This question is about intersection theory on the easiest (arithmetic) surface over $\mathbf{Z}$: $\mathbf{P}^1_{\mathbf{Z}}$.

Suppose we are given two distinct $\mathbf{Q}$-rational points $b_1$ and $b_2$ on the projective line $\mathbf{P}^1_{\mathbf{Q}}$. Let $P_1$ and $P_2$ be the natural sections of $P^1_{\mathbf{Z}}$ over $\mathbf{Z}$ corresponding to $b_1$ and $b_2$ given by the valuative criterion of properness. (The image of $P_1$ being just the closure of $b_1$ in $\mathbf{P}^1_{\mathbf{Z}}$.)

Can one determine if $P_1\cap P_2 = \emptyset$ just by looking at the generic fibre, i.e., at the points $b_1$ and $b_2$?

Example. The pairs $(b_1,b_2)$ given by $(0,1)$, $(0,\infty)$ or $(1,\infty)$ give sections such that $P_1\cap P_2=\emptyset$. (I can prove this by drawing a picture.)

What am I missing? I'm guessing it could happen that $P_1\cap P_2 =\emptyset$ after composing with an automorphism...