The Stacks project

Proof. This proof is the same as the proof of Lemma 93.12.3 but easier. We suggest the reader to skip the proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots , x_ n] \to B$. Let $S_ P \subset P$ be the inverse image of $S$. Set $P' = A'[x_1, \ldots , x_ n]$ and denote $S_{P'} \subset P'$ the inverse image of $S_ P$.

Suppose $A' \to B'$ is a flat lift of $A \to B$, in other words, $A' \to B'$ is flat and there is an $A$-algebra isomorphism $B = B' \otimes _{A'} A$. Then we can choose an $A'$-algebra map $P' \to B'$ lifting the given surjection $P \to B$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we find that $B'$ is a quotient of $P'$. In particular, we find that we can endow $B'$ with an $A'$-flat $P'$-module structure lifting $B$ as an $A$-flat $P$-module. Conversely, if we can lift $B$ to a $P'$-module $M'$ flat over $A'$, then $M'$ is a cyclic module $M' \cong P'/J'$ (using Nakayama again) and setting $B' = P'/J'$ we find a flat lift of $B$ as an algebra.

Set $C = S^{-1}B$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Elements of $C'$ which map to invertible elements of $C$ are invertible. We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. By the remark above these maps pass through localizations to give surjections $S_ P^{-1}P \to C$ and $S_{P'}^{-1}P' \to C'$ (for the second use Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $S_{P'}^{-1}P'$-module structure lifting $C$ as an $A$-flat $S_ P^{-1}P$-module. Conversely, if we can lift $C$ to a $S_{P'}^{-1}P'$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong S_{P'}^{-1}P'/\tilde J$ (using Nakayama again) and setting $C' = S_{P'}^{-1}P'/\tilde J$ we find a flat lift of $C$ as an algebra.

The syntomic locus of a morphism of schemes is open by definition. Let $J_ B \subset B$ be an ideal cutting out the set of points in $\mathop{\mathrm{Spec}}(B)$ where $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is not syntomic. Denote $J_ P \subset P$ and $J_{P'} \subset P'$ the corresponding ideals. Observe that $P' \to S_{P'}^{-1}P'$ is a flat ring map which induces an isomorphism $P'/J_{P'} = S_{P'}^{-1}P'/J_{P'}S_{P'}^{-1}P'$ by our assumption on $S$ in the lemma, namely, the assumption in the lemma is exactly that $B/J_ B = S^{-1}(B/J_ B)$. We conclude that our lemma is a consequence of Lemma 93.12.2 provided we can show that $B_ g$ lifts to an $A'$-flat $P'_ g$-module for $g \in J_ B$. However, the ring map $A \to B_ g$ is syntomic and hence lifts to an $A'$-flat algebra $B'$ by Smoothing Ring Maps, Proposition 16.3.2. Since $A' \to P'_ g$ is smooth, we can lift $P_ g \to B_ g$ to a surjective map $P'_ g \to B'$ as before and we get what we want. $\square$