Lemma 76.16.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ The following are equivalent:

1. $f$ is formally étale,

2. $f$ is formally unramified and the universal first order thickening of $X$ over $Y$ is equal to $X$,

3. $f$ is formally unramified and $\mathcal{C}_{X/Y} = 0$, and

4. $\Omega _{X/Y} = 0$ and $\mathcal{C}_{X/Y} = 0$.

Proof. Actually, the last assertion only make sense because $\Omega _{X/Y} = 0$ implies that $\mathcal{C}_{X/Y}$ is defined via Lemma 76.14.6 and Definition 76.15.5. 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 $X = X' \Leftrightarrow \mathcal{C}_{X/Y} = 0$ since after all by definition $\mathcal{C}_{X/Y} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$. $\square$

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