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Let $X$ be a complex projective variety, and let $B$ be an ample and globally generated line bundle on $X.$ By Example 1.2.9 in Lazarsfeld's book, the set $U$ of points $y\in X$ such that $B\otimes\mathfrak{m}_y$ is globally generated is an open subset, $\mathfrak{m}_y$ being the ideal sheaf of the closed point $y.$ However I do not understand if this set is also non-empty.

If $B$ is not very ample, in which case $U=X,$ I don't see why $U\neq\emptyset$: denoting by $\phi\colon X\rightarrow\mathbb{P}^N_{\mathbb{C}}$ the morphism determined by $B,$ (which is finite by ampleness,) if $y\in U,$ then $\phi$ separates points from $y$ and is unramified at $y.$ However $\phi$ is not injective (actually there exists a non-empty open subset in $\mathbb{P}^N_{\mathbb{C}}$ where fibers contain exactly $\deg(\phi)$ points), and the fibers containing a single point are those containing a point of total ramification.

Any help is greatly appreciated.

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1 Answer 1

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This set can be empty. Let $X$ be an elliptic curve, let $B=\mathcal{O}_X(2p)$ for some point $p\in X$. Then $B$ is ample and globally generated, but $\mathcal{O}_X(2p-q)\cong \mathcal{O}_X(2p)\otimes \mathfrak{m}_q$ is not globally generated for any $q\in X$: the space of global sections has dimension one by Riemann-Roch, so global generation would imply that there is a surjection $\mathcal{O}_X\to\mathcal{O}_X(2p-q)$. But a surjection of line bundles is an isomorphism, contradiction.

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