Timeline for Discrete vs. finitely generated subgroups of the adèles
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 22, 2017 at 15:37 | comment | added | YCor | 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. | |
| Dec 22, 2017 at 6:32 | comment | added | Venkataramana | 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. | |
| Dec 22, 2017 at 6:14 | comment | added | user113453 | Yes, I intended to ask something else. I have modified my question. Thanks | |
| Dec 22, 2017 at 6:12 | history | edited | user113453 | CC BY-SA 3.0 |
added 835 characters in body
|
| Dec 22, 2017 at 1:58 | review | Close votes | |||
| Dec 22, 2017 at 17:42 | |||||
| Dec 22, 2017 at 0:25 | history | edited | David Handelman | CC BY-SA 3.0 |
spelling of adèle, other typos
|
| Dec 21, 2017 at 22:30 | comment | added | paul garrett | 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? | |
| Dec 21, 2017 at 22:03 | comment | added | YCor | 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. | |
| Dec 21, 2017 at 21:30 | history | asked | user113453 | CC BY-SA 3.0 |