Linear equivalence vs algebraic equivalence of divisors on smooth projective surfaces

Question

Let $X$ be a smooth projective surface and $D_1, D_2$ be two divisors on $X$. Is it true that $D_1$ is linearly equivalent to $D_2$ if and only if $D_1$ is algebraically equivalent to $D_2$?

Answer 1

OK, let me write out an example to show that the answer to the OP's question is "no".

First, if $C$ is a smooth curve of genus $\geq 1$, then all divisors of the same degree are algebraically equivalent. On the other hand, the linear equivalence classes of these divisors form a $g$-dimensional abelian variety, called the Jacobian variety of $X$. In particular, if $p$ and $q$ are distinct points on $C$, then they are not linearly equivalent as divisors on $C$.

Now consider the surface $X=C \times \mathbf P^1$. The pullback map $\pi^*: Pic(C) \rightarrow Pic(X)$ is injective, so the divisors $\pi^*(p)$ and $\pi^*(q)$ are still not linearly equivalent, although they are evidently algebraically equivalent.