Lemma 37.8.6. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:
$f$ is formally étale,
$f$ is formally unramified and the universal first order thickening of $X$ over $S$ is equal to $X$,
$f$ is formally unramified and $\mathcal{C}_{X/S} = 0$, and
$\Omega _{X/S} = 0$ and $\mathcal{C}_{X/S} = 0$.
Proof. Actually, the last assertion only make sense because $\Omega _{X/S} = 0$ implies that $\mathcal{C}_{X/S}$ is defined via Lemma 37.6.7 and Definition 37.7.2. This also makes it clear that (3) and (4) are equivalent.
Either of the assumptions (1), (2), and (3) imply that $f$ is formally unramified. Hence we may assume $f$ is formally unramified. The equivalence of (1), (2), and (3) follow from the universal property of the universal first order thickening $X'$ of $X$ over $S$ and the fact that $X = X' \Leftrightarrow \mathcal{C}_{X/S} = 0$ since after all by definition $\mathcal{C}_{X/S} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$. $\square$
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Comment #927 by Jim on