**Proof.**
We will use without further mention that formation of the naive cotangent complex commutes with localization, see Section 10.134, especially Lemma 10.134.13. Note that $\Omega _{S/R}$ is a finitely presented $S$-module, see Lemma 10.131.15. Hence (2), (3), and (4) are equivalent by Lemma 10.78.2. It is clear that (1) implies the equivalent conditions (2), (3), and (4). Assume (2) holds. Writing $S_\mathfrak q$ as the colimit of principal localizations we see from Lemma 10.127.6 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 10.134.2. 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.
$\square$

## Comments (3)

Comment #8495 by Et on

Comment #8612 by amnon yekutieli on

Comment #9105 by Stacks project on