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.


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.

  • $\begingroup$ 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). $\endgroup$ Apr 28, 2010 at 5:31
  • $\begingroup$ 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.) $\endgroup$
    – naf
    Apr 28, 2010 at 6:00
  • $\begingroup$ 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 $G(\mathbb{R})$. $\endgroup$ Apr 28, 2010 at 6:10
  • 2
    $\begingroup$ @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. $\endgroup$
    – naf
    Apr 28, 2010 at 6:30
  • 5
    $\begingroup$ 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". $\endgroup$
    – BCnrd
    Apr 28, 2010 at 14:55

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