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Let $X$ be a smooth variety over $\mathbb{C}$, and let $D$ be a divisor on $X$. What is the condition on $D$ so that we can speak of a canonical section $s$ on $H^0(X,D)$ such that $D$ is the zero locus of $s$?

Thanks.

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What exactly is $H^0(X,D)$ for you? –  Jesko Hüttenhain Jan 30 '13 at 9:42
    
@minimax: do you count multiplicities in the zero locus ? What do you mean by canonical ? Any multiple of $s$ by a non-zero scalar with give the same zero locus (with or without multiplicities). –  user18119 Jan 30 '13 at 21:24
    
@JeskoHüttenhain: I just mean rational function $s$ such that $div(s)+D\geq 0$.... –  minimax Jan 31 '13 at 6:54
    
@QiL: For the multiplicity part, I am not sure. The reason I ask the question is sometimes I read book or paper, there are things like 'consider section $s$ correspond to effective divisor $D$', so I am wondering what is the $s$ that 'canonically' corresponds to $D$? –  minimax Jan 31 '13 at 6:57
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So this doesn't mean a unique section. The sentence can have two possible meanings: first, it can just say $s$ is a global section of the sheaf $O_X(D)$, or, less likely, a non-zero rational section $s$ of $O_X(D)$ (then the divisor of $s$ is $D$). –  user18119 Jan 31 '13 at 8:24

1 Answer 1

If I understand this correctly, the condition is for $\mathcal O(D)$ to be a globally generated sheaf of principal ideals. This is the case if and only if $D=-E$ is the negative of an effective cartier divisor $E$ which is locally given by a system $\{(U,s_U)\}_{U}$ with $s_U\in\mathcal O_X(U)$. In other words, $D$ is locally given by $\{(U,s_U^{-1})\}_{U}$. Note that since the $s_U\in \mathcal O_X(U)$ agree on intersections, you can glue them to a global section $s\in H^0(X,\mathcal O_X(-E)) \subseteq H^0(X,O_X)$.

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