 Let me recall what we did last time. So we consider it so-called perfectoid fields. And let me recall that a perfectoid field is a topological field k whose topology is induced by a non-discreet valuation of rank one. Such that the residue field is finite of characteristic p. And such that the Frobenius is surjective on k naught mod pi. So let me recall that k naught was the power-bonnet elements, which is a set of all x, absolute value at most one, using this rank one valuation. And in there we have the maximal ideal, which is also a set of topologically new potent elements. And it's also the set of all x of absolute value less than one. And last time we constructed a function called the tilting function from any such perfectoid field of characteristic p. And note that in characteristic p, the condition is basically that this complete non-archimedean field be perfect. And so some of the theory is much simpler over this characteristic p field. And we are trying to compare the two somehow. And so let me recall the description of k flat. So it's inverse limit over the piece power map r, for example, of k. And somehow if you consider truncated versions of the power-bonnet elements, then they become isomorphic. So pi and pi flat were chosen to be somewhat corresponding uniformizes. And so this was summarized somewhat in the following picture. So here's the spec k0 with its generic point and its special point and some infinitesimal sickening, which is the spectrum of this reduced thing. And then we have the tilt, which somehow starts the same. And then somehow goes on differently in the spectrum of k flat not. And last time I stated the following theorem that there is the tilting function l maps to a flat induces an equivalence between the finite etal extensions of k and of k flat. So that, in particular, the absolute gather groups of two fields will be canonically isomorphic. And implicit here is the statement that any finite extension of k will again be perfectoid. And the plan of proof for this was to use some almost mathematics. So recall that we defined the category of almost k0 modules as k0 a modules, which was the category of k0 modules, where you divide out by the six subcategory of m torsion objects. And this was possible because we are in a non-discrete valuation and hence the square of the maximal ideal is equal to itself. And then we defined notions of k0 a algebra and so on and so forth. And the refined version of the theorem is then that we have the following chain of equivalence of categories that somehow these finite etal extensions extend almost to the integral level. And some of them there's a lifting theorem stating that one can lift finite etal extensions over such, over a new potent elements. And then we can go to the other side because of this isomorphism of the truncated rings. And so part of this long equivalence of categories will generalize to a more general setup, which is the setup of perfectoid algebras. And these I will introduce now. So first over k, so a perfectoid k algebra is a Banach k algebra R. And then the first condition is of a topological nature saying that the ring of power bounded elements in R, it's always open but the condition is that it's bounded. And then comes sort of the perfectoid condition which says that reducing mod pi is surjective. As the Frobenius mod pi is surjective. And so let me remark that sort of in the classical setup of which a geometry will also consider Banach algebras. And this condition that the suffering of power bounded elements is open and bounded is precisely true for reduced algebras. But this condition will basically never be the case. So the second is then over the almost integral level. So there it's a flat and pyridically complete k not a algebra. A such that, again, while you look pi, the Frobenius should be surjective. But now we additionally require that the kernel is generated by the piece root of pi. So it was possible to somewhat choose these uniformisers to be equipped with a canonical system of p to the n roots. And the last point, also the reduced version, perfectoid k not a mod pi algebra is a flat k not a mod pi algebra a bar. And then we, again, put this condition about Frobenius, such that Frobenius uses such a nice morphism. OK? And then we have the following series of equivalents of categories. So let me denote by k perv, the category of perfectoid k algebras. And then we have k perv is equivalent to k not a perv. Then it's equivalent to k not a mod pi perv. And once we have this, then this will be just the same as on the other side again. And then we can live to the other side. So the proof splits into two parts. So first we have to prove this equivalent and then this. And at the first step, let me explain what some of the conditions imposed on such a perfectoid k not a algebra mean. So we put the condition as flat and pi at t complete. And so let me explain what this means in classical commutative algebra. So let m be a k not module, almost module. And recall that then we had to find this module m lower star, which was the homomorphisms in this almost category from k not a to m. And so this is a k not module. In fact, if you take the associate with almost module, we get back m. So the first part describes the condition when m is flat. So m is flat if and only if m lower star is flat, if and only if m lower star has no pi torsion. And let me recall that this thing here is called the module of almost elements of m. And for the following reasons, it's a good name. So assume m is the almost module associated to n. And n is flat, so flat k not module. Then the almost elements of m are described in the following It are the elements x and n revert pi, such that for all epsilon and m, the epsilon multiple really less than n. So some of it are not really elements of n itself, but as soon as you multiply them by something, in the maximum ideal, they become elements. So in this sense, almost elements. And so this basically explains the flatness condition. Now we want to explain what it means to be periodically complete. And there we have the following. So assume that m is flat. Then for all x and k not, one can compute what x m times lower star is. And it's just x and lower star, so that's OK. And secondly, one can wonder what m or x m lower star is. And there it will turn out that it does contain this thing. But in general, there are more elements. And I will give an example in a second. But it's not so bad, because for all epsilon and m, the image of m mod x epsilon m in m mod x m lower star is just this sub-module. And the last point is that, again, if m is flat, then m is periodically complete, if and only if m lower star is periodically complete. So in particular, the condition that this k not, almost algebra a, is flat and periodically complete can be rephrased as saying that a lower star, which is an actual k not module is periodically complete and flat. OK? And so let me first give you a remark about point three. So recall that, for example, if you want to compute k not mod pi lower star, then there was a general formula saying k not a mod pi. Lower star is a general formula saying that this is a set of morphisms from the maximal ideal into k not mod pi. So more generally, if you have any module, take its associated almost module and its almost elements, then this is computed by this formula. It takes more morphisms from the maximal ideal into there. And so this contains, as already mentioned last time, elements of the form sum of pi to the 1 over p to the i x i, where the x i are any elements in k not mod pi. Because as soon as you multiply by some small epsilon, this will just be a finite sum, which will then make sense. And so the problem is that these elements do not lie in k not mod pi, which is sub set of this in general. Cannot lift them to elements of k not. OK. But also we see that somehow these elements will not lift any further, because somehow there is this accumulation point of the exponents here. And they will not lift any further. And that's some of the reasons that here in 3 we get this last statement. And so let me give a proof of this. So for 1, first we check that if m lower star is flat, then m is flat. That's easy, because we call that m is flat means, by definition, that for all x k not modules and i bigger than 0, that's the tau i of x and m lower star is almost 0. But of course, if m lower star is flat, then this thing will be 0. That sort of this implication is clear. Then in the second implication is that if m is flat, then m lower star has no pi torsion. How do we do this? So first note that, again, using this formula for the almost elements, it's a set of homomorphisms from m into m lower star. And this description implies that this thing has no almost 0 elements, because m squared is equal to m again. So then we use that as a particular case of this condition, we use that the tau 1 of m lower star was k not mod pi, that this thing is almost 0. And this just translates into the condition that the kernel of multiplication by pi on m lower star is almost 0. But again, the kernel can have no almost 0 elements, because of this, and hence it follows that this thing is 0. And so which means that m lower star has no pi torsion. And then there is the last step saying that m lower star having no pi torsion implies m lower star flat. But of course, this is just a statement in standard commutative algebra and standard effect. So now for the second statement, so again, we use this formula that m lower star can be computed as the set of homomorphisms from the maximal ideal into n. And because n is flat, this embeds into homomorphisms from k into n invert pi. And how do we describe these homomorphisms that lie in the image? So homomorphisms are given by the element of 1, which is an element in n invert pi. And the condition is that as soon as you multiply by some epsilon in m, you have to get an element of n. But that's just what we claimed in the third point, if m is flat. Exactly, so we write x m lower star. We can write it as the set of elements in y in m lower star n invert pi, such that for all epsilon in m, some epsilon x times m lower star. And this can be rewritten as the set of all x times the set of all y in m lower star pi inverse such that for epsilon m epsilon times y lies in m lower star. But using the dot we proved in part two for the special case of n equal to m lower star, this is just x times m lower star. And because m maps to m lower star is left exact because it's right adjoined to something, we now get that m lower star module x m lower star lies in m module x m lower star. And so last assertion is left as an exercise. And for the last point, part four, we note first that both functors m maps to m lower star and n mapping to n almost at mid-left adjoins, giving the second case by the funcars that I have not yet introduced, which is m maps to m lower star, which is m lower shriek, which is m tensor m lower star. And hence commute with inverse limits. And so in one direction, we then get that if m is pi equal complete, then m lower star is the inverse limit of m mod pi to the n lower star. This commutes with inverse limit, so it's the inverse limit of m mod pi to the n lower star. And now we want to see that this is also the same as the inverse limit of m lower star mod pi to the n. So a priori would think that this might be only an inclusion because we only have this inclusion, but we require that somehow we have elements in this inverse limit. And so everything that occurs in this inverse limit has to lift to something larger. And then we use this statement here saying that then it's actualized in this sub-thing. And for the other direction, it's even simpler somehow. And so now we can prove some propositions. The first is that if we have a perfectoid k algebra, then phi induces, in fact, an isomorphism between r not mod pi reduced mod pi to the 1 over p, r not mod pi. So remember that we only required this phi to be surjective, but in all other rings, we required that we have such an isomorphism, and here it's automatic. And also, if one considers the almost algebra associated to r not a, then this is a perfectoid k not a algebra. And for the proof, just note that if x is an element of r not such where x is an r and r not such that its p's power is an r not, then it follows that such as this thing is power bounded, then obviously x divided by pi to the 1 over p is also power bounded, which gives the first statement. And for the other note that we know that r not is flat and pyridically complete by the condition that it's open and bounded, and hence by the previous lemma we say that the almost algebra is also flat and pyridically complete, and sort of the last condition also translates into the almost setup. And now we want to, so we can now pass from perfectoid k algebras to perfectoid k not almost algebras, and now we want to go the other way. That's the following lemma, so let a be a perfectoid k not a algebra, and let r be the algebra you get by inverting pi, and we want to see that this gives you a perfectoid k algebra. So first we give it a topology, so with topology making a lower star open and bounded. And then in fact, the power bounded elements of this thing will just be a lower star, and so in particular we see that r, we can reconstruct a by taking the almost algebra associated to r not, which means that the two functions will be inverse. And this also, then, by what's already on the blackboard implies that we get an isomorphism between a lower star mod pi to the 1 over p and a lower star mod pi, which might be surprising at first sight, because a priori we only required it to be an almost isomorphism, but it turns out that it's an isomorphism automatically, and r is perfectoid. And so let's prove this. So first we show that phi is injective. So first we know that phi, this map, is an almost isomorphism by assumption, meaning that the kernel and co-kernel are almost zero. And so the first step is to show that phi is, in fact, not only almost injective, but injective. So what does this mean? So we have some x in a lower star, such that x to the p is pi a lower star. And we want to prove that, in fact, x lies already in pi to the 1 over p a lower star. But almost injectivity says that this is at least true after multiplying by some epsilon. So for all epsilon and m, we see that epsilon times x has to be an element in pi to the 1 over p times a lower star. But this means that this x is almost an element of pi 1 over p times a. So it's in this set of almost elements. But then the last lemma tells us that this is just the same as what we have. So we get to the desired conclusion. And then next we need the following small lemma that if x is some element of r, such that its piece power lies in a lower star, then already x lies in a lower star. And for the proof, there exists some integer k, such that pi to the k over p times x lies in a lower star by definition of r. And if k is at least 1, then in particular we see that y to the p is pi a lower star. And by the injectivity that we have already proved, it follows that y lies in pi to the 1 over p times a lower star, which just says that also pi to the k minus 1 over p times x still lies in a lower star. And then we do induction to get the claim. And so next we prove that the power-bounded elements are just a lower star. And for this note that it's clear that a lower star is contained in the power-bounded elements, because all the powers of an element of a star will lie in a star. And a star is by definition a bounded subset of r. And for the other direction, assume that x be power-bounded. And then for all epsilon in the maximal ideal, epsilon times x is topologically new potent. And in particular, there exists some big n, such that epsilon times x to the p to the n will be in a lower star for n large enough. And this implies that epsilon times x is a lower star by the previous lemma. And hence, x again defines an almost element of a lower star. And hence, x itself, by the description of almost elements, is an a lower star. And so now we have to prove that we have the following equivalence. So in other words, we have to see that the a maps to a bar, which is a tensor for a mod pi. And so we have to see that a bar admits unique deformations. And for this, we will use a theory of the cotangent complex, the foundations of which are laid out in the Caesars of Euler-Z. And let me recall a little bit about this cotangent complex. So first in the classical commutative algebra setup somehow. So to any ring homomorphism, R to S, we associate the so-called cotangent complex, which is an element. You can consider it as an element in the derived category of S modules, whose homologies concentrated in degrees and non-positive degrees. And some properties of this are that first of the H, I guess it's more standard to use homological notation, H0 of this thing is just the differential of S over R. And for any R, S, T, one gets a triangle in the derived category of T modules connecting, now I have to get this right. L S over R pulled back from S to T, goes to the cotangent complex from T over R, goes to the cotangent complex of T over S. And so some of this extending the short exact sequence that one knows, somewhere from just omega 1. And let me say one word about the construction of this thing. So construction runs as follows. We will also some more needses. So first you use the dot-con equivalence saying that this derived category of S modules with this boundedness restriction is equivalent to the category of simplificial S modules, module of weak equivalence. And so instead of this thing in the derived category, we will construct some simplificial S module. And so the way we do this is to choose some simplificial resolution of S by 3R algebra. So meaning just a polynomial algebra and possibly infinitely many variables. And there's even a canonical way of doing this, if you like. And then once some are for such a polynomial algebra, it's easy to define what the associated cotangent complex should be. It should be just omega 1. And so one extrapolates from this case. So one sets L S over R to be the object in associated to omega 1 of this simplificial algebra, which then gives you a simplificial S dot module. And then you're getting this right. Yes. And then you change it over this S. And so why are we interested in this? It's because of the following theorem. So in deformation theory. So let R be some algebra, some ring. And I and R are an ideal, such that I squared is 0. And so set R0 to be R mod I. And let S be a flat R0 algebra. And we want to lift S to a flat R algebra. And we are asking, what's the obstruction to doing this is? And what are the possible ways of doing this? And that's answered by the following theorem, which says that there is an obstruction class in x2 of the cotangent complex of S0 over R0 with values in the ideal tensored up to S0. And the possible lifts are a torsor under x1 of the very same objects. And the automorphisms of one lift are x0 of the same objects. So in particular, for example, if the cotangent complex vanishes, then there is a unique lift. And it's sort of, there's no ambiguity whatsoever. And the second deformation result that we will need is that if we already have S and S prime to flat R algebras, and we are given a homomorphism from their reductions. So all the reductions to R0. And we are given a homomorphism from S0 to S0 prime, morphism of R0 algebras. And we want to lift this to a homomorphism from S2 S prime. And then there's the following that there is an obstruction class in x1 of the cotangent complex of S0 over R0 with values in S0 prime, tends to R0 i. And the possible lifts form a torsor under the group x0 of the same thing. So again, we see that if the cotangent complex vanishes, then also homomorphisms lift uniquely. And so it will be interesting for us to have a criterion for when the cotangent complex vanishes is the following very nice proposition that also appears. So I learned this from this book of Gabbard and Tramero. And I don't know whether this was observed before. So there's a following very easy criterion. So if R is the perfect Fp algebra, then the cotangent complex of R over Fp vanishes. So we will need a slightly enhanced version of this. So that R2 S be a morphism of Fp algebras. That R lower phi be the R algebra R via the homomorphism via the Frobenius. And in the same way, we define S phi. And we assume that the relative Frobenius 5S over R is an isomorphism even in the derived category between this and this. So this isomorphism in the derived sense just means that without the L, it's an isomorphism and all higher tor terms vanish. And then, again, the cotangent complex vanishes. And so for example, the first statement can be used to say that the ring of fit vectors of R adjusts a unique deformation of R to a flat and periodically complete Zp algebra, which is somehow much closer to how I think about the width vectors than the usual definition. And also, it's clear that the omega 1 of R over Fp is 0. Why is this? So if x and R and x is y to the p, then dx is equal to dy to the p. And that's p times something. And this is 0. And that's somehow the crucial identity behind this proposition here. And so we just have to go for a slightly more enhanced argument. And so this goes as follows. So we choose the simplification resolution. And so we also have sort of the relative Frobenius over R, which is a map from R phi tensor R S dot to S dot phi. And the assumption just says that this is a weak equivalent, or quasi-Somorphism, or whatever you call it. And so we have two resolutions of S phi by a free, simplificial S phi R phi algebra. Some of both sides of this give you such a resolution. And so it follows that they both compute the same cotangent complex, which means that R phi tensor R of the cotangent complex of S dot over R goes isomorphically to the cotangent complex of S dot S phi over R phi. But on the other hand, we can explicitly compute what does this thing here do on the differentials. So somehow if one part of this simplificial algebra is just some infinite polynomial algebra in general, then phi S k over R on differentials maps dx i to dx i to the p, which is again 0. So it follows that this very same map, which is a quasi-isomorphism, is also the 0 map. And in particular, both sides of this isomorphism have to vanish. And this is exactly what we wanted to prove. But that's somehow the same thing just considered from a different point of view as the cotangent complex of S over R. Now we have to use this vanishing of the cotangent complex. But first we have to sort of adapt the theory to the almost context. And this was also done in the book of Gabber and Tramiro. So they show the following that if R and S are k0 algebras, then if you consider the cotangent complex of S over R, but consider it in the almost way, then again, this is an object of derived category of S almost modules. And they show that this depends only on the morphism from R a to S a of k0 almost algebras. And so you get La a over b in the derived category of a modules for any morphism a to b of k0 almost algebras in this way. And in fact, Gabber and Tramiro define a slightly refined version of this cotangent complex, which actually is an actual module, somewhere over an actual ring. But we will only need this almost version. And then the previous theorems state true, theorems about deformations state true in the almost world. And we will then need sort of the following corollary of this vanishing of the cotangent complex, which is that if a bar is a perfectoid k0 a1 pi algebra, then the cotangent complex of a bar over this thing is 0. And one way to reduce it to the case that was already handled is sort of it's enough to represent a bar by a flat k0 pi algebra R such that Frobenius is an isomorphism from R pi to 1 over p was R. And again, because of these strange elements which appear in the, so someone could try this, but this does not work. But in this book, there's a different way of representing a normal string by an actual ring. So it's what they call a double lower shriek, which is you take a lower shriek, which was, I recall, maximal ideal tensor a1 star. This will, in general, be flat already, so that's better. So this thing might not even be flat. But it's not an algebra anymore of a k0 because it does not contain the identity. So we artificially sort of have to add the identity and sort of mod out m. And so this works. Don't want to get into the details here. And I guess one could also just do the same proof somehow. Or is it possible? Which is not that? Yes? OK. Well, somehow, if it's not a zero thing, then it will be facefully flat here somehow, so I guess. Well, maybe I should check again, but somehow. OK. And so as the corollary, we now get that the categories of k0a algebras, which provides us correctly last time, the perfectoid k0a algebras and the perfectoid k0a mod pi, algebras are equivalent. So that's for the proof. So we need to prove inductively that a bar lifts uniquely to a flat k0a mod pi to the n algebra, whatever I want to say, a bar n. And then a will just be the inverse limit of the a bar n. And to see this, we have to see that somehow, at each step, the cotangent complex will vanish. I want to add the almost thing here. And so we have an exact sequence. What does this go? Some easy verification, somehow that this lifts l a, a bar n over a mod pi to the n. And so this is 0, this is 0, and so this is also 0. OK. So finally, we get some of the equivalents of all of these categories. In particular, we get from perfectoid k algebras to perfectoid k flat algebras. But some of the way we did it was completely inexplicit. And it's nice to know what this equivalence looks like after all. So we want to compare this with this front-end construction. So let R be a perfectoid k algebra. And A be the associated k0almost algebra. And then we define A flat to be the inverse limit over Fabinius of A mod pi. And this is a k0, k flat, not A algebra, via k flat not A as the inverse limit over Fabinius of k not mod pi. And then we let R flat be the associated thing over k flat. And then we have the following proposition, that R flat is really the tilt of R. And we have some more properties somehow. R flat can also be described just as a set, as the inverse limit over the piece power map on R. And then also the ring of power bounded elements in the same way, A r star. In particular, one sees that the construction is independent of pi because the way we proved the equivalence we somehow implicitly chose some pi. But for example, one can see from this description that it doesn't matter. And somehow, again, we have the property that after reducing mod, the uniformizer, the two things come the same. And in particular, we get a continuous multiplicative map from R flat to R, which again is noted x maps to x sharp. And so the key point is just that somehow we have the series of equivalence of categories. k flat mod A mod pi to 12. And then we again lift to the other direction. And somehow, if we start with this R, then this maps to A and this maps to A bar, which is A mod pi. So that's all easy. And some of the key point is that this lifting here on characteristic p from the stuff mod pi flat to the stuff on the integral level, which we somehow did with this vanishing of the co-changing complex and which was completely inexplicit, can somehow, in this case, be made explicit just by taking the inverse limit of the fulvinous. Reasonably easy. Well, all the other verifications are basically standard. So maybe I should say how to get this map from R flat to R. So it's easy to see this. And then we have the projection map again here from the inverse limit over the piece power map. And in the same way that this was proved for fields, one checks that again there's an isomorphism here. And this then gives the desired map. It extends somewhere to the rational level. So one might wonder whether that's some or some or no, we have described the explicit function going from characteristic 0 to characteristic p. And one might wonder whether it's possible to go the other way and then you did this. So by some, the commerce function is given the following way. So some commerce function S maps to S sharp is S sharp. So it takes the ring of fit vectors of the power bounded elements in your characteristic p-ring. This becomes an algebra over the rate of fit vectors of k flat-node. And then in PAD code theory, you define this map theta from this ring of fit vectors to k. And then you tensile along this map theta. And so when we're drawing a picture, this would mean that we recall this picture somehow. And so we somehow went this way. And we showed that one can sort of invert both arrows. But this explicit function now goes somehow over this direction. This is the explicit function. So it does not go with a special point, but there is this whole surface somewhere in between which is some of the ring of fit vectors or a spectrum of the ring of fit vectors of this characteristic p-ring. And so you first extend over this whole surface and then specialize to a different point that somehow the explicit functions go as a circle. And it's possible to give a direct proof without using any almost mathematics that one gets an equivalence between these categories of perfected algebra, which is done by Kedler and here. So some of the independently came up with pretty much related ideas. And after I told them that there is this equivalence, I sort of saw that it's immediate to deduce it from their results. So let me give an example now. We have not seen any perfected algebra until now. So maybe that's one can, for example, take the following algebra. You take convergent power series in a certain number of variables, but you join all of the p-power roots of the variables as well. So by this ring, I mean you take the polynomial algebra over the ring of integers, complete it periodically and then invert pi. So this is perfectoid. And it still looks just the same. So still all flat. Very easy proof of this is to observe that first of the power-bounded elements are really just the PAD completion of this thing. And that phi is subjective on R0 mod pi. So we see that it's indeed a perfectoid algebra. And to see that the two are tilt of each other, it's enough to see that R0 mod pi is the same as R flat mod pi flat. So by how we proved the tilting equivalence, it's enough to show this isomorphism that is clear. Because it's just the same thing. And of course, you could sort of do the same example for any toric, for any monoids or for toric varieties. And finally, we want to use all of the stuff to finally prove this equivalence of Galois groups on the two sides. So we need to prove something about fields again. And that's the following lemma. So that if R is a perfectoid, k-algebra was tilt R flat. And somehow R flat is a topological algebra. And it makes sense to ask whether it's a perfectoid field. And that's precisely the case when R is a perfectoid field. And so proof runs as follows. So the rank voting valuation is necessarily given by equal to the spectral norm, which in our case can be defined as XR as the supremum overall, no the infimum, overall T inverse, where T is some unit such that T times X is an R0. And it's easy to see that spectral norm on the tilted side can be computed by taking the spectral norm of this sharp representative. And it's also easy to see that R is a perfectoid field if and only if this is multiplicative and R is a field. And because R flat is the inverse limit of R, it follows that if R is a field, then R flat is a field. And somehow it's also true that if this is multiplicative, then by this formula, this thing is multiplicative. So one gets one direction, and the other direction is also not hard. But so I leave it as an exercise. And so now we need to discuss the lifting of finite detail covers. So somehow how do finite detail covers? You want to show that if you have any perfectoid algebra in any of these settings and something finite at all over it, then again, it is perfectoid. And the starting point is the following proposition. Again, it's very nice everything that I need is already in the book, that if A bar is a perfectoid, K naught A mod pi algebra, and B bar is finite detail over A, then B bar again is a perfectoid. And for the proof, so that B bar flat over K naught A mod pi is clear, because it's flat over A bar and A bar is flat over this. So the hard part is to show that the following is true. So what's the following theorem? That we have this, and we have the Frobenius map here, and we have this thing, and this is a pullback diagram. So some other classical commutative algebra that might be well known, but some also extends to the almost setting. And once one has this, one easily sees that indeed the Frobenius does the expected thing. And so we get the following diagram. So we have some R corresponding to A corresponding to A bar, which is then the same as A bar flat. And then we lift this to some A flat, and then we lift to R flat. So let these be fixed somehow. And then we have all of these equivalents of categories. And so in the middle, we have some A bar finite at all, A bar flat finite at all. And we have just seen that there are these functions. And recall that last time I stated the theorem to the effect that one can uniquely lift finite at all covers over such algebras. So this is somewhere from last time. So in particular, we also get this function here. And so on also on the other side. So we have A finite at all. You'll have this function. And then these, of course, map to by inverting the uniformizer to finite at all algebras on the generic fiber. The first observation is that this diagram automatically implies the proposition that the outer upper arrows are fully faithful, because this here is an equivalence. And we get the same algebras here. So we have these inclusions. And later we will prove that both are equivalences. So this is known as faulting's almost purity theorem. I mean, so if I have B1, B2 in A finite at all say, then I'm interested in the set of homomorphisms A from B1 to B2. And so that's by the equivalence of categories. Yes, fully faithful. Yes. So if I have two such algebras which are finite at all over A, and I know that then they are also perfectoid. And so because for perfectoid algebras, I know that going to the generic fiber is fully faithful. I know that we have this equality. And this is just what we wanted to prove. And what we will do first is to prove this almost purity theorem somehow in characteristic p, which is much easier than in characteristic 0. So that's the following proposition. And again, it's already in the book of Gabel and Ramero. So that k be of characteristic p. So in order not to write flat everywhere, I sort of assume that from the start I'm in characteristic p and let R be a perfectoid k algebra and S finite at all over R. Then because S is a finite projective R module, it has a canonical topology. And then it makes sense to say that S is perfectoid. And on the almost integral level, it's still finite at all. So this means that this function here is fully faithful, essentially subjective. And there's the following extra thing that is useful, is that in fact, this thing is a uniformly finite projective R0A module. So in particular, there's a good notion of the rank of this extension. And so there's a good notion of speaking of the degree of some finite exhaled extension here. And so let me give the proof. So S, obviously, is a perfect k-banach algebra. But it does not yet mean that it's perfectoid because we have to show that the set of power-bounded elements is bounded. And so how do we do this? So let S0 in S be a finitely finite R0 subalgebra. So it's possible to choose one such that let S0 bot in S be the set of all x in S such that trace S over R x, say, is S0, S y is an R0. So here we use the trace form of S over R, which is a map from S to R. Last time introduced in the context of almost algebra, and of course, it exists also in usual commutative algebra. And so this is a perfect pairing. And then it's easy to see that S0 and S0 bot are open and bounded. And we claim that the power-bounded elements sit somewhere in between. So we let, what was my name for it, y is the integral closure of R0 in S. And then first of the trace form pairing on y takes values in R0. And it's also clear because this is a finite algebra that all elements in there are integral. And by this property, it follows that it's somewhere squeezed in between S0 and S0 bot. And so it follows that y is open and bounded. And but somehow using the arguments from the beginning, it's easy to see that the integral closure of y is almost the same as the power-bounded elements because if you have a power-bounded element and multiply it with some small epsilon, then it becomes topologically new potent, intense integral over this thing. And so it follows that S0 is open and bounded. So we know that this algebra S is perfectoid now. And now we have to see that there exists some n such that for all epsilon and m, there are maps to R0 to the n. And the other way around was composite equal multiplication by epsilon. For this, we'll imply then that S0 is a uniformly finite projective R0 module, S0A, which is part of being some of the assertion. And how do we construct this? So now it becomes a crucial argument. So let E and S10 or RS is the idempotent, showing that S over R is unremifed. And then there exists some big n such that pi to the n times E is an S0, 10 to the r0, S0. But because for being this projective, it follows that for all epsilon and m, in fact, epsilon times E is in this thing. So somehow, a priori, we only know that it's true that up to some power that it is unremifed, but then we can lift it almost to the integral level because for being this projective, it's really the crucial argument. And we write epsilon times E, some of the sum of xi tends to yi, where these xi and yi are in S0. And no, no, S0. So I'm not passing to almost stuff. So we have these. And if you think about it for a second, then you see that you can choose this n somehow uniform for all epsilon, somehow by taking one expression here and then taking just the piece roots out of it. And then we get maps from S0 to r0 to the n sending S to the tuple y1. And we get a map in the other direction sending r1 up to rn to the sum of ari, ri. And it's easy to see that they're composite multiplication by epsilon. So this way we see that S0a is such a nice module over r0a. And what's left to see is that S0a over r0a is unremifed, but we have already seen that E defines the required almost item potent somehow. So here we are. Yes, sorry. I shouldn't change notation from my notes. And so it follows that the diagram here extends here to an equivalent. And in particular it follows that there's also this arrow here. Or we have just proved that they are all perfectoid. And also another to extract at this step because to have the Frobenius. So somehow, once you have the Frobenius, somehow really not more p, but really on the ring itself, sort of you can really make this argument that if something is true up to some pi power, then it's really almost true already. This argument is much more difficult to make a perfectoid stuff in characteristic 0. And usually one has to work much harder to prove the same theorems. But somehow amazingly they all stay true. Yes, you can sort of, it's easy to make this explicit what it means. And I guess it's also a little bit stronger because you didn't assume that it's perfect. But I'll just say that it's complete. Good. It also follows that somehow in this picture all finite etal algebras are uniformly finite projectives in this picture. So it makes all settings sense to talk about the degree of some finite etal extension. And so as a final theorem for today, I want to finish the proof of this equivalence of Galois groups, which is to say that the only step left is to show that this very leftmost arrow there is an equivalent. And maybe first I collect something. So I want to collect sort of what we proved for perfectoid algebras in characteristics 0. So now assume that then there is a fully faceful function from the finite etal algebras of what's tilt to finite etal algebras over itself, which is inverse to the tilting function. And the essential image are those S and R finite etal such that S is perfectoid and S0A in this extents to the almost integral level. And somewhere in this case, if this is true, it's also true that S0A is uniformly finite projective over R0A. And then the final theorem is that it's a perfectoid field, then this function is in fact an equivalent. And for the proof, so we can assume that R is k. So we can just rename our k as this perfectoid field R. And so it has a tilt k flat. And so the idea is now that if k is algebraically closed, then we already proved this last time. And so we reduced to this algebraically closed case. So consider m, which is the algebraic closure of k flat hat. And then m is a perfectoid field. So it's perfect and complete. That's enough to see. And so in particular, it's a perfectoid k flat algebra. And so we can untilt it. We get an untilt m-sharp, which lives over k. And this is again a perfectoid field by the proposition that somehow tilting identifies the perfectoid fields. And it's algebraically closed by what we proved last time. And so for any finite extension L of k flat, so L is contained somewhat in an m, we can go in the other direction, get something over k. And we let n be the union over all L of these L-sharp things. So it's contained in m-sharp. And it's easy to see that this is a dense subfield. So because you know that modulo pi, some of the tilt, and this ring itself are the same. So modulo pi, you know that these are basically the same, and then it follows that they are dense. And then if some of such a field has algebraically closed completion, then it follows by Krasnar's lemma, that n itself is already algebraically closed. So in this setup, some are algebraically and separately closed are the same. And so what we get is that any finite extension F of k is contained in L-sharp, where in some L-sharp, where we can assume that L over k-sharp is some finite Galois extension. But tilting preserves automorphisms and degrees. So that's somehow the reason that I paid attention to this uniformly stuff, so to see that it really preserves the degrees. So it follows that L-sharp is again Galois. And so this F contains some L-sharp, this F contained in L-sharp corresponds to some subgroup of the Galois group of L-sharp over k, which is the same as the Galois group of L over k-flat. And so this corresponds to some F-flat over k-flat. And then we really have that this thing will give rise to some because we have the equivalent of categories. OK, and here we are. So we have proved this almost purity theorem for fields. And of course, I mean, it was a little bit, it would be easy, one could get easier proofs for this, but this somehow comes naturally also out of the theory. OK, that's it for today. Yes, so somehow, because you have this equivalent of categories, it's true that F-flat-sharp will be fixed by the subgroup. So it's contained in F, which are the invariants. And because they have the same degree again, it follows that they are equal. Yes, yes, yes, yes. So somehow you get this finite subgroup. This also exists on the other side because you have this equivalent of categories. And this, again, by Galois theory, gives you some finite extension. And if you untilt it, and by the equivalent of categories, the elements in there will be fixed under the subgroup. So it must be contained in F, but they have the same degree. And hence they are equal. Yes, yes, I need that to feel the sensilium, but it is, so somehow, a union of complete extensions.