Lemma 10.138.5. Let R \to S be a ring map. Let P \to S be a surjective R-algebra map from a polynomial ring P onto S. Denote J \subset P the kernel. Then R \to S is formally smooth if and only if there exists an R-algebra map \sigma : S \to P/J^2 which is a right inverse to the surjection P/J^2 \to S.
Proof. Assume R \to S is formally smooth. Consider the commutative diagram
\xymatrix{ S \ar[r] \ar@{-->}[rd] & P/J \\ R \ar[r] \ar[u] & P/J^2\ar[u] }
By assumption the dotted arrow exists. This proves that \sigma exists.
Conversely, suppose we have a \sigma as in the lemma. Let a solid diagram
\xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] }
as in Definition 10.138.1 be given. Because P is formally smooth by Lemma 10.138.4, there exists an R-algebra homomorphism \psi : P \to A which lifts the map P \to S \to A/I. Clearly \psi (J) \subset I and since I^2 = 0 we conclude that \psi (J^2) = 0. Hence \psi factors as \overline{\psi } : P/J^2 \to A. The desired dotted arrow is the composition \overline{\psi } \circ \sigma : S \to A. \square
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