The Stacks project

Proposition 10.138.8. Let $R \to S$ be a ring map. Consider a formally smooth $R$-algebra $P$ and a surjection $P \to S$ with kernel $J$. The following are equivalent

  1. $S$ is formally smooth over $R$,

  2. for some $P \to S$ as above there exists a section to $P/J^2 \to S$,

  3. for all $P \to S$ as above there exists a section to $P/J^2 \to S$,

  4. for some $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes S \to \Omega _{S/R} \to 0$ is split exact,

  5. for all $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes S \to \Omega _{S/R} \to 0$ is split exact, and

  6. the naive cotangent complex $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a projective $S$-module placed in degree $0$.

Proof. It is clear that (1) implies (3) implies (2), see first part of the proof of Lemma 10.138.5. It is also true that (3) implies (5) implies (4) and that (2) implies (4), see first part of the proof of Lemma 10.138.7. Finally, Lemma 10.138.7 applied to the canonical surjection $R[S] \to S$ (10.134.0.1) shows that (1) implies (6).

Assume (4) and let's prove (6). Consider the sequence of Lemma 10.134.4 associated to the ring maps $R \to P \to S$. By the implication (1) $\Rightarrow $ (6) proved above we see that $\mathop{N\! L}\nolimits _{P/R} \otimes _ R S$ is quasi-isomorphic to $\Omega _{P/R} \otimes _ P S$ placed in degree $0$. Hence $H_1(\mathop{N\! L}\nolimits _{P/R} \otimes _ P S) = 0$. Since $P \to S$ is surjective we see that $\mathop{N\! L}\nolimits _{S/P}$ is homotopy equivalent to $J/J^2$ placed in degree $1$ (Lemma 10.134.6). Thus we obtain the exact sequence $0 \to H_1(L_{S/R}) \to J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0$. By assumption we see that $H_1(L_{S/R}) = 0$ and that $\Omega _{S/R}$ is a projective $S$-module. Thus (6) follows.

Finally, let's prove that (6) implies (1). The assumption means that the complex $J/J^2 \to \Omega _{P/R} \otimes S$ where $P = R[S]$ and $P \to S$ is the canonical surjection (10.134.0.1). Hence Lemma 10.138.7 shows that $S$ is formally smooth over $R$. $\square$

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