The Stacks project

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) is quasi-isomorphic to a projective $S$-module placed in degree $0$. Hence Lemma 10.138.7 shows that $S$ is formally smooth over $R$. $\square$