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If $U\subseteq\mathbf{R}^n$ is an additive subgroup, discrete with respect to the induced topology, then $U$ is a finitely generated abelian group.

Question.

Given a discrete additive subgroup $U\subseteq\mathbf{A}_K^n$, with $\mathbf{A}_K$ the adèle ring of a number field $K$, what additional condition on $U$ forces $U$ to be a finitely generated abelian group?

Example.

(A) If $U = \mathcal{O}_K$, the ring of integers of a number field $K$, then $U$ is a finitely generated abelian group, and it is discrete both in $\mathbf{A}_K$ and in $\mathbf{R}\otimes_{\mathbf{Z}}\mathcal{O}_K\simeq\mathbf{R}^{[K:\mathbf{Q}]}$.

$\mathcal{O}_K$ satisfies the condition of being uniformly bounded with respect to the non-archimedean absolute values (it is contained in the $v$-adic unit ball in $K_v^n$ for every finite place $v$ of $K$).

(B) If $\mathcal{O}_K^{\times}$ is the unit group of $\mathcal{O}_K$, then the image $U$ of $\mathcal{O}_K^{\times}$ in the trace-zero hyperplane in $\mathbf{R}^{r_1 + r_2}$, an $r_1+r_2-1$ dimensional $\mathbf{R}$-vector space, is discrete.

$\mathcal{O}_K^{\times}$ also satisfies the condition of being uniformly bounded with respect to every non-archimedean absolute value.

The question may become:

Refined question.

Let $U$ be as in the question. Call $U_v$ the image of $U$ under the projection onto the $v$-factor $K_v^n$ of $\mathbf{A}_K^n$, $v$ a place of $K$.

Assume, in addition to discreteness of $U$, that there exists a constant $B>0$ such that $U_v$ is contained in the ball centered at zero and of radius $B$ in $K_v^n$, for every non-archimedean place $v$ of $K$.

Is $U$ a finitely generated abelian group?

Improve this question
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    $\begingroup$ If you don't include Archimedean places in adeles, the answer is trivially yes because the only discrete subgroup is $\{0\}$ (because any element generates an infinite subgroup dense in a compact group). If you include Archimedean places, the answer is clearly no since $K$ embeds as a discrete subring (and hence discrete subgroup) into its adeles. Unless I misunderstood the question. $\endgroup$
    – YCor
    Commented Dec 21, 2017 at 22:03
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    $\begingroup$ No: the field $k$ itself is a discrete subgroup of $\mathbb A_k$, but is not finitely-generated as an abelian group. Do you mean to ask a different question? $\endgroup$
    – paul garrett
    Commented Dec 21, 2017 at 22:30
  • $\begingroup$ Yes, I intended to ask something else. I have modified my question. Thanks $\endgroup$
    – user113453
    Commented Dec 22, 2017 at 6:14
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    $\begingroup$ For a non-archimedean place $v$, let $O_v$ be the ring of integers (an open compact subgroup of $K_v$. Your assumptions imply that the intersection $U'$ of $U$ with the open subgroup $(K\otimes {\mathbb R})^n\times \prod O_v^n$ has finite index in $U$. Since the product $\prod O_v^n$ is compact, it follows that the intersection $U'$ injects into a discrete subgroup of $(K\otimes {\mathbb R})^n$. Hence $U'$ is finitely generated and therefore so is $U$. I think this is an elementary exercise in the topology of adeles and not suitable as a question on math overflow. $\endgroup$
    – Venkataramana
    Commented Dec 22, 2017 at 6:32
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    $\begingroup$ Just a minor variation Venky's comment: a discrete subgroup of adeles (or more generally, of the direct product of $\mathbf{R}^n$ and a locally elliptic locally compact group $H$) is finitely generated if and only its projection to $H$ has compact closure. It's indeed an elementary exercise. $\endgroup$
    – YCor
    Commented Dec 22, 2017 at 15:37

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