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Can anyone help me with the following question? Let $X$ be a smooth, projective algebraic variety. Let $D$ be an effective divisor on $X$ and $m$ an integer.

Under which conditions there exists a line bundle $L$ such that $\mathcal{O}_X(D)=L^m$?

There is of course the obvious one: $m$ should divide de degree of $D$. Is that sufficient?

You can assume that $D$ has normal crossings but I don't think that matters for this particular question.

Thanks! (and Merry Christmas)

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(I just came across this old question. Even though the OP may no longer be interested, an answer might be useful to other users.)

First a slight correction: there is no such thing as "the degree" of $D$ if $X$ is not a curve. The obvious modification to your suggestion is to require that $m$ should divide the intersection number $D \cdot C$ for every curve $C \subset X$.

Here is an example to show this condition is not sufficient.

Let $X$ be a smooth surface in $\mathbf P^3$ of degree $d \geq 4$, and take $D$ to be a section of the hyperplane bundle $O_X(1)$. Then for all curves $C \subset X$, the intersection number $D \cdot C$ is divisible by $d$. But if $X$ is very general, the Noether–Lefschetz theorem says that $\text{Pic}(X)$ is generated by $O_X(1)$, so there is no appropriate line bundle $L$.

I doubt there is any sensible sufficient condition, in general.

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    @Ca$n$tlog: thanks for the counter-answer! – 2014-05-12