 So, let me start. So, I continue talking about perfectoid spaces, and again, let me give some short reminder. So, again, fix some perfectoid field. So, something like Qp, join all p-par roots of p to it and complete. And, of course, then we have this field I call k-flat, just given by the Fountain construction, taking the inverse limit, that was a p-spar map of k. In this case, this would be a similar field of our characteristic p. So, it takes a long series field, join all p-par roots of the uniformizer and complete. And so, in general, we have some uniformizer here, which would be something like t and some uniformizer on the other side, which would, in this case, be t. And then we had the perfectoid k-algebras, Banach k-algebra r, such that the power-bounded elements are open, are bounded. And the crucial condition is that the Frobenius is surjective, modulo pi. And then we had proved the tilting equivalents that the categories are equivalent. So, perfectoid k-algebras were equivalent to perfectoid k-flat-algebras. So, also this k-flat is perfectoid field and the same definition applies. And this just sends any such perfectoid algebra to its so-called tilt, which is r-flat, and it's just, again, the Fountain construction of taking the inverse limit over the piece-power map on r. And in particular, this gives us a map from r-til, r-flat to r, which is multiplicative, but not additive, which sends g just to something I call g-sharp, so it's just a projection to the first coordinate. That's automatic because of the, it's the power-bounded elements. So, if the piece-power is divisible by pi, then the thing itself will be divisible by pi to 1 over p, because you're considering the problem with elements. Right. Okay, so we have this, and then we went on and wanted to define spaces associated to such algebras, and the basic object that one associates spaces to in the SUBA theory was so-called affinoid algebras, and then we defined a perfectoid affinoid algebra is a pair rr-plus, where r is perfectoid, and r-plus is some open and integrally closed top ring of the power-bounded elements, and in most cases, in many cases, it's really just a set of power-bounded elements. And then also, such rr-plus, the category of such is also equivalent to the category of r-flat, r-flat-plus, where one way to get this r-flat-plus is again to just take this inverse limit over the piece-power limit. And then we looked at the spaces associated to them, so the so-called edict spectrum, which consisted of continuous variations on r, which are at most one on the sub-ring of sort of integral elements, and we sort of have the same thing on the other side. And then there is a map from here to here, sending x to x-flat, which is defined by requiring that if you evaluate this variation at this point, so this x-flat is a variation, which I denote in this way. So it should send any element of r-flat to some absolute value, and you just take the absolute value of this sharp represent, the sharp element, which will be an r, and then you can apply this variation here. And what we proved last time is that this, in fact, is in homomorphism, which identifies rational subsets. So the rational subsets were a certain natural basis for the topology. And there were certain pre-sheeps defined on this topological space, and we showed that for all rational subset u, which correspond to then under this homomorphism to the subset of the tilted space, it's true that this O x of u, O x plus of u is, again, perfectoid of phenoid, and its tilt is just given by evaluating the sheeps on the other side, the pre-sheeps. So some of the situation is that starting from any perfectoid algebra with this additional datum, we get a space which is the same on both sides, and in fact, we get pre-sheeps, which are also related under this procedure of tilting. And the first thing I want to prove today is that, in fact, this O x, and hence also O x plus is a sheaf, and that's a higher chromology of x vanishes, but even better, that the higher chromology of this integral sub-sheaf is m total, so it's almost zero, basically, for i bigger than zero. So as I did not insist that this r plus pk naught algebra, it does not, formally does not make sense to say almost zero at this point, because it's not a k-mode, not a module, but it's a very minor point. Okay? And so let me first do some preliminary, some first reductions. So the statement is equivalent to the following set for all covers by, finally, many rational subsets, y and x. The sequence O x, x, that's some perfectoid algebra, and then we can look at it at the almost integral level. So this is basically the same as O x plus of x, a. So this plus ring is almost the same as this ring of far-bonded elements, that's true in general, and then we can do the same thing here, and then we take the product. These are again rational subsets, and then so on. So this long exact sequence, this long complex, that this is a exact sequence of k-node a-modules. So that's some general nonsense, but because these O x of u, these algebras at the almost integral level, they are flat and periodically complete, and this is also equivalent that the same sequence is exact after reducing mod pi. Okay? But now, because we have proved that under the still thing somewhere, we get the structure sheet on the other side, this here is really just the same thing for the other side, modulo pi flat, and this is then really just the product of the O x flat of u i flat, 0 a modulo pi flat, and so on. And so it's enough to prove that this sequence, which now is in characteristic p, is exact, and so it follows that it's enough to do it in characteristic p. And so we have to prove that these pre-sheeps was vanishing in coromology, and so we do this by reduction, similar to the statements that are known in classical rigid geometry. So let me recall this, that S S plus ba reduced a finite k-algebra of topological finite type. So recall that this means that there exists some n such that this is a quotient, also in the topological sense, and that this S plus is really just the suffering of power bonded elements. So this thing will never be a perfect algebra, except in very trivial cases. But then we have the following results. So we have the theorem, which basically states a cyclicity theorem, and in full details it says the following. So first is that, in fact, because we assume that this algebra was reduced, this subset of power bonded elements really is just, it defines its topology, so it's open and bounded. And secondly, if u and x, which is bar S S plus, is a rational subset, then evaluating the thing, then we get a, again, a reduced sum of sub-sets of finite k-algebra of topologically finite type. And thirdly, and this is now finally the Tate-asciclicity theorem, all core model groups of this following sequence. So if x is covered by, finally, many rational subsets, then, so that's not quite the standard way of saying it, but it's equivalent. All core model groups of this complex are annihilated by a bounded power of pi. So usually one says that the complex without the zeros here is really on the rational level is exact, but additionally says that the transition maps are strict. And the strictness is equivalent by Banach's open mapping theorem to the fact that the common logic groups of this complex are annihilated by a certain power of pi. So I should say that this is basically Tate-asciclicity plus Banach's open mapping theorem. And now we want to sort of start from this case and deduce what we want. And so there's sort of an intermediate kind of perfectoid rings, which I call p-finite, so it's a following definition. It seems that the characteristic of Ksp says that a perfectoid definite k-algebra rr plus, and I don't know whether it's a really good name, I call it p-finite. If there exists some SS plus as above such that r plus is just take the perfection of your ring, so you take the direct limit over Fabrinius, S plus, and then complete periodically. And of course r has to be r plus invert pi. Okay, and then we have the following proposition. Assume that this is p-finite, and coming from, yes, it's also equal to r0. Assume that it's p-finite, and so coming from some of the SS plus. Then the first statement is that the space associated to rr plus is just the same as the space associated to, and the map goes in this direction. And this is an homeomorphism, which again identifies rational subsets. And secondly, for any rational subset, u and y, which then corresponds to some v and y, some of the situation is again the same that if you consider o x of u or x plus of u, then this is really just a completed perfection in the sense, as on the left, of o y of v o y plus of v. And finally, for all covers as above, somehow, by finitely many rational subsets, the sequence that we want to prove is exact is exact, so. Okay, and now this is easy. So, I mean, just basically as in the same way as you prove that the Fabrinius induces an isomorphism on spectra, it induces an isomorphism, so it's just the identity morphism. And somehow, it's also clear that if you take, I mean, you can define this at expectation also for non-complete things, and then you somehow general fact that taking a filtered direct limit corresponds to taking the filtered inverse limit on the spaces, so you get s. It's the inverse limit of this, but because all transition maps now are the identity, that's just this bar of ss plus, and it's a general effect due to Huber. Well, okay, so there's, so if you wouldn't consider topological rings, but usual rings, it's sort of in general true this statement. And then there's some small step showing that some of the continuous variations exactly correspond. So this, maybe you have to do by hand, but it's not difficult. Yeah, no, no, yeah, I'm sorry. I don't claim that there is a general statement. I don't claim that there is this general statement and this general entity, but in this case, it's okay. And, but in general, it's true that if you pass to the completion, then it won't change the space and also not the rational subset. Okay, and then for two, so this gives one, for two, it's enough to use in your universal property. So the completed perfection of this ring will be a perfectoid of k-algebra, and it's clear that it's the universal perfectoid of k-algebra with the diad property that some of this morphism factors over u. And for three, note that now, if you take the direct limit of our Frobenius of the sequence, or y of y, so we know that at each step, this complex is initiated by a power of pi, and if you take the limit of Frobenius, this power of pi will again shrink by p's power. So this is almost exact in the end. So again, the statement that if something is true up to a bounded p-power and you then iterate Frobenius long enough, then it will be almost true. And then we can complete. That's what we wanted to prove. And now the general case somehow follows by presenting any perfectoid of k-algebra and characteristic p as a direct limit, a complete direct limit of such p-finite algebras. And so let's do this. So again, assume that the characteristic of k-sp and any perfectoid of k-algebra are plus with a slight technical assumption that r plus is a k-not-algebra, which I will sort of ignore, is a completed direct limit, filtered direct, such that all r i plus r p finite. The second part says that in this case, the edict spectrum of r r plus really is again just the direct limit as the inverse limit of the edict spectra of these things. And again, this basically identifies rational subsets in the sense that any rational subset u in, let's call this x, these things x i, u and x is a pullback of a rational subset u i and x i for some i. And in this case, o x of u or x plus of u is the completed direct limit of all j which are bigger than i of o x j of u j or x j plus of u j. And just one thing I want to say is u j is a pre-image of u i and x j. And I think now comes the final part that, yes, we need the following technical statement that if v and x i is some quasi-compact open such that the image of x and x i is contained in v, then there exists some j such that also the image of this map will be contained in j. So this we will need to somehow show that if we have a cover of x, then each individual term of this cover can be seen as a pullback from a finite level, but we also need to see that somehow they cover the finite level and this will be due to this part of the lemma. And then finally we can show that the chief property is satisfied. So let's prove this. So in some sense the hardest part is part one. Difficult. So I mean we do just what we expect to do. So as for example one can present any ring as a filter direct limit of algebras of finite type. So for any subset i and r plus, we consider sub-algebra si and r given as the image of the quotient map from a polynomial algebra of finite mapping to r and that si plus and si is a power bond that elements will converge in power series somehow. So the coefficients go to zero. And somehow, so this gets the induced topology from, now this map is not strict, so si does not have the subspace topology of r. It has a quotient topology, I should say quotient, yes. So in this way it really becomes a, this with the power-bonded elements really becomes a state algebra of finite type. And it's reduced because it's a sub-ring here. So reduced a finite k-algebra of topologically finite type. Then we let ri, ri plus be this completion. And one can show that in this case then r r plus is really just the completed direct limit of these. Of course somehow because r r plus is perfect you really get the maps from ri into r. And you can check that somehow this plus ring really maps into this plus ring here. And okay, this gives some of this presentation. And for two we again note that this bar just of the direct limit of the ri and the direct limit of the ri plus. And here you sort of have to do the same sort of verification that the continuous variations exactly correspond. So you have this relation and then again the completion doesn't change anything. So it's the same as this edict spectrum of r r plus. I mean, yes, yes, yes, yes. So these are spectral space and these are spectral maps that will become important in a second. So part three is that again use universal property and part four is slightly interesting. So it's a general statement about spectral spaces and spectral maps that I want to sort of sketch. So let's call ai this complement xi minus v. So this is a spectral space because this v is a constructable subset and hence this complement constructable subset. And it's a general statement that's constructable subsets of spectral spaces against spectral and let aj be the preimage. Because all transition maps are spectral, this is again spectral and the assumption is that the inverse limit of the ai aj is empty. But if you consider the aj was a constructable topology, then this is a really a compact topological space. So quasi-compact ant house dwarf. Again it's a general statement about spectral spaces and because the transition maps are spectral it follows that if you give the aj this topology and the transition maps are continuous. The constructable topology. So now we have an inverse limit of compact topological spaces which is empty and hence it has to be empty at a finite level. Inverse limit of the space is empty and there's this general statement which is probably true enough that it's empty at a finite level. So you get the inverse limit. So just j in some finite set j, so j is finite. Aj is empty and then you take some k which is bigger than j. And at this index we check this because we take a filtered and direct limit, the thing will be empty. And finally we want to check the sheaf property and for this note that we can now take a conflict notation. So let's uk be a finite cover, write uk as pre-image of uk i in xi where i is large. And if you choose it very large then you can assume that it's the same i for all k. And then the union of the uk i kj covers xj for some j which is again larger than i. And so we may assume finally that the uk are pre-image of a cover uk i of some xi by renaming this jsi possibly. And then for all j bigger or equal to r we have that, so we know that the direct limit over the j bigger or equal to i of the sequence 0 maps to o x j of xj maps to the product over k of the o x j of uk j, am I getting it right? And so on this is exact because it's at each finite, each of these complexes is exact, intends to direct limit is exact and then complete gets the result statement. So maybe I should say two things. One might also try to prove this result by directly presenting any perfectoid algebra as a direct limit of algebras of finite type. And then you would also get some exact sequence like this but without the almost because you are not yet in the perfectoid situation. And such that at each step it's the comology is some of a little related by some power of pi. But in the direct limit this power might of course explode and you would have no control of what happens. And hence you sort of have to first go to the perfection with each algebra to see that it's almost 0 and then sort of this almost stuff cannot explode anymore and then you have it under control. And somehow again it's the proofs in characteristic p and then under the tilting procedure gets the result in characteristic 0 but I don't know of any direct way of proving that this structure pre-sheaf is actually a sheaf in characteristic 0. I am actually using what you already know. Okay, I am trying. Okay, so finally we have proved all the basic properties about a finite perfectoid space and finally we can define what a perfectoid space is. So what did we do? So for any perfectoid k algebra r r plus we have an affinuate edict space of r r plus over k. And we define that a perfectoid space of a k is an edict space of a k. So it's just a full subcategory of the category of edict spaces with the requirement that it's locally sort of affinuate perfectoid nature such that. And it is now a formality that we have again a tilting equivalent. So the category of perfectoid spaces over k is equivalent to the category of perfectoid spaces over k. Now in my first lecture I sort of defined what I mean by an edict space over k. So it's basically one which lives over spa k and z plus k naught integrally closed if you see it this way. So I only require that this, it's locally of the form spa r r plus where r r plus is an affinuate k algebra in the sense that this k is some k algebra with an open bounded subring and this r plus is as usual. So we have this tilting equivalent that now we have really kind of geometric spaces on both sides and eco characteristic and mixed characteristic and their theories are just equivalent. And so if x maps to x flat and then we have the following properties that the underlying topological spaces are the same. And that x is affinuate perfectoid meaning that it's of the form spa r r plus where r r plus is the perfectoid affinuate k algebra if and only if it's still this is so. And in this case, O x of x, O x plus of x is the tilt to the global sections on the other side. And it's really just a form of consequence of what we have proved. And we will need the following proposition, namely that fiber products exist in the category of edict perfectoid spaces. And this is remarkable in the sense because for usual edict spaces, one always has to impose a finite condition on one of the two arrows in order to get the existence of fiber products. Because some of our usual edict spaces, you always need some finite condition to ensure that this pre-sheaf is actually a sheaf. But now in our situation, we have this very general result that our pre-sheaf is a sheaf and just locally we have the following. If you consider the fiber product of spa r r plus over spa bp plus with spa cc plus is spa dd plus where d is a completion of a with c over b and it's easy to show that it's perfectoid again. And d plus inside is the integral closure of the image of a plus and the b plus. It's immediate to see that this has the desired universal property. I mean the hard part really is that somehow, I mean it's clear that somehow you would want to take this edict spectrum, but it's not somehow in general clear that again this O x, this pre-sheaf will again be a sheaf but as we again end up with a perfectoid ring it's okay. And so finally we can study the etal topology of perfectoid spaces and in particular also prove the almost purity theorem. And let me give the definition of what an etal and finally etal morphism is and then that's the break. Okay so in fact the first definition works quite in some general reality so that k be a non-archimedean field. We say that first we define what we mean by a finite etal morphism. So some of the problem with etal topology on perfectoid spaces is that we have no non-reduced perfectoid spaces. So we cannot define an etal morphism by using the infinitesimal lifting criterion and we have to do something else. And so now I explain the something else. So the morphism of a finite k-algebra is called finite etal. If b is the finite etal a-algebra just in the sense of usual commutative algebra and a plus is integral b plus is the integral closure. So in other words giving a finite etal algebra over some given a finite k-algebra is equivalent to giving a finite etal algebra over this rational thing over this a. Because some of the second component is uniquely determined. And then we say that the morphism x to y of etic spaces over k is finite etal if there exists a cover of y by some v i. Certain subsets v which are a finite open. So being equal to some spar a a plus such that f inverse of v is again affine rate. So some spar bb plus affine rate and such that the morphism of affine rate rings now is finite etal. So this is somehow as expected in a sense. And the last part may be unexpected namely we say that the morphism again of etic spaces over k is etal if for x and y now for how should I put the quantifiers x and x there exists open neighborhoods x and u contained in x and y is the image point of x. So this is supposed to be an element of some open subset v instead of y and a diagram. Now you have this map from u to v which is just this map restricted to you. So the image of u should be contained in v and require that we can complete this to a diagram where this map is finite etal and this is an open immersion. And then there's a following proposition which for classical rigid geometry probably have appeared first in the paper of De Jong and van der Poet that for etic spaces that are locally spar r r plus where r is strong in the Syrian. So somehow the kind of etic spaces that Huber considers. So Huber proved this sheaf property under the assumption that the ring is strongly in the Syrian. And for such etic spaces he studied the etal topology and showed that for such etic spaces this definition agrees with the infinitesimal criterion. And so this might look strange because for schemes it's wrong. So somehow an example would be that you take something like the f in line and over it you take some curve which somehow looks like this. But you remove some of these points but then if you take this point x here which has the same image as this point that you have taken out. Get some image point y and it's easy to see that for this x you cannot find such open neighborhoods because in any open neighborhood you would still somehow only remove some points here which would not help you. But somehow in the etic space picture you are somehow allowed to take some small neighborhood of x which looks like this and then you're fine. So somehow for general etic spaces I do not claim that these notions are well behaved but we will later see that they are well behaved for perfectoid spaces. But to see this we will need to use the almost purity theorem which we haven't proved yet. So somehow in the meantime we will need a slight and apparently stronger notion which is the following. We will now assume that k is a perfectoid field and we say that amorphism a plus to bb plus of perfectoid definite k algebras is strongly finite etal. If it's finite etal and the additional condition that it's still finite etal on the almost integral level is satisfied. That b not a is a finite etal. And then we get similar definitions of two and three just by inserting strongly everywhere. And so then we have the following. So recall that we had the following diagram. So if we have some algebra r and consider finite etal algebras and we showed that there's a fully faithful function from the finite etal covers on the almost integral level. And what we are requiring is that it lies in the essential image of this and we showed that under tilting these are identified and that also on the characteristic p side this is really an equivalent. So hence we see that f from x to y is strongly finite etal if and only if f flat from x flat to y flat is strongly finite etal. And it's also equivalent to the same map being just without the strongly. And what I will prove after the break now is that this really is an equivalent of categories actually. Okay and now the break. So we have these notions of finite etal morphisms and etal morphisms for perfectoid spaces. And some of the goal is finally to show this result and so let me first prove some lemmas. Some of which will not be directly needed for this but which are important anyway. So first we show that we can form some fiber products. So if you have an etal morphism and we pull it back by something it's again etal and so on. So let f from x to y be but this only a priori works under the assumption that it's strongly a finite etal. And any morphisms to y where x, y, z are perfectoid. And then this fiber product x times y, z to z this exists by this previous proposition. And this map is again strongly finite etal. And we have the additional property which will be important that if one considers the underlying topological space. Then this maps to the fiber product of the topological spaces and this map is subjective. And this somehow reduces to the finite case somehow. And so if everything is a finite and we are in the situation where this map from BB plus to AA plus is strongly finite etal. Then the fiber product is bar dd plus where d is just a changer BC. And d plus is integral closure of C plus. So in this case dd plus is strongly finite etal over C plus. And again we will have to reduce some statements and characteristics p to the p-finite case and then finally to the case of finite type somehow. And for this we have to need a similar result about pulling back etal maps between etic spaces which are of this form. Maybe I should give them a name. Let's call them for this talk locally in the Syrian etic spaces. So in the case that the characteristic of Ksp and f from x to y is a finite or just etal morphism of local Syrian etic spaces. No, I don't even. I have to find this for all the etic spaces now and it's okay. And g from z to y is any map such that z is perfectoid. Then this fiber product again exists x times y z exists is perfectoid. The map to z is finite etal or just etal. And again we have the property that the map from the topological space underlying the fiber product to the fiber product of the topological spaces is surjective. And again this reduces to the infinite case where everything is etal. So we have x again. So we have all of this. And so we require that the morphism from bb plus to aa plus is a finite etal. And then the fiber product is again of the form s and 2 somehow. So bc and d plus integral closure sum. And again it's untrue that dd plus is even strongly finite etal over k. And the proof is not difficult. So for 2 we just note that this fiber product a tensor bc is already complete. So it's finite projective over c and tends to already complete. And it's also true that d0a is a0a tensor d0a c0a. And so we know that this is finite etal over c0a and tends perfectoid. Because we require that we have something strongly finite etal. So this map really is finite etal. And well that's basically the content of part 2 for part 1. So somehow it's enough to do it for open emergence plus and for finite etal. So because somehow in the general case it's at least locally a composite of the 2. And so open immersion is clear somehow. And for finite etal we reduce to this case. Then part 4 was some of the analog of part, it's on the blackboard. So what we additionally have to use is the almost purity theorem in characteristic p. Which says that this fiber, this tensor product really is already perfectoid. And somehow it's finite etal over c and the almost purity theorem says that it's this automatically extends to the almost integral level. And rest as before. And also part 3 is s1. So again you can reduce to the case of an open immersion in the finite etal map. The finite etal map reduces to the affinity case and so on. So and now we want to reduce certain statements about finite etal maps and characteristic p to the case of finite type. So again we have to present everything somehow as a completed direct limit. And for this we need some statements about approximation of finite etal algebra so we have an atherian ring. So we have the following proposition. Which in the netherian case is an a-kicks is due to a-l-kick. And then to the non-netherian case generalized in the book of Gabel and Tramero. Which says that if A is a flat k-naught algebra. Such that A is a netherian along this ideal generated by pi. And let A hat be the pi at a separation of A. And then the finite etal algebras over the generic fiber for A. You can extend any such algebra to the completion. And the theorem says that this induces an equivalence of categories. And so we only need two properties about hensilieness. Namely that if A is already complete then it's hensilien. And the second property is that if A i are all hensilien along pi. Then also the direct limit is. So filter direct system say is hensilien. And so this gives us the following lemma, or corollary better to say. That if you have a filter direct system of complete k-naught algebras. And A B the completed direct image limit. Then the category of finite etal algebras over A. Is the two categorical direct limit of the category of finite etal algebras over the A i. So meaning that any finite etal algebra already comes from a finite level somehow. And any morphism of such also comes from a finite level. And any two morphisms become equal and so on. And you also need a second part. Namely that this part one above. So the characteristic of Ksp. And yes, sorry, sorry, sorry. Yes, of course we want to invert pi. I'm sorry. If r r plus is p finite. Coming somewhere from some SS plus. Then the finite etal algebras over r are just equivalent to the finite etal algebras over s. Some of the proof is now easy using what we have. That so finite etal covers and morphisms between them. And such are finitely presented objects. So that if we just take the direct limit of the A i. That this is automatic. And then this proposition, or maybe I should call it a theorem. Says that it's really the same as the finite etal algebras over this completed limit. And so this for the thing which was not called one. And for part two. Somehow this shows that the finite etal algebras over r are the two categorical direct limit. Of the algebras of p to the nth roots of s. But because Frobenius induces an equivalence of the category of finite etal algebras. This then just the same as finite etal algebras over s. So the next step is the following proposition. If we have a strongly finite etal morphism. Perfectoid spaces. Then we require that locally on why we can find some affinoid subset such that the pre-image is again affinoid. And we get a finite etal morphism of affinoid algebras. But in fact, one would expect that this is true for any open affine. And this is in fact true. Then for all open affine. Perfectoid. V inside of y, the pre-image, u and x is again perfectoid. And the morphism which you get on rings. So the morphism from o y of v, o y plus of v to o x of u, o x plus of u is strongly finite etal. And for the proof, all the statements translate directly under the tilting procedure to the other side. So we may assume that is the characteristic of Ksp. And now we want to reduce to the same statement which we know for the stuff of finite type. So we want to sort of show that we can pull back this finite etal cover from a finite level. So we can assume that y is really just this open affinoid. So that v is y which is some spar r r plus the affinoid perfectoid. And we can write r r plus as a completed direct limit p finite stuff. We claim that there exists some i and some xi over yi finite etal. So yi is this bar of r i r i plus finite etal such that x is just a pullback of xi to y. And to justify this claim, we know that locally this is possible. So we can find a finite cover of this y by some rational subset such that in each rational subset we can find such finite etal algebra due to the corollary part one. So the gluing is also possible on a finite level because the spaces are quite separated. So meaning that somehow the gluing condition can also be seen at a finite level which also is somehow due to the corollary part one. Because any morphism of finite etal algebra is also defined on a finite level and the core cycle condition is also satisfied on a finite level. Because this is a two-categorical direct limit also says that if you have two morphisms which gives the same inverse limits and they already are the same at a finite level. So again this is due to the corollary part one. And then you can glue this and get something finite etal over this space. So we may assume that r r plus is in fact from the beginning p finite because if we know that here the global sections are finite etals then we know that the fiber product really is just what we get on rings. So we have some SS plus giving rise to this by perfection and then let's call this y naught. So y naught is the r y which is the bar of r r plus. And now we claim that there exists some x naught which is finite etal such that we have this fiber product and the proof is the same using now part two of the corollary. Now this SS plus is reduced of finite type. No, I could just glue some space. I mean locally I have some space and then can I just glue them together? Yes, yes, yes. So finally it's enough to prove our claim for x zero to y zero because we know how to compute the fiber product if this is affinoid and finite etal over this and so that's a classical statement in rigid geometry. Okay, so now we have this and okay now we can with the same method we can prove some other statements. So it follows now that in particular finite etal covers strongly, finite etal covers of bar r r plus are the same as strongly finite etal covers, what covers strongly finite etal over it. So it's clear that somehow anything strongly finite etal covers something here and now we prove that conversely any such space which is strongly finite etal over it really comes from some such algebra. And for the same method of proof one shows that the following proposition, assume that again the characteristic of Ksp and assume that f from x to y is a etal morphism of perfectoid spaces. Then for any x, there exists a diagram that x is in u mapping to v via this restriction of f to u and what to say this, a pullback diagram as this where this is etal and u naught and v naught are locally Nusserian. Locally any etal map between perfectoid space and characteristic p comes as a pullback of an etal map between stuff of finite type and there's a corollary and this is for general K. Strongly finite etal maps are open, strongly etal maps and the composite of two strongly etal maps is against strongly etal. Somehow to prove these statements it's again enough to do it in characteristic p and then using this procedure you would use to the statements for eddy spaces of finite type where you know that they are true. But finally we can come to the big theorem. R R plus be a perfectoid phenol K algebra. So here again K is of any characteristic and x be the eddy spectrum and was tilled. And the first statement is just collecting what we know first. So for any affineurate perfectoid subset open u and x we have a fully faithful function so we have that the finite etal covers of u and what we have proved it's the same as the finite etal covers of this on the almost integral level and this embeds fully facefully into the category of finite etal covers of OX of u. And we would like to know that this is an equivalence and that's summer part two so this is an equivalent. What is strongly finite etal? It's the finite etal covers of u in the sense defined I mean so and by what I've written on the right it's ah sorry strongly yes yes yes yes yes yes yes strongly I'm sorry. And somehow then collecting this this means that for all perfectoid K algebras R we have that finite etal R algebras really so I should say it's differently. And any I over R which is finite etal we know that first of S is again perfectoid on the almost integral level it's still finite etal and it's a technical statement that in fact it's uniformly finite projective or not a module. So this is where the generalization of faulting is almost pure T. Okay so we know that part one is true and we also know that part three is an immediate consequence of part two. So it's really about proving part two. And the idea is now that we somehow are allowed to prove this locally so we can localize some of them on the perfectoid space and then glue some of our finite etal cover that's somehow the idea and so somehow the idea is that locally such a on the perfectoid space is basically given by this perfectoid field at this point and for perfectoid fields we already know the equivalence. And so let's carry this out. So we may assume that of course it use X and so fix S over R finite etal. So the first claim is the following. For X and X there exists some rational subsets say a rational neighborhood of X and a strongly finite etal cover V of U which gives rise to the algebra S tensor over R with O X of U in. So somehow we can spread out this finite etal algebra for finite etal algebras and the claim is that locally somehow this is in the image of this function which we have in one. And for the proof recall first that the completed residue field is really just the completion of O X of U over our rational subsets and then taking the completion. So this was this strange behavior of attic spaces that if you take the completed local ring somewhere then you are left with the residue field. But that's very good in fact. So this implies by what we know that the finite etal algebras over this are just the two categorical direct limit of the finite etal algebras of O X of U. On the other hand we have proved the tilting equivalence for fields. So we know that this is the same as finite etal algebras over the tilt and we also know that the residue this tilted thing is also just the residue field at the tilted point. And then again we know that this is the two categorical direct limit over all X and U of O X flat of U flat finite etal. So somehow this means that locally we can we have our finite etal algebra and locally we can tilt it to the other side because we already can do it for fields. And so somehow in a sense you should imagine that somehow we start with this finite etal cover in characteristic zero. Locally at each point we can tilt it to the other side then on the other side we glue everything together. We have the almost purity theorem there and we deduce the almost purity theorem in characteristic zero. So we can find V flat over U flat finite etal such that the tilt V to U gives this thing over X but then over some U prime smaller than U prime which is the fiber product to U prime will give O X of U prime tensor RS. So first we know somehow that we can find something on here which somehow after pulling back will give the correct thing but then tilting it back to characteristic zero we get another finite etal cover which agrees on this residue field and hence on a small neighborhood. So this justifies some of the claim. So we know that locally we can find them and so what we get is that there exists the finite cover because our space is quasi-compact X, the union of Ui with Ui rational and Vi to Ui finite etal gives somehow the finite etal algebra O X of Ui tensor RS and because this puncture in one is fully faithful this glued to a finite etal, strongly finite etal Did I say strongly? No. Strongly finite etal morphism. I have no Y yet. So F from Y to X. Now the previous proposition shows that Y is again a finite so it's some spa of A A plus where A A plus is strongly finite etal over R R plus and so finally it's enough to show that this ring which we get on global sections is really just the algebra that we started with but that's just no photos from the sheaf properties. So we have the sheaf property for O Y which gives us the exact sequence that the global sections can be calculated by taking O X on these subsets but there they are just given by these tensor products by construction and this stays true on subsets and similarly for O X we have an exact sequence which says that R is this, I and J, O X, I intersect U J and now we're basically done because S is flat over R and the second sequence tender R S is just the first sequence and we see that really A is equal to S. And here we are.