# Line bundles which are fiber-wise algebraically trivial.

$$\DeclareMathOperator{\Pic}{Pic}$$ Let $$f: M \to B$$ be a smooth morphism of smooth varieties over $$\mathbb C$$ such that the natural map $$\mathcal O_B \to f_* \mathcal O_M$$ is an isomorphism (i.e. $$f$$ is proper with connected fibers). I know that the following holds[1, Lemma 2.7]:

If $$L \in \Pic(M)$$ is such that for all $$b \in B$$, the restriction $$L_b = L|_{f^{-1}(b)}$$ is trivial, then $$L \cong f^* L'$$ for some line bundle $$L' \in \Pic(B)$$.

Does a similar statement hold for algebraic equivalence? So assume $$L_b$$ is algebraically equivalent to zero for all $$b$$, which means that $$0 = c_1(L_b) \in H^2(f^{-1}(b), \mathbb C).$$ Is it true that $$L$$ is algebraically equivalent to $$f^* L'$$ for some $$L' \in \Pic(B)$$?

Equivalently, using the cited fact above, we may ask if there is a line bundle $$L'' \in \Pic^0(M)$$ such that $$L_b \cong L''_b$$ for all $$b \in B$$. Then $$L \otimes L''^\vee \cong f^* L'$$ for some $$L' \in \Pic(B)$$, since $$L \otimes L''^\vee$$ is trivial on the fibers of $$f$$.

[1] Kleiman, The Picard Scheme

No. A counterexample is given by the Poincaré line bundle on the product $$A \times A^\vee$$ for any abelian variety $$A$$.