This is a basic question. Let $G$ be a finite group, $M$ a finitely generated $\mathbb Z[G]$-module so that the $\mathbb Z_p[G]$-module $M_p$ is free for all prime numbers $p$, i.e. is locally free.

Here $\mathbb Z_p[G]=\mathbb Z_p\otimes_\mathbb Z\mathbb Z[G]$ and $M_p=\mathbb Z_p\otimes_\mathbb ZM$ where $\mathbb Z_p$ is the $p$-adic completion of $\mathbb Z$.


1: How do we show that $\mathbb Q\otimes_\mathbb ZM$ is a free $\mathbb Q[G]$-module of well defined rank?

If the rank of $M$ is defined as the $\mathbb Q[G]$-rank of $\mathbb Q\otimes_\mathbb Z M$ then

2: How do we show that it's also the rank of $M_p$ over $\mathbb Z_p[G]$, for all $p$?

Thank you for your help.

  • 3
    $\begingroup$ 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). $\endgroup$ – Alex B. Apr 29 '16 at 23:15
  • 1
    $\begingroup$ 2. Just look at rank over $\mathbb{Z}_p$, respectively over $\mathbb{Q}$. $\endgroup$ – Alex B. Apr 29 '16 at 23:17
  • $\begingroup$ Even more concretely, if $G$ acts trivially on $\mathbb{Z_p}\otimes_\mathbb{Z} M$ then $G$ necessarily acts trivially on $M$. $\endgroup$ – Lior Silberman May 6 '16 at 23:01

The module $M$ becomes free over $\mathbb{Q}_p$ for some (and every) $p$. Let us write $\psi$ for the character of $M$. Then it follows that $\psi(g)=n\delta_{1,g}$ where $n$ is the dimension of $M\otimes_{\mathbb{Z}}\mathbb{Q}_p$. But this implies already that $M$ is free over $\mathbb{Q}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.