I have a question regarding additive subgroups of $\mathbb Q^n$. Suppose we have such a guy $M$, and suppose we know it is pointwise finitely divisible. That is, for every $m \in M$, there are only finitely many integers $n$ such that $m/n \in M$. Does it follow that $M$ is finitely generated?
Some thoughts:
It is well-known that a subgroup of $\mathbb Q^n$ is finitely generated if and only if it is discrete, i.e. the denominators are "globally bounded", whereas the given property only says each element has bounded denominators. Thus it seems weaker, but I haven't been able to find an example of an $M$ with this property and which is not finitely generated.
If $n = 1$, the statement is true: we may assume that $1 \in M$, and one can then show that $1/N = \text{min}\{m \in M \ | \ nm = 1 \text{ for some }n \geq 1\}$ generates $M$.
By the above point, an equivalent way to state the question is this: suppose $M \subset \mathbb Q^n$ is a subgroup whose intersection with any line through the origin is finitely generated. Does it follow that $M$ is finitely generated?
