# divisors and powers of line bundles

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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And even for curves, this is not sufficient (take $D$ equal to the sum of two points). –  Cantlog May 12 '14 at 12:03
@Cantlog: Indeed. I started to write that as my answer, but then the details got a little messier than I cared to deal with... –  user64687 May 12 '14 at 12:07
Suppose $P_1+P_2\sim 2Q$ with $P_1\ne P_2$. Let $f$ be a rational function whose divisor is $P_1+P_2-2Q$. It defines a degree $2$ morphism from the curve to the projective line. So if the curve is not hyperelliptic we are already done. If the curve is hyperelliptic, then $P_2$ must be the conjugate of $P_1$ under the hyperelliptic involution. So this happens rarely. –  Cantlog May 12 '14 at 12:12
@Cantlog: that's a nice argument. I was worried about the possibilty that $L$ is not effective, but I suppose one might as well assume that $C$ has genus 1. –  user64687 May 12 '14 at 12:15
Oops, you are right, I completely missed this case. Then for curves the answer is positive because the Jacobian is $m$-divisible. –  Cantlog May 12 '14 at 13:05