We will add more here as needed.
Lemma 69.23.1. (For a more general version see More on Morphisms of Spaces, Lemma 76.35.1). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian. The following are equivalent
$f$ is finite, and
$f$ is proper and $|X_ k|$ is a discrete space for every morphism $\mathop{\mathrm{Spec}}(k) \to Y$ where $k$ is a field.
Proof. A finite morphism is proper according to Morphisms of Spaces, Lemma 67.45.9. A finite morphism is quasi-finite according to Morphisms of Spaces, Lemma 67.45.8. A quasi-finite morphism has discrete fibres $X_ k$, see Morphisms of Spaces, Lemma 67.27.5. Hence a finite morphism is proper and has discrete fibres $X_ k$.
Assume $f$ is proper with discrete fibres $X_ k$. We want to show $f$ is finite. In fact it suffices to prove $f$ is affine. Namely, if $f$ is affine, then it follows that $f$ is integral by Morphisms of Spaces, Lemma 67.45.7 whereupon it follows from Morphisms of Spaces, Lemma 67.45.6 that $f$ is finite.
To show that $f$ is affine we may assume that $Y$ is affine, and our goal is to show that $X$ is affine too. Since $f$ is proper we see that $X$ is separated and quasi-compact. We will show that for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. This implies that $H^1(X, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemmas 69.15.1 and 69.5.1. Then it follows that $X$ is affine from Proposition 69.16.7. By Lemma 69.22.8 we conclude that the stalks of $R^1f_*\mathcal{F}$ are zero for all geometric points of $Y$. In other words, $R^1f_*\mathcal{F} = 0$. Hence we see from the Leray Spectral Sequence for $f$ that $H^1(X , \mathcal{F}) = H^1(Y, f_*\mathcal{F})$. Since $Y$ is affine, and $f_*\mathcal{F}$ is quasi-coherent (Morphisms of Spaces, Lemma 67.11.2) we conclude $H^1(Y, f_*\mathcal{F}) = 0$ from Cohomology of Schemes, Lemma 30.2.2. Hence $H^1(X, \mathcal{F}) = 0$ as desired. $\square$
As a consequence we have the following useful result.
Lemma 69.23.2. (For a more general version see More on Morphisms of Spaces, Lemma 76.35.2). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\overline{y}$ be a geometric point of $Y$. Assume
$Y$ is locally Noetherian,
$f$ is proper, and
$|X_{\overline{y}}|$ is finite.
Then there exists an open neighbourhood $V \subset Y$ of $\overline{y}$ such that $f|_{f^{-1}(V)} : f^{-1}(V) \to V$ is finite.
Proof. The morphism $f$ is quasi-finite at all the geometric points of $X$ lying over $\overline{y}$ by Morphisms of Spaces, Lemma 67.34.8. By Morphisms of Spaces, Lemma 67.34.7 the set of points at which $f$ is quasi-finite is an open subspace $U \subset X$. Let $Z = X \setminus U$. Then $\overline{y} \not\in f(Z)$. Since $f$ is proper the set $f(Z) \subset Y$ is closed. Choose any open neighbourhood $V \subset Y$ of $\overline{y}$ with $Z \cap V = \emptyset $. Then $f^{-1}(V) \to V$ is locally quasi-finite and proper. Hence $f^{-1}(V) \to V$ has discrete fibres $X_ k$ (Morphisms of Spaces, Lemma 67.27.5) which are quasi-compact hence finite. Thus $f^{-1}(V) \to V$ is finite by Lemma 69.23.1. $\square$
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)