4
$\begingroup$

The sources known to me (Knutson's Algebraic Spaces and Pascual-Gainza's Ampleness criteria for algebraic spaces) define a line bundle $L$ on an algebraic space $X$ (over a base scheme $S$) to be ample if a tensor power of $L$ is very ample, i.e. the pullback of $i^*\mathcal{O}(1)$ given a factorization $X \xrightarrow{i} \mathbb{P}^n_S \to S$. The existence of an ample line bundle shows then obviously that $X$ is a scheme.

Alternatively, one could define a line bundle $L$ on $X$ to be ample as in the scheme case: We call $L$ ample if for every quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$ of finite type, there exists an integer $n_0$ such that $\mathcal{F}\otimes L^{\otimes n}$ is generated by global sections for every $n\geq n_0$. As usual one can show in this case that the non-vanishing loci of global sections of tensor powers of $L$ form a basis of topology for the underlying space $|X|$ of $X$, at least if $X$ is noetherian.

My question is now the following:

When does the existence of an ample line bundle in my sense imply the existence of a very ample line bundle? When can one deduce from the existence of an ample line bundle that the algebraic space is schematic?

At least the latter question seems to have a positive answer in the case that $X$ is noetherian and normal (and quasi-separated and of finite type over $S$). It is known that in this case $X$ is the coarse moduli space of $Y/G$ for a scheme $Y$ and a finite group $G$ (see Is every algebraic space the quotient of a scheme by a finite group? ). The morphism $f: Y \to X$ is indeed finite. Thus, $f^*L$ is ample. Thus, $Y$ is quasi-projective. Hence, $X$ is a scheme.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.