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Section 12 The Hodge-Tate comparison map

Reference.

[18], lecture V.
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

Definition 12.1.1.

Let E be a (not necessarily commutative) graded ring. We say that E is graded commutative if
ab=(1)mnba(aEn,bEm).

Proof.

The multiplication map on K is given by a morphism KALKK in D(A). We may directly read off the multiplication in n0Hn(K) 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 ARing, a differential graded algebra over A (also known as a A-dga) is a complex (E,d) of A-modules in which E is also equipped with the structure of a (not necessarily commutative) graded A-algebra subject to the signed Leibniz rule
dn+m(ab)=dn(a)b+(1)nadm(b)(aEn,bEm).
We say that an A-dga (E,d) is commutative if E is graded commutative. We say that it is strictly commutative if it is commutative and moreover a2=0 for any a of odd degree. (This last condition is redundant if E is 2-torsion-free.)
The prototypical example of this definition is the following construction.

Definition 12.2.2.

Let AB be a morphism in Ring. The de Rham complex
(ΩB/A,ddR)=(BΩB/A1ΩB/A2),
in which ΩB/Ai=BiΩB/A1, is a strictly commutative A-dga with multiplication given by the wedge product.
The universal property of the module of Kähler differentials can be reinterpreted as follws.

Proof.

See [18], Lecture V, Lemma 3.3.

Definition 12.2.4. The completed de Rham complex.

For I a finitely generated ideal in ARing and R a derived I-complete A-algebra, we may define the module of completed Kähler differentials Ω^R/A1 as the derived I-completion of the usual module ΩR/A1. If A is derived I-complete and R is the derived I-completion of a finitely generated A-algebra, then Ω^R/A1 is a finitely generated A-module.
Now suppose that A is derived I-complete. Then the completed de Rham complex Ω^R/Ai is a strictly commutative A-dga, and in Lemma 12.2.3, if E is derived I-complete, then η extends uniquely to a map Ω^B/AE of A-dgas.

Subsection 12.3 Construction of the Hodge-Tate comparison map

Definition 12.3.1.

Let (A,I) be a prism and let R be an A-algebra (writing A=A/I). For MModA and n an integer, define the Breuil-Kisin twist M{n}=MA(I/I2)n; note that this makes sense even if n<0 because I/I2 is an invertible A-module (from the definition of a prism).
For n0, consider the exact sequence
0In+1OΔ/In+2InOΔ/In+2InOΔ/In+10
of OΔ-modules on (R/A)Δ, then take a connecting homomorphism to obtain the Bockstein differential
βI:Hn(ΔR/A){n}Hn+1(ΔR/A){n+1}.
It will follow from Lemma 12.3.2 that these indeed form the differentials in a complex (H(ΔR/A){},βI).
As per Definition 11.3.1, the object ΔR/AD(A) carries the structure of a commutative ring object over A. From Lemma 12.1.2, we deduce that the graded group n0Hn(ΔR/A){n} carries the structure of a commutative A-dga. It is also strictly commutative, but this requires some extra verification; see Lemma 12.3.4.
Suppose now that R is derived p-complete. Then the universal property of the completed de Rham complex gives us a morphism of A-dgas
(12.1)ηR:(Ω^R/A,ddR)(H(ΔR/A){},βI).
To see that the Bockstein differentials are indeed the differentials of a complex, we make the following general observation.

Proof.

Consider the commutative diagram in Figure 12.3.3 in which the rows are exact.
Figure 12.3.3.
By applying MAL to the terms and comparing the two rows, we see that the map βn factors as
Hn(MALIn/In+1)Hn+1(MALIn+1/In+3)Hn+1(MALIn+1/In+2)
where the first map is the connecting homomorphism obtained from the upper row of Figure 12.3.3. By applying MAL to the exact sequence
0In+2/In+3In+1/In+3In+1/In+20
we deduce that the composition
Hn+1(MALIn+1/In+3)Hn+1(MALIn+1/In+2)βn+1Hn+2(MALIn+2/In+3)
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 p=2. For a more conceptual approach, see Proposition 15.3.2.

Proof.

We may assume p=2 as otherwise this follows from ordinary commutativity; this will allow us to use the universal formula (for a,b in any δ-ring)
(12.2)δ(ab)=δ(a)δ(b)+b(ab).
Using Lemma 5.2.5, we may also reduce to the case where I=(f) with fA distinguished (this is mostly just to simplify notation).
We use the fact that (R/A)Δ contains a weakly final object (F,IF) which moreover is f-torsion-free (Proposition 11.6.5) to compute Hodge-Tate cohomology using the cosimplicial ring (F,d)as per Remark 11.6.7. Lift η(t)F/IF to TF0. Let U,VF1 and X,Y,ZF2 be the images of T under the various maps F0F1 and F0F2 in the cosimplicial ring F, so that
d0(T)=UV,d1(U)=XY+Z.
Since UVF1 vanishes modulo f (the reductions of U and V modulo f are the two images of t) and F is f-torsion-free, the unique element αF1 with UV=fα is also a cocycle. Tracing through the construction of the Bockstein differential, we see that βI(t) equals the image of α in H1(F/fF), so we need to check that the latter squares to zero.
Multiplying by f again, we may instead check that UV squares to zero in H2(f2F/f3F). The square is represented by (XY)(YZ)f2F2; we will check that this is the boundary of f2δ(α)f2F1. To begin, note that ϕ(α) is a cocycle because ϕ commutes with the differential in F. Hence on one hand,
δ(UV)=δ(fα)=f2δ(α)+ϕ(α)δ(f)
and so
d1(δ(UV))=d1(f2δ(α)).
On the other hand, by (12.2),
δ(UV)=δ(U)δ(V)+V(UV)=d1(δ(T))+V(UV)
and so
d1(δ(UV))=d1(V(UV))=Y(XY)+Z(YZ)Z(XZ)=(XY)(YZ).
(Compare [18], Lecture V, Lemma 5.4.)

Subsection 12.4 The Hodge-Tate comparison theorem

Proof.

Proof.

By Proposition 6.5.3, we may view R as the p-completion of a smooth A-algebra; consequently, Ω^R/A1 is a finite projective R-module, so the completed de Rham complex consists of finite projective R-modulse. We may thus deduce the claim from Theorem 12.4.1.

Example 12.4.3. The Hodge-Tate isomorphism in q-de Rham cohomology.

Take (A,I)=(Zp[[q1]],([p]q)). We identify A=A/I with Zp[ζp] via qζp.
Take R=AX±, i.e., the p-adic completion of the Laurent polynomial ring A[X±], so that
R=iZ^AXi.
We will show that
ΔR/A(AX±qAX±dXX)iZ^(AXi[i]qAXidXX)
where [i]q=(qi1)/(q1) is the q-analogue of i, where the hat denotes the (p,[p]q)-completion. We now distinguish between the case where i0(modp), in which case [i]q maps to a unit in A, and the case where i0(modp), in which case [i]q maps to zero in A. Thus reduction modulo I yields a quasi-isomorphism
ΔR/AkZ^(AXkp0AXkp).

Exercises 12.5 Exercises

1.

Give an example of a Z-dga which is commutative but not strictly commutative.