Lemma 37.6.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open such that $f(U) \subset V$. If $f$ is formally unramified, 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 37.6.1. If $f$ is formally ramified, then there exists at most one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Hence clearly there exists at most one such morphism 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Your email address will not be published.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)