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)


(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.

  • 1
    $\begingroup$ And even for curves, this is not sufficient (take $D$ equal to the sum of two points). $\endgroup$ – Cantlog May 12 '14 at 12:03
  • $\begingroup$ @Cantlog: Indeed. I started to write that as my answer, but then the details got a little messier than I cared to deal with... $\endgroup$ – user64687 May 12 '14 at 12:07
  • $\begingroup$ 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. $\endgroup$ – Cantlog May 12 '14 at 12:12
  • $\begingroup$ @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. $\endgroup$ – user64687 May 12 '14 at 12:15
  • $\begingroup$ Oops, you are right, I completely missed this case. Then for curves the answer is positive because the Jacobian is $m$-divisible. $\endgroup$ – Cantlog May 12 '14 at 13:05

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