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Let $X,X'$ be normal, proper and flat $\mathbb{Z}$-schemes. Write $X_{\mathbb{Q}}$ for the generic fiber of $X$, i.e. $X_{\mathbb{Q}}:=X \times_{Spec (\mathbb{Z})} Spec (\mathbb{Q})$.

Let us assume that there is an isomorphism $X_{\mathbb{Q}}\cong X'_{\mathbb{Q}}$ of $\mathbb{Q}$-schemes. The following fact seems to be well known:

There exists a finite set of prime numbers $\Sigma=\{p_1,...,p_{n}\} \subset \mathbb{Z}$ such that the above (generic) isomorphism extends to an isomorphism of $\mathbb{Z}(\Sigma^{-1})$-schemes (where $\mathbb{Z}(\Sigma^{-1}):=\mathbb{Z}[p_{1}^{-1},...,p_{n}^{-1}]$), i.e. we have

$$X \times_{Spec(\mathbb{Z})} Spec(\mathbb{Z}(\Sigma^{-1})) \cong X' \times_{Spec(\mathbb{Z})}Spec(\mathbb{Z}(\Sigma^{-1}))$$

Can anyone give me a reference for this statement or, if it seems reasonable, even a (sketch of a) proof?

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up vote 6 down vote accepted

The morphism $X_{/\mathbb Q} \to X'_{\mathbb Q}$ ultimately involves some explicit polynomials, which have only finitely many primes in the denominators of their coefficients, and so can be extended over $X_{\mathbb Z(\Sigma^{-1})}$ for a sufficiently large (but finite) set of primes $\Sigma$.

Likewise for the map in the other direction (enlarging $\Sigma$ if necessary). Since the composite of the resulting two morphisms is the identity generically, it it will have to the be the identity (by flatness, say).

If you want a more formal argument, you could do the following: let $f: X_{\mathbb Q} \to X'_{\mathbb Q}$ be the given morphism, and form its graph $\Gamma_f \subset (X \times_{\mathbb Z} X')_{\mathbb Q}$. Now closed up $\Gamma_f$ in $X\times_{\mathbb Z} X'$. The result is a closed subscheme $Z$ of the product, and the projection of $Z$ onto $X$ is an isomorphism generically (i.e. over Spec $\mathbb Q$) (since over $\mathbb Q$ this is just the projection of $\Gamma_f$ onto $X_{\mathbb Q}$, and the characteristic property of a graph is that this projection is an isomorphism).

Now check that a morphism between two finite type flat $\mathbb Z$-schemes which is an isomorphism over $\mathbb Q$ is an isomorphism over some open subset Spec $\mathbb Z(\Sigma^{-1})$ of Spec $\mathbb Z$. Thus, over $\mathbb Z(\Sigma^{-1})$ we see that the projection $Z_{\mathbb Z(\Sigma^{-1})} \to X_{\mathbb Z(\Sigma^{-1})}$ is an isomorphism, and so $Z_{\mathbb Z(\Sigma^{-1})}$ is the graph of some extension $f'$ of $f$ to $X_{\mathbb Z(\Sigma^{-1})}$.

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Dear @Matt E, thank you for your answer! I have two points: First I am unfortunately not quite able to check the assertion in your last paragraph (the one which starts with "Now check..."), which I think is because I don't really know much about flatness. Second, your first sentence is quite astonishing to me as well and I'd love to learn more about its background. Put precisely, how can you describe an isomorphism $X_{\mathbb{Q}}\cong X'_{\mathbb{Q}}$ using polynomials?? – Nils Matthes Nov 1 '12 at 22:04

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