Lemma 72.12.2. Let $S$ be a scheme. Let $f : X \to Y$ be a proper morphism of algebraic spaces over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. There exists an open subspace $V \subset Y$ characterized by the following property: A morphism $Y' \to Y$ of algebraic spaces factors through $V$ if and only if the pullback $\mathcal{L}'$ of $\mathcal{L}$ to $X' = Y' \times _ Y X$ is ample on $X'/Y'$ (as in Divisors on Spaces, Definition 69.14.1).

**Proof.**
Suppose that the lemma holds whenever $Y$ is a scheme. Let $U$ be a scheme and let $U \to Y$ be a surjective étale morphism. Let $R = U \times _ Y U$ with projections $t, s : R \to U$. Denote $X_ U = U \times _ Y X$ and $\mathcal{L}_ U$ the pullback. Then we get an open subscheme $V' \subset U$ as in the lemma for $(X_ U \to U, \mathcal{L}_ U)$. By the functorial characterization we see that $s^{-1}(V') = t^{-1}(V')$. Thus there is an open subspace $V \subset Y$ such that $V'$ is the inverse image of $V$ in $U$. In particular $V' \to V$ is surjective étale and we conclude that $\mathcal{L}_ V$ is ample on $X_ V/V$ (Divisors on Spaces, Lemma 69.14.6). Now, if $Y' \to Y$ is a morphism such that $\mathcal{L}'$ is ample on $X'/Y'$, then $U \times _ Y Y' \to Y'$ must factor through $V'$ and we conclude that $Y' \to Y$ factors through $V$. Hence $V \subset Y$ is as in the statement of the lemma. In this way we reduce to the case dealt with in the next paragraph.

Assume $Y$ is a scheme. Since the question is local on $Y$ we may assume $Y$ is an affine scheme. We will show the following:

If $\mathop{\mathrm{Spec}}(k) \to Y$ is a morphism such that $\mathcal{L}_ k$ is ample on $X_ k/k$, then there is an open neighbourhood $V \subset Y$ of the image of $\mathop{\mathrm{Spec}}(k) \to Y$ such that $\mathcal{L}_ V$ is ample on $X_ V/V$.

It is clear that (A) implies the truth of the lemma.

Let $X \to Y$, $\mathcal{L}$, $\mathop{\mathrm{Spec}}(k) \to Y$ be as in (A). By Lemma 72.12.1 we may assume that $k = \kappa (y)$ is the residue field of a point $y$ of $Y$.

As $Y$ is affine we can find a directed set $I$ and an inverse system of morphisms $X_ i \to Y_ i$ of algebraic spaces with $Y_ i$ of finite presentation over $\mathbf{Z}$, with affine transition morphisms $X_ i \to X_{i'}$ and $Y_ i \to Y_{i'}$, with $X_ i \to Y_ i$ proper and of finite presentation, and such that $X \to Y = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$. See Limits of Spaces, Lemma 68.12.2. After shrinking $I$ we may assume $Y_ i$ is an (affine) scheme for all $i$, see Limits of Spaces, Lemma 68.5.10. After shrinking $I$ we can assume we have a compatible system of invertible $\mathcal{O}_{X_ i}$-modules $\mathcal{L}_ i$ pulling back to $\mathcal{L}$, see Limits of Spaces, Lemma 68.7.3. Let $y_ i \in Y_ i$ be the image of $y$. Then $\kappa (y) = \mathop{\mathrm{colim}}\nolimits \kappa (y_ i)$. Hence $X_ y = \mathop{\mathrm{lim}}\nolimits X_{i, y_ i}$ and after shrinking $I$ we may assume $X_{i, y_ i}$ is a scheme for all $i$, see Limits of Spaces, Lemma 68.5.11. Hence for some $i$ we have $\mathcal{L}_{i, y_ i}$ is ample on $X_{i, y_ i}$ by Limits, Lemma 32.4.15. By Divisors on Spaces, Lemma 69.15.3 we find an open neigbourhood $V_ i \subset Y_ i$ of $y_ i$ such that $\mathcal{L}_ i$ restricted to $f_ i^{-1}(V_ i)$ is ample relative to $V_ i$. Letting $V \subset Y$ be the inverse image of $V_ i$ finishes the proof (hints: use Morphisms, Lemma 29.36.9 and the fact that $X \to Y \times _{Y_ i} X_ i$ is affine and the fact that the pullback of an ample invertible sheaf by an affine morphism is ample by Morphisms, Lemma 29.36.7). $\square$

## 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)