 So today I want to speak about finally about perfectoid spaces and so let me recall a little bit about what we did. So first off we defined the notion of a perfectoid field is a topological field k which is complete with respect to a non-discreet valuation of rank one such that the residue field has positive characteristic p and the Frobenius is surjective from what you look by p. And over such fields so as an example you can take say something like the completion of qp to which you're drawing all the p-paral roots of p for example. And over such fields we consider the following type of topological algebras these were complete algebras they were Banach algebras algebra r with the topological condition that r0 which is a set of power bounded elements is bounded subset so it's always open but it should be bounded so comparing with classical rigid geometry that this would mean that the ring is reduced and also in this case it implies that the ring will be reduced and the crucial condition is that again so here pi is a certain uniformizer so that we chose somehow in this business of when I discussed perfectoid fields that this map is surjective okay and what we had proved last time basically was that if you introduce the field which I call k-flat which can be yes somehow yes you need that one way to say this it's absolute value is bigger than the absolute value of p if you take the inverse limit of the piece power map of k this can be given a natural ring structure that's a construction due from to fountain and then this is a perfectoid field of characteristic p where it just means that it's perfect and complete and to any ring r we associated to any perfectoid k-algebra r we associated another perfectoid k-flat algebra r-flat which can be described in the same way as the inverse limit over the piece power map of r and so this is a perfectoid k-flat algebra and again in characteristic p perfectoid so the condition that phi from r0 surjective is equivalent to the condition that r is perfect so in this case it's probably offset so so get messaging characteristic p a perfectoid k-algebra k-flat algebra is a one of k-flat algebra such that the power bound elements are open and bounded and the ring is perfect so it's something that is very easy to write down what it is and so it's amazing so-called tilting theorem tells you that the two categories are equivalent so a category of perfected k-algebra which I denote by k-perf is equivalent to the category perfected k-flat algebra by a procedure by which you can translate some of the deep phenomena that the current characteristic 0 to phenomena in equal characteristic p which are much easier to understand that's some other way to think about it yes so some of morphisms are continuous morphisms yes okay and so today we want to associate spaces to such objects and recall that somehow from the first talk that we defined the notion of a tate k algebra is a topological k-algebra again call it r such that there exists an open and bounded subring r0 and r and so in particular we see that perfectoid k algebras are tate so that's nice and then we will consider the following type of aphinoid algebras perfectoid aphinoid k-algebra so again it's a pair rr plus where r is perfectoid and r plus in our notice of meant integrally closed so it's just an aphinoid k-algebra such that the ring r is perfectoid and so two such rings we associated this topological space x which is called the adic spectrum of rr plus which is a set of all continuous variations such that for all functions in this plus ring the variations at most one and we take them up to equivalence and on this we defined pre-shefs oax and oax plus which somehow on global sections would give you back r and r plus because here we are in a complete situation okay and so the first lemma we need today is that we also have an equivalence of categories between perfectoid aphinoid k-algebras which are the objects that perfectoid spaces of a k are built out of they are the same as the same thing over the other side so mapping rr plus to r flat r flat plus and so somehow what we need to see is that we can identify these open and integrally closed subrings and for this note that first that necessarily these elements are an r plus because they are topologically new potent and because r plus is open some power of them will be in there and because it's a really close they will themselves lie in there and then we have the set of allowed r plus and r not is in bijection with integrally closed subrings of r not modulus of maximal ideal and we call that somehow under this procedure of tilting you somehow had an isomorphism between this thing and this thing where somehow here pi and pi flat are somehow corresponding uniformisers in these two fields and so in particular you also have said r not mod the maximal ideal is the same thing as on the other side and hence this thing here really is the same as integrally closed subrings of r flat not mod and flat which again you can then identify with the possible r plus r flat pluses in r flat so this way you get the desired identification between such pairs and now I can formulate the theorem that we want to prove today and it says following so let r r plus be a perfectoid fenoid k algebra with tilt of flat of that plus and let's consider the associated spaces x is an x spectrum of r r plus and also x flat okay and of course we also have all these structure sheaths on on these spaces structure pre-sheaths are priori and the first part says is that we have a canonical homeomorphism which identifies rational subsets so recall that the topology on these spaces was defined by choosing a so-called rational subsets as the basis for the topology and I will recall the description in a second and it's very easy to write down the map in one direction namely if x maps to x flat under this correspondence then I should say that summer in general there's a map from r flat to r by projecting to the last coordinate here and this map I denote by f maps to f sharp because it sort of goes in the other direction and then the formula is given by assigning the valuation sending a function f to the valuation of this function f sharp yes so yeah I mean if you identify the topological spaces in the rational subsets here exactly the rational subsets here so in both directions somehow if I have one here then the image will be one here and one in the very first version of my preprint sort of I've only proved it in one direction but it's true in both and I can now give a proof okay so the second part uses some on its formulation the first part and it says it for any rational u and x corresponding then under this correspondence to some u flat and x flat rational you can form or x of u or x plus of u and by definition of the structure sheaves pre-sheafs this was some in a finite algebra and the claim is that it's a perfectoid affinoid key algebra was tilled given by sort of what we get by evaluating the pre-sheafs on the other side so somehow this relation sheifies between r r plus and it's tilled and then the third part is that in fact these things are really sheaves and this implies that also these other things are sheaves and lastly one would like to have an analog of tase i-psychlicity theorem saying that the higher chromology groups vanish and this is true so the h i x o x is 0 for i bigger than 0 but in this perfectoid situation it's even better namely the statement which is a statement about the generic fiber somehow extends to the slightly generic fiber in the sense that if you take the plus sheaf then this will be m torsion for positive i which is something that is not true in classical rigid geometry so this could there's a theorem stating that the non-trivial theorem stating that this is annihilated by some bounded power of the uniformizer but it might be somehow large okay and the proof will occupy us at least for today and so let me give an outline of the proof first so first one deals with one and two and proves them completely and then afterwards one proves three and four completely and so so the first step will be to show a weak form of part two but with the additional precision that we know explicit an explicit description of almost description of x plus a few which will be important and then the second step is some of the crucial step so the problem is somehow that the map f maps to f sharp from r to flat to r is far from subjective so some in the definition of some of this flat valuation here on the other side somehow you only see the valuations on these sharp elements and somehow we want them to determine everything some of the valuation but this map is very far from being subjective in general and so but still we need some kind of approximation lemma that we can approximate any function by a function of the form f sharp and so that would be the following approximation lemma any f and r can be approximated by some g sharp where g of that such that the following holds for all x and x the absolute value of f is at least the same as the absolute value of g sharp except if possibly of both are very small so you cannot exactly approximate the function but at least sort of its absolute value can be approximated in a nice way and then after you have this it's easy to conclude with part and as this will take some time I will not yet sketch the outline of the other two parts and instead let's start with the proof of one and two so what's easy to see is that this map from x to x flat it's well-defined but somehow you need to check that this actually defines the valuation on on the swing of flat but that's not so difficult so we need to see that it's continuous and for this first recall that a rational subset is something of the form u given by some f's and the g subset of so it's a set of all x and x such that's the absolute value of the f is at most the absolute value of g where the fi are certain elements of r that generate r is an ideal and g is also some element of r and at this point we will need the following remark that we can we can always assume that all the f's and the g's are in r plus for example or in r naught just by multiplying them by a large power of the uniformizer and we can also assume that fn is just some power of pi and some natural number because there exists n such that pi to the n is in r plus f1 plus fn as a subset of r because if you invert pi then they will really generate r so they exist some such n and then we can just add this pi to the n to the set of f's and this last condition will be sort of automatically satisfied so in particular we sort of see that on rational subsets the absolute value of g of x is at least so it's non-zero so bounded from below okay and now how does this imply that this map is continuous if we take a rational subset of the tilt with these conditions then the preimage is you sharp given in the form that it's a rational subset associated to all these sharp elements instead of x so that somehow this is correct as a set it's clear somewhere from the definition and what is we have to check is that it's really a rational subset and for this we have to see that the f-sharp still generate r but somehow if fn was some power of the uniformizer then fn sharp will just be this pi to the n and then already this one element generates generates r so that's okay I'm not completely yeah I'm not sure I would expect yes but some I haven't checked this is true that if they did not generate r then there's a point probably yes even if they are not Syrian ah okay yeah yes yes yeah then one could sort of do this yeah okay I must admit that I just haven't checked but sort of this way you get your way through at least and so now we come sort of the first part of this proof so the weak form of two so here we start with some rational subset expression flat rational and we assume these conditions so that all of them some fact enough that they are power bounded and that fn is just some power of the uniformizer and so let u-sharp in x be the pre-image and now we want to get this description and so let r0 so we want to define a perfectoid ring and for this somehow you have to join all of these p power roots but the nice thing about these elements that we have to join anyway is that they come equipped with canonical p power roots somewhere because they come as sharp elements so so let's this be the pieatic completion the ring generated by all of these elements inside of r where you invert g-sharp and then and maybe I should give this monster your name so let's just be a then the almost algebra associated to a is a perfectoid k0 almost algebra the second part is that o x of u-sharp is again a perfectoid k algebra and recall that perfectoid k-algebras were equivalent to such perfectoid k0 almost algebras and so was associated almost algebra which was defined in this way just really being this thing we have somehow not only described the structure sheet now but we have sort of almost described it on the integral level in this way so I mean our priority are somehow some passing to the integral closure stuff that we have to do but sort of it's enough to join just the p power roots of these elements to get almost the integral closure of the whole thing and the last part is that or x of u-sharp tilt to or x-flat the second part implies that this independent of the presentation almost because this thing we know is independent of any choices so let me I will recall the definition of this structure pre-chief maybe it will become clear as this algebra yes but after passing to the almost algebra it will not depend because of part 2 and that's what I will prove let me just proceed to prove it okay so so first we do part one in case that's a characteristic of ksp so recall that this implies that k-flat is just k and so on so I can ignore all the flats and sharps and so on and so it's clear that a is flat and pyridic complete and to show is that the map this map is an almost isomorphism okay and for this because look at the polynomial so this clearly is a perfectoid so here is this property is true and this surjects on to the algebra that we're looking at the others so this is surjective so so ti to the i over p to the m sent to fi over g 1 over p to the m so this is surjective and the kernel contains the ideal i which is generated by all elements of the form ti to the i over p to the m times g minus f1i and the claim is that the induced map to this thing is almost injective so it's an almost isomorphism and how do we prove this so so first it's an isomorphism after inverting pi because in this case we also invert fn and and g because g divides fn and once you have divided we have inverted g the ti really just become the quotient and we get see that it's an isomorphism and so that f be in the kernel then there exists some k such that pi to the kf is an i and so in particular p to the jf to the p to the j is an i because it's just a mortal pull of this one but this implies that this thing itself is an i because i is perfect by its definition and so we see that it's almost injective because sort of any element in the kernel lies already almost in this ideal i by which we have divided okay and so as it's clear that if you consider now this other ring which is almost the same then the statement that we want is literally true that if you reduce mod pi or 1 over p and mod pi that you get the same thing this is true get the claim next we prove that 1 implies 2 and for general k so let me recall one definition of the structure sheath so so first we take something some open and bounded suffering and in our case we can just take the power bounded elements because in our case they are open and bounded and then take this thing which is a pi at a completion of the subalgebra and then or x of u will just be this thing and then you invert pi okay and we have to see a summer that we get that we can also express this in terms of our algebra a so so what we have is that if we are only adding these elements and of course we get a suffering of this one where we join all the people roots but I claim that it's this is also contained in pi to the minus n times m times this thing and this is because f in here can be written as the sum of terms in 1 over g sharp to the n times are not why I don't invert so much because somehow there may be some denominators in was sort of a p power in the denominator of g sharp but sort of you can take the closest integer to this and then the additional part will just have sort of for every f just a little bit less than 1 times g sharp in the denominator but 1 over g sharp to the n where is equal to pi to the minus n times m times f n sharp over g sharp n is yeah no n is correct to the n and again this also lies in the subalgebra and so we only have this denominator left okay and so this means that because sort of they differ only by bounded power of the uniformizer we can pass to the pi at the completions and they still have the same behavior on the pi at the completions and then inverting pi we get the same thing so we get that o x of u sharp is just a where you invert pi and so this means that this thing here really is the perfect to a k-algebra associated to a and finally we can do the general case so the rest of the proof so so in general it's still true that a is flat and pilotically complete and we still need to see that so what's enough to see is that if I reduce a mod pi or the almost algebra that's also enough then I get the same thing for the other side so I can write down the similar algebra a flat which is just defined as our flat not where you invert all the fi at all at all the fi over g's and the p power roots then you get in such an algebra so let me write this down it's off that not because if this is true then because we know the result for the other side we know that here Frobenius is so objective his kernel almost generated by the piece root of the uniformizer and hence is true so this gives them part one hence also part two and this equation then also entails that the algebra which you have in characteristic zero is really just the tilt of the algebra and characteristic p so we only have to see that this is true and for this so we still have this thing mapping to a where here i is generated by this ti to the one over p to the m times g sharp and also the additional input that we need is that we can consider the tilt of or x flat of u or x flat plus of u because you have already proved the resulting characteristic p this is a perfect or k-algebra so it's the until the fact that algebra and so we can tilt to a until to a perfect to it affinuate k-algebra and then we have the diagram that sparse s plus maps to this bar of which is just u inside of x flat and you have this diagram and from this diagram it follows that this thing factors over you so some we need to get the comparison between this characteristic p-ring to this characteristic zero ring and this somehow we will get this so but the universal property of of the structure she if we get a map from o x of u sharp or x plus of u sharp to s s plus and now we have the following diagram so we have the following composite maps so we have are not when join all of these things so this perfectoid so let's do it with the almost algebra's this maps to a almost it's clear that these elements will map to or x of u sharp and this maps to further to s zero sirk sharp so this is a map of perfectoid k-sirk a algebra's was tilt exactly the single characteristic p mapping to the composite map I mean somehow we don't know that intermediate things are already perfect okay not a algebra's but we know that the two outer things are and the tilt is given by this map some of this is the ideal that occurs so this is the same as a flat almost by what we have proofed in characteristic p and so this is an almost also the same as o x flat of u flat yes yes so that's bad notation so by a flat I denoted the ring that is constructed in a similar way so maybe I should just denoted by B sorry I think it was also on this blackboard I think that okay so okay and so what would you get do we get from this so we know that as this is an isomorphism so as the composite because these rings are just the same some of from the explicit description you see that sort of what you mod out is just the same module the uniformisers and so this sort of corresponds to the composite map to a almost modular pi going further to s not a what you look hi so as an aside as not a is equal to be what you do no no sorry sorry sorry no there's a tilt so sorry yeah so it's a composite map but this thing here is subjective and hence both isomorphisms and hence this here is an isomorphism and as this is just by definition the tilt of B we get to the result okay so some of B was the thing on the integral level whose almost algebra would give this thing yes a little bit tricky but yes yes yes so the power-bounded elements sort of the problem in this proof is that a priori after a past irrational subset the power-bounded elements could get very large somehow so they I don't know they might not even be sort of bounded sort of in their ring anymore and sort of one has to show that they are do not get too large and so it's so difficult to do this if the perfecto at field has characteristic zero and sort of the proof is by first proving it in characteristic P and then somehow using this resulting characteristic P in some way to bound this structure sheaf here and so where you bound this is to use this universal property of the structure sheaf and the fact that we already know what happens in characteristic P to get some bound here on the structure sheaf which is given by this map I don't know one so I mean it would be nice to have a proof which does not first do it in characteristic P and then translates so I don't know one but I haven't sort of tried for a long time to do this I mean but somehow often the case that in order to prove something about perfecto at algebra and characteristic zero you first have to prove the analogous result in characteristic P and then use it somewhere in the proof now we have finished this description of the structure sheaf on certain rational subsets and the next thing we want to do is prove such an approximation lemma and first we prove this an approximation lemma sort of for hyper surfaces in P and more or less that's how you should think of the following statement so it goes as follows so as our perfecto at algebra we take just the polynomial algebra so this is the one of a P graded algebra and and the lemma says following for any f in the power bounded elements which can also be described just as k0 and then so as those power series which have integral coefficients and any c greater equals to 0 and epsilon greater than 0 there exists some g in r flat not which is also just k flat not and again you take power series in this number of variables so in this G depends on C and epsilon such that for all x in the space associated to R we have the following inequality that difference between f and g sharp at this number x it's not too large and the way it's bounded is in the following way so this means that if f is large this can also be comparably large but sort of it's less than the absolute value of f and if the absolute value of f gets small then also the absolute value of g gets small with an absolute bound somehow given in terms of this c I remark about this and there's something that I forgot to say I assume that f is homogeneous of some degree and then you can also choose this g to be homogeneous of the same degree so d is some number so the condition so if epsilon less than 1 condition implies the following statement that for all x we have that the maximum of the absolute value of f at x and the absolute value of pi to the c is the same as the same for g sharp so somehow if one of them is large then the other one is large of the same absolute value but sort of if they're small they're both small but sort of can maybe different less than one so I mean that epsilon will always be less than one but okay and so this also be a rather technical proof but that will be the most technical proof sort of this lecture series so so let me try to get all epsilon's right so we fix epsilon greater than zero and assume that it's less than one and we can also do I assume this yes that epsilon is in fact a rational number with only p powers in the denominator we also fix f and we proved by induction the following statement induction on c the following there exists some epsilon that will depends on our c which is positive and some gc in this ring homogeneous of degree d such that for all x we have the following inequality I make it only in the homogeneous case so the homogeneity will sort of play a role in the proof somehow what I would like to have in the end is only one minus epsilon but sort of in the course of the proof we will always lose in the induction step some small epsilon so we have to add some small epsilon of c at each step sort of to keep going and then what we have so okay and at each step we will increase c a little bit and for this we need to fix source the increment which will increase c so we induct from c to c prime which is c plus a and c is equal to zero now yes if c is equal to zero we can do the following then yes it's a little bit tricky to get all the constants right and all the quantifiers but again we can just choose any lift g of f bar in which is the same thing as on the other side and and then we can even choose epsilon c equal to epsilon because it will then be that the absolute value of f of x minus g sharp zero of x is always at most the absolute value of pi and this is less than equal to pi times maximum f of x okay and so now we go from so we go from c to c prime which is c plus a and we may assume that this epsilon of c that we have is between I mean we can just make it smaller if we like so we can assume that it's between zero and epsilon and also again that it's a rational number with only p in the denominator okay so now we have set up our induction now comes a real argument so somehow in the first step we just took some thing which is equal to the thing that we started with modulo pi so as a first approximation that's good and now sort of there's this subset where both of them are at most pi and now we want to sort of in this rational subsets get a better approximation so we now we look at the subset where both of them are at most pi so that you see in x flat which is the rational subset where the absolute value of gc of x is at most at pi to the c then you see sharp which is an x bar r not is given by gc sharp of x is at most it's pi flat pi to the c and by this remark here we know that this is just the same locus where the absolute value of f is at most pi to the c so and so the induction assumption implies that if you consider the element h which is the difference of f and gc sharp which is the thing that we want to make small then we know that on this rational subset on this rational subset the difference is always divisible by this number because on this subset this maximum that we have will be always pi to the c and then this is just what this inequality here says that for all points in this open subset the absolute value of this thing divided by this is at most one in tensors license plus algebra and okay and the previous lemma tells us what this algebra is at least almost so we know that o x of u sharp the almost algebra here is given by taking r naught and adding gc sharp over pi flat no pi to the c and all its p power roots it it is true but I don't want to use it but sort of they are almost the same because this lies between the power bounded elements and the maximum ideal times the power bounded elements yes but sort of it's not clear that this is true on the whole sheath somehow it's I assume that sort of on the global sections I just take this r plus to be equal to r naught I may take this bar of r r naught I mean sort of it's this r plus I just choose r naught it's one possibility but it's not clear that this is true then for the whole sheath that also on this rational subset some of this o x plus of u sharp is the same as o x of u sharp power bounded elements but it's almost true because for any such open equation where you close suffering it lies between the power bounded elements the maximum ideal times the power bounded elements so as far as the almost serious concern there's no difference okay and because h is homogeneous h lies almost in the pi at a completion of the direct sum overall I zp inverse 0 is equal to i is equal to 1 that's this four factor and then we can sort of take gc sharp over pi to the c to some power I and then we can multiply this with some element of degree equal to d minus d i to get something of degree d so some of the ideas that now on this smaller rational subspace we have this difference h so we can write h as the sum of over such I of so here we choose some smaller epsilon which we need to do because we don't know that it itself lies in there but any small multiple lie in there so after we decrease this epsilon of c by a little bit we will can write it in this form times the sum of our i of gc sharp over pi to the c to the i times certain elements are i where the r i of degree d minus d i and they go to 0 and so now we have to sort of you just take some naive approximation of this thing on this rational subset and add it to our gc to get a better approximation so we set so we choose s i in after that search such that s i mod pi sharp is equal to r i mod pi and also such that they go to 0 and we set gc prime to be gc and to this we add something that makes our approximation better at least on the special subset and now we have to check that it satisfies the desired inequality um so we want to verify the following claim that for all x and x the following is true we have this factor c plus 1 minus epsilon plus epsilon of c prime and then the maximum of the absolute value of f at x and yes um so the worst denominator we can have is pi sharp to the c here and this is cancelled out by this factor here because 1 minus epsilon plus epsilon of c prime will be positive while this i looking back here it's between 0 and 1 and this precisely where we have the needs of homogeneity because if we didn't know that it was a homogeneous element we wouldn't get sort of this we could get arbitrary i's here but knowing that it's homogeneous element we know that the power of gc sharp which we can get is bounded by 1 because the higher powers will give elements of too large degree so so there are two cases case one is that the absolute value of f at x is larger than pi to the c so we are outside of of this rational subset where we constructed our element in the way that it might be a better approximation and so we need to see that sort of we didn't destroy the good properties of our function out so on as a complement of this rational subset and so it's enough to check that all the terms that we added sort of do not contribute anything bad so that okay times gc over pi flat to the c to the i times s i sharp at x is at most pi to the 1 minus epsilon plus yes exactly times f of x so the maximum this case is taken on by times f of x so we have this and there should be no c here yes okay so we need to check this but the absolute value of s i sharp at x is less than 1 and ignoring this term you get the maximum for i equal to 1 because this quotient here has norm bigger than 1 and i should also say that in this case yes this is equality so this means that for i equal to 1 we cancel this c we get 1 minus epsilon plus epsilon of c prime which is what we want to have and we get the absolute way of gc sharp which is exactly the absolute way of f of x so we see that on this rational subset outside of this rational subset we didn't do anything bad and now we have to check that also inside we did improve the approximation and here it's enough to show that this difference is bounded by this term because in any case this will be larger than the maximum that we have there and for this again it's enough that f minus gc prime sharp is an element of pi to the c plus 1 of o x plus well it's almost in pi to the c plus 1 times the power bounded elements as so this means that it's also almost in the integral elements and because c plus 1 is greater than c prime plus 1 minus epsilon plus epsilon of c prime which follows from all the choices that we made somehow in pulling all in getting our induction started so it's enough to check this but this really just falls directly from the construction of gc prime because somehow we precisely chose it to sort of make the approximation better on this rational subset so okay this gives this approximation lemma now that we have it we can sort of finish the proof of part one and two namely now we get the following nice corollaries so now that r r plus b any perfectoid f not k algebra was tilled half flat half that plus and x again and x flat and they're at x spectra and it turns out that somehow the case that we handled above is somehow the universal case in some sense so we can reduce it for any other situation so for any f and r c and epsilon there exists some gc epsilon and r flat such that for all x and x the very same inequality is true so we get such an approximation lemma for all possible perfectoid algebras and this allows us then to uh no there's some intermediate steps so for all x and x the completed residue field is perfectoid well i ignore homogeneity because some of them are not rated anymore so if we would apply this in the previous case we would only get something weaker of course because here we don't claim anything more anymore about homogenous elements i will prove this in a second um and the third part is that this map from x to x flat which we have is a homomorphism identifying rational subsets okay let's prove this for one so just by multiplying f by a large by a large pi power we may assume that f is in r naught or even in r plus and we can also assume that c is some integer and then we can write some can always sort of approximate f by the element of the g from g sharp up to some uh up to pi and okay then we go on right to the c times g c sharp plus some gc plus one where all the g naught up to gc are some elements of r flat plus and or maybe it's better called fc plus one um is in r plus yes sorry and we can assume in fact that this error term fc plus one is just zero because looking at the inequality that we want to prove it won't disturb it and we have some map from k up to gc to r in r sending g i over one over p to the m to g i over one over p to the m sharp and then f is the image of t naught plus p pi t one plus pi squared t two and so on pi to the ctc which is an homogeneous element and now we apply the previous result to this homogeneous element in this algebra okay um so now we have this general approximation lemma and uh first let's now prove part two of the characteristic of ksp in this case i should have stated this in the first lecture they said if x is r plus can in fact be any fv node k algebra and x is some point of x then the pi added completion of i mean you have this local ring or xx plus and the funny thing is that if you complete it which you would think does not sort of it's still some of a complete local ring but in fact the opposite it's true it's equal to just the completion of the residue field and for the proof just note that the kernel of this projection map um from here to here to the residue field it's also the kernel on the rational stuff and this is pi divisible so somehow if you have a function which vanishes at your point then the function over pi will be integral on a smaller subset and it will still lie in the kernel and so if you complete sort of the whole kernel sort of vanishes because you take separated completion so this means in our context that that uh well this is almost the same so this is a on a is a completion of the direct limit so x and u rational of perfectoid k algebras k not a algebras and so it follows that this itself is perfectoid k not a algebra and hence this is a perfectoid k algebra and it's also this also in general too it's a non-archimedian field and so it this means that it's a perfectoid field so first read we now have to we want to show that the map from x to x flat that this is homeomorphism identifying rational subsets and so what that u and x be rational in fact u if you can is again is of the form u f 1 up to f n over g okay and then in fact u is equal to v sharp where v v is a rational subset let's call this f 1 flat to f n flat with v flat in x flat where the fi flat and g flat form approximations in the sense of one uh of f 1 to f n and g so we know that we can approximate these functions by functions of the form something sharp and if you do it well enough then it's easy to check that the rational subsets here in fact uh the same so this means that uh any rational subsets of x comes as a preimage of some rational subset of x flat and because x is t naught so you can separate points it follows that the map from x to x flat must be injective because if two points two points would be contracted to one point then there could be no rational subset separating them because any such comes as a preimage from here which either contains both points or none of them okay and uh finally we have to see that it's subjective so so the valuation factor is as you have some ring of that then it maps to this completed residue field and then on this you have some valuation to some gamma union zero and so we know that this is a perfectoid k algebra so by proving part two in characteristic p and hence we can tilt the map and get a map from r to some untilled thank you uh to some perfectoid field l and we have checked in the first lecture about perfectoid fields that in fact the continuous variations of perfectoid fields are identified under tilting so we can also tilt this valuation and this gives so we call this just a variation at some point x get and then this gives us a preimage x of well i shouldn't call this x from the x from the beginning so let's call this so this proves that um this is a bijection and we have proved that the preimage of a rational subset is a rational subset and the other way around so it identifies rational subsets and hence is a homomorphism and and the previous lemma now shows that o x of u is in fact perfectoid k algebra for all u and x rational we first proved this only if our u sort of comes as a preimage of rational subsets but now we know that they are the same so we know uh that this is true in general and then also the proof of two two works in general now so what we have done some hours in total now is that we have taken a perfectoid algebra and we have associated it to a space and sort of can now consider a perfectoid algebra somewhere as a family of perfectoid fields and this point of view will then then later show turn out to be very useful in proving this almost purity theorem but you're using some of the general case to a case of field summer by doing it locally around each point okay and let's also collect some other corollaries of what we have proved so we call that we also want to claim the following so if you are an axis rational and corresponding to some u flat and x flat then we want to see that this tilts to x flat of u flat and recall that we already know that both of these are perfectoid if you know k algebras because we have checked that this integral thing here really is a perfectoid k algebra and the proof is not very easy we can characterize so we have a universal property describing or x of u or x plus of u now it describes it among we have a universal property describing it among all of u k algebras but in particular it describes it among perfectoid of u k algebras and then you can tilt this universal property and you get the universal property characterizing hence we know that this has to tilt to this and this finishes the proof of parts one and two of the theorem and what's left to prove is that the structure pre-sheeps on this thing are in fact really structure-sheeps so there and that they satisfy an i-cyclicity theorem and yeah it doesn't make sense to start with this now so