The well known Dirichlet's unit theorem states that the unit group of a maximal order in a quadratic number field is finitely generated of rank blah blah blah. I think it's pretty naive to expect a most radical generalization to hold, namely:

Generalized Dirichlet unit theorem. The unit group of an associative unital ring with finitely generated additive group is finitely generated.

Is there any concrete counterexample?

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    $\begingroup$ It's even finitely presented. Because it's an arithmetic group. I'll post details later. $\endgroup$
    – YCor
    Jul 5, 2018 at 20:44
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    $\begingroup$ Related: Kahn proved that if $X$ is a reduced scheme of finite type over $\operatorname{Spec} \mathbb Z$, then $\Gamma(X,\mathcal O_X)$ is finitely generated. He gives counterexamples if the reducedness assumption is dropped (e.g. $\mathbb Z[x][\varepsilon]/(\varepsilon^2)$), but those are not finite over $\operatorname{Spec} \mathbb Z$ (only of finite type). See Bruno Kahn, Sur le groupe des classes d'un schéma arithmétique. $\endgroup$ Jul 5, 2018 at 21:01
  • $\begingroup$ In this case when the underlying abelian group is torsion-free it yields a complete argument (in this setting Borel-Harish-Chandra is enough; Kahn probably relies on it in general). Possibly there's some additional argument needed to tackle the case with torsion. $\endgroup$
    – YCor
    Jul 5, 2018 at 22:38
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    $\begingroup$ If you were already aware of the use of B-HC in this question (and thus the torsion-free case) it would have been useful to mention it from the beginning. $\endgroup$
    – YCor
    Jul 6, 2018 at 8:04
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    $\begingroup$ Related: math.stackexchange.com/questions/2445562/… $\endgroup$ Jul 7, 2018 at 8:44

2 Answers 2


Let $A$ be a ring that is finitely generated as an abelian group. Let $I$ be the subgroup of torsion elements. By finite generation, $I$ is finite. On the other hand, $I$ is a two-sided ideal of $A$, and $A/I$ is a ring that is torsion-free as an abelian group. By Borel and Harish-Chandra, $(A/I)^\times$ is finitely-generated. Since $I$ is finite, the map $A^\times \to (A/I)^\times$ is an isogeny (i.e. has finite kernel and cokernel, see Theorem 1.3 in this paper), and therefore $A^\times$ is finitely generated.

  • $\begingroup$ Great. Rather it's Lemma 3.6 in this paper that we need (the general case makes use of it). It says that if $A\to B$ is a surjective ring homomorphism with finite kernel, then the group homomorphism $A^\times\to B^\times$ is also surjective (clearly it has finite kernel). $\endgroup$
    – YCor
    Jul 6, 2018 at 11:23
  • $\begingroup$ @YCor Yes, you are right. I think the general case is worth knowing about, so I will leave the reference like this if you don't mind! $\endgroup$
    – Aurel
    Jul 6, 2018 at 11:27
  • $\begingroup$ I'd be curious how the assumption that $I$ is finite can be relaxed (for the surjectivity statement). At least, $I$ of finite length as left $A$-module is enough? $\endgroup$
    – YCor
    Jul 6, 2018 at 12:54
  • $\begingroup$ Not sure what the weakest assumption possible is. I don't have time to think about it now, but I will ask Alex if he has an idea. $\endgroup$
    – Aurel
    Jul 6, 2018 at 17:41
  • $\begingroup$ May I ask someone to be more specific about which bit of the work of Borel and Harish-Chandra is being applied in the comment of @YCor and in this answer? $\endgroup$
    – Lee Mosher
    Jul 12, 2018 at 18:36

At Lee Mosher's request, here is how Borel-Harish-Chandra is applied, which is the bulk of the argument (with some additional work in the case with torsion as described in Aurel's post).

Let $A$ be a finitely generated (associative unital) ring whose underlying additive group is free abelian of finite rank. From left multiplication we get a representation of $A$ on $B=A\otimes_\mathbf{Z}\mathbf{R}$, and thus an injective ring homomorphism $i:A\to M=\mathrm{End}(B)\simeq M_d(\mathbf{R})$, where $\mathrm{End}$ means as real vector space; it maps into $E=E(\mathbf{R})$, the set of endomorphisms as algebra.

Inside $E\times E$, consider the set $G$ of pairs $(x,y)$ such that $xy=1$. This is a Zariski closed submonoid of $E\times E$, and can be identified to the automorphism group of $B$; it lies inside $M_{2d}(\mathbf{R})$. For $x\in A^\times$, consider $j(x)=(x,x^{1})$. Then the group homomorphism $j$ maps injectively $A^\times$ into $G(\mathbf{Z})$. We claim that this is surjective. Indeed, consider $(x,y)\in E(\mathbf{Z})\times E(\mathbf{Z})$ with $xy=1$. Then $x$ preserves the lattice of integral points, i.e., $x$ induces an automorphism of $A=B(\mathbf{Z})$. [Note: we had to introduce $E\times E$ instead of $E$ to describe $G$ as a closed subgroup, which is necessary to apply Borel-Harish-Chandra.]

Thus we have $A\simeq G(\mathbf{Z})$. Borel and Harish-Chandra precisely proved (Annals of Math, 1962) that $G(\mathbf{Z})$ is finitely generated (and even finitely presented, as mentioned in Borel's ICM proceeding of 1962) for every $\mathbf{Q}$-defined subgroup of $\mathrm{GL}_k$.

[Remark: $G(\mathbf{Z})$ is possibly not a lattice in $G(\mathbf{R})$; however $G(\mathbf{Z})=H(\mathbf{Z})$ for some $\mathbf{Q}$-defined normal subgroup $H$ of $G$, such that $G^0/H^0$ is a $\mathbf{Q}$-split torus, and $H(\mathbf{Z})$ is a lattice in $H(\mathbf{R})$.]

  • $\begingroup$ (Note: it's rather a Zariski-closed submonoid of $E\times E^{\mathrm{op}}$, but this does not matter) $\endgroup$
    – YCor
    Jul 13, 2018 at 8:07

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