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This question is from Liu's book on Algebraic Geometry. Let $\mathcal{L}$ be an invertible sheaf on a scheme X. Suppose that X is an algebraic variety over a field k and that there are n points of X that are not basepoints for $\mathcal{L}$ . Show that if k is either infinite or with cardinality greater than n, there exists a section $s \in H^0(X,\mathcal{L})$ that generates $\mathcal{L}$ at these n points that are mot basepoints.

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Hint:If $x$ is not a basepoint for $\mathcal{L}$, then the global sections that do not generate $\mathcal{L} _{x}$ form a proper subspace $V_{x} \subseteq H^{0}(X, \mathcal{L})$. If there is no section that generates $L _{x_{i}}$ for $1 \leq i \leq n$, then $\cup _{i} V_{x_{i}} = H^{0}(X, \mathcal{L})$. Is this possible? – Piotr Pstrągowski Mar 30 '13 at 15:53
Piotr: I suspect not, but I am not sure why. Is it analogous to where a vector space over an infinite field is not a union of finitely many proper subspaces? How could I see that the global sections are a vector space over k? (Probably easy?) – Heidar Svan Mar 30 '13 at 17:06
Well, normally when people say invertible sheaf, they assume that the sheaf in question is an $\mathcal{O}_{X}$-module. Thus, $\mathcal{L}(X)$ is an $\mathcal{O}_{X}(X)$-module and a $k$-vector space in particular. And yes, this is completely analogous to the proof that over an infinite field a vector space is a not a finite union of proper subspaces. The proof will also work in this case. – Piotr Pstrągowski Apr 1 '13 at 12:30

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