**Proof.**
Assume $R_0 \to A_0$ syntomic, in particular a local complete intersection (More on Algebra, Lemma 15.32.5). Choose a presentation $A_0 = R_0[x_1, \ldots , x_ n]/J_0$. Set $C_0 = \text{Sym}^*_{A_0}(J_0/J_0^2)$. Note that $J_0/J_0^2$ is a finite projective $A_0$-module (Algebra, Lemma 10.134.16). By Lemma 16.3.1 the ring map $A_0 \to C_0$ is smooth and we can find a presentation $C_0 = R_0[y_1, \ldots , y_ m]/K_0$ with $K_0/K_0^2$ free over $C_0$. By Algebra, Lemma 10.134.6 we can assume $C_0 = R_0[y_1, \ldots , y_ m]/(\overline{f}_1, \ldots , \overline{f}_ c)$ where $\overline{f}_1, \ldots , \overline{f}_ c$ maps to a basis of $K_0/K_0^2$ over $C_0$. Choose $f_1, \ldots , f_ c \in R[y_1, \ldots , y_ c]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$ and set

\[ C = R[y_1, \ldots , y_ m]/(f_1, \ldots , f_ c) \]

By construction $C_0 = C/IC$. By Algebra, Lemma 10.134.11 we can after replacing $C$ by $C_ g$ assume that $C$ is a relative global complete intersection over $R$. We conclude that there exists a finite projective $A_0$-module $P_0$ such that $C_0 = \text{Sym}^*_{A_0}(P_0)$ is isomorphic to $C/IC$ for some syntomic $R$-algebra $C$.

Choose an integer $n$ and a direct sum decomposition $A_0^{\oplus n} = P_0 \oplus Q_0$. By More on Algebra, Lemma 15.9.11 we can find an étale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$ and a finite projective $C'$-module $Q$ such that $Q/IQ$ is isomorphic to $Q_0 \otimes _{A_0} C/IC$. Then $D = \text{Sym}_{C'}^*(Q)$ is a smooth $C'$-algebra (see More on Algebra, Lemma 15.9.13). Picture

\[ \xymatrix{ R \ar[d] \ar[rr] & & C \ar[r] \ar[d] & C' \ar[r] \ar[d] & D \ar[d] \\ R/I \ar[r] & A_0 \ar[r] & C/IC \ar[r]^{\cong } & C'/IC' \ar[r] & D/ID } \]

Observe that our choice of $Q$ gives

\begin{align*} D/ID & = \text{Sym}_{C/IC}^*(Q_0 \otimes _{A_0} C/IC) \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes _{A_0} C/IC \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes _{A_0} \text{Sym}_{A_0}^*(P_0) \\ & = \text{Sym}_{A_0}^*(Q_0 \oplus P_0) \\ & = \text{Sym}_{A_0}^*(A_0^{\oplus n}) \\ & = A_0[x_1, \ldots , x_ n] \end{align*}

Choose $f_1, \ldots , f_ n \in D$ which map to $x_1, \ldots , x_ n$ in $D/ID = A_0[x_1, \ldots , x_ n]$. Set $A = D/(f_1, \ldots , f_ n)$. Note that $A_0 = A/IA$. We claim that $R \to A$ is syntomic in a neighbourhood of $V(IA)$. If the claim is true, then we can find a $f \in A$ mapping to $1 \in A_0$ such that $A_ f$ is syntomic over $R$ and the proof of (1) is finished.

Proof of the claim. Observe that $R \to D$ is syntomic as a composition of the syntomic ring map $R \to C$, the étale ring map $C \to C'$ and the smooth ring map $C' \to D$ (Algebra, Lemmas 10.134.17 and 10.135.10). The question is local on $\mathop{\mathrm{Spec}}(D)$, hence we may assume that $D$ is a relative global complete intersection (Algebra, Lemma 10.134.15). Say $D = R[y_1, \ldots , y_ m]/(g_1, \ldots , g_ s)$. Let $f'_1, \ldots , f'_ n \in R[y_1, \ldots , y_ m]$ be lifts of $f_1, \ldots , f_ n$. Then we can apply Algebra, Lemma 10.134.11 to get the claim.

Proof of (2). Since a smooth ring map is syntomic, we can find a syntomic ring map $R \to A$ such that $A_0 = A/IA$. By assumption the fibres of $R \to A$ are smooth over primes in $V(I)$ hence $R \to A$ is smooth in an open neighbourhood of $V(IA)$ (Algebra, Lemma 10.135.16). Thus we can replace $A$ by a localization to obtain the result we want.
$\square$

## Comments (1)

Comment #3751 by slogan_bot on