Let $D$ and $C$ bet an irreducible effective divisor and an irreducible curve on $\mathbb{P}^1 \times \mathbb{P}^2$, respectively. Assume that both $D$ and $C$ are mapped surjectively onto $\mathbb{P}^1$ via the first projection. What can we say about $D$ and $C$ if we suppose moreover that $D\cap C = \emptyset$?
My geometric intuition says that this cannot happen unless $D=\mathbb{P}^1 \times \gamma$, where $\gamma \subset \mathbb{P}^2$ is a curve and $C=\mathbb{P}^1\times \{q\}$ where $q\notin \gamma$. Maybe this can be proved via intersection theory?
Edit: I think that a possible sketch of proof goes as follows:
$D$ is linearly equivalent to $\mathcal{O}_{\mathbb{P}^1\times \mathbb{P}^2}(a,b)$ with $a,b\geq 0$. We note that $a=0$ or $b=0$, since otherwise $D$ is ample and hence $D\cdot C > 0$. On one hand, If $b=0$ then $D$ is the pullback of a divisor on $\mathbb{P}^1$ and hence it is a plane (since it is irreducible) of the form $\{pt\}\times \mathbb{P}^2$. On the other hand, $C$ surjects onto $\mathbb{P}^1$ via the first projection, so the latter case is impossible. We have therefore that $a=0$ and hence $D$ is the pullback of a divisor on $\mathbb{P}^2$, i.e., $D\cong \mathbb{P}^1 \times \gamma$, where $\gamma\subseteq \mathbb{P}^2$ is a curve. Moreover, the projection formula gives $0=C\cdot \operatorname{pr}_2^*\ell = (\operatorname{pr}_2)_* C \cdot \ell$, where $\ell$ is a line on $\mathbb{P}^2$ and thus $C$ is of the form $\mathbb{P}^1 \times \{q\}$, where $q\notin \gamma$.
Any opinions? Thanks a lot in advance!