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.

Is it true that if $U\subseteq\mathbf{A}_K^n$ is an additive subgroup, with $\mathbf{A}_K$ the adèle ring of a number field, $U$ discrete with respect to the induced topology, then $U$ is a finitely generated abelian group?

For example, 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 $\mathbf{R}\otimes_{\mathbf{Z}}\mathcal{O}_K\simeq\mathbf{R}^{[K:\mathbf{Q}]}$. 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.

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?

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

Is it true that if $U\subseteq\mathbf{A}_K^n$ is an additive subgroup, with $\mathbf{A}_K$ the adéle ring of a number field, $U$ discrete wrto the induced topology, then $U$ is a finitely generated abelian group?

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

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.

Is it true that if $U\subseteq\mathbf{A}_K^n$ is an additive subgroup, with $\mathbf{A}_K$ the adèle ring of a number field, $U$ discrete with respect to the induced topology, then $U$ is a finitely generated abelian group?

For example, 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 $\mathbf{R}\otimes_{\mathbf{Z}}\mathcal{O}_K\simeq\mathbf{R}^{[K:\mathbf{Q}]}$. 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.