Timeline for Rank of a locally free $\mathbb Z[G]$-module
Current License: CC BY-SA 3.0
6 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
May 6, 2016 at 23:01 | comment | added | Lior Silberman | Even more concretely, if $G$ acts trivially on $\mathbb{Z_p}\otimes_\mathbb{Z} M$ then $G$ necessarily acts trivially on $M$. | |
May 6, 2016 at 20:55 | answer | added | Ehud Meir | timeline score: 2 | |
Apr 30, 2016 at 14:00 | history | edited | YCor |
edited tags; edited tags
|
|
Apr 29, 2016 at 23:17 | comment | added | Alex B. | 2. Just look at rank over $\mathbb{Z}_p$, respectively over $\mathbb{Q}$. | |
Apr 29, 2016 at 23:15 | comment | added | Alex B. | 1. Clearly, if $M_p$ is free (for just a single prime $p$!), then $\mathbb{Q}_p\otimes M=\mathbb{Q}_p\otimes_{\mathbb{Q}}(\mathbb{Q}\otimes M)$ is a free $\mathbb{Q}_p[G]$-module, whence $\mathbb{Q}\otimes M$ is free (it is a very general fact that if $K$ is a field, and two modules over $K[G]$ become isomorphic after extending the field of scalars, then they are already isomorphic over the smaller field). | |
Apr 29, 2016 at 22:05 | history | asked | eddie | CC BY-SA 3.0 |