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!



Relatively numerically trivial divisor
Hi,


So the result is true and a proof has been nicely explained to me by Yoshinori Namikawa.


First, let's recall the following "classical" fact (whose proof can be found as lemma $7.7$ in the book *Fourier-Mukai and Nahm transforms in Geometry and Mathematical Physics*, for instance):


**Lemma** :
Let $f : Y \rightarrow X$ be a proper morphism of varieties such that that 


$$ \mathrm{R} f\_\* \mathcal{O}\_X = \mathcal{O}\_Y. $$ 


Let $E$ be a Cartier divisor on $Y$. Then $E$ is the pull back of a Cartier divisor on $X$ if and only if for all $x \in X$, there is a neighborhood $U$ of $x$ in $X$ such that $E$ restricted to $f^{-1}(U$) is trivial.


Let $x \in X$, and let $U$ be a contractible neighborhood of $x$ in $X$. Since $\mathrm{R} f\_\* \mathcal{O}\_{f^{-1}(U)} = \mathcal{O}\_U$, the exponential exact sequence shows that:
$$ \mathrm{Pic}(f^{-1}(U)) = H^2(U,f^{-1}(U), \mathbb{Z}).$$


Let $F\_x$ be the reduced scheme underlying $f^{-1}(x)$. As the open $U$ is contractible, we have $H^2(f^{-1}(U), \mathbb{Z}) = H^2(F\_x, \mathbb{Z})$. So we only have to prove that $E$ is homologically trivial on $F\_x$. Denote by $\mathrm{Pic}^h(F\_x)$ (resp. $\mathrm{Pic}^{\tau}(F\_x)$ the group of homologically trivial line bundles (resp. numerically trivial line bundles) on $F\_x$. Since $F\_x$ is reduced, a theorem of Matsusaka (see Kleiman : *Towards a numerical theory of ampleness* for a proof) tells us that the quotient group:
$$ \mathrm{Pic}^{\tau}(F\_x)/\mathrm{Pic}^h(F\_x)$$
is of finite order. Hence some multiple of $E$ is homologically trivial on $F\_x$, which conclues the proof.


