Consider a minimal smooth conic bundle $S$ of dimension two. Assume that there are two curves $C,F$ on $S$ such that $C^2 < 0$ and $F^2 = 0$. Let $D$ be a pseudoeffective divisor on $S$ such that $D\sim\_{\mathbb{Q}} aC+bF$ (numerical equivalence) where $a,b\in\mathbb{Q}$.


Is it true then that $mD$ must be an integral effective divisor for some $m\geq 1$?



Pseudoeffective divisors on surfaces
If I understand the question correctly, then the Picard number of $S$ is 2 and hence the existence of the curve $F$ implies that the cone of effective curves is closed, so every pseudoeffective divisor is actually effective.


**Added to answer Friedrich's question in the comments:** Let $C\subseteq S$ be a(n effective) curve with negative self-intersection. First of all, as the intersection number of different irreduicble components is non-negative, at least one of the irreducible components of $C$ must have negative self-intersection and hence we might as well assume that $C$ is irreducible.


However, then $C$ is non-negative on every irreducible curve other than itself. Intersecting with $C$ is a linear functional on the space of curves, and by the previous observation the hyperplane defined by $C\cdot (\ \ )=0$ separates $C$ from all other irreducible curves. This implies that $C$ must be on the boundary of the effective cone. (In particular, this also means that on a surface with Picard number $2$, there can be at most two irreducible curves with negative self-intersection. An example with two is a general degree 4 surface in $\mathbb P^3$ containing a conic. I leave it for the reader to figure out what the two curves with negative self-intersection are...)


