Lemma 15.37.9. Let $R$, $S$ be rings. Let $\mathfrak n \subset S$ be an ideal. Let $R \to R'$ be a ring map. Set $S' = S \otimes _ R R'$ and $\mathfrak n' = \mathfrak nS$. If
the map $R \to R'$ embeds $R$ as a direct summand of $R'$ as an $R$-module, and
$R' \to S'$ is formally smooth for the $\mathfrak n'$-adic topology,
then $R \to S$ is formally smooth in the $\mathfrak n$-adic topology.
Proof. Let a solid diagram
as in Definition 15.37.1 be given. Set $A' = A \otimes _ R R'$ and $J' = \mathop{\mathrm{Im}}(J \otimes _ R R' \to A')$. The base change of the diagram above is the diagram
with continuous arrows. By condition (2) we obtain the dotted arrow $\psi ' : S' \to A'$. Using condition (1) choose a direct summand decomposition $R' = R \oplus C$ as $R$-modules. (Warning: $C$ isn't an ideal in $R'$.) Then $A' = A \oplus A \otimes _ R C$. Set
Then $J' = J \oplus J''$ as $A$-modules. The image of the composition $\psi : S \to A'$ of $\psi '$ with $S \to S'$ is contained in $A + J' = A \oplus J''$. However, in the ring $A + J' = A \oplus J''$ the $A$-submodule $J''$ is an ideal! (Use that $J^2 = 0$.) Hence the composition $S \to A + J' \to (A + J')/J'' = A$ is the arrow we were looking for. $\square$
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