Lemma 76.21.7. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

$f$ is formally étale,

$H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = H^0(\mathop{N\! L}\nolimits _{X/Y}) = 0$.

Lemma 76.21.7. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

$f$ is formally étale,

$H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = H^0(\mathop{N\! L}\nolimits _{X/Y}) = 0$.

**Proof.**
Assume (1). A formally étale morphism is a formally smooth morphism. Thus $H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = 0$ by Lemma 76.21.6. On the other hand, a formally étale morphism if formally unramified hence we have $\Omega _{X/Y} = 0$ by Lemma 76.14.6. Conversely, if (2) holds, then $f$ is formally smooth by Lemma 76.21.6 and formally unramified by Lemma 76.14.6 and hence formally étale by Lemmas 76.19.4.
$\square$

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