MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I think this is of relevance, though it may not answer everything. Once you have local-global, I suspect what you desire, should follow. 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
@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
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,

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...).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.