Lemma 10.138.7. 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 the sequence

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

of Lemma 10.131.9 is a split exact sequence.

Proof. Assume $S$ is formally smooth over $R$. By Lemma 10.138.5 this means there exists an $R$-algebra map $S \to P/J^2$ which is a right inverse to the canonical map $P/J^2 \to S$. By Lemma 10.131.10 we see that the sequence is split.

Assume the exact sequence of the lemma is split exact. Choose a splitting $\sigma : \Omega _{S/R} \to \Omega _{P/R} \otimes _ R S$. For each $\lambda \in S$ choose $x_\lambda \in P$ which maps to $\lambda$. Next, for each $\lambda \in S$ choose $f_\lambda \in J$ such that

$\text{d}f_\lambda = \text{d}x_\lambda - \sigma (\text{d}\lambda )$

in the middle term of the exact sequence. We claim that $s : \lambda \mapsto x_\lambda - f_\lambda \mod J^2$ is an $R$-algebra homomorphism $s : S \to P/J^2$. To prove this we will repeatedly use that if $h \in J$ and $\text{d}h = 0$ in $\Omega _{P/R} \otimes _ R S$, then $h \in J^2$. Let $\lambda , \mu \in S$. Then $\sigma (\text{d}\lambda + \text{d}\mu - \text{d}(\lambda + \mu )) = 0$. This implies

$\text{d}(x_\lambda + x_\mu - x_{\lambda + \mu } - f_\lambda - f_\mu + f_{\lambda + \mu }) = 0$

which means that $x_\lambda + x_\mu - x_{\lambda + \mu } - f_\lambda - f_\mu + f_{\lambda + \mu } \in J^2$, which in turn means that $s(\lambda ) + s(\mu ) = s(\lambda + \mu )$. Similarly, we have $\sigma (\lambda \text{d}\mu + \mu \text{d}\lambda - \text{d}\lambda \mu ) = 0$ which implies that

$\mu (\text{d}x_\lambda - \text{d}f_\lambda ) + \lambda (\text{d}x_\mu - \text{d}f_\mu ) - \text{d}x_{\lambda \mu } + \text{d}f_{\lambda \mu } = 0$

in the middle term of the exact sequence. Moreover we have

$\text{d}(x_\lambda x_\mu ) = x_\lambda \text{d}x_\mu + x_\mu \text{d}x_\lambda = \lambda \text{d}x_\mu + \mu \text{d} x_\lambda$

in the middle term again. Combined these equations mean that $x_\lambda x_\mu - x_{\lambda \mu } - \mu f_\lambda - \lambda f_\mu + f_{\lambda \mu } \in J^2$, hence $(x_\lambda - f_\lambda )(x_\mu - f_\mu ) - (x_{\lambda \mu } - f_{\lambda \mu }) \in J^2$ as $f_\lambda f_\mu \in J^2$, which means that $s(\lambda )s(\mu ) = s(\lambda \mu )$. If $\lambda \in R$, then $\text{d}\lambda = 0$ and we see that $\text{d}f_\lambda = \text{d}x_\lambda$, hence $\lambda - x_\lambda + f_\lambda \in J^2$ and hence $s(\lambda ) = \lambda$ as desired. At this point we can apply Lemma 10.138.5 to conclude that $S/R$ is formally smooth. $\square$

Comment #1808 by Peter Johnson on

Around line 4 of the proof, P/J^2 is a direct sum of R-modules, but in general will only be a split extension of R-algebras.

Comment #3206 by Dario Weißmann on

Typo: The map $S\to P/J^2$ is a right inverse to $P/J^2 \to S$.

And why not invoke lemma 10.130.10 from there?

Comment #3310 by on

A rare change which shortens the Stacks project. Thanks very much. Change is here.

Comment #4297 by Bogdan on

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