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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!

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up vote 8 down vote accepted

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).

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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 $S$? – Daniel Loughran Jul 19 '10 at 8:42

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.

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Thanks for your help. I think that I have what I need now to prove my lemma.. – Daniel Loughran Jul 19 '10 at 16:32
@Laurent : Salut, et bienvenu ! – Chandan Singh Dalawat Jul 20 '10 at 9:36
@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! – quim Nov 20 '15 at 11:04

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