Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a group scheme of finite type over $\mathbf{Z}$. Must $G(\mathbf{Z})$ be finitely presented?

(The question is inspired by a not yet successful attempt to answer a question of Brian Conrad.)

A few special cases:

  1. If $G$ is the Néron model of an abelian variety over $\mathbf{Q}$, then a positive answer amounts to the Mordell-Weil theorem (combine with restriction of scalars to get the full Mordell-Weil theorem).

  2. If $G$ is the restriction of scalars of $\mathbf{G}_m$ from the ring of integers of a number field down to $\mathbf{Z}$, then a positive answer follows from Dirichlet's unit theorem.

share|improve this question

1 Answer 1

It follows from Theorem 6.12 of Borel and Harsh-Chandra, "Arithmetic subgroups of algebraic groups", that $G(\mathbb{Z})$ is finitely generated if $G$ is affine. Perhaps one can combine this with Chevalley's theorem to deduce finite generation in the general (not necessarily affine) case.

EDIT (Added idea for proof of general case; see also Torsten Ekedahl's comment below)

EDIT (Proof completed (assuming $G$ is separated) and simplfied using comments of BCnrd below)

As discussed in the comments below, we may assume $G$ is flat and we may also assume it is connected. By Chevalley's theorem, there is an affine subgroup scheme $H_{\mathbb{Q}}$ of the generic fibre $G_\mathbb{Q}$ of $G$ such that the quotient $G_{\mathbb{Q}}/H_\mathbb{Q}$ is an abelian variety. Let $H$ be the Zariski closure of $H_{\mathbb{Q}}$ in $G$ with the reduced induced structure. Then $H$ is a closed subgroup scheme of $G$. By a theorem of Raynaud (see comment of BCnrd below for the reference) $H$ is also affine.

We have an incusion of groups

$G(\mathbb{Z})/H(\mathbb{Z}) \subset G(\mathbb{Q})/H(\mathbb{Q}) \subset (G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$.

Since $G_{\mathbb{Q}}/H_{\mathbb{Q}}$ is an abelian variety, by the Mordell-Weil theorem $(G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$ is a finitely generated abelian group, hence so is $G(\mathbb{Z})/H(\mathbb{Z})$. Since $H(\mathbb{Z})$ is finitely generated by the Borel-Harish-Chandra theorem, it follows that $G(\mathbb{Z})$ is finitely generated.

share|improve this answer
I think it's a bit more subtle. For cocompact lattices, I think you still need to use Weil's paper, and, as I recall, one of the reasons Kazhdan defined Property (T) was to prove that nonuniform lattices were finitely generated (a group with Property (T) is compactly generated, so discrete implies finitely generated). –  Matthew Stover Apr 28 '10 at 5:31
Well, I am not an expert and am only quoting the theorem; here the question is one about arithmetic groups so one does not have to deal with general (non-arithmetic) lattices. (If one embeds G in $GL_{n,Z}$ as closed subgroupscheme then if I am not mistaken, $G(\mathbb{Z})$ is the same as the $G_Z$ in their theorem.) –  ulrich Apr 28 '10 at 6:00
Well, first of all, $G(\mathbb{Z} \otimes \mathbb{R})$ might not be reductive, as in the examples given in the question. Perhaps I'm missing something, but I don't see why the argument works even in the semisimple setting. It is a lattice, which is not accidentally analogous to Dirichlet's unit theorem, but I still don't see how you get to finite generation for a generic discrete subgroup of finite covolume in $G(\mathbb{R})$. –  Matthew Stover Apr 28 '10 at 6:10
@Torsten Ekedahl. There is no reductivity assumption in the theorem I quoted (Theorem 6.5 deals with the reductive case.) I admit to not having read the proof so perhaps I am still missing something. –  ulrich Apr 28 '10 at 6:30
Let's assume $G$ separated. A $G/H$ is not needed to prove finite generation, since $G(\mathbf{Z})/H(\mathbf{Z})$ is subgroup of $G(\mathbf{Q})/H(\mathbf{Q})$, a subgroup of (finitely generated) Mordell-Weil group of ab var. $G_{\mathbf{Q}}/H_{\mathbf{Q}}$. A more serious point is to prove $H$ is affine! Since finite type with affine generic fiber, it is affine over $\mathbf{Z}[1/N]$ for some $N$. To prove affine is problem over $\mathbf{Z}_{(p)}$ for each $p|N$. Now use SGA3 result of Raynaud (needs sep'tdness) that appears with proof as Prop. 3.1 of Prasad-Yu paper "Quasi-reductive groups". –  BCnrd Apr 28 '10 at 14:55

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.