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*}
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
Lemma 12.1.2.
For \(A \in \Ring\text{,}\) let \(K^\bullet\) be a commutative \(A\)-algebra object in \(D(A)\text{.}\) Then \(\bigoplus_{n \geq 0} H^n(K^\bullet)\) carries a natural graded ring structure, with respect to which it is graded commutative.
Proof.
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.
Lemma 12.2.3. Universal property of the de Rham complex.
Let \((E^\bullet, d)\) be a graded commutative \(A\)-dga supported in degrees \(\ge 0\text{.}\) Let \(\eta\colon B \to E^0\) be a map of \(A\)-algebras such that for each \(x \in B\text{,}\) the element \(y = d(\eta(x)) \in E^1\) satisfies \(y^2 = 0\) (note that this is automatic if \(E^\bullet\) is strictly commutative). Then \(\eta\) extends uniquely to a map \(\Omega^\bullet_{B/A} \to E^\bullet\) of \(A\)-dgas.
Proof.
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.
Lemma 12.3.2.
Let \(I\) be an invertible ideal of a ring \(A\) (e.g., the principal ideal generated by a non-zerodivisor). Given \(M^\bullet \in D(A)\text{,}\) let
\begin{equation*}
\beta^n\colon 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+2})
\end{equation*}
be the connecting homomorphism in the exact sequence obtained by applying \(M^\bullet \otimes^L_A *\) to the sequence
\begin{equation*}
0 \to I^{n+1}/I^{n+2} \to I^n/I^{n+2} \to I^n/I^{n+1} \to 0.
\end{equation*}
Then the composition \(\beta^{n+1} \circ \beta^n\) vanishes for all \(n\text{.}\)
Proof.
Consider the commutative diagram in Figure 12.3.3 in which the rows are exact.
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.
Lemma 12.3.4.
For any \(t \in R\text{,}\) the class \(\beta_I(\eta(t)) \in H^1(\overline{\Prism}_{R/A})\) squares to zero in \(H^2(\overline{\Prism}_{R/A})\text{.}\)
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 \(\delta\)-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
Theorem 12.4.1. Hodge-Tate comparison theorem.
Let \((A,I)\) be a bounded prism. Let \(R\) be a \(p\)-completely smooth \(\overline{A}\)-algebra. Then the Hodge-Tate comparison map (12.1) is an isomorphism.
Proof.
Corollary 12.4.2.
With notation as in Theorem 12.4.1, the object \(\overline{\Delta}_{R/A} \in D(R)\) is perfect.
Proof.
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
1.
Give an example of a \(\ZZ\)-dga which is commutative but not strictly commutative.
