Let $X$ be a rational elliptic surface over $\Bbb{C}$.
It is well known that $-K_X$ is linearly equivalent to a general fiber $F$.
I'm trying to prove/disprove the following: let $D$ be an effective, nef divisor on $X$ such that $D^2=D\cdot F=0$, then $D$ is numerically equivalent to $nF$ for some $n\in\Bbb{Z}$.
In one of the comments on this question, someone claims that a consequence of Hodge index theorem is that if $A,B$ are nef and $A\cdot B=0$, then $A,B$ are numerically proportional.
If this is true, then I have what I need. Since $F,D$ are nef and $D\cdot F=0$, then $[D]=n[F]$ in $\text{Num}(X)$. And I wouldn't even have to use $D^2=0$.
But I can't see why the claim in the comment is true. I can only see that if $A^2>0$, then by Hodge Index Theorem $B^2\leq 0$ with equality only if $[B]=0$ in $\text{Num}(X)$. But since $B$ is nef, $B^2\geq 0$, so ideed $[B]=0$. By symmetry, $B^2>0$ would also imply $[A]=0$. How do I get $A,B$ numerically proportional?
As observed by Mohan, it can certainly happen that $D$ is equivalent to a rational multiple of a general fiber (simply consider the case that we have multiple fibers). Here is a simple argument that shows that $D$ is always numerically a rational multiple of a general fiber (without using the HIT). It is clear that $D$ is vertical as it does not intersect a general fiber. Now, let $f:X\to Y$ be the given fibration. Subtracting the pullback of points $P_i\in Y$ with appropriate coefficients $p_i$ from the base we have that $−(D−f^*(∑p_iP_i))$ is effective but contains no fiber. If $D−f^*(∑p_iP_i)=0$ we are done, otherwise, pick a curve $C$ contained in a fiber and intersecting its support, then $−(D−f^∗(∑p_iP_i))\cdot C>0$ which is impossible as $C\cdot f^*(∑p_iP_i)=0$ and $D$ is nef.