Notes on prismatic cohomology

If $N$ is derived $t$-complete, then the cone $F$ of $N\to N/t$ is complete for the $t$-adic filtration, and the $\varphi$-action is topologically nilpotent because $\varphi (t)=t^p\in t^{2}B\text{.}$ From this we see that $F^{\varphi =1}=0\text{,}$ proving the claim.

For convenience, we take colimits in the ambient category $D(B)$ (in contrast to the original statement).

For (1), we first verify that the two assertions are equivalent. Consider a diagram $\{(M_i,{\varphi }_i)\}$ in $D_{\mathrm{comp}}(B[F])\text{.}$ Let $F$ be the cone of the map from ${\mathrm{colim}}_i{M}_i$ to its derived $t$-completion; note that $F$ is uniquely $t$-divisible, so $F$ is also the cone of the map obtained after inverting $t$ on both sides. Now both statements of the lemma are equivalent to the vanishing of $F^{\varphi =1}\text{,}$ and hence to each other.

For (2), using (1) it suffices to check that each map $(M,\varphi )\stackrel{\varphi }{\to }(M,\varphi )$ induces an isomorphism upon applying either ${\bullet }^{\varphi =1}$ or $(\bullet [t^{-1}]{)}^{\varphi =1}\text{,}$ both of which are clear. (Compare [18], Lecture IX, Proposition 1.2.)

This formally reduces to the case $n=1\text{.}$ In this case, apply Proposition 22.2.3 with $B=A/p,t=d$ where $d$ is a generator of $I$ (Theorem 7.2.2).

Let $R\to V$ be a morphism in $\mathbf{Ring}$ with $V$ a eudoxian valuation ring. The image of $p$ in $V$ is then one of the following.

It suffices to check descent for an arc${}_p$-covering $R\to S$ (so in particular both rings are derived $p$-complete). By Lemma 22.3.3, $R\to S\oplus R/p\oplus R[1/p]$ is an arc-covering. Since derived $p$-completion followed by inverting $p$ kills both $(R/p)$-modules and $R[1/p]$-modules, we may deduce the claim from arc-descent for étale cohomology (Theorem 22.3.4). (Compare [21], Corollary 6.17.)

If $V$ is a $p$-complete AIC valuation ring, then $\mathrm{Frac}V$ is a perfectoid field (Definition 8.3.1). We may thus deduce the claim from Lemma 8.3.3.

Let $\mathrm{♯}:V^{\mathrm{♭}}\to V$ be the multiplicative map obtained by composing the constant lift $[\bullet ]:V^{\mathrm{♭}}\to W(V^{\mathrm{♭}})$ with the quotient map $W(V^{\mathrm{♭}})\to V\text{.}$ It is customary to write the image of $x$ under $\mathrm{♯}$ as $x^{\mathrm{♯}}$ rather than $\mathrm{♯}(x)\text{.}$

Suppose that $V$ is a valuation ring. If $\mathrm{Frac}V$ has characteristic $p\text{,}$ then $V=V^{\mathrm{♭}}$ and there is nothing more to check; we may thus assume that $\mathrm{Frac}V$ has characteristic 0. Since $V$ is an integral domain, the $p$-power map on $V$ is injective; hence for $x\in V^{\mathrm{♭}}\text{,}$ $x^{\mathrm{♯}}=0$ if and only if $x=0\text{.}$ This in turn implies that $V^{\mathrm{♭}}$ is an integral domain (if $xy=0$ then $x^{\mathrm{♯}}{y}^{\mathrm{♯}}=0$) and that the principal ideals of $V^{\mathrm{♭}}$ are totally ordered with respect to inclusion (if $x^{\mathrm{♯}}$ is divisible by $y^{\mathrm{♯}}\text{,}$ then the ratio admits a coherent sequence of $p$-power roots and so is itself in the image of $\mathrm{♯}$). Hence $V^{\mathrm{♭}}$ is a valuation ring.

Conversely, suppose that $V^{\mathrm{♭}}$ is a valuation ring. Again, we may assume that $p\ne 0$ in $V\text{;}$ since $V^{\mathrm{♭}}$ is an integral domain, we may apply Exercise 8.5.3 to deduce that $V$ is $p$-torsion-free. Since $V^{\mathrm{♭}}$ is a local ring, so are $W(V^{\mathrm{♭}})$ and its quotient $V\text{.}$ Choose $\varpi \in V$ as per Lemma 8.2.3.

We need to show that given any two nonzero elements $x,y\in V\text{,}$ one is a multiple of the other. By dividing by powers of $\varpi$ as needed, we may reduce to the case where $x$ and $y$ have nonzero images in $V/\varpi$ and hence in $V/p=V^{\mathrm{♭}}/d\text{.}$ Since $V^{\mathrm{♭}}$ is a valuation ring, after possibly swapping terms we can write $x=yz+pu$ for some $z,u\in V\text{.}$ Similarly, we can write $\varpi \equiv yw+pv$ for some $w,v\in V\text{.}$ Since $V$ is classically $\varpi$-complete, $1-(p/\varpi )v$ is a unit in $V\text{;}$ hence $\varpi$ is divisible by $y\text{,}$ as then is $p\text{.}$ Consequently, $x=yz+pu$ is also divisible by $y\text{,}$ as desired.

From the proof of Lemma 22.4.2, we see that $\mathrm{♯}$ induces an injective map from the value group of $V^{\mathrm{♭}}$ to that of $V\text{.}$ To prove surjectivity, we must check that every element $x$ of $V$ has an associate in the image of $V^{\mathrm{♭}}\text{.}$ As in the proof of Lemma 22.4.2, we may prove this by first dividing by a suitable power of $\varpi$ to ensure that $x\not\equiv 0(\mathrm{mod}\varpi )\text{,}$ then showing that in this case $x$ is an associate of $[\bar{x}]$ where $\bar{x}\in V/p$ is the image of $x\text{.}$

By Lemma 22.4.2, $V$ is a valuation ring if and only if $V^{\mathrm{♭}}$ is; it thus remains to show that if $V^{\mathrm{♭}}$ is not AIC, then neither is $V\text{.}$ We may assume that $V$ has characteristic 0, as otherwise there is nothing to check.

Let $R$ be the integral closure of $V^{\mathrm{♭}}$ in a nontrivial finite Galois extension of its fraction field with Galois group $G\text{.}$ By Theorem 7.3.5, $V^{\prime }=V{\otimes }_{W(V^{\mathrm{♭}})}W(R)$ is a lens with $V^{\prime \mathrm{♭}}\cong R$ and, by Lemma 22.4.2, again a valuation ring.

By construction, $G$ acts on $V^{\prime \mathrm{♭}}$ with fixed subring $V\text{.}$ By functoriality, $G$ also acts on $V^{\prime }\text{;}$ since $V^{\prime }$ is of characteristic $0\text{,}$ we can we can see by averaging over the group action that the fixed subring $V^{\prime G}$ is equal to $V\text{.}$

By the Artin and Dedekind lemmas in Galois theory, we see that $\mathrm{Frac}{V}^{\prime }$ is a finite Galois extension of $\mathrm{Frac}V$ of degree $\mathrm{\#}G>1\text{.}$ This proves that $V$ is not AIC.

Let $R^{\mathrm{♭}}\to V$ be a map to a eudoxian valuation ring, which we may assume is perfect. If $d$ maps to a unit in $V\text{,}$ then the map extends to $R^{\mathrm{♭}}[d^{-1}]\text{.}$ Otherwise, we may replace $V$ with its $d$-adic completion; by Theorem 7.3.5, the map $R^{\mathrm{♭}}\to V$ corresponds to a map $R\to V^{\mathrm{♯}}=A{\otimes }_{R}W(V)$ whose target is a lens and (by Lemma 22.4.2) a $p$-complete eudoxian valuation ring. Since $R\to S$ is an arc-covering, we get an extension to a map $S\to V^{\prime }$ for some eudoxian valuation ring $V^{\prime }$ containing $V^{\mathrm{♯}}\text{,}$ which we may take to be $p$-complete and AIC and hence a lens (Lemma 22.4.1). Then $S^{\mathrm{♭}}\to V^{\prime \mathrm{♭}}$ gives the desired extension. (Compare [25], Proposition 8.9.)

Write $R$ as the slice of a perfect prism $(A,I)\text{.}$ By applying Corollary 21.1.8 to the arc-covering from Lemma 22.5.1, then taking derived $d$-completions (which kills all terms involving $R^{\mathrm{♭}}[d^{-1}]$), we deduce the stated result. (Compare [25], Proposition 8.9.)

We first observe that both sides of (22.1) admits descent for the arc$p$-topology: for the left-hand side this is Theorem 22.3.4, and for the right-hand side it follows from Corollary 22.2.4 and Theorem 22.5.2. (Compare [25], Corollary 8.10.)

Using arc${}_p$-descent and the Artin-Schreier-Witt exact sequence (Proposition 22.1.1), we obtain a map (from left to right in (22.1)) of the desired form. To check that it is an isomorphism, we may apply arc${}_p$-descent again: using a v-covering as in Example 20.3.8, we may reduce to a case where $R=\prod _i{R}_i$ is a product of $p$-complete AIC valuation rings (and in particular a lens, by Lemma 22.4.1). Note that by Theorem 23.1.1, each ring $R_i^{\mathrm{♭}}$ is an AIC valuation ring.