TITLE: Numerically equivalent divisors on a surface
QUESTION [8 upvotes]: What is an example of divisors on a surface which are numerically equivalent but not linearly equivalent? Can a curve be numerically equivalent to 0?
I don't have a good intuition for this. In projective space, any two curves intersect. In some less nice surfaces, some curves don't intersect. But can you have a curve which intersects no other curve?

REPLY [7 votes]: For your first question, consider a ruled surface $S$ over a curve $C$. If the genus of $C$ is no less than $1$, then any two different fibers are numerically equivalent but not linearly equivalent. In fact, for any smooth variety $X$, each pair of algebraically equivalent divisors are linearly equivalent if and only if $Pic^0(X)$  is trivial, since $NS(X)=Pic(X)/Pic^0(X)$. Notice that algebraically equivalence is finer than numerically equivalence, so $Pic^0(X)$ trivial is a necessary condition for each pair of numerically equivalent divisors to be linearly equivalent.
For your second question, any curve $C$ cannot be numerically equivalent to $0$ in any (nonsingular) projective variety $X$. For the reason, choose an embedding of $X$ to the projective space, and denote the hyperplane section of $X$ by $H$, then $C\cdot H=\deg C>0$, which means $C$ cannot be numerically trivial. Notice that every nonsingular complete surface is projective, so any curve on a surface cannot be numerically trivial.
However, if $X$ is nonsingular, complete but not projective, it may happen that the sum of two different curves may be numerically trivial (see Hironaka's example -- Example 3.4.1 in Appendix B of Hartshorne's "Algebraic Geometry").