Lemma 10.137.12. 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 finite free $S_\mathfrak q$-module,

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

4. $H_1(L_{S/R})_\mathfrak q = 0$ 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 10.134, 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. $\square$

Comment #8495 by Et on

why does the zariski freeness of the module follow from Lemma 10.127.6? I feel like a more delicate argument is needed here (albeit a standard one)

Comment #8612 by amnon yekutieli on

In the lemma, presumably the whole discussion is about the naive contangent complex NL_{S/R}. But the text goes back and forth between that and the full cot cmplx L_{S/R}.

Granted, these complexes have the same H_1 (or H^1, as later, in \ref{https://stacks.math.columbia.edu/tag/08R6/cite}), but it might be better to clarify.

Comment #9105 by on

@#8495: The lemma states that the category of finitely presented modules is the colimit of the categories, so if the colimit is free then it is free at some stage. So I think this is easy enough to leave as is.

@#8612: This is because of the somewhat "bad" choice of writing $H_1(L)$ as a shorthand for $H_1(NL)$. See Definition 10.134.1. I guess we should have used the classical $\Gamma_{B/A}$ and less confusion would have been caused. Sorry, but unless more people complain I am going to leave this as is.

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