Direct summand conjecture and perfectoid Abhyankar lemma: an overview
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Transcript: English(auto-generated)
00:05
Good morning and good evening to everybody. So thank you very much for the invitation
00:26
to speak here. Let me take this opportunity to thank also Ofer Gabor who is here for a very helpful list of criticisms and corrections on a premature version of my work. Let me
00:43
start with an introduction. So A will be a commutative ring with a unit and B will be
01:07
a finite extension ring throughout the talk. Finite means that it is finitely presented as
01:22
a module. This is in the notarian case, you mean? No, no. It will soon be in the notarian case, I start like this. No, the definition of the finite extension in general is finitely generated. Yes, for me it will be this. If you like, you can take a notarian ring. For this part
01:49
of the blackboard it will be this definition. So let us consider the following exact sequence.
02:05
So exact sequence of finitely presented A modules. And the question of the day is when
02:23
this sequence splits. First I state a simple fact about the meaning of the splitting of
02:43
this sequence. The following are equivalent. So this sequence splits or in ideal theoretic terms for any ideal of A. Any ideal of A is contracted, so when you consider Ib intersected
03:15
with A is just I. And you also ask similarly for every polynomial extension of A, for
03:33
T1. So in other words, ideals are universally contracted. Another formulation which is
03:45
equivalent and more category minded, if we consider the base change between A modules and B modules, so this is just a product with B over A, this should be faithful.
04:15
Okay, so some examples where this condition is satisfied. For instance, when B is flat
04:26
over A, then because you have this finite condition and it is an inclusion, it is faithfully flat. And then B over A is flat, but it's also finitely presented, so it's projective,
04:47
so you have the splitting. So this is somehow a trivial case in this story.
05:00
Another very special case is when A is a normal domain and the degree, the degree is just the dimension of the vector space which you get by tensoring B with the fraction
05:20
field of A, is invertible in A, then this also implies that the sequence splits using divided trace. So now the direct sum and conjecture, due to Hoekstra's direct sum
05:59
and conjecture, is that this sequence splits whenever A is a notarian regular ring.
06:30
So a word about this condition, it is a rather sharp condition. In fact, if you try to weaken it a bit, you will find some counterexamples. For instance, a counterexample
06:51
in a non-regular case. So in characteristic two, I copy from upstairs, so I think it's
07:02
that one, T1 cubed plus T2 cubed inside, so this is A, this is B, so the index here is just four, so this condition is not satisfied, this is not flat, and this is in fact not
07:24
split, so not split, so the exact sequence is not split, whereas this is a normal complete intersection ring. Excuse me, when you don't have the extra term with the cubes, I think the index is
07:46
four. So when you have the cube, you also have two, right? Yes, yes, yes. So it's an example to show that already for surfaces, in positive characteristics,
08:10
you cannot weaken so much this important condition. So in this talk, I didn't plan
08:21
to talk really about the motivations of this conjecture. Let me say at least a few words. It came out of a complex of conjectures in commutative algebra called the homological
08:41
conjectures, which have their source in work by Peskins, Pirro, and Uxter, and other people, which deal with problems with local intersections, numbers, and so on. And another source is some problematic of descent by Olivier, Renaud, Grison, et cetera, where
09:15
in their paper, the conjecture is not stated explicitly, but it's somehow implicit. For
09:32
example, the paper by Renaud Grison, that any injective integral extension of notarian
09:42
ring descends flatness of modules. So the conjecture is not stated like this in your paper, but it's stated that this conjecture, so this statement implies what I said. So this is another direction, another source.
10:05
You mean they state the problem that whether it descends flatness, so they prove in the finite case and they state it... Exactly. For integral and non-finite case, it's a non-trivial problem. And then there was another paper showing
10:21
that the direct sum and conjecture implies this, yes? No, no, it's implicit in their paper that direct sum and conjecture implies that you can pass from finite to integral extension and have a flatness descent. There is an
10:40
argument in the paper which shows this. OK, so let me state a few facts about this conjecture. First, there are some which are mainly due to upstairs, except for one.
11:06
So one can reduce to the case where A is a complete local or regular domain and where
11:24
B is a domain. And so now there are some cases. So in characteristic 0, then you can
11:40
apply a second argument here using a divided trace. So this is OK, the conjecture is true. In characteristic p, this is more surprising, equal characteristic 0. In characteristic p,
12:06
it's also true, but it is more tricky. So let me give a short proof due to upstairs himself. Let us introduce, so this is a subring of A, p to the n between brackets, generated
12:29
by the p to the nth power. And so you have the ideal generated by the pnth to the nth. It's
12:54
just now, but it will gradually decrease. So please ask again and again. So it is contained
13:02
in p to the n. So when you take the intersection, you see that this is 0. I'm just proving this fact. So now, because B is finite over A, there is certainly a linear form, which
13:25
is not 0. And so there is some element of B over which lambda B is not 0. You can translate by B. And so you get another linear form, which has the property that
13:44
lambda 1 is not 0. And so since you have these properties, there exists an index n such that lambda 1 is not in m to the p of n. So now you use the fact that A is regular,
14:09
and also that it is a complete local regular domain, which says that this implies that Frobenius is finite flat. Yes, yes, thank you. Residual field. One can assume this
14:36
without problem. So this means that A is finite free over p to the n. Then this property
14:57
implies by Nakayama that this thing is a part of a basis of A over p to the n.
15:12
Now then you can take a linear form, mu A p to the n linear, A p to the n, which maps this lambda
15:30
1 to 1 in A p to the n. Now when you restrict mu lambda to B lambda to the n, p to the n,
15:46
sorry, this gives you a retraction B p to the n to A p to the n, which is the inclusion here is just a section of this mu lambda. So you solve the problem, not for A and B,
16:04
but for A p to the n, B p to the n. But by transport of structure, by the nth iterate of Frobenius, you get it. So Frobenius minus p to the n gives you the section. So that's the
16:30
difficult case of mixed characteristics of p. Well, in fact, one can reduce to,
16:45
so by Cohen structure, this is something of this form, but one can even assume that it is un-rambified. Now this is part of a long paper of Oster in 1980s. Exactly, this uses some, yes.
17:01
It uses three key things, the canonical element and... Exactly, this reduction is non-trivial, but it will not be very important for the sequence. So it was, essentially this case was open, so this was done in,
17:20
here, this was done in 1973, in fact, and this remained open, except that in dimension three, so it's solved by Heitmann, very long paper.
17:45
And in dimension two, it is very easy. Yes, yes, yes, it's very easy, yes, essentially because you reduce to flatness.
18:03
So the aim of this talk is to explain proof proposed at the beginning of August, I think, of this case, hence of the general case of the Dirac-Sommer conjecture
18:27
in characteristic 0p, using some perfectoid techniques. So I would like first to explain the strategy.
18:45
So maybe it's still one small thing, which will be useful. So to say that the conjecture is true, it means that under certain circumstances, this sequence splits, and in fact to have this splitting in the characteristic,
19:01
the mixed characteristic case, it is enough to split, this is also remarked by Uchster, a kind of Mittag-Leffler argument, enough to split this sequence mod p to the n for all n,
19:21
or maybe n because there is another n here. So it's enough to do it modulo something. Or even mod powers of the maximal ideal, it's the same argument. Yes, yes, I have it, yes, you are right. In fact, one first proved what you said.
19:45
Okay, let me explain the strategy.
20:27
So let us start with this proof in characteristic p. You see that first you have an analog of Frobenius here in a mixed characteristic for A.
20:44
So in the sequel, I should write it here, in the sequel, I consider this case. So you have Frobenius here, no problem.
21:03
But on B, maybe there is no Frobenius, so you cannot imitate the proof. There is no meaning of anything like p to the n, so this proof is hopeless to generalize. So let us try another kind of argument in characteristic p.
21:25
So back to characteristic p for a while, very short while, but in the separable case. So this is very special.
21:43
And also for simplicity, when A is, I replace this v by something characteristic p of the form, let's say, k, perfect field of characteristic p, t0. And then I have the other variables, so t1 and so on.
22:03
So this is characteristic p analog of this question. And I assume that this is separable. Then one can consider, instead of applying Frobenius, one can apply a kind of inverse
22:23
of Frobenius, so take a perfect closure. So 1 over p infinity, and then 1 over p infinity. So this is still separable.
22:42
And what is the advantage here to pass to the perfect closure is that the trace has better properties. So here, if you take the trace in the sense before, so the trace fraction field of A,
23:06
then it is not subjective, anything like this, so you cannot really hope to use the trace to get a retraction of this inclusion. But here, at the top level, at infinity, there is more hope,
23:26
because the trace of B infinity is a radical ideal, so it is itself 1 over p infinity. And so, in particular, it is not contained in ideal generated by the t i's.
24:04
But this extension is in fact face fully flat, but even free. Free over the t i's are some indices, where the indices are in 1 over p,
24:22
and less than 1, between 0 and 1. And then it means that there exists some m i, such that the projection on the m i factor of some,
24:41
and there exists some B, in B, 1 over p infinity of B, is essentially 1 of the trace. So you can use this element, so you embed B to B, 1 over infinity,
25:07
then here you take the map B prime to the trace of B prime B, and this goes to A, 1 over p infinity, and then here I take the projection for
25:23
this m i over A. This sends 1 to 1, so it is a retraction of the embedding of A into B. Somehow, the idea of using the trace still works, provided you go to the perfect closure.
25:45
So now this argument can suggest some analog in mixed characteristics,
26:04
at least under some condition, which is somehow analog of the probability, I don't know. So I come back to this case, A, a ring of formal power series of a discrete valuation ring,
26:24
and this special case of mixed characteristics, which was studied by Bart, maybe it was published two years ago, I'm not sure, which is the case where
26:47
when you invert p, you get an eta, so a finite eta extension. So in that case,
27:07
one can, instead of this A, 1 over p infinity, one can replace A by, so one can define, let's say, A infinity to be something which you got by,
27:25
so get by ramifying all the variables, and maybe p itself, that's fine, and B infinity,
27:45
which replaces this thing, will be the normal closure, or integral closure, of A infinity, of,
28:09
which you get when you invert p. Yes, this is the same as formal power series, because in A,
28:30
okay, so now the the key properties which replace this argument of trace is Faltin's
28:50
almost purity, which says that when you have a thing like this finite eta extension, I should say that you get also that A infinity 1 over p inside B infinity 1 over p is again finite eta,
29:11
of course, yes, and then there is an extra property. Why I consider this one? It's not a
29:25
perfect thing, but it is perfect to read, in the sense that I have the property that when you went out by p, you can let Frobenius act, and Frobenius is surjective here.
29:41
So Faltin's almost purity says that in this situation, you can almost remove the 1 over p, so B infinity is, so p 1 over p infinity almost finite eta over A infinity. In particular,
30:07
it is almost faithfully flat. So now you have the following diagram,
30:25
which is analog to this diagram, A, you put it here, A, A infinity, B infinity, B. Okay, so here this is faithfully flat, in fact three,
30:49
faithfully flat, and here it is almost faithfully flat. So this implies in particular that when you consider the extension not for A, but for A infinity and
31:13
B tensor over A infinity, this splits almost. Now, one needs some device to remove the almost.
32:09
Device is a simple lemma, says that if R is a local ring, so a local
32:38
material ring, and S is a faithfully flat extension, maybe not material,
32:55
M, a finitely generated R module, which is such that
33:06
M tensor with S is killed by some idempotent ideas. Let me call it J, where J is J square,
33:21
an ideal in S. And you assume that this idea intersected with R is not zero, then M is zero. So this condition says that M is almost zero in the sense of I, or rather M.
33:42
When S is almost zero, then you can deduce that M is zero. This is very simple, but it's crucial, it's the way you tie together the Newtonian world with the non-Newtonian world of all this perfect reasoning. So you apply this to R is A, and M is
34:16
the A module generated by the extension class given by the exact sequence. This S is just A
34:34
infinity, and what I wrote there is that A infinity, you multiply it by any fractional power of P,
34:52
I'm sorry, A tensor one, one over A infinity, is zero.
35:00
And so the conclusion by this lemma is that E is zero, so star splits. So this is a slight rewriting of the proof suitable for my extension to the general case.
35:21
So now I would like to keep a little bit this line of thought, but attack the general case. So maybe before I do this, let me make a comment. Of course, one could elaborate a little bit this proof and weaken the assumptions. For instance, instead of assuming
35:48
there that you really have a netile extension after inverting P, you can ask for some logarithmic version or relative logarithmic version, and it can be done. But what you cannot
36:03
do is to hope that by some miraculous resolution of singularity, you will get the general case by this kind of argument. Because splitting properties, this exact sequence star splits,
36:23
you cannot descend this property by blowing up, for instance. So this is one case where I think any hope to use logarithmic geometry in this context is vain.
36:45
So one has to attack the... So really the case where you have some ramification and especially some analog of Felting's almost purity theorem in a ramified case.
37:04
But in some sense, the assumption that it was regular maybe could be changed to some assumption where just the almost purity still exists. When you have a tower and the almost purity exists. So it looks like the regularity in this case is a little bit...
37:22
In fact, I use even the... You use it to reduce to this case, but if I start with A where I have a tower and almost purity, I cannot play the same game. Exactly. It's not important. This is not... Just I like to have concrete things. You can take some powers of the TI's in a regular ring.
37:47
Here you have something. You have canonical coordinates. So let me explain what one does in general. So I was explaining the strategy, so general case.
38:09
The general case one needs one floor more. So consider as before, this A infinity is the same. B, B infinity, but I have to consider something more. B infinity, infinity hat. So there is
38:31
one floor more. Well, this is obtained by adding, adjoining some roots of the discriminant.
38:41
So this is very much in the spirit of Abiancar's lemma. I will speak about this effectively Abiancar's lemma. So this A infinity, infinity hat is a completion of A infinity, G over P infinity, just like this, where G is a discriminant. By this, I mean an element of
39:11
A plus B A such that B over 1 over PG is finite et al over A over 1 over PG.
39:33
So it turns out that... So what is this thing? So it's very... So it contains A infinity,
39:43
PG, which is just the ring which you get by adjointing the roots of G and completing periodically. But then what I do is I invert P and take the periodically integral element.
40:01
So there are more elements. In fact, it's very difficult to describe this. Very difficult to describe. In fact, even without any variable for n equals 0, it's very difficult to describe
40:21
explicitly this, at least at finite level. So this is what we need. And the B infinity,
40:43
infinity is almost what you guess. So it's complete integral closure of this A infinity, infinity of B tensor over A in infinity 1 over PG.
41:17
So maybe I will not try to comment on this complete thing. So we are not in the notarian
41:24
case. So in the perfectoid world, the notion of integral closure is not a good notion. It's much better to deal with complete integral closure and these kind of things. This was an old concept introduced by Krull, but a little bit forgotten.
41:43
So now I discuss analog in this context, this ramified context of fighting almost purity. So this is the perfectoid. So I keep the same notation.
42:39
So there's one. And in this case, it says the following, that this extension is
42:56
so it's first this p1 over p infinity almost finite eta after inverting G.
43:10
N is G p1 over infinity almost finite eta mod p to the m for n m.
43:30
Of course, the natural thing, the simplest thing would say that it is simply G p1 over infinity almost finite eta, but I cannot prove this at the moment for some
43:41
technical reason. So and in fact, this is part of the game which says that in fact, this ring is perfectoid, integral perfectoid. And this, which means if you prefer that
44:02
when you mod p, you take four menus, this is subjective. And this is almost the same for b, except that it is almost subjective. So this is almost perfectoid. Maybe it's not perfectoid. So in this context, where one deals with almost mathematics, not in the usual case,
44:25
for some, the usual case is when you take some non-disquiet valuation ring. Here, one has to deal maybe for the first time with really a more general case of almost
44:41
mathematics. Fortunately, the book by Gabor Hamaro is written in the complete generality, so it's really useful here. And in this context, well, there are some technicalities. Let me, so this is really in the spirit of Abirankar's lemma. So to get
45:04
some kind of almost subtleness, instead of inverting just discriminant, you ramify it. Discriminant here is PG. So I cannot say much about this. This is rather technical,
45:24
but it uses the perfectoid space. So I consider the perfectoid, so the principal perfectoid space attached to a infinity, infinity one over p.
45:44
And I consider the complement of the tubular neighborhood of divisor G equals to zero. So this is rational domain, bigger than p to the G. And then it
46:07
turns out that, so I know, so it's p to the J for various J, J is one, two, and so on. So I consider that. And the point is that this ring are in fact the limit of some localization,
46:34
so p to the J. Well, this is the ring of functions bounded by one on those domains.
46:56
So this is a kind of a Riemann, a perfectoid version of Riemann extension theorem,
47:02
and you have the same for b. But now here you have a perfectoid algebra of Scholz.
47:24
But you are in a situation where you avoid the discriminant. So you can apply Faltin's purity or some variance, like Liu and Scholz, to those extensions. And so you have
47:42
something almost finite et al. The problem is to go to the limits. It's not so easy. So I use a Galois technique here, which is I think very useful and simple, to show that something is almost finite is not easy. And one way is to show that it is Galois,
48:11
because to show that it is Galois is just an equation. It's not a property, so to speak. So this is a standard thing in a computative algebra. If you have a
48:22
finite extension and a group, a finite group, which acts on S such that the invariants are R, so you can write a map, canonical map, from S tensor S to the product of S by the element of the group, which you imagine. This is like in classical Galois theory. And if this map is an
48:50
finite et al, S is finite et al over R. So you get the finiteness for free. You just have to
49:03
prove that something is an isomorphism. But if it is an almost isomorphism, you get that it is almost finite et al. So this is a nice way. Now to have such an isomorphic, you have to prove that certain idempotent are in this algebra, but you know the Feltings theorem that they are
49:26
in all this algebra, and they are all compatible, so they go to the limit. By the way, this kind of thing can simplify the proof of Feltings' almost priority itself. The key point is just to prove
49:44
B, that if you have R perfectoid, and if you know that S is perfectoid, and the extension is finite et al, when you invert P, then you can get almost for free that the finite et alness,
50:06
almost finite et alness, reducing to the Galois case, because here you know that, so let me do it, so let's revisit, if I have time, yes maybe, let's revisit
50:24
Feltings' almost purity, so a small proposition, so S over A, R, finite et al,
50:41
extension of perfectoid algebras over some perfectoid field, then if you take the integral element, so S over R is almost finite et al. So there is a standard reduction to the Galois case
51:13
using a torsion, so you reduce to the Galois case, and then now you apply this
51:28
trick, so what you know is S tensor S over R is an isomorphic S, and then you take the integral element here, but it is known that integral, so at least if you take the completion
51:51
integral element is almost the same as the completed tensor product, so this is the basic
52:02
properties of perfectoid algebra. So you can conclude except for this completion, here there should be no completion in the definition of the Galois. So the trick is to reduce mod p, to say that this is okay, or mod p n for any n, and when you have something finite et al,
52:22
or almost finite et al, for mod p n, for all n, you can go to the limit, this is some variant of Grothendieck equivalence of Markable, but as Ofer-Gaber pointed out to me, you cannot do this in the general situation, where you deal with this nasty ideal, because p is not contained
52:43
in this ideal. So here you also need to get the full force of all things good, you have to prove it, you start from algebraic finite et al, but it is not perfectoid. You have to prove that S is perfectoid, so this is one part of this. Yes, but for instance in Kedlaja's
53:02
use approach, they prove first this, and then they use all staff robot rings to get the other thing. I claim that this small proposition dispense you of all these robot rings things. Scholz has another way of doing, he proves both things simultaneously. So let me explain
53:26
now maybe the second part. So this gives you here, so I imitated this part, so at least I have, this is almost faithfully flat, at least mod p to the m for all n.
53:43
So I have a question, I forgot a little bit about what you, so you have this, there are several possible completion and the order which is important in this, so when you construct, you add g1 over p to the n, you can edit algebraically just there, and then you could pass the limit over n, periodically complete, but now this one is not
54:11
uniform, is it, so there is a concept of uniform Banach algebra. Yes, this one,
54:21
you have to take this one, so this is the unit ball for the spectral norm, this one is not good. Yeah okay, now is this one, so is it the case that the norm on this is equivalent at least to the spectral norm or even not so? In this simple case, it is really a spectral norm, it is a spectral norm that I consider.
54:50
When you take just the periodic completion, before what you crossed out, so what do you take, the completion is over what, after 1 over p,
55:01
where is the completion in the formula, because it depends? Well, let me take it like this, yes, I think it makes no difference in this case, but in any case I mean this, I mean something complete.
55:21
Okay, with this definition it looks like okay, but now the fact that it contains the actual completion is not completely clear unless you prove that the periodic is something that there are no elements of spectral norm zero in this Banach algebra.
55:42
Yes, I think it's okay, but I don't need this, you can forget about this thing, which is the wrong thing to consider, so I really consider a perfectoid algebra. And you don't prove that the thing before is uniform, I mean it's not, I forgot now it's not. That one? Yes.
56:00
It is not. I do not know. No, no, no, it is not. Okay, this is what I... No, no, it is absolutely not, this one thing is much bigger than this thing. So okay, let us to finish, so if you word about the other part, so you know that this is free,
56:22
so face fully flat, what you need is to know some flatness, then you can copy the argument, which is maybe still there, no, or there, no maybe not, but with the extensions. So let me explain something like this, so this is the last but important point,
56:50
this is not the most difficult part, but this may be the most surprising, because these extensions, as I said, are a little bit mysterious at finite level.
57:03
So this is the last point, adjoining roots of G. So the theorem is that this algebra is p to the 1 over p infinity almost face fully flat
57:42
over a infinity. So once you have this, you can copy exactly the argument I told you before, but just there is one formula. So this is a little bit surprising, because as I said, if you consider, for instance,
58:03
if you consider, just you add some maybe square of G to A, and invert p and take the element, so it's not known, it was not known,
58:27
that whether it is flat over A. Do you do it for a given n or all of them?
58:41
This is a no, for n even for n equal 2, it's not clear, I think. So in fact it is maybe, there are some counterexamples, in fact, I think, not exactly for, so maybe, to make counterexamples, maybe I should also invert,
59:02
take square root of p, then maybe it's not flat, and the diarangement conjecture would imply that at least there is a splitting, and it was not even known, but here, going to infinity makes things simpler.
59:27
So the idea, so I just have to say what is the idea, the idea is to add one variable t, and look at the perfectoid, so this is a geometric idea, like another case,
59:44
the perfectoid space attached to A infinity t, so you have this perfectoid space, and in it, you have
01:00:00
divisor, if I may say, t equals to g, and you consider small tubular neighborhood. So t minus g s equals to p to the i. Then the key device in the proof is to reinterpret this algebra, which is constructed
01:00:28
by adjoining roots and roots of g and then completing and taking some integral closure, the ring which you get when you invert p, then it can be considered as a completed direct limit
01:00:51
of rings or functions. So these are functions bounded by one on
01:01:10
this domain. So what you have to prove is that these rings are in fact almost facefully flat over
01:01:30
infinity. So you still have to take the unit ball in those things? Yes, my notation is very vague. You have to invert the p and take the ball.
01:01:45
Yes, by this I mean I invert p and take the unit ball, which amounts geometrically to say that I take all functions bounded by one geometrically. So then one has to use the fact
01:02:05
that those things are perfect to it. Scholz's approximation argument to replace this t minus g by something which admits roots and then to use another proposition of Scholz, which describes
01:02:22
explicitly those localizations, at least at the almost level. So the point is that by the perfectoid theory, you can describe those rings going to infinity, adjoining infinitely
01:02:42
many roots of g, almost explicitly by a double co-limit, completed co-limit, and then at the end you can prove by some computation this plays a role of constant here. So it's reasonable that you get without too much effort, almost faceful flatness.
01:03:02
And the equal sign is precise or up to almost? Because your function bounded by one, when you extend it to a taller neighborhood, maybe the bound will be slightly worse. So maybe this equality is... No, no, it's okay like this. It's okay. This is something of this kind and this is a completed co-limit of something
01:03:27
of this kind. I mean... So when you just apply to the unit ball, the limit of unit balls, when you complete it, you get at least something which will be essentially the same.
01:03:44
I'm asking about subjectivity, because the function is... I mean even without taking unit ball, if there are perfectoid algebras, you invert P. I say that this one is the co-limit in the category of perfectoid algebra of this one.
01:04:02
And then I take... But it's a project of a co-limit in the category of perfectoid algebra, as you say. You take the unit ball for the spectral norm in the co-limit. I don't think it is true on the nose that it is the completed direct limit
01:04:21
of the unit balls. I think that it is true up to almost, because you cannot... In general, if you take something in the... It can be approximated by something in some finite stage whose norm is very close to being less than or equal to one, but not exactly, maybe. I'm not sure. And so it is not...
01:04:43
But it's enough for your purposes. It's enough and I think in this case this works anyway, but we can discuss. If you like, you can put an almost and this is enough for my purpose, but I think this is really true as it stands, but we can discuss.
01:05:03
So thank you. Okay, so we will take maybe a few questions from Tokyo, then from Beijing. So we start by talking.
01:05:21
Okay, so Tokyo. So is there some other question from Tokyo? Sorry, no question from Tokyo. Sorry, no question from Tokyo, from Beijing.
01:05:45
A question from Beijing. Stupid question. Stupid question. So usually in the perfectoid story you see some tilt occurring, but here we saw no tilt. So is there...
01:06:07
So is it hidden? Hidden? Yes, it is hidden, yes. It's hidden everywhere. For instance, everywhere. So when I state that somehow this property here, for instance,
01:06:28
this is proved by tilting. So many things are proven by tilting. I just... Yes.
01:06:42
No, tilting is very useful in all these stories. No more questions from Beijing. Okay, thank you. So questions from Pence? Yes, so in your abstract you also mentioned something about Bitcoin, Macaulay...
01:07:02
Thanks for the question. I had no time to mention it. So I will return to this original motivation. I mentioned the homological conjectures, can be just a little bit more precise. So in fact, this is a basic problem in
01:07:26
commutative algebra that not all rings are coin Macaulay. So it's so nice to have coin Macaulay rings. But at least you can try to map a non-coin Macaulay ring to some coin
01:07:46
Macaulay ring. It's such a way, a local ring, such a way that the maximal ideal of the original ring is not mapped to zero. Okay, in general it's hopeless to expect that you
01:08:02
do this in the notarian world, but you can expect some big coin Macaulay algebra, which contain, not necessarily content, but which are, such that any local notarian ring is mapped to them. So this is in fact a conjecture by Huxley. I think that
01:08:21
they always exist. And it is slightly, this implies a direct sum and conjecture, and this is only slightly stronger. What I meant in the abstract is that by combining these kind of techniques with some tools due to Huxley, one can prove this
01:08:47
existence of big coin Macaulay algebras through this conjecture. In fact, the only remaining open case is a case of mixed characteristic, and that's what I did in my
01:09:01
second paper, the pre-published version, so there is a section on this. So there is still a stronger conjecture, which implies all of the homological conjecture, but it's still open, but I think it's very accessible. It would be that
01:09:21
these big coin Macaulay algebras, which now exist, should be weakly functorial, so that if you have a local map of local rings, you should be able to construct big coin Macaulay algebra in order to make a certain square commute.
01:09:43
This I can almost do, except that there is a very nagging problem related to this algebra, that I don't know that this algebra, which is just, let me recall, this is just essentially you take this v and you add all this p,
01:10:04
this t as 1 over p infinity, and the g 1 over p infinity, and you take, you complete somehow. So this is the same, so you can also consider
01:10:23
g minus 1 over p infinity of this ring, and is this an isomorphism? I don't know. So this is just the intersection, by definition, of all these modules, all k,
01:10:44
but in fact this is a ring, and this is, I don't know in fact, this is a very simple question, but it's very nagging, so if this is the case, or at least if you can choose, in any case,
01:11:00
g is such that this is the case, I think one can prove with the same techniques, weak functoriality as well. Other questions? So maybe just because of your question, the almost purity in the beginning is really motivated by computing et al cohomology, so the new case you prove of this almost purity,
01:11:26
does it have some consequences on et al cohomology? Could you compute more, or could you simplify a little bit some parts of Fenton's computation of almost et al cohomology? I understand your question, I can't really
01:11:46
answer by a very concrete answer, but I can say something in this direction. Up to now, in order to apply Fenton's almost purity, you had to have coordinates,
01:12:01
you have to extract some roots of coordinates, so somehow you need to be in a toric situation. So you have to, yes, but it's very, it's annoying because it's difficult to match and you have to reduce the toric situation.
01:12:22
Here, this gives you a way to construct almost perfect to it envelops to any, let's say, rigid algebra by doing, by using notar, how do you say, how do you say this, notar,
01:12:49
normalization, so you have an affine algebra, an affine algebra, you take notar normalization, so you start with the big, which is an affine or affine algebra, you take notar normalization
01:13:23
and you have a discriminant here, so this is over p edicts, and then you can play this game, construct a infinity and the b infinity, and this one contains b, it is almost perfectoid,
01:13:45
this is a way to construct a lot of perfectoid or almost perfectoid algebra without reducing to the toric situation, so in that sense maybe it is useful to compute
01:14:01
cohomology without reducing to small affine subsets. Thank you, so if there is no other questions, we thank the speaker again.