Lemma 93.12.3. Let $A' \to A$ be a surjective map of Noetherian rings with nilpotent kernel. Let $A \to B$ be a finite type flat ring map. Let $\mathfrak b \subset B$ be an ideal such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is syntomic on the complement of $V(\mathfrak b)$. Then $B$ has a flat lift to $A'$ if and only if the $\mathfrak b$-adic completion $B^\wedge $ has a flat lift to $A'$.
Proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots , x_ n] \to B$. Let $\mathfrak p \subset P$ be the inverse image of $\mathfrak b$. Set $P' = A'[x_1, \ldots , x_ n]$ and denote $\mathfrak p' \subset P'$ the inverse image of $\mathfrak p$. (Of course $\mathfrak p$ and $\mathfrak p'$ do not designate prime ideals here.) We will denote $P^\wedge $ and $(P')^\wedge $ the respective completions.
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 = B^\wedge $ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is complete with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (Algebra, Lemma 10.97.10). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through completions to give surjections $P^\wedge \to C$ and $(P')^\wedge \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^\wedge $-module structure lifting $C$ as an $A$-flat $P^\wedge $-module. Conversely, if we can lift $C$ to a $(P')^\wedge $-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^\wedge /\tilde J$ (using Nakayama again) and setting $C' = (P')^\wedge /\tilde J$ we find a flat lift of $C$ as an algebra.
Observe that $P' \to (P')^\wedge $ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^\wedge /\mathfrak p'(P')^\wedge $. 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 $g \in \mathfrak p'$. 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. $\square$
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