Notes on prismatic cohomology

Let $(A,I)$ be the perfect prism with $A/I\cong R$ and let $R^{\mathrm{♭}}$ be the tilt. Then $R^{\mathrm{♭}}$ is perfect and $R/p\cong R^{\mathrm{♭}}/p\text{,}$ so $R/p$ is semiperfect. This proves (1).

For (2), apply Theorem 7.2.2 to write $I=(d)$ with $d\in A$ a distinguished element. By Lemma 7.1.2, we have $d=[a_0]-pu$ for some $a_0\in R^{\mathrm{♭}}$ and some unit $u\in A^{\times }\text{.}$ We may then take $\varpi$ to be the image of $[a_0]$ to obtain (2).

To check (3), we show that $\sqrt{pR}=\bigcup _n({\varpi }^{p^{-n}})\text{.}$ Since the left side contains the right side, it suffices to observe that the quotient $\bar{R}=R/\bigcup _n({\varpi }^{p^{-n}})$ is itself perfect, and hence reduced.

If $R=A/I$ is a lens, then all of the stated conditions follow directly from Theorem 7.2.2 and $R/p$ is semiperfect by Lemma 8.2.3. Conversely, suppose that these conditions hold; we will show that for $A=W(R^{\mathrm{♭}}),I=\ker ({\theta }_R)\text{,}$ the pair $(A,I)$ is a perfect prism with $A/I\cong R\text{.}$ Since ${\theta }_R$ induces a surjective map mod $p\text{,}$ it is in fact surjective, so $A/I\cong R\text{.}$ The ring $A$ is classically $p$-complete; we may check that it is classically $(p,I)$-complete by checking that $A/p\cong R^{\mathrm{♭}}$ is classically $I$-complete, which is straightforward.

At this point, we must show that $I$ admits a distinguished generator. To this end, choose $x,v\in A$ lifting $\varpi ,u$ and put $g=pv-x^p\in \ker (I)\text{.}$ The series expansion of $x$ has zero coefficient of $p\text{;}$ since $v$ is a unit (because $A$ is $I$-local), we deduce from Lemma 7.1.2 that $g$ is distinguished. Since $I$ is principal, we may apply Lemma 5.2.1 to deduce that $g$ is in fact a generator.

Conversely, suppose that the given conditions hold. It will suffice to show that the kernel of ${\theta }_R$ is principal, as then we can apply Proposition 8.2.5. We first show that the kernel of the (surjective) Frobenius map $R/p\to R/p$ is generated by $\varpi \text{.}$ Given $x\in R$ with $x^p\in pR\text{,}$ write $x^p={\varpi }^{p}y$ for some $y\in R\text{.}$ Then $(x/\varpi {)}^p=y\in R$ and so $x\in \varpi R\text{.}$

Now the Frobenius on $R/p$ factors as $R/p\to R/\varpi \to R/p$ where the first map is the canonical projection and the second map is an isomorphism. Since this composite is surjective, the image of $\varpi$ in $R/p$ admits a compatible system of $p$-power roots ${\bar{\varpi }}^{p^{-n}}\text{.}$ By induction on $n\text{,}$ the kernel of ${\varphi }^n$ on $R/p$ is generated by ${\bar{\varpi }}^{p^{-n}}\text{.}$ Hence the kernel of ${\bar{\theta }}_R:R^{\mathrm{♭}}\to R/p$ is generated by the element $\bar{\varpi }$ of $R^{\mathrm{♭}}$ corresponding to the coherent sequence $({\bar{\varpi }}^{p^{-n}}{)}_n\text{.}$ Since both $W(R^{\mathrm{♭}})$ and $R$ are $p$-torsion-free and classically $p$-complete, the kernel of ${\theta }_R$ is generated by any element of the kernel lifting $\bar{\varpi }\text{,}$ and in particular is principal. (Compare [18], Lecture IV, Proposition 2.10.)

If $K$ is of characteristic $p\text{,}$ then ${\mathfrak{o}}_K/p={\mathfrak{o}}_K$ is reduced and semiperfect, hence perfect. We thus assume hereafter that $K$ is characteristic 0; we may then check the conditions of Proposition 8.2.6.

It is clear that ${\mathfrak{o}}_K$ is classically $p$-complete and $p$-normal (since it is integrally closed), and by hypothesis ${\mathfrak{o}}_K/p$ is semiperfect. Since $K$ is not discretely valued, we can choose an element $x\in {\mathfrak{o}}_K$ of positive valuation such that $x^p$ divides $p\text{.}$ Since ${\mathfrak{o}}_K/p$ is semiperfect, there exists $y\in {\mathfrak{o}}_K$ such that $y^p\equiv p/x^p(\mathrm{mod}p)\text{;}$ put $\varpi =xy\text{.}$ Then ${\varpi }^p/p\equiv 1(\mathrm{mod}{x}^p)$ and so $u={\varpi }^p/p$ is a unit in ${\mathfrak{o}}_K\text{.}$

See [80] or [107] (or other references as given in [81], Remark 2.1.8).

We first check that $J^{\prime }$ is radical. If $x\in R$ with $x^p\in J^{\prime }\text{,}$ then $x^{p}J=0\text{;}$ since $R$ is perfect, it follows that $xJ=0$ and so $x\in J^{\prime }\text{.}$

We next check that $J+J^{\prime }$ is radical. The ideal $J+J^{\prime }$ is the kernel of $R\to R/J{\otimes }_{R}R/J^{\prime }\text{;}$ the target is a colimit of perfect rings and hence is itself perfect.

The square in question is already a pushout square at the level of groups, hence also at the level of rings. To check that it is a pullback square, we must check that $J\cap J^{\prime }=0\text{.}$ To this end, choose $x\in J\cap J^{\prime }\text{;}$ since $x\in J^{\prime }$ we have $xJ=0\text{,}$ but since $x\in J$ this implies $x^2=0$ and finally $x=0$ because $R$ is perfect.

We first show that the square is a pullback. By Theorem 7.2.2 we can write $I=(d)$ with $d$ distinguished. By Lemma 7.1.2 we can write $d=[a_0]-pu$ with $a_0\in R^{\mathrm{♭}}\text{,}$ $u\in A^{\times }\text{.}$ Consider the perfect ideals $J=(a_0^{p^{-\mathrm{\infty }}})$ and $J^{\prime }=R^{\mathrm{♭}}[J]$ of the perfect ring $R^{\mathrm{♭}}\text{.}$ The square Figure 8.4.5 consists of $p$-torsion-free, $p$-adically complete rings and its reduction modulo $p$ is the pullback square from Figure 8.4.2; hence by devissage it is a pullback square.

Since $d$ is a non-zerodivisor in each of the rings in Figure 8.4.5 by Lemma 7.1.2, we may reduce modulo $d$ to get another pullback square (Figure 8.4.6). Let $S^{\prime }$ be the top right and bottom right entry of the new square.

Since both $d$ and $p$ are non-zerodivisors on $S^{\prime }\text{,}$ by Exercise 8.5.3 we have $(S^{\prime }/(d))[p]\cong (S^{\prime }/(p))[d]=(R^{\mathrm{♭}}/J^{\prime })[d]\text{.}$ The latter vanishes because the element $d=a_0$ of $R^{\mathrm{♭}}/J^{\prime }$ is a non-zerodivisor (by Lemma 6.4.2). We deduce that $S^{\prime }/(d)$ is $p$-torsion-free, and so the surjection $R\to S^{\prime }$ from the top row factors through $R/R[p^{\mathrm{\infty }}]=S\text{.}$ As in the previous paragraph, we may identify the bottom right entry with $S^{\prime }/\sqrt{p{S}^{\prime }}\text{.}$

Let $K$ be the kernel of $R\to S^{\prime }\text{.}$ Since Figure 8.4.6 is a pullback square, $K$ embeds into $W(R^{\mathrm{♭}})/J)/d=\bar{R}$ and hence is $p$-torsion. Hence $K\subseteq R[p^{\mathrm{\infty }}]$ and so the induced map $R/R[p^{\mathrm{\infty }}]=S\to S^{\prime }$ is injective. Since it is also surjective (because $R\to S^{\prime }$ is) it is an isomorphism; this proves that Figure 8.4.4 and Figure 8.4.6 are the same square, and hence the former is a pullback square.

To conclude, note that the first numbered assertion is included in Lemma 8.2.3; the second and third assertions follow from the fact that Figure 8.4.4 is now a pullback square; and these in turn imply that the square is a pushout. (Compare [18], Lecture IV, Proposition 3.2.)

By Proposition 8.4.3, we may reduce to the cases of a perfect ring of characteristic $p$ and of a $p$-torsion-free untilted ring. In the former case, it is evident that any perfect ring is reduced. in the latter case, let $R$ be the lens in question. Apply Lemma 8.2.3 to choose an element $\varpi \in R$ such that ${\varpi }^p=pu$ for some unit $u\in R^{\times }\text{.}$ It will suffice to check that any $x\in R$ with $x^p=0$ vanishes, or (because $R$ is $p$-adically separated) that any such $x$ is divisible by ${\varpi }^n$ for any positive integer $n\text{.}$ We prove this by induction starting with the base case $n=0\text{.}$ Given that $x={\pi }^{n}y$ for some nonnegative integer $n$ and some $y\in R\text{,}$ we have ${\varpi }^{np}{y}^p=0$ and hence $y^p=0$ because $R$ is $p$-torsion-free. By Lemma 8.2.3 again, the kernel of the Frobenius on $R/p$ is generated by $\varpi \text{;}$ hence $y\in \varpi R$ and $x\in {\varpi }^{n+1}R\text{.}$ (Compare [18], Lecture IV, Corollary 3.3.)