These are my live-TeXed notes for Professor Bhargav Bhatt's Eilenberg lectures *Geometric aspects of p-adic Hodge theory* at Columbia, Fall 2018.

Notice that Bhargav has written up his own notes (read those!) I keep the live-TeXed notes here for the sole purpose of reinforcing my understanding.

09/05/2018

##
Introduction

The goal of this series to explain prismatic cohomology, which unifies various cohomology theories in -adic geometry.

###
Motivation

The motivation comes from the classical de Rham comparison theorem.

(de Rham, Serre)
Let

be a compact complex manifold. Then

Notice that RHS is the cohomology of the holomorphic (or algebraic, hence the name "Serre") de Rham complex of

, which depends on the geometry of

, while LHS only depends on the topology of

. Explicitly, the isomorphism is given by integration along cycles and each homology class

defines an obstruction to integrating

-forms on

.

RHS is rather computable (e.g., by Macaulay) using the defining equation of , which helps one to obtain topological information about . Conversely, the comparison shows that deforming the complex structure without changing the underlying topology still keeps holomorphic invariants.

How to see mod

cohomology classes (or equivalently,

-torsion classes) on

"geometrically"?

At least when

is algebraic, these give obstructions to integrating

-forms on "

".

###
Global statements

Let be a smooth closed subvariety, whose defining equations live in . Assume that is still smooth mod (i.e., good reduction outside ).

(Bhatt-Morrow-Scholze)
In this set-up, for any

. We have

Here

is the associated complex analytic manifold,

is the reduction of

mod

, and

is the algebraic de Rham cohomology.

If

. Then

has no

-torsion.

This gives an *algebraic* way to control the torsion in singular cohomology.

Conversely, torsion in singular cohomology forces de Rham cohomology to be larger, as in the following example.

Suppose

is an Enriques surface in characteristic 2. Then

has the form

for some Enriques surface

in characteristic 0.

is always a quotient of K3 surface by a fixed point-free involution, so has fundamental group

. So the inequality in Theorem

2 implies that

even though

(this recovers an example of Illusie, W. Lang).

The strategy of the proof of Theorem 2 is to consider a cohomology theory valued in -modules (an example of *prismatic cohomology*), such that

- There is an identification .
- There is an injective map
Notice that a) + b) clearly implies Theorem 2.

###
Local structure of prismatic cohomology

Fix a prime for the rest of the semester ( is a very good prime for computation in prismatic cohomology!) Set . Let a "Frobenius lift" on from . Let be the ideal defining the "diagonal" of . Let be (the -adic completion of) a smooth -algebra. Write , and , where is an algebraic closure of . Let be the algebraic de Rham complex of .

Let

be the

-adic completion of

. Then

is a free

-module of rank 1 generated by

, and higher exterior powers vanish. So we find

09/17/2018

##
-rings

Last time we tried to make the point that prismatic cohomology is a deformation of de Rham cohomology (traditionally crystalline cohomology is an example of such a deformation). We would like to stress that it is a good idea to also carry along a lift of the Frobenius with the deformation. This leads to the notion of -rings (Joyal, Buium, Borger...)

Let be a (commutative) ring with a map such that . Then for each , we have . The notion of -ring is trying to remember (rather than ). Notice that if is -torsion-free, then is a function of .

A

*-ring* is a pair

where

is a ring,

is a map of sets satisfying

- ,
- ,
- .

- If is a -ring, then gives a ring map lifting .
- If is -torsion-free, then a) gives a bijection between the -structures on and endomorphisms lifting on .

For example, given

, let us check

is additive:

which is equal to

as desired.

¡õ
- There is a unique -ring structure on , given by = identity. Then . One can check this is the initial object in -rings.
- Let . Then for any polynomial , the map gives a -structure on .
- Let be a perfect field of characteristic . Then the ring of Witt vectors (the unique -adically complete and -torsion-free ring lifting ) with (induced by on ), is a -ring.
- Let -algebra. The any ring map gives a -ring on (as we can write down by dividing ).
- Let (not -torsion-free). Then there exists a unique -structure on with (the existence would not be true without quotienting ). So there can be -torsion in -ring. However, cannot kill 1 as the following lemma shows.

There is no nonzero

-ring

where

for some

(think:

is a "

-derivation", which lowers the order of

-adic vanishing by 1).

Say

, and

is a

-algebra with a

-structure. Apply

to

we obtain

. On the other hand

Hence

and

is the zero ring.

¡õ
An element

in a

-ring

has

*rank 1* if

. This terminology is motivated by

-theory where

comes from the Adams operator and when

,

is what the Adams operator does on line bundles.

###
The category of -rings

Reference: C. Rezk, *Etale extensions of -rings*.

(Truncated Witt Vectors)
For any ring

, we define a new ring

as follows: as a set

, with addition

and multiplication

Using these formulas, it is easy to check the following lemma.

Let

be a

-ring. Then the map

is a ring map lifting the identity after the restriction map

And conversely, any section of the restriction map

gives a

-structure on

.

The category of

-rings has all limits and colimits, and they are computed on the underlying rings (i.e., limits and colimits commutes with forgetful functor to rings).

It is easy to check for limits (embed the limit into a product and check component-wise). For colimits, say

is a diagram of

-rings. They give maps

compatible in

. We obtain a map

. There is also a canonical map

(as

is a functor). One can check the composition gives a

-structure on

.

¡õ
The free

-ring

on a variable

is given by

and

.

Consider

say sending

(or any other polynomial). Pushing-out along

we obtain a

-ring

in which

. However, it is hard to explicitly compute

, as free

-rings are of infinite type.

Let

be a

-ring. Let

be a multiplicative subset such that

. Then

has a unique

-structure compatible with that of

.

###
Perfect -rings

Now we discuss a property that is special to the category of -rings.

A

-ring

is

*perfect* if

is an isomorphism.

Perfect -rings are essentially perfect algebras in characteristic :

The key lemma to prove this theorem is the following:

If

is a

-adically complete

-ring and

is a

-torsion element, then

. In particular, perfect

-rings are

-torsion-free.

Apply

to

, we obtain

Since

is

-adically complete, we know that

is a unit. To prove

if suffice to show that

. But

Here the last equality is due to

.

¡õ
09/24/2018

We recall the following standard construction.

Let

be a perfect

-algebra. There exists a unique multiplicative map, the

*Teichmuller lift*,

splitting the projection

. Given

, we define

to be

, where

is some lift of

. One can check this is well-defined using essentially the binomial identity:

Any

can be written uniquely as a power series

One simply write

to be the image of

in

,

to be the image of

and so on.

¡õ
These 's are called the *Teichmuller coordinates* of (though these coordinates does not respect the ring structure).

###
Distinguished elements

For a commutative ring , we write to be the Jacobson radical of . A more useful way to think about it is that it consists of "small" elements We will *always* assume that (so all other prime numbers are invertible in ).

An element

in a

-ring

is

*distinguished* (or

*primitive*) if

(this terminology dates back to Fontaine). Since

is a ring map, we know that if

is distinguished, then

is also distinguished.

A distinguished element can be thought of as a "deformation" of .

- (crystalline cohomology) . Then is distinguished as . In fact, is distinguished in any -ring .
- (-de Rham cohomology) (). Then is distinguished. In fact, consider the map . This a -map. Moreover, is a unit if and only if is a unit (as is topologically nilpotent). Now , so is a unit, hence is a unit. The intuition here is that specializing to allows one check to an element is distinguished more easily.
- (Breuil-Kisin cohomology) (). Then is distinguished. One can check this using the specialization . (here is even "smaller" than as contracts ).
- (-cohomology) -adic completion of (). Then is distinguished (from b)).

In all these examples we know how to deform the de Rham cohomology from to , but the constructions are different (at least three constructions). Our goal is to uniforms these different constructions.

Let

be a distinguished element in a

-ring. Assume

. Then

is also distinguished for any unit

.

Notice that

the first and third terms are in

(as

and

are), and the second term is a unit (as

is distinguished), we know that

is also a unit.

¡õ
(Irreducibility)
Assume

is distinguished. Assume

. Then

is distinguished and

is unit. (In particular, in a

-ring there can not be a square root of

).

Notice

Again the first and third terms are in

. So the second term

is a unit, and hence

is a unit and

is distinguished.

¡õ
Assume

. Then

is distinguished if and only if

. In particular, the condition that

is distinguished only depends on the ideal

.

Notice

. So if

is distinguished, then

. Conversely, write

. We need to show that

is a unit. This is equivalent to that

is a unit in

, or equivalently

. We will proceed by contradiction. We may assume that

(after localizing along the locus where

is not a unit). Then

for some

. So

Notice that RHS is distinguished as

and so

is a unit. Thus the irreducibility lemma (Lemma

10) implies that

is distinguished, a contradiction.

¡õ
###
Digression: derived completions

A prism will be a -ring together with an ideal that is locally cut out by distinguished elements (but not necessarily globally). It is also more convenient to assume that it is "complete" along the ideal. We thus need a good notion of completion (for non-noetherian rings).

Let

and

. An

-complex

is

*derived -complete* if for any

,

is 0 in

(here

is the right derived functor of the inverse limit functor). This is equivalent to

where

acts on

by

. Notice that the completion here is less naive (using the noetherian ring

instead of

). See notes for the rest of the assertions.

- All derived -complete -complexes form a triangulated subcategory closed under product. It has a left adjoint .
- is derived -complete if and only if each (put in degree 0) is derived -complete.
- All derived -complete -modules form an abelian subcategory of all -modules.
- (Derived Nakayama) Assume is derived -complete. Then if and only if .

10/08/2018

##
Prisms

Recall (Lemma 11):

A

*-pair* consists of a

-ring

and an ideal

.

Prisms will be the objects of the prism site.

A map

of prisms (a ring map

sending

into

) is

*(faithfully) flat* if

is (faithfully) flat. Here flat means that the target complex has cohomology only in degree 0 and this cohomology is flat over the source.

A prism

is called

*perfect* if

is perfect;

*crystalline* if

;

*bounded* if

has bounded

-torsion. Every prism we will encounter will be bounded.

- Any -torsionfree and -adically complete -ring gives a prism .
- Perfect prisms = perfectoid rings.

We don't know a natural example of a prism of where is locally principal but not principal (though abstract examples exist). The following lemma shows that is not far from principal.

Let

be a prism. Then

is principal and any generator is a distinguished element. In particular,

(exercise).

Write

, where

. Then one can check that

is a generator of

. The key is to use the irreducibility lemma (Lemma

10) for distinguished elements (see notes).

¡õ
Let

be a map of prisms. Then

is an isomorphism (so a map of prisms is determined on the underlying

-rings).

Both sides are locally generated by distinguished elements. Use the irreducibility of distinguished elements locally on

.

¡õ
##
Perfect prisms

A commutative ring

is

*perfectoid* if

for a perfect prism

.

- Let be a perfect and -adically complete -ring, and . Then for a perfect -algebra and (Theorem 4). So any perfect -algebra is perfectoid.
- with is a -torsionfree -ring, and is generated by a distinguished element. So is a perfectoid ring.

Let

be a perfect

-algebra. Let

. Then

(

-torsion in

) is equal to

for any

. (The latter is killed by a small power of

, known as an

*almost zero* module).

Suppose

such that

for some

. So

for any

. By reducedness we obtain

for any

.

¡õ
Let

be a perfect prism. Then

. In particular, perfect prisms are bounded.

See notes.
¡õ

The functor

gives an equivalence between the category of perfect prisms and the category of perfectoid rings.

. Then

(a "2-dimensional" object), and

(all nilpotents in

are killed when passing to the perfection).

Write

and

. (a) is clear since Frobenius on

is surjective and

is a quotient of

. For (b), write

, where

. So we may write

, where

. Thus in

, we have

. We may then take

(the Teichmuller map is multiplicative). For (c), we claim that

. It suffices to show that

is reduced (equivalently, perfect). This follows from the general fact that if

is a perfect

-algebra, and

an ideal. Then

.

¡õ
- If , are maps of perfectoid rings. Then is perfectoid (no Tor even without flatness assumptions in the perfectoid world!)
- If is perfectoid, then is perfectoid (and -torsionfree), and is a pullback diagram. So one can build any perfectoid ring using a -torsionfree perfectoid ring and a perfect -algebra . In particular, is
*reduced*.

10/15/2018

##
The prismatic site and the prismatic cohomology

We will let be a "base" prism. Assume is generated by a distinguished element and is bounded.

- (crystalline) Let be any -torsionfree, -complete -ring and .
- (Breuil-Kisin) Let with and , where is any Eisenstein polynomial (e.g. ).
- () Let be a perfectoid ring and .
- (-de Rham) Let with , and .

Let be a formally smooth -algebra (e.g., the -adic completion of a smooth -algebra). Our goal is to construct an object such that

- gives differential forms on relative to .
- is related to etale cohomology of .

The

*prismatic site* of

relative to

consists of prisms

over

together with a map

over

. (Notice the direction of the map is different from the crystalline site). Pictorially we have

We will write it as

.

Define functors

and

on

given by sending

to

(an

-algebra over

) and

(an

-algebra) respectively. (In the perfectoid case

and

).

- Let . Then is the category of prisms over . In particular, it has a initial object and hence the prismatic cohomology will be simply .
- Let (the -adic completion of ). Then there is no initial object. There exists formally smooth lift of together with a -structure on (e.g., and . Then . (One can also do the same for any ).

The

*prismatic cohomology* of

is defined to be

It admits a Frobenius action

(as the category

admits a Frobenius action).

The

*Hodge-Tate cohomology* of

is defined to be

It no longer has the Frobenius action (but is linear over the larger ring

).

These both are commutative algebra objects.

Let

. Then

.

##
The Hodge-Tate comparison

Let be a map of commutative rings. We have an *algebraic de Rham complex* We view it as a strictly commutative differential graded -algebra:

- (graded commutative) .
- (strictly graded commutative) if is odd (which only matters in characteristic 2).

(Universal property of the de Rham complex)
Let

be a graded commutative

-dga. Assume we have a map

of

-algebras. Further assume that for any

,

squares to 0. Then there exists a unique extension of

to graded commutative

-dgas,

We have a Bockstein exact sequence of sheaves on , Thus we obtain a *Bockstein differential* In this way we obtain a graded commutative -dga . We also have (by -algebra structure). One can also show that for any , squares to 0. Thus by the universal property of the de Rham complex, we obtain a map from the de Rham complex to the Hodge-Tate complex:

(Hodge-Tate comparison)
is an isomorphism (as genuine complexes). In particular,

And so

can be represented by a perfect complex (this is the source of finiteness in the global situation).

How to compute cohomology of categories? It turns out to be extremely simple.

Let

be a small category. Let

be the category of presheaves on

. Then

is the derived functor

of

Assume there exists

that is

*weakly final* (i.e.,

for any

). Assume that

has finite nonempty product. Then

is calculated by

So in order to compute the cohomology of the prismatic complex, it suffices to find a weakly final object and compute its self-products.

Let

be a

-pair over

. Then there exists a universal map

to a prism over

.

Pure category theory.
¡õ

We denote this universal by (think: the universal prism where becomes divisible by ).

has finite nonempty coproducts.

Let

and

be two objects in

. Set

. We have two maps

via

(and similarly for

), which are not necessarily the same. ; Let

and let

, then the two maps becomes the same in

. Then

works.

¡õ
10/22/2018

The goal today is to sketch a proof of the Hodge-Tate comparison theorem, which is divided into two steps.

- If (characteristic case), prove the crystalline comparison for (before reduction).
- Use the Cartier isomorphism when to deduce the Hodge-Tate comparison theorem in general.

###
Reminder on crystalline cohomology

Fix a -torsionfree ring , and a smooth -algebra .

(Divided power envelope)
Let

be a (ind)-smooth

-algebra. Fix a surjection

over

. (think: embed

into a smooth affine space.) Let

. The

*divided power envelope of * is defined to be

(

-adic completion), where

. (This is the characteristic

analogue of allowing integration of power series in

.)

One can check that there exists an induced surjection with kernel having divided powers. The map is universal with this property (hence the name envelope).

Let

,

, and

. Then

is the divided power polynomial ring. One can check that

has the cohomology of a point (

,

). This is the characteristic

analogue of the Poincare lemma.

Choose a surjection

with

(ind-)smooth over

. Let

a cosimplicial

-algebra. We have an induced surjection

with kernel

for any

. In this way we obtain an ideal

. We define

(think: geometrically embed a singular space into powers of a smooth affine space and look at its tubular neighborhoods). Define

(de Rham-crystalline comparison)
If

is a smooth lift of

to

. Then

In particular, the de Rham complex on LHS is independent of the lift

.

There exists a canonical quasi-isomorphism

(Cartier isomorphism)
Let

be the Frobenius twist of

. So the absolute Frobenius

factors as

, where the relative Frobenius

is

-linear. Notice that

is an

-linear complex (by the Leibniz rule we have

for

). We have a map

and also have a Bockstein differential

So by the universal property (Lemma

17), we obtain a map of strictly graded commutative dgas,

(Cartier)
This map is an isomorphism.

###
Relating divided powers to -structures

Let

be a

-torsionfree

-rings. Let

be an (ind-)smooth

-algebra over

. Let

such that

is a non-zero-divisor on

. Then

Here LHS means formally adjoining in

-rings, and the RHS is the divided power envelope.

First consider the free case

over

. We have a pushout diagram

Notice that the left vertical arrow is faithfully flat, and hence the right vertical arrow is faithfully flat, thus

is

-torsionfree.

The next goal is to identify with . First, by we obtain that hence . To check that , by induction it suffices to show that if is any -torsionfree -ring, and such that , then . In fact, The second term has the same -adic valuation as , so it remains to show that the first term is in . Indeed, by assumption that we have and thus

Finally, one can show that (key fact: , where is a -adic unit).
¡õ

When

, reduce to the previous lemma. In general, do a base change argument.

¡õ
10/29/2018

###
Crystalline comparison for prismatic cohomology

Let be a -torsionfree complete -ring. Then is a bounded prism. Let be a smooth -algebra.

There exists a canonical isomorphism

of commutative algebra objects in

, compatible with the Frobenius action on both sides.

Let us explain why there is a canonical morphism from LHS to RHS.

Choose a free -ring together with a surjection . Recall that by definition . Since is now chosen to have a -structure, we know that is a cosimplicial -algebra over . On the other hand, the kernel of has divided powers. Consequently, we have This the *key* relation between the -structure and divided power structure and gives a commutative diagram Notice that the lower right part of the diagram gives a cosimplicial object in the prismatic site . Thus we obtain a canonical map So by adjunction we obtain a map One can check this is an equivalence by explicitly computing both sides using Cech-Alexander complexes and using Lemma 20 to identify -structures and divided powers structures.

###
Hodge-Tate comparison

Let be a bounded prism. Let be a formally smooth -algebra. Now let us sketch a proof of the Hodge-Tate comparison (Theorem 6).

- Deduce the Hodge-Tate comparison from the crystalline comparison (Theorem 9) and the Cartier isomorphism (Theorem 8) when (and worry about the Frobenius twist when is not perfect).
- In general, set . Then the irreducibility lemma for distinguished elements (Lemma 10) implies that in . We look at the base change along . Using Lemma 20, we may deduce the theorem from the previous case .

For any formally smooth

-scheme

, one can define its

*prismatic cohomology* , compatible with passing to affine open subsets

(i.e.,

).

##
Extension to the singular case

###
Non-abelian derived functors

Our next goal is to explain how to extend the prismatic cohomology to the singular case. We will need to derive the functor of Kahler differentials, which is a functor from rings to modules. Since the category of rings is not abelian, we need some formalisms on non-abelian derived functors.

Let be a commutative ring. Let be the category of finitely generated polynomial -algebras. Consider a functor e.g., . To derive , we view as "projective" objects in .

###
Cotangent complexes

(Cotangent complexes)
Define the cotangent complex

to be

.

###
Derived de Rham cohomology

Fix a ground ring of characteristic .

Define the derived de Rham cohomology

to be

.

In char 0, for an affine space the functor is a constant functor so there is no higher derived de Rham cohomology. But in char , even for an affine space has a lot of cohomologies (given by the Cartier isomorphism). The following property helps us to control the derived de Rham cohomology.

(Derived Cartier isomorphism)
For any

, there exists an increasing exhaustive filtration (

*conjugate filtration*)

on

equipped with canonical isomorphisms

where

. More precisely, we have a lift of

into the filtered derived category

.

11/12/2018

If

is smooth, then

.

For

smooth, we have

.

¡õ
The moral here is that if we have a functor whose cohomology can be described in terms of differential forms, then its derived functor should also has a description in terms of differential forms. We will realize this idea for prismatic cohomology as well.

###
Derived prismatic cohomology

Let be a bounded prism. Let be a formally smooth -algebra. We constructed the prismatic cohomology together with an action of the Frobenius . Moreover, we have the Hodge-Tate comparison

The

*derived prismatic cohomology* is obtained by deriving

where

is the

-adic completion of

, and

is the derived category of

-complete

-complexes. Define

which is the same as the derived functor of

(as the non-abelian derived functor commutes with filtered colimits, in particular with

).

One can check that if is the -adic completion of a polynomial -algebra. We have the following derived version of the Hodge-Tate comparison.

For any

, we have an increasing exhaustive filtration

on

such that

in

.

From now on by abuse of notation we will write .

- is a commutative algebra object.
- is a commutative algebra object.

##
Perfections in mixed characteristic

Let be a perfect prism (e.g., , where ). Let be a -complete -algebra. Our next goal is to construct a "perfectoidization" of . It may not be perfectoid, but better be thought of as a "derived perfectoid ring". We will realize this using the prismatic cohomology of .

First let us look at the case of *characteristic *, where we already know what should be.

Let

be a perfect field of characteristic

. For any

-algebra

, define

. The map

is the universal map from

to a perfect

-algebra.

Let

. Then

.

Let

be a

-algebra, the prefection of the derived de Rham cohomology

(where

is induced by the Frobenius on

) identifies with

via the projection

.

Reduce to

= a polynomial

-algebra, and use

is zero for any

.

¡õ
Now we can do the same thing for prismatic cohomology.

Let

,

. Let

be a

-algebra. The perfection

identifies with

via the natural map

.

Use the fact that

kills

. There is an extra subtlety: the Frobenius for the prismatic cohomology comes from the objects on

, while the Frobenius on differential forms comes from the ring

in characteristic

. Nevertheless one can check they are the same.

¡õ
We have a natural identification of

(

-completion) and

.

Now let us come to the *mixed characteristic* case. Let be a perfect prism. Let be an -algebra. We use the prismatic cohomology to define the perfectoidization of .

- (-completion).
- (-completion).

- is a commutative algebra object, and is an isomorphism on it.
- is a commutative algebra object.

- If , then .
- If is already perfectoid (e.g., ), then (check: , the unique lift of from to ).
- Let . Let (a torus). We use the fact that (more on this later) Then one can compute that where and for any . Notice that when specializing to the term in degree one of this two-term complex is zero and we obtain a complex with one term. However, note that the element (= the image of ) in degree one is not a co-boundary, and also nonzero in (= nonzero mod ). Therefore is genuinely derived (with nontrivial higher cohomology).

11/19/2018

Two examples of perfect prisms to keep in mind today:

- .
- (especially for etale comparison).

- (there is no cohomology in negative degree; reason: power operation in homotopy theory).
- If (i.e., in degree 0), then is a perfect prism, and is a perfectoid ring.
- is independent of the choice of .
- The functor commutes with faithfully flat base change .

As an application, we now reprove one of the key lemmas in André's recent breakthrough on the direct summand conjecture.

(André)
Let

be a perfectoid ring. Let

. Then there exists

-completely faithfully flat map

of perfectoid rings such that

has a compatible system of

-power roots

in

.

Let

be a perfectoid ring and

an ideal. Let

. Then

lives in degree 0, and

is surjective.

Let

. Let

and

. I don't know how to describe

explicitly (or even write down a nontrivial element in the kernel explicitly).

##
The etale comparison

Let be a perfect prism. Let be a -complete -algebra. Assume is finitely generated over and is bounded.

There exists a canonical isomorphism

Our next goal is to prove Theorem 11. To simplify notation, assume . The main steps are:

- Reduce from to (most interesting step).
- Reduce to semiperfectoid (quotient of perfectoid; its prismatic cohomology lives in degree 0).
- Reduce to perfectoid (for which the theorem was known before: dates back to e.g. Fontaine-Wintenberger in 70's).

Step (a) A *continuity property* inspired by topological cyclic homology.

Let

be an

-algebra and

. Define the category of Frobenius modules

and a subcategory

such that

is

-complete. The colimits in

are computed by the

-completion of the usual colimit.

- The fixed point functor commutes with colimits.
- The fixed point functor commutes with colimits.
- Let , then the perfection induces an isomorphism on .

The upshot is that to prove the Theorem 11, it suffices to show the same thing for (after the perfection).

Step (b) Let be a polynomial algebra. We take the perfectoid ring . We have a faithfully flat map . Let be the Cech nerve of (i.e., ). Then each is *semiperfectoid*.

For more general , choose topological generators to get a surjective map . Repeating the previous construction we obtain , where each is semiperfectoid.

Now the strategy is to reduce Theorem 11 for to the same thing for .

- (descent for etale cohomology) (true for any torsion contractible sheaves).
- (descent for prismatic cohomology) and the same for and (the proof uses tools developed in Akhil Mathew's undergraduate thesis).

11/26/2018

Step (c) Let be semiperfectoid and be its perfectoidization. By Corollary 11, we know that lives in degree 0 and hence is perfectoid. We claim that both sides of the desired isomorphism do not change when replacing by . For the RHS, we claim that In fact both reduce to mod (by the Hodge-Tate comparison and the definition of ), and hence the claim is true by the derived completeness of both. For the LHS, one shows that induces an isomorphism of associated -sheaves (which we have not covered) and one gets (by a theorem) that they have the same etale cohomology for the generic fiber.

Therefore it remains to prove the etale comparison when is perfectoid. In this case, by a (classical) theorem, we have (in modern terminology, this can be proved using the pro-etale site of perfectoid spaces and the exactness of the Artin-Schreier sequence). Since is perfectoid, we have and hence . Thus we have proved the etale comparison when is perfectoid.

In particular, the -etale cohomology of lives in only two degrees (in fact, by the same etale comparison theorem with nontrivial coefficient systems, one can show that is a ).

Now let us come back to prove our main application.

Let

be complete and algebraically closed. Let

be the valuation ring (which is perfectoid) with residue field

. Let

be a proper smooth formal scheme. Then

Let

be the perfect prism associated to

. So

. The map

induces a map

, and a map of perfect prisms

. The prismatic complex

is obtained by gluing

for all open

. Similarly, we have the prismatic complex

. We have an base change isomorphism (by the Hodge-Tate comparison)

(Notice that since

is a formal scheme,

as topological spaces and we can identify sheaves on them).

Let , and similarly define . We claim that is a *perfect complex*, i.e., represented by a finite complex of finite free modules. To prove the claim, it suffices to show that . This is true by the Hodge-Tate comparison: has a filtration whose graded pieces are given by , which is perfect since is *proper* and *smooth*.

Let with fraction field and residue field . Then by the semicontinuity for finitely presented -modules, we obtain

The RHS is given by which is given by (the Frobenius of twist) by the crystalline comparison for prismatic cohomology.

For the LHS: by the etale comparison we obtain We then apply the following linear algebra lemma to to get the desired inequality. (In fact, using the -de Rham complex, we will see that is an isomorphism for and hence the dimension of the LHS is *equal* to the dimension of -etale cohomology of the generic fiber ).
¡õ

Let

be an algebraically closed field of characteristic

. Let

and

. Then

is injective, and moreover an isomorphism when

is an isomorphism.

##
The -de Rham complex

As an analogue of crystalline cohomology being the de Rham cohomology of a lift, we would like to compute on the nose (as a genuine complex).

We will work with with , and . Let .

(Aomoto, Jackson)
Let

. We define its

*-de Rham complex* (depending on the choice of

)

where we define the

-linear map

Notice that the definition makes sense since

. Notice that

. Therefore

on the nose. We have a similar construction for

.

Let

be a formally smooth

-algebra. A

*framing* of

is a formally etale map

We call such

a

*framed pair*.

(-de Rham complex for framed pairs)
Let

be a framed pair. We obtain a formally etale map

of

-algebras. For

, we have an automorphism

of

such that

if

and

if

. Since

, we obtain a unique automorphism

of

extending it (by the formal smoothness of

). We define

and the

*-de Rham complex* to be the Koszul complex

12/03/2018

Namely,

.

Calculus.
¡õ

(Scholze)
There is a symmetric monoidal functor

from formally smooth

-algebras to

, equipped with natural isomorphisms

for each choice

of framing. In particular,

is a commutative algebra in

, and each

is an

-algebra.

Our next goal is to prove Conjecture 1. It is hard to prove from first principle: it is already not clear how to write down the endomorphism of the -de Rham complex for the simple change of variable .

##
The -crystalline cohomology

Our goal is to construct a *-crystalline site* whose cohomology is computed by -de Rham complexes.

Write with . Let . Let . Observe that the ideals and define the same topology (one is contained in the power of the other and vice versa).

The main new ingredient is the notion of *-PD thickenings*.

- is a -PD pairing, and is the initial such pair. More generally for any -completely flat -algebra over , we obtain a -PD pair .
- If in , then condition (c) is equivalent to that for any , for any (the usual PD-structure requires , but it turns out to not affect anything).

Let

be a

-torsionfree

-algebra over

. Let

such that

. Then

If

, this lemma saying that if

, then

.

Reduce to the universal case, where

is

-flat. We would like to show that

In

, we have

(think:

and

are "transverse" to each other), so it suffices to show that

By definition,

applying

we obtain

and

by assumption.

¡õ
Take

, where

, and

form a regular sequence.

¡õ
Let

,

, and

with

,

. Then one can compute that

contains

And they form a topological basis over

. (This was the motivating example for inventing the

-crystalline site).

(-crystalline site)
Let

be a formally smooth

-algebra. Define

to be the category of

-PD pairs

such that

, and make it into a site via indiscrete topology (so presheaves=sheaves). Let

be the sheaf

. Define the

*-crystalline cohomology* to be

To compute , we use the Cech-Alexander complexes for (compare Theorem 7). Choose a surjection with kernel such that is a -completion of a free -algebra over . We obtain a cosimplicial -algebra over with ideal such that . Taking -PD envelopes gives a cosimplicial object of , Category theory (Lemma 18) then implies that is given by .

( specialization)
There exists a canonical isomorphism

- where (Theorem 7).
- is computed by (by Proposition 12 (c)), which also computes LHS by the what we just said.
¡õ

12/10/2018

##
-crystalline comparison and -de Rham comparison

Let be formally smooth over . Let . Notice that we have the following commutative diagram The left vertical column corresponds to the -crystalline cohomology, and the right vertical column corresponds to the prismatic cohomology. The twist is needed as is not a distinguished element but is.

(-crystalline comparison for prismatic cohomology)
There exists a natural isomorphism

To get a map

, we need to show that for each

, we get an object

. By the definition of

-PD thickening, we have a commutative diagram

Linearizing along

we obtain the desired object in the prismatic site. To check this map is an isomorphism, we check after

and use the crystalline comparison for prismatic cohomology.

¡õ
Our next goal is to relate to . In the classical characteristic setting this relation is saying that the crystalline-de Rham comparison does not depend the choice of the lift. One can basically add everywhere in the proof for the classical case, and it works.

(-de Rham complexes for -PD envelopes)
Let

be a formally smooth

-algebra, formally etale over

(which is given a unique

-structure satisfying

). Assume there is a surjection

and

is the

-PD envelope of

. By the following lemma, we can extend the

-de Rham complex

to

(from

to

).

Each

-derivative

extends uniquely to

.

Recall that

. It suffices to show that each

extends uniquely to an automorphism of

congruent to

mod

. Using the universal property of

, we need to show that for each

, we have

and

In fact,

for some

, and so

Dividing by

, we obtain the desired result.

¡õ
(-crystalline-de Rham comparison)
Let

be a framed

-algebra. Then there exists a natural isomorphism

Let

be the unique lift of

to

with coordinates

. Let

be the Cech nerve of

. We have a surjection

with kernel

, and

is a cosimplicial

-algebra over

computing

. To relate it to

, we use the bicomplex

. Notice that

is computed by the first row of this bicomplex and

is computed by the first column of this bicomplex. To show they are quasi-isomorphic, we combine:

- All horizontal maps give quasi-isomorphisms of the columns (reduce mod and use the Poincare lemma: adding free variables does not change the cohomology). So the bicomplex totalizes to the first column.
- All rows except first column are acyclic (write down explicit combinatorial contracting homotopy). So the bicomplex also totalizes to the first row.
¡õ

##
Prismatic cohomology via topological Hochschild homology

Finally, let us mention how to recover from the topological Homology homology . The characteristic 0 story is classical (Quillen, Connes, Tsygan...).

Let

be a map of (commutative) rings. Define the

*Hochschild homology* to be

(i.e., derived self-intersection of the diagonal in

).

We observe:

- .

(Hochschild-Kostant-Rosenberg)
If

is smooth, then

as graded

-algebras. Moreover, if

(characteristic 0), then we have a natural decomposition (a lift to the derived category)

To bring the de Rham differential in, one uses the following observation of Connes: has an -action. To see this use ( is obtained by gluing two 's at two points). Then induces a map which is the de Rham differential under the previous theorem.

Define the

*periodic homology* to be the Tate cohomology of

(a periodic version of

),

Assume

is smooth, and

. Then

where

has degree 2.

Now let us come to the prismatic story. Let be a perfect prism. Let be a smooth algebra over .

For any ring

, define the

*topological Hochschild homology* (point: tensoring over the sphere spectrum

gets rid of the factorials in the denominators). It also has an

-action, and we similarly define the

*topological periodic homology* to be the Tate cohomology of

.

(Bokstedt, Hesselholt)
Let

be perfectoid. Then

where

has degree 2 (if tensor over

instead of the sphere spectrum, one gets the divided power algebra), and

.

(Bhatt-Morrow-Scholze)
Let

be smooth over a perfectoid

. Then there is a filtration

on

with