Lemma 67.52.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-finite and separated. Let $Y'$ be the normalization of $Y$ in $X$. Picture:
Then $f'$ is a quasi-compact open immersion and $\nu $ is integral. In particular $f$ is quasi-affine.
Proof. By Lemma 67.51.1 the morphism $f$ is representable. Hence we may apply Lemma 67.52.1. Thus there exists an open subspace $U' \subset Y'$ such that $(f')^{-1}(U') = X$ (!) and $X \to U'$ is an isomorphism! In other words, $f'$ is an open immersion. Note that $f'$ is quasi-compact as $f$ is quasi-compact and $\nu : Y' \to Y$ is separated (Lemma 67.8.9). Hence for every affine scheme $Z$ and morphism $Z \to Y$ the fibre product $Z \times _ Y X$ is a quasi-compact open subscheme of the affine scheme $Z \times _ Y Y'$. Hence $f$ is quasi-affine by definition. $\square$
Your email address will not be published.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)