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Let $X$ be a normal, projective surface (or more generally a variety) over $\mathbb{C}$ (i.e., $X$ is irreducible). Fix a polarisation on $X$. I am looking for examples of rank one, degree zero (degree under the fixed polarisation) reflexive sheaf $F$ such that there exists an integer $r>0$ for which there is no rank one reflexive sheaf $G$ for which $G^{\otimes r}=F$. Is there some general strategy to produce such examples? Moreover, what happens if $F$ is an invertible sheaf?

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    $\begingroup$ What is your definition of "degree zero sheaf"? For a smooth quadric hypersurface $X$ in $\mathbb{P}^3$ that is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$, do you consider $\text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^*\mathcal{O}(-1)$ to have degree zero? $\endgroup$
    – Jason Starr
    Commented Oct 14, 2018 at 14:33
  • $\begingroup$ @JasonStarr My definition of degree is the same as the one in Huybrects-Lehn page 13. I have edited the question to fix a polarisation at the beginning. If I understand correctly, your example is of degree zero. Moreover, if $X$ is smooth, then the Picard variety is an abelian variety so, any invertible sheaf has an $r$-th root. So, my question is not interesting if $X$ is smooth. $\endgroup$
    – Chen
    Commented Oct 14, 2018 at 14:44
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    $\begingroup$ "Moreover, if $X$ is smooth, then the Picard variety is an abelian variety ..." This is not correct. According to your definition, the group of isomorphism classes of degree zero, rank one reflexive sheaves on a quadric hypersurface $X$ in $\mathbb{P}^3$ is a free cyclic group generated by the class of $\text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^*\mathcal{O}(-1)$. It is not the set of $k$-points of an Abelian variety. It is not divisible. The example I wrote in my comment above is a counterexample for every integer $r>1$. $\endgroup$
    – Jason Starr
    Commented Oct 14, 2018 at 14:52
  • $\begingroup$ @JasonStarr I am missing something. I want $k$-points of $Pic^0(X)$ to be "degree zero" invertible sheaves. $\endgroup$
    – Chen
    Commented Oct 14, 2018 at 14:58
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    $\begingroup$ "I am missing something." With the definition that you wrote above, certainly every $k$-point of $\text{Pic}^0(X)$ does give a degree zero invertible sheaf. However, with the definition that you wrote above, there are also degree zero invertible sheaves that are not algebraically equivalent to the structure sheaf. For instance, the invertible sheaf that I have twice listed in the comments above has degree zero with respect to the fixed polarization, yet it is not algebraically equivalent to zero. $\endgroup$
    – Jason Starr
    Commented Oct 14, 2018 at 15:02

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