Lemma 10.131.10. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Moreover, assume that there exists an $R$-algebra map $S' \to S$ which is a right inverse to $S \to S'$. Then the exact sequence of $S'$-modules of Lemma 10.131.9 turns into a short exact sequence

$0 \longrightarrow I/I^2 \longrightarrow \Omega _{S/R} \otimes _ S S' \longrightarrow \Omega _{S'/R} \longrightarrow 0$

which is even a split short exact sequence.

Proof. Let $\beta : S' \to S$ be the right inverse to the surjection $\alpha : S \to S'$, so $S = I \oplus \beta (S')$. Clearly we can use $\beta : \Omega _{S'/R} \to \Omega _{S/R}$, to get a right inverse to the map $\Omega _{S/R} \otimes _ S S' \to \Omega _{S'/R}$. On the other hand, consider the map

$D : S \longrightarrow I/I^2, \quad x \longmapsto x - \beta (\alpha (x))$

It is easy to show that $D$ is an $R$-derivation (omitted). Moreover $x D(s) = 0$ if $x \in I, s \in S$. Hence, by the universal property $D$ induces a map $\tau : \Omega _{S/R} \otimes _ S S' \to I/I^2$. We omit the verification that it is a left inverse to $\text{d} : I/I^2 \to \Omega _{S/R} \otimes _ S S'$. Hence we win. $\square$

Comment #1086 by Nuno Cardoso on

Typo: It should be "assume that there exists an $R$-algebra map $S' \to S$ which is a right inverse to $S \to S'$" instead of "assume that there exists an $R$-algebra map $S' \to S$ which is a right inverse to $S' \to S$".

Comment #8559 by on

The readers might find interesting to know the following: In Matsumura's Commutative Algebra (not to be confused with Commutative Ring Theory), Ch. 10, (26.I), in Theorem 58 it is stated a sufficient and equivalent condition for the map $I/I^2\to\Omega_{S/R}\otimes_S S'$ from Lemma 10.131.9 to have an $S'$-linear retraction (namely, that there is an $R$-algebra map $S'\to S/I^2$ which is a right inverse to the canonical map $S/I^2\to S'$). It is immediate to verify that "there exists an $R$-algebra map $S'\to S$ which is a right inverse to $S\to S'$" implies the condition between parentheses.

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