Section 12 The Hodge-Tate comparison map
Reference.
In this section, we formulate our first application of prismatic cohomology, the Hodge-Tate comparison theorem. The proof will be sketched in Section 15.
Subsection 12.1 Graded commutativity for graded rings
Lemma 12.1.2.
For let be a commutative -algebra object in Then carries a natural graded ring structure, with respect to which it is graded commutative.
Proof.
The multiplication map on is given by a morphism in We may directly read off the multiplication in and its properties (associativity, distributivity over addition) to obtain the graded ring structure. It remains to check graded commutativity; this follows from the Koszul sign rule appearing in the Alexander-Whitney construction (see [83], tag 00P4). We will see this concretely in Remark 13.2.3.
Subsection 12.2 The de Rham complex
Definition 12.2.1.
For a differential graded algebra over (also known as a -dga) is a complex of -modules in which is also equipped with the structure of a (not necessarily commutative) graded -algebra subject to the signed Leibniz rule
We say that an -dga is commutative if is graded commutative. We say that it is strictly commutative if it is commutative and moreover for any of odd degree. (This last condition is redundant if is -torsion-free.)
The prototypical example of this definition is the following construction.
Definition 12.2.2.
The universal property of the module of Kähler differentials can be reinterpreted as follws.
Lemma 12.2.3. Universal property of the de Rham complex.
Let be a graded commutative -dga supported in degrees Let be a map of -algebras such that for each the element satisfies (note that this is automatic if is strictly commutative). Then extends uniquely to a map of -dgas.
Proof.
See [18], Lecture V, Lemma 3.3.
Definition 12.2.4. The completed de Rham complex.
For a finitely generated ideal in and a derived -complete -algebra, we may define the module of completed Kähler differentials as the derived -completion of the usual module If is derived -complete and is the derived -completion of a finitely generated -algebra, then is a finitely generated -module.
Now suppose that is derived -complete. Then the completed de Rham complex is a strictly commutative -dga, and in Lemma 12.2.3, if is derived -complete, then extends uniquely to a map of -dgas.
Subsection 12.3 Construction of the Hodge-Tate comparison map
Definition 12.3.1.
Let be a prism and let be an -algebra (writing ). For and an integer, define the Breuil-Kisin twist note that this makes sense even if because is an invertible -module (from the definition of a prism).
For consider the exact sequence
As per Definition 11.3.1, the object carries the structure of a commutative ring object over From Lemma 12.1.2, we deduce that the graded group carries the structure of a commutative -dga. It is also strictly commutative, but this requires some extra verification; see Lemma 12.3.4.
Suppose now that is derived -complete. Then the universal property of the completed de Rham complex gives us a morphism of -dgas
To see that the Bockstein differentials are indeed the differentials of a complex, we make the following general observation.
Lemma 12.3.2.
Let be an invertible ideal of a ring (e.g., the principal ideal generated by a non-zerodivisor). Given let
be the connecting homomorphism in the exact sequence obtained by applying to the sequence
Proof.
Consider the commutative diagram in Figure 12.3.3 in which the rows are exact.
By applying to the terms and comparing the two rows, we see that the map factors as
where the first map is the connecting homomorphism obtained from the upper row of Figure 12.3.3. By applying to the exact sequence
we deduce that the composition
vanishes. Combining these two observations proves the claim. (Compare [117], tag 0F7N.)
To check strict commutativity, we make an explicit computation. Remember that there is nothing to check here unless For a more conceptual approach, see Proposition 15.3.2.
Lemma 12.3.4.
Proof.
We may assume as otherwise this follows from ordinary commutativity; this will allow us to use the universal formula (for in any -ring)
Using Lemma 5.2.5, we may also reduce to the case where with distinguished (this is mostly just to simplify notation).
We use the fact that contains a weakly final object which moreover is -torsion-free (Proposition 11.6.5) to compute Hodge-Tate cohomology using the cosimplicial ring as per Remark 11.6.7. Lift to Let and be the images of under the various maps and in the cosimplicial ring so that
Since vanishes modulo (the reductions of and modulo are the two images of ) and is -torsion-free, the unique element with is also a cocycle. Tracing through the construction of the Bockstein differential, we see that equals the image of in so we need to check that the latter squares to zero.
Multiplying by again, we may instead check that squares to zero in The square is represented by we will check that this is the boundary of To begin, note that is a cocycle because commutes with the differential in Hence on one hand,
and so
On the other hand, by (12.2),
and so
(Compare [18], Lecture V, Lemma 5.4.)
Subsection 12.4 The Hodge-Tate comparison theorem
Theorem 12.4.1. Hodge-Tate comparison theorem.
Let be a bounded prism. Let be a -completely smooth -algebra. Then the Hodge-Tate comparison map (12.1) is an isomorphism.
Proof.
Corollary 12.4.2.
Proof.
By Proposition 6.5.3, we may view as the -completion of a smooth -algebra; consequently, is a finite projective -module, so the completed de Rham complex consists of finite projective -modulse. We may thus deduce the claim from Theorem 12.4.1.
Example 12.4.3. The Hodge-Tate isomorphism in -de Rham cohomology.
We will show that
where is the -analogue of where the hat denotes the -completion. We now distinguish between the case where in which case maps to a unit in and the case where in which case maps to zero in Thus reduction modulo yields a quasi-isomorphism
Exercises 12.5 Exercises
1.
Give an example of a -dga which is commutative but not strictly commutative.