Timeline for Comparing lower central series and augmentation ideal completions
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2010 at 21:24 | comment | added | Torsten Ekedahl | Your stated isomorphism is not true. To the left you have only finite sums, to the right you have some infinite sums. | |
| Sep 14, 2010 at 6:47 | comment | added | user2529 | Beriefly the question is like this. Let G be a group. Is the $I$ adic completion of the group ring ${\mathbb{Z}}/p[G]$,denoted by $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ is isomorphic to ${\mathbb{Z}}/p[\varprojlim_{q}G/\gamma_qG]\cong\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG]$? | |
| Sep 12, 2010 at 12:15 | comment | added | Torsten Ekedahl | Yes, I mean exactly what you propose. I have to admit that I am a little bit queasy about the non-fg case but I do not see the contradiction you suggest, could you be a bit more specific? I am not suggesting (if that is what you are saying) that $\varprojlim_q \mathbb Z/p[G/\gamma_qG] = \mathbb Z/p[G^p]$. | |
| Sep 12, 2010 at 5:33 | comment | added | user2529 | To clarify: By completion of the group ring in the $I$-adic topology do you mean $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ or something else? Thanks. | |
| Sep 12, 2010 at 5:21 | comment | added | user2529 | it seems there is a contradiction. If the group $G$ is not abelian and not finitely generated, then $\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG] \rm{not}\cong \varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$. However, for the same $G$, consider $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ as the completion of the group ring $\mathbb{Z}}/p[G]$ in the $I$-adic topology, it is isomorphic to $\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG]$. | |
| Sep 11, 2010 at 6:01 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |