**Proof.**
By Lemma 10.131.10 applied to $R \to S \to R$ we see that $I/I^2 = \Omega _{S/R} \otimes _{S, \sigma } R$. Since by definition of a smooth morphism the module $\Omega _{S/R}$ is finite locally free over $S$ we deduce that (1) holds. If $I/I^2$ is free, then choose $f_1, \ldots , f_ d \in I$ whose images in $I/I^2$ form an $R$-basis. Consider the $R$-algebra map defined by

\[ \Psi : R[[x_1, \ldots , x_ d]] \longrightarrow S^\wedge , \quad x_ i \longmapsto f_ i. \]

Denote $P = R[[x_1, \ldots , x_ d]]$ and $J = (x_1, \ldots , x_ d) \subset P$. We write $\Psi _ n : P/J^ n \to S/I^ n$ for the induced map of quotient rings. Note that $S/I^2 = \varphi (R) \oplus I/I^2$. Thus $\Psi _2$ is an isomorphism. Denote $\sigma _2 : S/I^2 \to P/J^2$ the inverse of $\Psi _2$. We will prove by induction on $n$ that for all $n > 2$ there exists an inverse $\sigma _ n : S/I^ n \to P/J^ n$ of $\Psi _ n$. Namely, as $S$ is formally smooth over $R$ (by Proposition 10.138.13) we see that in the solid diagram

\[ \xymatrix{ S \ar@{..>}[r] \ar[rd]_{\sigma _{n - 1}} & P/J^ n \ar[d] \\ & P/J^{n - 1} } \]

of $R$-algebras we can fill in the dotted arrow by some $R$-algebra map $\tau : S \to P/J^ n$ making the diagram commute. This induces an $R$-algebra map $\overline{\tau } : S/I^ n \to P/J^ n$ which is equal to $\sigma _{n - 1}$ modulo $J^ n$. By construction the map $\Psi _ n$ is surjective and now $\overline{\tau } \circ \Psi _ n$ is an $R$-algebra endomorphism of $P/J^ n$ which maps $x_ i$ to $x_ i + \delta _{i, n}$ with $\delta _{i, n} \in J^{n -1}/J^ n$. It follows that $\Psi _ n$ is an isomorphism and hence it has an inverse $\sigma _ n$. This proves the lemma.
$\square$

## Comments (2)

Comment #2161 by Daniel Smolkin on

Comment #8338 by Et on