The Stacks project

First proof. This proof is a cheat. Namely, if $B$ has a flat lift $B'$, then taking the henselization $(B')^ h$ we obtain a flat lift of $B^ h$ (compare with the proof of Lemma 93.8.8). Conversely, suppose that $C'$ is an $A'$-flat lift of $(B')^ h$. Then let $\mathfrak c' \subset C'$ be the inverse image of the ideal $\mathfrak b^ h$. Then the completion $(C')^\wedge $ of $C'$ with respect to $\mathfrak c'$ is a lift of $B^\wedge $ (details omitted). Hence we see that $B$ has a flat lift by Lemma 93.12.3. $\square$

Second 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 $P^ h$ and $(P')^ h$ the respective henselizations. We will use that taking henselizations is functorial and that the henselization of a quotient is the corresponding quotient of the henselization, see More on Algebra, Lemmas 15.11.16 and 15.12.7.

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 $C = B^ h$ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is henselian with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (by More on Algebra, Lemma 15.11.9 and the fact that the kernel of $C' \to C$ is nilpotent). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through henselizations to give surjections $P^ h \to C$ and $(P')^ h \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^ h$-module structure lifting $C$ as an $A$-flat $P^ h$-module. Conversely, if we can lift $C$ to a $(P')^ h$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^ h/\tilde J$ (using Nakayama again) and setting $C' = (P')^ h/\tilde J$ we find a flat lift of $C$ as an algebra.

Observe that $P' \to (P')^ h$ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^ h/\mathfrak p'(P')^ h$ (More on Algebra, Lemma 15.12.2). 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. $\square$