
Lemma 10.135.12. Let $R \to S$ be of finite presentation. Let $\mathfrak q$ be a prime of $S$. The following are equivalent

1. $R \to S$ is smooth at $\mathfrak q$,

2. $H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega _{S/R, \mathfrak q}$ is a projective $S_\mathfrak q$-module, and

3. $H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega _{S/R, \mathfrak q}$ is a flat $S_\mathfrak q$-module.

Proof. We will use without further mention that formation of the naive cotangent complex commutes with localization, see Section 10.132, especially Lemma 10.132.13. It is clear that (1) implies (2) implies (3). Assume (3) holds. Note that $\Omega _{S/R}$ is a finitely presented $S$-module, see Lemma 10.130.15. Hence $\Omega _{S/R, \mathfrak q}$ is a finite free module by Lemma 10.77.4. Writing $S_\mathfrak q$ as the colimit of principal localizations we see from Lemma 10.126.6 that we can find a $g \in S$, $g \not\in \mathfrak q$ such that $(\Omega _{S/R})_ g$ is finite free. Choose a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I$. We may work with $\mathop{N\! L}\nolimits (\alpha )$ instead of $\mathop{N\! L}\nolimits _{S/R}$, see Lemma 10.132.2. The surjection

$\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _ R S \to \Omega _{S/R} \to 0$

has a right inverse after inverting $g$ because $(\Omega _{S/R})_ g$ is projective. Hence the image of $\text{d} : (I/I^2)_ g \to \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _ R S_ g$ is a direct summand and this map has a right inverse too. We conclude that $H_1(L_{S/R})_ g$ is a quotient of $(I/I^2)_ g$. In particular $H_1(L_{S/R})_ g$ is a finite $S_ g$-module. Thus the vanishing of $H_1(L_{S/R})_{\mathfrak q}$ implies the vanishing of $H_1(L_{S/R})_{gg'}$ for some $g' \in S$, $g' \not\in \mathfrak q$. Then $R \to S_{gg'}$ is smooth by definition. $\square$

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