Notes on prismatic cohomology

This is a direct consequence of Theorem 22.5.2. Compare [25], Corollary 8.10.

The hypothesis on $S\to S^{\prime }$ implies the hypothesis of Lemma 25.2.2, so Figure 25.2.3 is a pullback square in $D(S)\text{.}$ In particular, the cones of the two rows are isomorphic in $D(S)\text{.}$ The bottom row consists of two objects which by construction are $J$-almost zero (Corollary 24.3.6), so its cone is also $J$-almost zero; hence the top row is a $J$-almost isomorphism.

We may assume from the outset that $p\in J\text{.}$ Suppose first that $S$ admits an action by a finite group $G$ such that $R\to S$ is a $J$-almost $G$-Galois cover. Note that this hypothesis is preserved by a $p$-completely flat base extension, as it can be checked modulo $p$ thanks to derived Nakayama (Remark 6.6.6); moreover, all of the conclusions can also be checked after such a base extension. By Theorem 19.4.4, we may thus assume that $R$ is absolutely integrally closed. By Lemma 19.4.2, we can then find generators $f_1,\dots ,f_r$ of $J$ such that $R\to S$ splits outside $V(f_i)$ for $i=1,\dots ,r\text{.}$ Since being a $J$-almost isomorphism is equivalent to being an $(f_i)$-almost isomorphism for $i=1,\dots ,r\text{,}$ we may reduce to the case where $J=(f)$ and we have an $R$-algebra isomorphism $S[f^{-1}]\cong \prod _{i\in I}R[f^{-1}]$ for some finite index set $I\text{.}$ Put $S^{\prime }=\prod _{i\in I}R$ and let $S^{\prime }$ be the integral closure of $R$ in $S[f^{-1}]\text{;}$ we then have maps $S\to S^{″},S^{\prime }\to S^{″}$ to which we may apply Corollary 25.2.5. This allows us to equate all of the desired assertions about $S$ to the corresponding statements about $S^{\prime }\text{,}$ which are self-evident.

Assume next that $R\to S$ has constant degree $r$ outside $V(J)\text{.}$ By Lemma 24.4.5, we can find an $S_r$-Galois covering of $\mathrm{Spec}(R)\setminus V(J)$ which is an $S_{r-1}$-Galois covering of $\mathrm{Spec}(S)\setminus V(J)\text{.}$ Using Corollary 24.4.4, we may reduce to the previous case.

Now consider the general case. In this case, $\mathrm{Spec}(R)\setminus V(J)$ can be partitioned as a finite union $\underset{i}{⨆}{U}_i$ of closed-open subsets, on each of which the degree of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ is constant. For each $i\text{,}$ let $R_i$ be the image of $R$ in $H^0(U_i,\mathcal{O})$ and let $R_{i,\mathrm{lens}}$ be the lens coperfection of $R_i\text{.}$ The map $R\to \prod _i{R}_{i,\mathrm{lens}}$ satisfies the condition of Lemma 25.2.2, so we may reduce to the previous case.

Exercise (Exercise 25.6.1), or see [25], Lemma 10.12.

By passage to filtered colimits, we may assume that $R\to S$ is module-finite and finitely presented. By Lemma 19.3.4 and Lemma 19.3.5, we know that ${\mathbf{\Delta }}_{S/A,\mathrm{perf}}$ is concentrated in degrees $\ge 0\text{,}$ and everything will follow once we show that it is also concentrated in degrees $\le 0\text{.}$ For this, we fix a sequence $g_1,\dots ,g_n$ as in Lemma 25.3.1 and induct on $n\text{.}$

For the base case $n=1\text{,}$ the map $R\to S_{\mathrm{red}}$ is finite étale, and so $S_{\mathrm{red}}$ is a lens. By arc-descent for lenses (Theorem 22.5.2), $S_{\mathrm{lens}}\to S_{\mathrm{red},\mathrm{lens}}$ is an isomorphism.

For the induction step $n>1\text{,}$ the induction hypothesis (and arc-descent) implies that $(S/g_1{)}_{\mathrm{lens}}$ is concentrated in degrees $\le 0\text{;}$ by Lemma 24.3.7, it is enough to check that $S_{\mathrm{lens}}$ is $g_1$-almost concentrated in degrees $\le 0\text{.}$ By Remark 25.3.2, $R[g_1^{-1}]\to S_{\mathrm{red}}[g_1^{-1}]$ is a finite étale covering.

Let $S^{\prime }$ be the integral closure of $R$ in $S_{\mathrm{red}}[g_1^{-1}]\text{.}$ By Lemma 25.2.2, the map $S_{\mathrm{lens}}\to S_{\mathrm{lens}}^{\prime }$ is a $g_1$-almost isomorphism, so it will be enough to check that $S_{\mathrm{lens}}^{\prime }$ is $g_1$-almost concentrated in degrees $\le 0\text{.}$ But this may be deduced from Theorem 25.2.6 by approximating $S^{\prime }$ with module-finite, finitely presented $R$-algebras. (Compare [25], Theorem 10.11.)

Combine Theorem 25.2.6 with Theorem 25.3.3.

By Theorem 25.3.3, $S_{\mathrm{lens}}$ and $(S/J{)}_{\mathrm{lens}}$ are both lenses concentrated in degree $0\text{.}$ By Corollary 19.4.6, the map $S_{\mathrm{lens}}\to (S_{\mathrm{lens}}/J{S}_{\mathrm{lens}}{)}_{\mathrm{lens}}=(S/J{)}_{\mathrm{lens}}$ is surjective. This proves the claim: the cone is actually concentrated in degrees -1 and 0.

By derived Nakayama (Proposition 6.6.2), this reduces to the corresponding statement after reduction modulo $I\text{,}$ which is Lemma 25.4.1.

Combine Corollary 25.4.2 with the étale comparison theorem (Theorem 22.6.1).

This reduces to Corollary 25.4.3 using the “method of the trace” ([117], tag 03SH) as in [25], Theorem 11.1.

It suffices to check that $R/p^n\to S/p^n$ splits in ${\mathbf{Mod}}_{R/p^n}$ for every $n\text{,}$ as then an application of the Artin-Rees lemma shows that $R\to S$ splits (see [16], Lemma 5.3). That is, we must show that the boundary class $\alpha \in {\mathrm{Ext}}_R^1(S/R,R)$ vanishes modulo $p^n$ for all $n\ge 2\text{.}$

Choose a nonzero element $f\in R$ such that $R[f^{-1}]\to S[f^{-1}]$ is finite étale, and define the ideal $J=(p,f)R_2\text{.}$ By Theorem 25.3.4, $R_2/p^n\to S_{2,\mathrm{lens}}/p^n$ is almost finite étale for the context $(R_2/p^n,J_{\mathrm{lens}}/p^n)\text{.}$ Consequently, ${\alpha }_2/p^n$ is $(J_{\mathrm{lens}}/p^n)$-almost zero; in particular, it is killed by $(pf{)}^{p^{-m}}\in R_2$ for all $m\ge 0\text{.}$ (Note that $f^{p^{-m}}$ makes sense in $R_2$ because the latter is AIC; this is why we didn’t stop at $R_1\text{.}$ Also, we are using that $S_2$ maps to $S_{2,\mathrm{lens}}$ but not any closer relationship between these two objects.)

Now suppose that $\alpha /p^n\ne 0$ for some $n\ge 2\text{.}$ By Krull’s intersection theorem ([117], tag 00IP), $pf\notin ({\mathrm{Ann}}_{R/p^n}(\alpha /p^n){)}^{p^m}$ for $m\gg 0\text{.}$ Since $R\to R_2$ is $p$-completely faithfully flat, we also have $pf\notin ({\mathrm{Ann}}_{R_2/p^n}(\alpha /p^n){)}^{p^m}\text{;}$ but this contradicts the previous paragraph. This conclusion yields the desired result. (Compare [16], Theorem 5.4.)