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Let $X$ be an arbitrary scheme.

I think one calls $X$ a characteristic zero scheme if all the residue fields of $X$ are of characteristic zero, for any point $x$ on $X$.

If $X$ is a scheme over $\mathbb Q$, then of course it is a scheme of characteristic zero.

My question is: does also the converse hold? And how does one make this rigorous?

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Assuming that $X$ has characteristic zero, you have to construct a morphism $X\to\mathrm{Spec}\:\mathbb Q$. At the level of topolgical spaces, there are no choices to be made. Can you see what can be done at the level of structure sheaves? –  Mariano Suárez-Alvarez Nov 23 '11 at 17:45
    
It might be worth noting that $X$ may not actually have any $\mathbb{Q}$-points. This may not be confusing to other people, but confused me for awhile. If $E$ is a genus $1$ curve of characteristic zero that has a $\mathbb{Q}(\sqrt{2})$-point but no $\mathbb{Q}$-point, then people might say that $E$ is defined as an elliptic curve over $\mathbb{Q}(\sqrt{2})$ but is not defined over $\mathbb{Q}$. As the argument below shows, this does not mean that it is not a scheme over $\mathbb{Q}$. Ignore this if that was already obvious to you. –  Matt Nov 23 '11 at 20:02
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up vote 3 down vote accepted

Yes, you only have to show that any non-zero integer $n$ is invertible in $O_X(X)$ (then the universal property of localization induces a homomorphism from $\mathbb Q \to O_X(X)$ hence to the sheaf $O_X$). For all $x\in X$, as $n$ is no zero in $k(x)$, it is invertible in $O_{X,x}$, hence invertible in an open neighborhood of $x$. The local inverses glue into a global inverse by the uniqueness of the inverse.

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