Notes on prismatic cohomology

Let \((A, I)\) be the perfect prism with \(A/I \cong R\) and let \(R^\flat\) be the tilt. Then \(R^\flat\) is perfect and \(R/p \cong R^\flat/p\text{,}\) 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 \(a_0 \in R^\flat\) 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 \(\overline{R} = R / \bigcup_n (\varpi^{p^{-n}})\) 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 \(A = W(R^\flat), I = \ker(\theta_R)\text{,}\) 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 \(A/p \cong R^\flat\) 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 \(\overline{\varpi}^{p^{-n}}\text{.}\) By induction on \(n\text{,}\) the kernel of \(\phi^n\) on \(R/p\) is generated by \(\overline{\varpi}^{p^{-n}}\text{.}\) Hence the kernel of \(\overline{\theta}_R\colon R^\flat \to R/p\) is generated by the element \(\overline{\varpi}\) of \(R^\flat\) corresponding to the coherent sequence \((\overline{\varpi}^{p^{-n}})_n\text{.}\) Since both \(W(R^\flat)\) and \(R\) are \(p\)-torsion-free and classically \(p\)-complete, the kernel of \(\theta_R\) is generated by any element of the kernel lifting \(\overline{\varpi}\text{,}\) and in particular is principal. (Compare [18], Lecture IV, Proposition 2.10.)

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

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

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

We next check that \(J+J'\) is radical. The ideal \(J+J'\) is the kernel of \(R \to R/J \otimes_R R/J'\text{;}\) 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 \(J \cap J' = 0\text{.}\) To this end, choose \(x \in J \cap J'\text{;}\) since \(x \in J'\) 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 \(a_0 \in R^\flat\text{,}\) \(u \in A^\times\text{.}\) Consider the perfect ideals \(J = (a_0^{p^{-\infty}})\) and \(J' = R^\flat[J]\) of the perfect ring \(R^\flat\text{.}\) 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 \(S'\) be the top right and bottom right entry of the new square.

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

Let \(K\) be the kernel of \(R \to S'\text{.}\) Since Figure 8.4.6 is a pullback square, \(K\) embeds into \(W(R^\flat)/J)/d = \overline{R}\) and hence is \(p\)-torsion. Hence \(K \subseteq R[p^\infty]\) and so the induced map \(R/R[p^\infty] = S \to S'\) is injective. Since it is also surjective (because \(R \to S'\) 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.)