Let $X$ be an integral, regular, projective scheme of dimension $2$, flat over $\operatorname{Spec }\mathbb Z$. I have a couple of questions about horizontal and vertical divisors on $X$:
- Is it always possible to find two horizontal divisors on $X$ with no intersection points? I'm tempted to say yes, but I'm not able to prove it formally.
- Given an irreducible horizontal divisor $D$ is it always possible to find a vertical divisor $E$ linearly equivalent to $D$?
Many Thanks
I think the answer for the first question is yes, because horizontal divisors correspdoning to the closure of points in the generic fiber (see Liu). So if you pick different points, they will give you different closures. I think horizontal divisors and vertical divisors are almost never equivalent.