The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.136.16. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = S \otimes _ R R'$. Then $R \to S$ is formally smooth if and only if $R' \to S'$ is formally smooth.

Proof. If $R \to S$ is formally smooth, then $R' \to S'$ is formally smooth by Lemma 10.136.2. To prove the converse, assume $R' \to S'$ is formally smooth. Note that $N \otimes _ R R' = N \otimes _ S S'$ for any $S$-module $N$. In particular $S \to S'$ is faithfully flat also. Choose a polynomial ring $P = R[\{ x_ i\} _{i \in I}]$ and a surjection of $R$-algebras $P \to S$ with kernel $J$. Note that $P' = P \otimes _ R R'$ is a polynomial algebra over $R'$. Since $R \to R'$ is flat the kernel $J'$ of the surjection $P' \to S'$ is $J \otimes _ R R'$. Hence the split exact sequence (see Lemma 10.136.7)

\[ 0 \to J'/(J')^2 \to \Omega _{P'/R'} \otimes _{P'} S' \to \Omega _{S'/R'} \to 0 \]

is the base change via $S \to S'$ of the corresponding sequence

\[ J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0 \]

see Lemma 10.130.9. As $S \to S'$ is faithfully flat we conclude two things: (1) this sequence (without ${}'$) is exact too, and (2) $\Omega _{S/R}$ is a projective $S$-module. Namely, $\Omega _{S'/R'}$ is projective as a direct sum of the free module $\Omega _{P'/R'} \otimes _{P'} S'$ and $\Omega _{S/R} \otimes _ S {S'} = \Omega _{S'/R'}$ by what we said above. Thus (2) follows by descent of projectivity through faithfully flat ring maps, see Theorem 10.94.5. Hence the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0$ is exact also and we win by applying Lemma 10.136.7 once more. $\square$


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