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Section 12 The Hodge-Tate comparison map


[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 \(E^\bullet\) be a (not necessarily commutative) graded ring. We say that \(E^\bullet\) is graded commutative if
\begin{equation*} ab = (-1)^{mn}ba \qquad (a \in E^n, b \in E^m). \end{equation*}
The multiplication map on \(K^\bullet\) is given by a morphism \(K^\bullet \otimes_A^L K^\bullet \to K^\bullet\) in \(D(A)\text{.}\) We may directly read off the multiplication in \(\bigoplus_{n \geq 0} H^n(K^\bullet)\) 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 \(A \in \Ring\text{,}\) a differential graded algebra over \(A\) (also known as a \(A\)-dga) is a complex \((E^\bullet, d)\) of \(A\)-modules in which \(E^\bullet\) is also equipped with the structure of a (not necessarily commutative) graded \(A\)-algebra subject to the signed Leibniz rule
\begin{equation*} d^{n+m}(ab) = d^n(a)b + (-1)^n a d^m(b) \qquad (a \in E^n, b \in E^m). \end{equation*}
We say that an \(A\)-dga \((E^\bullet, d)\) is commutative if \(E^\bullet\) is graded commutative. We say that it is strictly commutative if it is commutative and moreover \(a^2 = 0\) 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 \(A \to B\) be a morphism in \(\Ring\text{.}\) The de Rham complex
\begin{equation*} (\Omega^{\bullet}_{B/A}, d_{\dR}) = \left( B \to \Omega^1_{B/A} \to \Omega^2_{B/A} \to \cdots \right), \end{equation*}
in which \(\Omega^i_{B/A} = \wedge^i_B \Omega^1_{B/A}\text{,}\) 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.
See [18], Lecture V, Lemma 3.3.

Definition 12.2.4. The completed de Rham complex.

For \(I\) a finitely generated ideal in \(A \in \Ring\) and \(R\) a derived \(I\)-complete \(A\)-algebra, we may define the module of completed Kähler differentials \(\widehat{\Omega}^1_{R/A}\) as the derived \(I\)-completion of the usual module \(\Omega^1_{R/A}\text{.}\) If \(A\) is derived \(I\)-complete and \(R\) is the derived \(I\)-completion of a finitely generated \(A\)-algebra, then \(\widehat{\Omega}^1_{R/A}\) is a finitely generated \(A\)-module.
Now suppose that \(A\) is derived \(I\)-complete. Then the completed de Rham complex \(\widehat{\Omega}^i_{R/A}\) is a strictly commutative \(A\)-dga, and in Lemma 12.2.3, if \(E^\bullet\) is derived \(I\)-complete, then \(\eta\) extends uniquely to a map \(\widehat{\Omega}^\bullet_{B/A} \to E^\bullet\) 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 \(\overline{A}\)-algebra (writing \(\overline{A} = A/I\)). For \(M \in \Mod_{\overline{A}}\) and \(n\) an integer, define the Breuil-Kisin twist \(M\{n\} = M \otimes_{\overline{A}} (I/I^2)^{\otimes n}\text{;}\) note that this makes sense even if \(n \lt 0\) because \(I/I^2\) is an invertible \(\overline{A}\)-module (from the definition of a prism).
For \(n \geq 0\text{,}\) consider the exact sequence
\begin{equation*} 0 \to I^{n+1} \calO_\Prism/I^{n+2} \to I^n \calO_{\Prism}/I^{n+2} \to I^n \calO_{\Prism}/I^{n+1} \to 0 \end{equation*}
of \(\calO_{\Prism}\)-modules on \((R/A)_{\Prism}\text{,}\) then take a connecting homomorphism to obtain the Bockstein differential
\begin{equation*} \beta_I\colon H^n(\overline{\Prism}_{R/A})\{n\} \to H^{n+1}(\overline{\Prism}_{R/A})\{n+1\}. \end{equation*}
It will follow from Lemma 12.3.2 that these indeed form the differentials in a complex \((H^\bullet(\overline{\Prism}_{R/A})\{\bullet\}, \beta_I)\text{.}\)
As per Definition 11.3.1, the object \(\overline{\Prism}_{R/A} \in D(\overline{A})\) carries the structure of a commutative ring object over \(\overline{A}\text{.}\) From Lemma 12.1.2, we deduce that the graded group \(\bigoplus_{n \geq 0} H^n(\overline{\Prism}_{R/A})\{n\}\) carries the structure of a commutative \(\overline{A}\)-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 \(\overline{A}\)-dgas
\begin{equation} \eta_R^\bullet\colon (\widehat{\Omega}^\bullet_{R/\overline{A}}, d_{\dR}) \to (H^\bullet(\overline{\Prism}_{R/A})\{\bullet\}, \beta_I).\tag{12.1} \end{equation}
To see that the Bockstein differentials are indeed the differentials of a complex, we make the following general observation.
Consider the commutative diagram in Figure 12.3.3 in which the rows are exact.
Figure 12.3.3.
By applying \(M^\bullet \otimes^L_A \star\) to the terms and comparing the two rows, we see that the map \(\beta^n\) factors as
\begin{equation*} H^n(M^\bullet \otimes^L_A I^n/I^{n+1}) \to H^{n+1}(M^\bullet \otimes^L_A I^{n+1}/I^{n+3}) \to H^{n+1}(M^\bullet \otimes^L_A I^{n+1}/I^{n+2}) \end{equation*}
where the first map is the connecting homomorphism obtained from the upper row of Figure 12.3.3. By applying \(M^\bullet \otimes^L_A *\) to the exact sequence
\begin{equation*} 0 \to I^{n+2}/I^{n+3} \to I^{n+1}/I^{n+3} \to I^{n+1}/I^{n+2} \to 0 \end{equation*}
we deduce that the composition
\begin{equation*} H^{n+1}(M^\bullet \otimes^L_A I^{n+1}/I^{n+3}) \to H^{n+1}(M^\bullet \otimes^L_A I^{n+1}/I^{n+2}) \stackrel{\beta^{n+1}}{\to} H^{n+2}(M^\bullet \otimes^L_A I^{n+2}/I^{n+3}) \end{equation*}
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 \(p=2\text{.}\) For a more conceptual approach, see Proposition 15.3.2.
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 \(\delta\)-ring)
\begin{equation} \delta(a-b) = \delta(a) - \delta(b) + b(a-b).\tag{12.2} \end{equation}
Using Lemma 5.2.5, we may also reduce to the case where \(I = (f)\) with \(f \in A\) distinguished (this is mostly just to simplify notation).
We use the fact that \((R/A)_{\Prism}\) 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 \((F^\bullet, d^\bullet)\)as per Remark 11.6.7. Lift \(\eta(t) \in F/IF\) to \(T \in F^0\text{.}\) Let \(U,V \in F^1\) and \(X,Y,Z \in F^2\) be the images of \(T\) under the various maps \(F^0 \to F^1\) and \(F^0 \to F^2\) in the cosimplicial ring \(F^\bullet\text{,}\) so that
\begin{equation*} d^0(T) = U-V, \qquad d^1(U) = X - Y + Z. \end{equation*}
Since \(U-V \in F^1\) vanishes modulo \(f\) (the reductions of \(U\) and \(V\) modulo \(f\) are the two images of \(t\)) and \(F^\bullet\) is \(f\)-torsion-free, the unique element \(\alpha \in F^1\) with \(U-V = f\alpha\) is also a cocycle. Tracing through the construction of the Bockstein differential, we see that \(\beta_I(t)\) equals the image of \(\alpha\) in \(H^1(F^\bullet/fF^\bullet)\text{,}\) so we need to check that the latter squares to zero.
Multiplying by \(f\) again, we may instead check that \(U-V\) squares to zero in \(H^2(f^2 F^\bullet/f^3 F^\bullet)\text{.}\) The square is represented by \((X-Y)(Y-Z) \in f^2 F^2\text{;}\) we will check that this is the boundary of \(f^2 \delta(\alpha) \in f^2 F^1\text{.}\) To begin, note that \(\phi(\alpha)\) is a cocycle because \(\phi\) commutes with the differential in \(F^\bullet\text{.}\) Hence on one hand,
\begin{equation*} \delta(U-V) = \delta(f\alpha) = f^2 \delta(\alpha) + \phi(\alpha) \delta(f) \end{equation*}
and so
\begin{equation*} d^1(\delta(U-V)) = d^1(f^2 \delta(\alpha)). \end{equation*}
On the other hand, by (12.2),
\begin{equation*} \delta(U-V) = \delta(U) - \delta(V) + V(U-V) = d^1(\delta(T)) + V(U-V) \end{equation*}
and so
\begin{equation*} d^1(\delta(U-V)) = d^1(V(U-V)) = Y(X-Y) + Z(Y-Z) - Z(X-Z) = (X-Y)(Y-Z). \end{equation*}
(Compare [18], Lecture V, Lemma 5.4.)

Subsection 12.4 The Hodge-Tate comparison theorem

By Proposition 6.5.3, we may view \(R\) as the \(p\)-completion of a smooth \(\overline{A}\)-algebra; consequently, \(\widehat{\Omega}^1_{R/A}\) 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 \((A, I) = (\ZZ_p\llbracket q-1 \rrbracket, ([p]_q))\text{.}\) We identify \(\overline{A} = A/I\) with \(\ZZ_p[\zeta_p]\) via \(q \mapsto \zeta_p\text{.}\)
Take \(R = \overline{A}\langle X^{\pm} \rangle\text{,}\) i.e., the \(p\)-adic completion of the Laurent polynomial ring \(\overline{A}[X^{\pm}]\text{,}\) so that
\begin{equation*} R = \widehat{\bigoplus_{i \in \ZZ}} \overline{A} X^i. \end{equation*}
We will show that
\begin{align*} \Delta_{R/A} &\cong \left( A \langle X^{\pm}\rangle \stackrel{\nabla_q}{\to} A\langle X^{\pm}\rangle \frac{dX}{X} \right)\\ & \cong \widehat{\bigoplus_{i \in \ZZ}} \left( A X^i \stackrel{[i]_q}{\to} AX^i \frac{dX}{X} \right) \end{align*}
where \([i]_q = (q^i-1)/(q-1)\) is the \(q\)-analogue of \(i\text{,}\) where the hat denotes the \((p, [p]_q)\)-completion. We now distinguish between the case where \(i \not\equiv 0 \pmod{p}\text{,}\) in which case \([i]_q\) maps to a unit in \(\overline{A}\text{,}\) and the case where \(i \equiv 0 \pmod{p}\text{,}\) in which case \([i]_q\) maps to zero in \(\overline{A}\text{.}\) Thus reduction modulo \(I\) yields a quasi-isomorphism
\begin{equation*} \overline{\Delta}_{R/A} \cong \widehat{\bigoplus_{k \in \ZZ}} \left( \overline{A} X^{kp} \stackrel{0}{\to} \overline{A} X^{kp} \right). \end{equation*}

Exercises 12.5 Exercises


Give an example of a \(\ZZ\)-dga which is commutative but not strictly commutative.