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The Stacks project

Lemma 10.138.2. Let R \to S be a formally smooth ring map. Let R \to R' be any ring map. Then the base change S' = R' \otimes _ R S is formally smooth over R'.

Proof. Let a solid diagram

\xymatrix{ S \ar[r] \ar@{-->}[rrd] & R' \otimes _ R S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[u] \ar[r] & R' \ar[r] \ar[u] & A \ar[u] }

as in Definition 10.138.1 be given. By assumption the longer dotted arrow exists. By the universal property of tensor product we obtain the shorter dotted arrow. \square


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