Lemma 15.36.9. 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

1. the map $R \to R'$ embeds $R$ as a direct summand of $R'$ as an $R$-module, and

2. $R' \to S'$ 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

$\xymatrix{ S \ar[r] & A/J \\ R \ar[u] \ar[r] & A \ar[u] }$

as in Definition 15.36.1 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

$\xymatrix{ S' \ar[r] \ar@{-->}[rd]^{\psi '} & A'/J' \\ R' \ar[u] \ar[r] & A' \ar[u] }$

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

$J'' = \mathop{\mathrm{Im}}(J \otimes _ R C \to A \otimes _ R C) \subset J' \subset A'.$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).