bad reduction of a rational surface

Question

Let $X$ be the blowing-up of $\mathbb{P}^2_{\mathbb{Q}}$ at four rational points on a line. Can one show that $X$ has bad reduction at 2? Or does $X$ secretly has good reduction at 2?

What one can be sure is that the closures in $\mathbb{P}^2_{\mathbb{Z}}$ of any such four points cannot be disjoint over $(2)\in{\rm Spec\ }\mathbb{Z}$, since each rational line in $\mathbb{P}^2_{\mathbb{F}_2}$ contains only 3 rational points. If (one of) the cross ratio of the four points is $a/b$ with ${\rm gcd}(a,b)=1$, the same consideration suggests that $X$ has bad reduction at $p$ iff $p\mid a$ or $p\mid b$ or $p\mid(a-b)$.

Answer 1

You are completely right that for any 4 rational points on the projective line, two have equal reductions mod two. However as ulrich points out, this does not create a singularity of the underlying surface.

The reason is that if we blow up $[1,0,0], [0,1,0],[1,1,0]$ in $\mathbb P^2_{\mathbb Z}$, say, we can then blow up, not the inverse image of the copy of $\operatorname{Spec} \mathbb Z$ defined by $[1,-1,0]$, but instead its strict transform. The strict transform can be defined as the closure in the blown-up surface of the relevant rational point. From this it is easy to see that the strict transform is isomorphic to $\operatorname{Spec} \mathbb Z$ and that the surface is smooth at it, so there is no problem blowing it up to obtain a smooth surface.

Explicitly, we can see that the strict transform of $[1,-1,0]$ intersects $[1,1,0]$ in the direction given by $\frac{ [1,-1,0]- [1,1,0]}{2} = [0,-1,0]$ - i.e. at the intersection of the exceptional divisor and the strict transform of the line that both points lie on.