Let $X$ be a projective variety over $\mathbb{Z}$, and suppose that $X$ has everywhere good reduction. Let $Y$ be the blow-up of $X$ at an integral point.


Then is it the case that $Y$ also has everywhere good reduction?


The example situation that I have in mind is the following (my main motivation is del Pezzo surfaces).


Take $X= \mathbb{P}^2$ over $\mathbb{Z}$. This clearly has good reduction everywhere. Next let $Y$ be the blow-up of $\mathbb{P}^2$ at the integral point $(0:0:1)$. This can be realised as the subvariety of $\mathbb{P}^2 \times \mathbb{P}^1$ (with variables $x\_0,x\_1,x\_2$ and $y\_1,y\_2$) given by the equation $x\_1 y\_2 = x\_2 y\_1$. Then $Y$ has everywhere good reduction (at least if my caluclations are correct). 


I am curious to know if this happens after successively blowing up more integral points to obtain other del Pezzo surfaces. Note however that I am not claiming that all del Pezzo surfaces have everywhere good reduction!


Thanks in advance!



Good reduction and blow-ups
It is true if the integral point $T$ is actually a section (as in your example), because you then blow-up a smooth scheme $X\to {\rm Spec}(\mathbb Z)$ along a smooth center $T\simeq {\rm Spec}(\mathbb Z)$. In general, as $T$ is flat over $\mathbb Z$, the fiber $Y\_p$ of $Y$ at a prime $p$ is the blow-up of $X\_p$ along $T\_p$. At a $p$ ramified for $T\to {\rm Spec}(\mathbb Z)$, $T\_p$ is non-reduced and in general $Y\_p$ is not smooth. 


As an example, take $X=\mathbb P^2={\rm Proj}\ \mathbb Z[x,y,z]$ and $T=V\_+(x, y^2-2z^2)$. Then $Y$ has singular fiber at $2$. 


[**Edit**] Sorry, I was a little to optimistic on the compatibility of the blowing-up of $X$ with the base change $X\_p\to X$. However the conclusion is the same. Suppose for simplicity that the generic fiber of $X$ is geometrically connected. Let $p$ be any prime number and let $(X\_p)'$ be the blow-up of $X\_p$ along $T\_p$. Then we have a canonical closed immersion $(X\_p)'\to Y\_p$ which commutes with $(X\_p)'\to X\_p$ and $Y\_p\to X\_p$. Suppose now that $Y$ is smooth, then as $X\_p$ and $Y\_p$ are connected and smooth of the same dimension and $(X\_p)'\to X\_p$ is birational, $(X\_p)'\to Y\_p$ is an isomorphism. Hence $(X\_p)'$ must be smooth too. But in general this is not the case as $T\_p$ is not necessarily reduced (in the above example
$(X\_2)'$ is a normal singular surface). 


