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Let $X$ be a scheme of finite type over $\mathbb{Z}$. Let $R$ be the ring of algebraic integers. My intuition is that $X(R)$ is practically always infinite.

More specifically, suppose that $X$ is faithfully flat over $\mathbb{Z}$, of relative dimension $\geq 1$, and the generic fiber is geometrically irreducible. Is that enough to guarantee infinitely many algebraic integer points?

This question is inspired by this one; I have no application in mind.

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I think this is of relevance, though it may not answer everything. Once you have local-global, I suspect what you desire, should follow. math.uga.edu/~rr/ArithAllAlgInt.pdf In this note we will establish two theorems about Diophantine equations over the ring of all algebraic integers $O$. Let $V$ be a geometrically irreducible affine variety over $K$. The first theorem is a general local-global principle: $V$ has points over $O$ iff it has points over all the local $O_v$. The second theorem applies the first to get: Hilbert's Tenth problem has a positive solution over $O$. –  Junkie Oct 10 '11 at 4:05
    
If $X$ is flat over $\mathbf{Z}$ and the generic fibre is geometrically irreducible, does it follow that $X\to \mathrm{Spec} \mathbf{Z}$ is surjective (since $\mathrm{Spec} \mathbf{Z}$ is a Dedekind scheme)? Moreover, as you probably know, if $X$ is proper over $\mathbf{Z}$, $X(R) = X_K(K)$, where $R=O_K$. Now, I'm pretty sure that $X_K(K)$ is infinite for some field $K$, but maybe you can find a curve $X_K/K$ and field extensions $L_1,L_2,\ldots$ of $K$ of increasing degree with $X_K(L_i)$ finite. Why would this not be possible? Anyway, it's the non-proper case which probably interests you. –  Taicho Oct 10 '11 at 6:36
    
@Taicho. NB: If X is a (say, projective, flat model of) a projective curve of genus $g\geq 2$ over $\mathbf Q$, then $X_K(K)$ is finite for any algebraic number fields, by Faltings's Theorem (Mordell conjecture). –  ACL Oct 10 '11 at 7:10
    
Don't you want to assume that $X$ is affine (with maybe some extra conditions) ? Note that if $X$ is proper over $\bf Z$ then $X(R)=X(\bar{\bf Q})$ and clearly $X(\bar{\bf Q})$ is infinite if the dimension of the generic fibre of $X$ is positive. –  Damian Rössler Oct 10 '11 at 11:04
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@Taicho: perhaps $Spec(\mathbf{Z}[X,Y]/(2XY-1))$ is an illustrative example. If I didn't make a mistake, this is flat over $\mathbf{Z}$ but not faithfully flat, and clearly has no integral points. –  Kevin Buzzard Oct 10 '11 at 20:07

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up vote 12 down vote accepted

This is basically true, in view of a density theorem due to Robert Rumely (Arithmetic over the ring of all algebraic integers, J. reine u. angew. Math. 368, 1986, p. 127-133). It relies on Rumely's capacity theory, and his extension of the theorem of Fekete-Szegö.

For a generalization, and an algebraic proof, see also Laurent Moret-Bailly, Groupes de Picard et problèmes de Skolem. II. Annales scientifiques de l'École Normale Supérieure, Sér. 4, 22 no. 2 (1989), p. 181-194. (Numdam, http://www.numdam.org/item?id=ASENS_1989_4_22_2_181_0)

The hypothesis of Moret-Bailly's Theorem is that $X$ be irreducible, surjective and of positive relative dimension over ${\rm Spec}\mathbf Z$, and that its generic fiber be geometrically irreducible. Then, he proves that $X$ has $\overline{\mathbf Z}$-points which can be chosen arbitrarily close to a given $p$-adic point (end even more...).

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Antoine, I think $X$ should also be quasi-projective. –  Qing Liu Oct 10 '11 at 12:00
    
No, this is not a necessary hypothesis in Moret-Bailly's Theorem. In fact, he later proved similar results for Artin stacks. (see Problèmes de Skolem sur les champs algébriques, Compositio Math., 125, 1-30 (2001). –  ACL Oct 10 '11 at 13:59

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