7
$\begingroup$

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!

$\endgroup$

2 Answers 2

8
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Thanks for your answer but I just wanted to clarify: In general if $X \to S$ is a smooth morphism of schemes, then the blow-up $Y$ of $X$ at an $S-$valued point is also smooth over $S$? $\endgroup$ Jul 19, 2010 at 8:42
9
$\begingroup$

More generally, if $X\to S$ is flat and finitely presented, and if $T$ is a closed subscheme of $X$ which is a relative local complete intersection over $S$, then the blow-up $Y$ of $T$ in $X$ is flat over $S$ and commutes with every base change $S'\to S$. This is just because the powers of the defining ideal $I$ of $T$ are all flat and commute with base change, and $Y$ is by definition Proj($\bigoplus_{n\geq0}I^n$).

This applies in particular if $X$ and $T$ are both smooth over $S$. In this case, compatibility with base change implies that the geometric fibers of $Y\to S$ are blow-ups of smooth subvarieties in smooth varieties, hence smooth. Since $Y$ is flat over $S$, it is smooth.

$\endgroup$
3
  • $\begingroup$ Thanks for your help. I think that I have what I need now to prove my lemma.. $\endgroup$ Jul 19, 2010 at 16:32
  • $\begingroup$ @Laurent : Salut, et bienvenu ! $\endgroup$ Jul 20, 2010 at 9:36
  • $\begingroup$ @Laurent : Is it really needed that T is a relative local complete intersection? It seems to me that flatness of T and X already give flatness of all powers $I^n$ and quotients $I^n / I^{n+1}$, and so the commuting with base change. Is that not so? Thanks! $\endgroup$
    – quim
    Nov 20, 2015 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.