Timeline for Cartier Divisor generated by Global Sections

6 events

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Nov 28, 2017 at 23:56	comment	added	user267839		@Mohan: Do you mean that it's clear by using surjectivity of $ H^0(X, \mathcal{O}_X(a)) \to H^0(X, \mathcal{F})= k(a)$ and the fact that $\mathcal{F}$ globally generated? So for every $x \in X$ there exist a $l \in k(a)$ such that $l_x \neq 0$ in $\mathcal{F}_x$ which provides a preimage $s \in H^0(X, \mathcal{O}_X(a))$ such that $s_x \neq 0$ in $\mathcal{O}_X(a))_x$ by commutativity of section-stalk diagram? Or do you mean it in other way?
Nov 28, 2017 at 11:34	vote	accept	user267839
Nov 28, 2017 at 6:43	answer	added	gdb		timeline score: 6
Nov 28, 2017 at 5:12	comment	added	gdb		Do you assume that your curve is smooth? Otherwise a point $a$ might not be a Cartier divisor. For example, take $a$ to be a singular point of a nodal cubic.
Nov 28, 2017 at 1:25	comment	added	Mohan		You have too many issues. If $X$ is not projective, it is affine and thus every quasi-coherent sheaf is globally generated. If it is projective and since it has a $k$-rational point, easy to check $H^0(X,\mathcal{O}_X)=k$. What exactly do you mean an invertible element in $H^0(\mathcal{O}_X(a))$? It is a vector space, not a ring. Also, I hope you have realized that $\mathcal{F}=k(a)=k$. Rest should be clear.
Nov 28, 2017 at 0:16	history	asked	user267839	CC BY-SA 3.0