Hi,

Let $f : X \rightarrow Y$ a projective morphism of quasi-projective algebraic varieties over $\mathbb{C}$. Assume that $X$ is smooth, that $Y$ is normal and that:

$$\textbf{R} f_* \mathcal{O}_X = \mathcal{O}_Y $$

(here $ \textbf{R} f_* $ denotes the derived push-forward of $f$). Let $E$ be a Cartier divisor on $X$ such that $E$ is numerically trivial on every fiber of $f$. Is there an integer $m$ such that $mE = f^* F$ for some Cartier divisor $F$ on $Y$ ?

In case $Y$ is a point, I think it is true, because $E$ numerically trivial implies that $mE$ is a deformation of $\mathcal{O}_X$ for some $m>>0$. One concludes with the equality: \begin{equation*} T_{\mathrm{Pic}^0(X), \mathcal{O}_X} = H^1(X,\mathcal{O}_X), \end{equation*} where $T_{\mathrm{Pic}^0(X), \mathcal{O}_X}$ is the tangent space to the neutral component of the Picard scheme at $\mathcal{O}_X$.

I guess that if $f$ is flat, the same argument should work with the "relative Picard scheme". However I am interested in the case where $f$ is not flat.

Many thanks in advance!

trivialon every fiber. Does it follow that $E = f^* F$ for an invertible sheaf $F$ on $Y$?" $\endgroup$ – Piotr Achinger Mar 29 '13 at 3:18