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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
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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