Lemma 37.8.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open subschemes such that $f(U) \subset V$. If $f$ is formally étale, so is $f|_ U : U \to V$.
Proof. Consider a solid diagram
\[ \xymatrix{ U \ar[d]_{f|_ U} & T \ar[d]^ i \ar[l]^ a \\ V & T' \ar[l] \ar@{-->}[lu] } \]
as in Definition 37.8.1. If $f$ is formally ramified, then there exists exactly one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since $|T'| = |T|$ we conclude that $a'(T') \subset U$ which gives our unique morphism from $T'$ into $U$. $\square$
Post a comment
Your email address will not be published.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)