharp. So it has the same valuation, so it doesn't quite change anything. But it has the advantage that then pi sort of has a canonical system of p to the n th roots. So given as first taking the p to the nth root of pi flat, which exists because this lives in a perfect field and then taking the lift. Now we have a uniformizer with p power roots, which is at some points quite convenient. And so conversely, if you have some uniformizer with a system of p power roots, then you get an element pi flat given by pi pi 1 over p and so on in this inverse limit in k flat, which is the inverse limit over the piece power map of k. So somehow giving this uniformizer and characteristic p is the same thing as giving 1 and characteristic 0 with these p power roots. And let me also just shortly sketch an example. So you can take, for example, k as a completion of this thing. And then k flat will be the completion of some of the similar things in characteristic p, taking the long series field over fp and then adjoining all of these p power roots of t. And so in this case somehow we can take pi equal to p with its given p to the nth root. Then pi flat will somehow be this t. And we can also somehow see this isomorphism after reduction somehow. Because in this case, k naught mod p is just, you take zp, adjoin all of these p power roots, and reduce mod p. But now this has become an fp algebra. So we can also consider this as fp. And somehow rename this p as t. And then adjoin all of these p power roots of t, modulo t, which then is just k flat not mod t. So that's somehow what's more or less going on in general. And so some of the picture is sort of as follows. So you can imagine somehow the spectrum of k, not somehow it has a generic point. So this is spec k. And it has a special point, which is somehow the spectrum of k naught modulo the maximal ideal. But then sort of this special point here has an infinitesimal sickening, which is the spectrum of k naught mod p. And somehow all of this somehow morally lifts some on the line. And so now the tilting procedure is somehow you have on the one hand this picture with this generic point and this sickened special point here for this character 6-0 version. And then some of the characteristic p version starts to look the same, but then tilts some of the other way. So that's somehow the spectrum of k flat not. And somehow this is what I call the tilt. They would sit in a larger object indeed. So you could consider the ring of bit vectors of your field of this thing. And this would somehow give something two-dimensional, which will sort of fill out the space. OK. And so as one preparation for later, when we will consider perfectoid spaces, it will be useful to compare the valuations, the continuous valuations of the field and its tilt. So we have the following proposition. Let k be a perfectoid field. So it was tilt k flat. Then the set of continuous valuations of k, some bijection with the set of continuous valuations on k flat by sending some variation here to the variation flat, which is given by f of, yes. Yes, yes, yes. Always more your equivalence. But sort of if I take the equivalence classes, then sort of there's always exactly one representative where this is true. Yes, yes, yes. Yes. And to sum up, at this point, these absolute values are meant to denote a general valuation on k and not just a series of rank one valuation. And so a quick proof of this is, for example, that continuous valuations correspond to open valuation sub-rings k plus contained in k naught, which in turn correspond to valuation sub-rings of the residue field k naught modulo is a maximal ideal. But now this is the same as on the other side. And then you go back to continuous valuations on k. And this in turn will then correspond to continuous valuations on k flat. And you can check by hand that somehow this map really is the one that sort of comes out of this abstract formalism. And so the main theorem is the following generalization of the theorem of Fontaine and Winter-Berger. And so the following, if k is perfectoid and L over k is the finite extension, then also L is perfectoid, somehow with a natural topology of L as the finite k vector space. And then it makes sense to claim the following, that tilting L maps to L flat induces an equivalence of categories between the category of finite extensions of k and finite extensions of k flat, preserving degrees. So in particular, sort of you get this isomorphism of algebra groups. And let me sort of first sketch how this will be proved. So maybe there's a more direct way, but sort of I incorporate some parts of the proof somewhere, then in the more general framework, which will be of use anyway. So the first step is to show that 1 is true in characteristic p, that's clear. Then the second part will be to invert the tilting function, to get a function from k flat algebras to k algebras, which preserves some of the property. So somehow from perfectoid fields, at least, over k flat to the similar category over k. And it will turn out that sort of this can be done in great generality. So for what I will talk about in the next talk for so-called perfectoid k algebras. And I'll say a little more about this also in a minute. And so we get sort of a functor in the other way. So for any finite extension of k flat, we can produce one of k. So get some fully-faceful functor, finite et al over k flat to finite et al over k. Yes, yes, yes, I will use some almost things there. And so now we have to show that this is essentially surjective, which is in some sense the hardest part always. So somehow to show that the hardest part always is to show that something which is finite et al in characteristics 0 really comes via this inverted tilting procedure from something characteristic p. And some of the starting point to get to start it will be the following proposition. And for this we use some of the following proposition that if k is a perfectoid field with tilt k flat, then if k flat is already algebraically closed, then so is k. So somehow, at least in the algebraically closed case, we can sort of get this conclusion. And the proof is just similar by an explicit approximation argument. So we take some polynomial, this polynomial ring, and we need to show that this has a 0. And so we may assume that the Newton polygon is a line. And we may also assume that the absolute value of a naught is 1, because the value group of k is the same as the value group of its tilt that somehow follows from this part 2, that for any pi we can find a uniform water pi flat. And this is a q vector space. And so somehow by rescading this a naught, we can, by rescading x, we can make this absolute value 1, if you like. And then we choose some polynomial q of x. So somehow, again, x is d plus b times d minus 1 x to d minus 1 and so on plus b naught. On the other side, such that p mod pi is the same as q mod pi flat, which now live in the sort of the same ring k naught mod pi x equals 2. And then choose some root of this. And then we define a new polynomial, which is p of x plus y sharp. And then this y sharp gives sort of the first approximation to a solution. And then the constant term has absolute value, s and pi inverse at pi. And then we continue with this algorithm and we converge to a solution. OK? Yes? OK. And so now our proof will make use of almost mathematics. And let me, before the break, already start a little with this and explain what it has to do with the proof of the theorem that the Galois groups are isomorphic. So the idea behind this almost mathematics is due to faultings. And I will make sort of use of this book of Gabber and Tramero where they lay the foundations in really great detail. So we fix the perfectoid field k. And so we have this maximal ideal m inside of k naught. And so the idea is to neglect m torsion everywhere. So for example, if m is a k naught module, then an element is called almost 0 if it's annihilated by m. And we call m almost 0 if m times m is 0. So or in other words, if all of its elements are annihilated by m. And then the basic fact is that the full subcategory of almost 0 objects is sick. So the only non-trivial part of the seminal, which is also not difficult, but if you have a short exact sequence of k naught modules and the two outer terms are annihilated by m, then we want to prove that also the thing in the middle is annihilated by m. So you cannot start with something almost 0 and then build up something larger out of it. And well, OK, so in general, it's, of course, true that m squared times m is 0. But we are in this favorable situation of a nondiscrete valuation. So the square of the maximal ideal is still the maximal ideal. And so we get that m times m is 0, OK? And so we get a sequence of localization functions. So we can start with the category of k naught modules. So this is something, some integral structure on something also. And then, of course, we can pass to k modules, which you get somewhere by localizing along our p-par total objects. And in between, somehow, you can only forget about the almost 0 objects. And so this is some of the generic fiber. And so this is something strictly in between. So one might call it a slightly generic fiber. Or if you think about it from the other direction, it's somewhat an almost integral structure. And so the philosophy somehow is that for perfectoid objects and properties of the generic fiber extend to the slightly generic fiber. And so in characteristic pieces, for example, as follows, if something is true over k, then this implies that there is some big number n such that it's true over on the integral level up to pi to the n total. So maybe you have to invoke some finiteness of the objects involved or something like this at this point. But then because the Frobenius is bijective, the same statement will just literally hold true modulo pi to the n over p torsion. And then you go on. And then somehow it's true up to m, which is the ideal generated by all n over p to the m torsion. So it's almost true, in some sense. So that's somehow the basic argument. And so in our situation, when we use this in the following way, we will prove the following series of equivalences of categories. So you start with something finite etal over k. And if this philosophy is true, then this should extend. So maybe I should give this function here a name. So somehow this m maps to m almost, which is a, so this category is by definition the category of k not a module, so almost k not modules. So somewhere you can take this almost version of your power-bound elements and then consider something finite etal over this. In some almost sense, I have to define what this is. But then there will be all your theorems stating that finite etal algebras in this almost sense also lift over nil-potent. So you can also reduce modulo pi and then take the finite etal covers. But now this thing is just the same thing as on the other side, which then, again, you can lift back to the other side. So somehow, again, in this strange picture that somehow you have k and you have k flat, somehow you start with something here. This extends all the way to the almost integral level. So you somehow have to cut off, in the sense, this close point here. And then it lifts over here. But then you can lift it back to the other side. I don't need this ramification theory of deeply ramified fields now. I mean, that somehow the idea to reduce to this case of algebraic closed fields is due to Ketlaya somehow. Ketlaya does very similar stuff. And he has a different way of proceeding, which avoids almost mathematics, but instead sort of needs some tedious calculations with width vectors. So I find it less conceptual somehow as argument. OK, so let me continue doing some almost mathematics for the rest of the talk. So the definition is that the category of almost K naught modules is somehow K naught a modules, which is the category of K naught modules divided by the almost zero ones. As it is a six subcategory, this makes sense. And so we get a localization function from K naught modules to K naught a modules, m mapping to ma. So in particular, any almost K naught module can be represented by an actual K naught module. And one may wonder if one somehow has two K naught modules, how to compute the homomorphisms as almost modules between the associated almost modules. And there's a very simple formula for this. It's just homomorphisms from m tens, so as K naught modules from m tens are m to n. So somehow the idea is that there is some calculus of fractions. So somehow you can, any morphism from ma to na is represented by some morphism. So first you replace m by some far the isomorphic one, and then you map to n. Whereas this is an almost isomorphism, meaning that the kernel and co-kernel are almost zero. And then it turns out that m prime, which is m tens or m, is the final object for such m prime. So somehow instead of any m prime, you can just take m tens or m, which is a map to m, is an almost isomorphism. And well, I don't know. I never know what the final and initial is sort of here. Sorry. Initial, yes, it's probably true. And so the map to m is an almost isomorphism and any other thing, so it maps to any other one. And so this then implies the statement here. So OK. So you can somehow explicitly say what the homomorphisms in this almost category are. And in particular, we see that for all almost modules, the set of homomorphisms from x to y is a K naught module. And so somehow it has more structures than just somehow you can define than the set of almost homomorphisms from x to y just as being the associated almost module, which then gives an internal hom, but somehow there is more structure to it in the hom set. OK. And so this category of almost K naught modules has all the good formal properties of the category of modules over a ring. So technically, this category is an abelian tensor category where one defines kernels, co kernels, and tensor into the unique way compatible with definition for K naught modules. So for example, if you take the tensor product of two almost modules, then it's just the almost module associated to the tensor product for any two actual modules. OK. And it's also true that somehow the usual adjunction between tensor product and almost homomorphisms is true. So we have that's almost for L, M, N, almost modules. We have that homomorphisms from L into the almost homomorphisms are the same as homomorphisms from L, tensor, M. So in particular, because it's in the BN tensor category, we can define the notion of almost K naught algebras, denoted K naught A algebras. And so it's clear how you find them. So they are some module plus a multiplication map from A tensor A to A plus all the other data. Just satisfying the usual axioms. And then if A is an almost K naught algebra, then you also have to define the notion of A modules. So these are some almost modules, almost K naught modules plus a multiplication map from A plus all the other data and conditions that you have. Sorry. And localization gives a function from K naught algebras to K naught A algebras and also from if R, say maps to R almost. And then we get also a function from R modules to our A modules, M mapping to MA. And so for modules of a K naught, it was the case that they are all represented by actual modules. And one would like to know that some of the same is true now also for algebras and so on. So some of that one can actually represent any almost K naught algebra by some actual algebra. And this does work. So somehow, because of the following proposition, so there is a right adjoint from almost modules back to K naught modules denoted to write adjoint to somewhere N mapping to N A, function M maps to M lower star, which is given by taking the homomorphisms as K naught A modules from K naught A into, this is the so-called module of almost elements. And this induces also functions from K naught A algebras to K naught algebras. So somehow, on this thing, if A is in almost K naught algebra, then on a lower star, one can define a natural structure of K naught algebra. And also from A modules then to a lower star modules. And there's one thing I should say. So the adjunction from M lower star A mapping back to M, this is an isomorphism. So somehow, this M lower star gives you a sort of canonical representative of your almost module. So this is just an almost isomorphism. Yes? It means you see the one page? Yes. Yes, one can sort of, if M is in K naught module, then you can form the almost module. And then it's almost elements. And by definition, it's the same as this. And now we use this proposition telling us what the homomorphisms are. And so it's the same as homomorphisms of K naught modules, now from this maximal ideal into M, which is also the reason that one calls these almost elements, these elements of this module, because somehow an element would be given by a map from K naught into here. But this is not really a map from K naught, but only from the maximal ideal. So you don't actually have an element of M, but for every small epsilon in M, we have sort of the epsilon multiple of this element. So let's just say this. So any K naught A algebra comes by a localization from K naught algebra and same for modules. OK, so now we have in this almost setting somebody find what algebras and modules of algebras and so on are, and now we want to define some more notions from commutative algebra in this almost context. So yes, the following definition or proposition. So let A be any almost K naught algebra. Then for example, you want to define what flat modules are. So name module M is flat if X maps to M tends to AX is exact. And so what does this mean concretely? So if R is an A algebra, that's a K naught algebra, and N is an R module, then the associated almost module N A is a flat RA module if and only if the following condition is satisfied that for all R modules X and for all i bigger than 0, the tor i of N and X, I mean, if it was flat, then it should be 0. But if it's almost, in this sense, almost flat, then this should only be almost 0. So you can play the same game with projective modules. But for some, you can define sort of tensor products now also for A modules. So one way to define it would be to take sort of first the representatives and then go back to the almost module. And then there's the definition of what projective is. But for some reason, one adds the adjective almost projective. And the reason is the following that somehow there's a notion of a projective module in any useful category. But this notion of projectivity in this category sense is not very useful for a reason that I will explain afterwards. So we say it's almost projective. If the function X maps to the almost homomorphisms from SA modules from M to X, if this is exact. And we have the same criterion. So if R is a k naught algebra and N is some R module, then NA is an almost projective RA module if some of the same thing is true for X. So for all X, which are R modules, and i bigger than 0, the X i of R modules. This is to be an R, I guess, of which way N to X is almost 0. And then we also need some finiteness conditions on the modules. And these are defined in the following way. So we start in this case with, so the easiest way to define it is to start with an actual algebra so that R be a k naught algebra and N be some R module. And then the RA module, NA, is said to be almost finitely generated, respectively almost finitely presented as the following condition is satisfied. If for all epsilon in the maximal ideal, there exists a finitely generated, respectively finitely presented R module M epsilon and a map F epsilon from epsilon to M, whose kernel and co-kernel are annihilated by epsilon, killed by epsilon. Some taking condition. And let me give some examples for this. So first an example showing that the notion of projectivity, categorical notion of projectivity is bad. So even k naught, A is not a projective. Sorry, yes, I wanted to write N's everywhere. Because it's consistent now. And so some are implicit and three is a claim that this only depends on the RA module NA. It's not a projective k naught A module. So somehow you could, for example, look at the map. This map, this should be so objective if it's projective. And this can be identified just with k naught. And so this is identified with the homomorphisms from M into k naught mod pi. And so for any elements x i and k naught mod pi, i greater than running through the positive integers, the sum pi to the 1 over minus 1 over p to the i x i exists in the right-hand side. Because if you multiply this by some epsilon, then this will become a finite sum. But you cannot lift this element somehow in general to an element in k naught. Yes, so this is an example. I mean, you can show that it's in no case projective, I think. Because you show that if p is projective categorically, then p lower shriek will be a projective k naught module, which would then have to be free and so on. But anyway. So at least in this non-sphereically complete case, this shows that k naught, a is not projective. But of course, it's almost projective because somehow you can arbitrarily well approximate any element on the right-hand side. And the other thing I want to give is an example. So for this almost finitely generated business, so say k is the completion of qp and you join lots of p power roots of p. And so p is not 2. And it's my favorite example. And else you join a square root of p to it. And then l naught a is a almost finitely presented k naught a module. And this is done as follows. So for any integer m, we have some map from k naught plus p to the 1 over 2 times p to the m, k naught into l naught. So somehow, if you would invert p, this would become an isomorphism. Because l is just a vector space a direct sum of the two. And integrally, this is not quite an isomorphism. So it's an injective. And the co-kernel is annihilated by this thing. And then somehow this gives you some module which works for any given epsilon. But it's somehow not the case that you can find one module which works for all epsilon. So that's the reason that this definition there is somehow slightly twisted. OK. And we have the following proposition that, again, a is some almost k naught algebra and m is some a module. Then m is almost finitely presented and flat, if and only if it's almost finitely generated and almost projective. So some are given an almost analog of the classical theorem in commutative algebra. And in order to save some chalk, I want to sort of for this lecture series, just call such modules finite projective. Because they somehow, at least for our purposes, just form the correct analog of finite projective modules, except for one slight problem is that in general, such modules do not have a very good theory of ranks. So let me give some additional property under which they do have a nice theory of ranks. So we say that m is uniformly almost finitely generated. If definition above, there exists some n such that an epsilon can be chosen to be generated by at most n elements. So for example, in this case, this is true just because we can take n is equal to 2. And then the proposition and say m is uniformly finite projective if, well, what you guess is true. So it's finite projective plus almost uniformly almost finitely generated. And then there's the proper, let's call it theorem. It's in the book of Gabbard and Tremul, everything that I talked about here in almost mathematics is in this book of Gabbard and Tremul, which says that if m is a uniformly finite projective a module, then there exists a decomposition of a into a product such that each somehow, if you pull back this m now to each of these some more components of a, then this has rank i by which the following is meant. If you take the i's exterior power of mi, then this is invertible. And the i plus first power is 0. And l is invertible if l tender the almost homomorphisms from l to a is just a. So this is supposed to say that for these uniformly finite projective modules, there's a good notion of rank, which one can use. And so recall that we were basically interested in the theory of finite etal algebras over something, so that's what we will come to now. So we will define what an etal algebra is. So again, that a, b, a k0 algebra and b, b and a algebra. Then in particular, we get some meaning, for example, an algebra in the category of almost modules. That's one way to say it. So we get a multiplication map mu to b. And then b is said to be unremifed over a. So in the classical picture, you would say that the diagonal is a closed embedding. And so in particular, there should be some idempotent, which somehow describes this closed embedding. And so in this context, it's just an almost element. So if there is an idempotent over the fiber product, satisfying the expected property. So it's an idempotent under the multiplication map. It's mapped to one. And in some words, if you multiply it with a kernel of mu, then you get 0. And b is a tall over a if unremifed and flat. Yes, so maybe it's saying so. b is a algebra is either an algebra in this category, a modules, or it's a k0 algebra plus a map of k0 a algebras from a to b. Some of the two points of you are equivalent. And so the notion that we are really interested in is the notion of finite data algebra. So we say that b is finite at all over a if at all, and b is a finite projective a module. So that's some very abstract stuff, but we have to go through this once. And so there's an equivalent characterization of this condition of being finite at all in terms of trace maps. So let me recall this. So let a be some k0 algebra, and p be some finite projective a module. Then first of all, there's a dual module, which is pdual, which is a set of almost homomorphisms of a modules from p to a. And then it's true that if you take the dual module twice, then you get again p, so p is reflexive. And it's also true that the endomorphism algebra over p, if you then, I mean, this is again an actual real set, not just something almost. But if you take the associated almost thing, then it's just again given by the tensor product of p and p dual, and in particular there's a trace map, what was the notation, trace of p over a to a. And then the following definition, and b, some a algebra, such that b is a finite projective a module. And now some of you are interested in the question of whether this b now actually finds a finite eta a algebra, so where does this map say eta? And for this we define the trace form between their a, b as the composition of the multiplication map and b maps to the trace of multiplication by b, which defines an endomorphism of. Then one can show that this map is finite eta, if and only if this trace form is a perfect pairing, saying that it induces an isomorphism from b to its dual. Well, so finally we have defined all of these notions in almost commutative algebra. And I recall that we wanted to prove the following series of equivalence of categories that goes on to the other side. And in particular there should be some theorem stating that one can lift such finite eta algebras, and that's given by the following theorem, also in the book of Gabba and Tramiru, which, for example, says the following. If you have a k naught a algebra, so for example, just k naught a, such that a is flat, and pi adequately complete, i.e., i, a isomorphic to the inverse limit of quotients by pi to the n. Then b maps to b tensor a, a mod pi induces an equivalence between the finite eta algebras over a and over its reduction mod pi. And later we will also need that any such algebra b, which is finite eta algebras over a, is again flat and pi adequately complete. So the theorem takes care of this isomorphism here. And sort of what we still have to show then is that this here is an isomorphism, and on the other side as well. OK. Now let me just, so it's now basically finished this part. Let me just give a short preview of next time. So over k, yes, over k naught. I mean, by definition, I mean, that's basically clear because it's flat over a and a is flat over k naught a. So it's also flat over b is, again, flat over k naught a, because it's transitive, but it's not upper or clear that it's also pi adequately complete. So some other picture analyzes to perfectoid k algebras. So the definition is as follows. Again, some on the outer side, we have some classical commutative algebra objects. So perfectoid k algebra is a Banach k algebra r, such that r naught, which is the set of power bounded elements, in r, that this is open and bounded. And the crucial condition that phi is surjective on r naught mod pi. And again, in characteristic p, some of this last condition just says that r is perfect. And some way, we again want to have some change of chain of equivalences, which goes from one side to the other, whereas this first going from generic to slightly generic phi, but then reducing mod pi, then going to the other side, shifting to the other side, and again going to the generic fiber. So we need all these categories in between. And then the perfectoid k naught a algebra is a flat, pi adequately complete k naught a algebra a, such that phi induces an isomorphism between a reduced mod pi to the 1 over p and a mod pi. Yes, but this would not be equivalent to the other categories. So I will just approve you some of the next time. And here's a flat k naught a pi r a algebra a bar, such that this is satisfied. And then we get a chain of equivalences from some perfectoid algebras over k. Let me just write them as k perv. k naught a perv. And then there's some of the strange phenomenon that a perfectoid algebra somewhere lifts uniquely somewhere from mod pi to this whole level. And then we again have just this equality, because these are the same rings. And then going back to the other side, k naught a perv. And sum of any, so if sum of r corresponds to a, corresponds to a bar here, then we also get maps from the finite detail algebras here. And sum of this theorem just stated says that they are also the same as here. And then again, we can sort of go to the other side and flat finite detail, flat finite detail. And sum of what's not so clear is that these two maps here, they are fully faithful, but it's not so clear that they are isomorphisms. And then what's relatively easy is to show that it's in characteristic piates. It's an isomorphism. And then some of, in Falking's almost purity theorem just says that we'll say that this is almost always an isomorphism. It's an equivalence. And that's somehow what we will then prove in a long series of reductions by finally reducing it to the case of algebraic closed fields that was handled today. And so that's it for today.