Lemma 15.40.3. Let $A \to B$ be a local homomorphism of Noetherian local rings. Assume $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. Then $A \to B$ is flat.

Proof. We may assume that $A$ and $B$ a Noetherian complete local rings by Lemma 15.37.4 and Algebra, Lemma 10.97.6 (this also uses Algebra, Lemma 10.39.9 and 10.97.3 to see that flatness of the map on completions implies flatness of $A \to B$). Choose a commutative diagram

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

as in Lemma 15.39.3 with $R \to S$ flat. Let $I \subset R$ be the kernel of $R \to A$. Because $B$ is formally smooth over $A$ we see that the $A$-algebra map

$S/IS \longrightarrow B$

has a section, see Lemma 15.37.5. Hence $B$ is a direct summand of the flat $A$-module $S/IS$ (by base change of flatness, see Algebra, Lemma 10.39.7), whence flat. $\square$

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