Notes on prismatic cohomology

To prove (2), it is enough to show that $(A/d)[p^2]=(A/d)[p]\text{.}$ That is, given $f,g\in A$ with $p^{2}f=gd\text{,}$ we must have $pf\in dA\text{.}$ Since $gd\in p^{2}A\text{,}$ we have $\delta (gd)\in pA$ and hence $\varphi (g)\delta (gd)\in pA\text{.}$ Rewriting this as $\delta (d)g^p\varphi (g)+\delta (g)\varphi (gd)\text{,}$ we see that $\delta (d)g^p\varphi (g)\in pA\text{.}$ Since $A$ is $(p,d)$-local, $\delta (d)$ is a unit in $A\text{,}$ so $g^p\varphi (g)\in pA$ and so $g^{2p}\in pA\text{.}$ Because $A/p$ is reduced, this implies $g\in pA\text{;}$ since $A$ is $p$-torsion-free, this implies that $pf\in dA$ as desired. (Compare [25], Lemma 2.34.)

By Lemma 5.3.6, the ideal $I$ is principal and any generator $d$ of $I$ is a distinguished element. By Lemma 2.2.8, $A$ is $p$-torsion-free.

Since $A$ is $p$-torsion-free and derived $p$-complete, it is also classically $p$-complete by Lemma 6.4.2. By the previous paragraph, it is also classically $(p,I)$-complete.

By Proposition 3.3.6, $A\cong W(A/p)\text{.}$ By Lemma 7.1.2, any generator $d$ of $I$ is a non-zerodivisor. By Lemma 6.4.3, $(A/p)[I^{\mathrm{\infty }}]=(A/p)[I]\text{.}$ By Lemma 7.1.2, $(A/I)[p^{\mathrm{\infty }}]=(A/I)[p]\text{.}$

Let $A^{\prime }$ be the coperfection of $A\text{;}$ by Lemma 2.2.8, $A^{\prime }$ is $p$-torsion-free. Let $A^{″}$ be the classical $p$-completion of $A^{\prime }\text{;}$ by Lemma 6.4.2, $A^{″}$ is also the derived $p$-completion. By Exercise 2.5.8, $A^{″}$ can be canonically promoted to a $\delta$-ring over $A\text{.}$ Now Proposition 3.3.6 implies $A^{″}\cong W(A^{″}/p)\text{.}$

For each positive integer $n\text{,}$ we may now argue as in the proof of Theorem 7.2.2 that the derived $I$-completion of $A/p^n$ coincides with the classical completion. Consequently, if we take $A^{‴}$ to be the classical $(p,I)$-completion of $A^{″}$ (or equivalently of $A^{\prime }$), then $A^{‴}$ also equals the derived $(p,I)$-completion of either $A^{\prime }$ or $A^{″}\text{.}$ By Exercise 2.5.8, $A^{‴}$ can be canonically promoted to a $\delta$-ring over $A^{″}\text{.}$ Again, Proposition 3.3.6 implies $A^{‴}\cong W(A^{‴}/p)\text{.}$

At this point, $(A^{‴},I{A}^{‴})$ is a prism (the conditions on the ideal $I{A}^{‴}$ are implied by the corresponding conditions on $I$) and $A^{‴}$ is universal for maps of $A$ to derived $(p,I)$-complete $\delta$-rings. Thus the proof is complete. (Compare [18], Lecture IV, Lemma 1.3 or [25], Lemma 3.9.)

Everything will follow once we construct a natural isomorphism $A\cong W({\bar{A}}^{\mathrm{♭}})\text{.}$ By Theorem 7.2.2, it will suffice to construct a natural isomorphism $A/p\cong {\bar{A}}^{\mathrm{♭}}\text{.}$

By Theorem 7.2.2, $I$ admits a generator $d$ which is a distinguished element. By definition, we have $\bar{A}/p=A/(p,d)\text{.}$ For each positive integer $n\text{,}$ the $n$-fold Frobenius $A/(p,d)\to A/(p,d)$ identifies with the canonical map $A/(p,d^{p^n})\to A/(p,d)$ compatibly with $n\text{,}$ so the limit $\underset{\varphi }{lim}\bar{A}/p$ gets identified with $\underset{\varphi }{lim}A/(p,d^{p^n})\text{.}$ The latter is naturally isomorphic to $A/(p)$ because the latter is clasically $d$-complete (Lemma 6.4.3).

It will suffice to explain how to recover $A$ and $I$ functorially from $\bar{A}\text{.}$ Since $\bar{A}$ is in the essential image of the functor, $\varphi :\bar{A}/p\to \bar{A}/p$ is surjective and so ${\bar{A}}^{\mathrm{♭}}\to \bar{A}/p$ is surjective. Consequently, ${\theta }_A:W({\bar{A}}^{\mathrm{♭}})\to \bar{A}$ is also surjective. We can now reconstruct the diagram of Figure 7.3.4 to recover $A=W({\bar{A}}^{\mathrm{♭}})$ and $I=\ker (A\to \bar{A})\text{.}$