Let $C$ be a separated irreducible reduced curve which is quasi-finite over $\mathrm{Spec}\: \mathbb Z$. Is it necessarily affine i.e. $\mathrm{Spec}\: \mathcal O$ where $\mathcal O$ is an order in a number field? What if we look at curves faithfully flat over $\mathrm{Spec}\: \mathbb Z$?

Crossposted from MSE: https://math.stackexchange.com/questions/3098577/what-are-arithmetic-curves

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    $\begingroup$ Every finite morphism of schemes is affine, by definition. Thus, if the target of the morphism is affine, then also the domain of the morphism is affine. When you write that $C$ is finite over $\text{Spec}\ \mathbb{Z}$, do you mean that the morphism is quasi-finite rather than finite? $\endgroup$ – Jason Starr Feb 3 at 19:37
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    $\begingroup$ @JasonStarr Yes, that's precisely what I mean, I'll edit the question. $\endgroup$ – Anna Abasheva Feb 3 at 22:05
  • $\begingroup$ Is "curve" understood to mean that fibers over SpecZ are curves (in the usual sense of varieties) over the residue fields? $\endgroup$ – Qfwfq Feb 3 at 22:57
  • $\begingroup$ @Qfwfq From her example of SpecO, I'd suppose that she means a scheme of dimension one (rather than relative dimension one). $\endgroup$ – WhatsUp Feb 3 at 23:03

The answer is yes.

Let $C$ be an integral separated quasi-finite $\mathbb{Z}$-scheme. Then, by Zariski's Main Theorem, there is an open immersion $C\subset D$ with $D$ an integral scheme, and a finite morphism $D\to \mathrm{Spec} \mathbb{Z}$. Since $D\to \mathrm{Spec} \mathbb{Z}$ is finite, there is a finitely generated subring $A\subset \overline{\mathbb{Q}}$ and an isomorphism $D\cong \mathrm{Spec} A$.

It is well-known that open subsets of Dedekind schemes are affine.

Let $D'\to D$ be the normalization of $D$. The pull-back $C':=C\times_D D'$ is open in Spec $D'$, and hence affine. Since $C'$ is affine and $C'\to C$ is finite, it follows that $C$ is affine.


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