**Proof.**
Let $J \subset B$ be the kernel of $B \to A/I$ so that $B/J = A/I$. By Algebra, Lemma 10.137.3 the sequence

\[ 0 \to I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0 \]

is split exact. Thus $\overline{P} = J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a finite projective $A/I$-module. Choose an integer $n$ and a direct sum decomposition $A/I^{\oplus n} = \overline{P} \oplus \overline{K}$. By Lemma 15.9.11 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A$-module $K$ which lifts $\overline{K}$. We may and do replace $A$ by $A'$. Set $B' = B \otimes _ A \text{Sym}_ A^*(K)$. Since $A \to \text{Sym}_ A^*(K)$ is smooth by Lemma 15.9.13 we see that $B \to B'$ is smooth which in turn implies that $A \to B'$ is smooth (see Algebra, Lemmas 10.135.4 and 10.135.13). Moreover the section $\text{Sym}^*_ A(K) \to A$ determines a section $B' \to B$ and we let $B' \to A/I$ be the composition $B' \to B \to A/I$. Let $J' \subset B'$ be the kernel of $B' \to A/I$. We have $JB' \subset J'$ and $B \otimes _ A K \subset J'$. These maps combine to give an isomorphism

\[ (A/I)^{\oplus n} \cong J/J^2 \oplus \overline{K} \longrightarrow J'/((J')^2 + IB') \]

Thus, after replacing $B$ by $B'$ we may assume that $J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a free $A/I$-module of rank $n$.

In this case, choose $f_1, \ldots , f_ n \in J$ which map to a basis of $J/(J^2 + IB)$. Consider the finitely presented $A$-algebra $C = B/(f_1, \ldots , f_ n)$. Note that we have an exact sequence

\[ 0 \to H_1(L_{C/A}) \to (f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to 0 \]

see Algebra, Lemma 10.132.4 (note that $H_1(L_{B/A}) = 0$ and that $\Omega _{B/A}$ is finite projective, in particular flat so the Tor group vanishes). For any prime $\mathfrak q \supset J$ of $B$ the module $\Omega _{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega _{B/A}$ is finite projective and because $\Omega _{B/A} \otimes _ B B/J$ is free of rank $n$ (see Algebra, Lemma 10.77.2). By our choice of $f_1, \ldots , f_ n$ the map

\[ \left((f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2\right)_{\mathfrak q} \to \Omega _{B/A, \mathfrak q} \]

is surjective modulo $J$. Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism (this is because by Nakayama's Algebra, Lemma 10.19.1 the map is surjective and then for example by Algebra, Lemma 10.15.4 because $((f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2)_{\mathfrak q}$ is generated by $n$ elements the map is injective). Thus $H_1(L_{C/A})_{\mathfrak q} = 0$ and $\Omega _{C/A, \mathfrak q} = 0$. By Algebra, Lemma 10.135.12 we see that $A \to C$ is smooth at the prime $\overline{\mathfrak q}$ of $C$ corresponding to $\mathfrak q$. Since $\Omega _{C/A, \mathfrak q} = 0$ it is actually étale at $\overline{\mathfrak q}$. Thus $A \to C$ is étale at all primes of $C$ containing $JC$. By Lemma 15.9.4 we can find an $f \in C$ mapping to an invertible element of $C/JC$ such that $A \to C_ f$ is étale. By our choice of $f$ it is still true that $C_ f/JC_ f = A/I$. The map $C_ f/IC_ f \to A/I$ is surjective and étale by Algebra, Lemma 10.141.8. Hence $A/I$ is isomorphic to the localization of $C_ f/IC_ f$ at some element $g \in C$, see Algebra, Lemma 10.141.9. Set $A' = C_{fg}$ to conclude the proof.
$\square$

## Comments (2)

Comment #2170 by JuanPablo on

Comment #2199 by Johan on