Lemma 37.13.6. 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 37.13.6. 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.**
A formally étale morphism is formally smooth and hence we have $H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = 0$ by Lemma 37.13.5. On the other hand, we have $\Omega _{X/Y} = 0$ by Lemma 37.8.6. Conversely, if (2) holds, then $f$ is formally smooth by Lemma 37.13.5 and formally unramified by Lemma 37.6.7 and hence formally étale by Lemmas 37.11.4.
$\square$

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