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Let $X$ be a smooth projective variety with Picard number 1 over $\mathbb{C}$. Let $F$ be a coherent sheaf on $X$ such that $c_1(F)$ is algebraically trivial, and hence numerically trivial. Also rank $F=0$.

So $F$ is supported on a proper closed subscheme of $X$. Can $F$ be supported on a divisor, or is the codimension of Support $ F\geq 2$.

Since the Picard number of $X$ is 1, any non-zero effective divisor is ample. Since $c_1(F)$ is numerically trivial, if $F$ is supported on a divisor, it is non-effective. Is this possible?

Note, I posted this question on MathSE, but have not received any answers.

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    $\begingroup$ Your hypothesis that $F$ has rank $0$ and $c_1(F)$ is numerically trivial is enough to conclude that $F$ cannot be supported on a divisor. (The Picard number 1 hypothesis is not necessary for this.) $\endgroup$
    – naf
    Commented Oct 24, 2017 at 11:13
  • $\begingroup$ @ulrich, thank you for the clarification. It would be helpful if you can elaborate a bit. $\endgroup$
    – user52991
    Commented Oct 24, 2017 at 11:14
  • $\begingroup$ @ulrich, I have been trying to understand what you said. Do we have that - if we have a numerically trivial divisor $D$, then it cannot be the support of any sheaf. Is it because the $D$ is not effective? $\endgroup$
    – user52991
    Commented Oct 24, 2017 at 16:09
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    $\begingroup$ Yes, the point is that on a projective variety a nonzero effective divisor is never numerically trivial. $\endgroup$
    – naf
    Commented Oct 25, 2017 at 4:56

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