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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)