Lemma 33.15.3. Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_ K$ is quasi-projective over $K$ for some field extension $K/k$, then $X$ is quasi-projective over $k$.
Proof. By definition a morphism of schemes $g : Y \to T$ is quasi-projective if it is locally of finite type, quasi-compact, and there exists a $g$-ample invertible sheaf on $Y$. Let $K/k$ be a field extension such that $X_ K$ is quasi-projective over $K$. Let $\mathop{\mathrm{Spec}}(A) \subset X$ be an affine open. Then $U_ K$ is an affine open subscheme of $X_ K$, hence $A_ K$ is a $K$-algebra of finite type. Then $A$ is a $k$-algebra of finite type by Algebra, Lemma 10.126.1. Hence $X \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type. Since $X_ K \to \mathop{\mathrm{Spec}}(K)$ is quasi-compact, we see that $X_ K$ is quasi-compact, hence $X$ is quasi-compact, hence $X \to \mathop{\mathrm{Spec}}(k)$ is of finite type. By Morphisms, Lemma 29.39.4 we see that $X_ K$ has an ample invertible sheaf. Then $X$ has an ample invertible sheaf by Lemma 33.15.1. Hence $X \to \mathop{\mathrm{Spec}}(k)$ is quasi-projective by Morphisms, Lemma 29.39.4. $\square$
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Your email address will not be published.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)