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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 $\mathbb Z$.

Then

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.

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    $\begingroup$ 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.
    Commented Apr 29, 2016 at 23:15
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    $\begingroup$ 2. Just look at rank over $\mathbb{Z}_p$, respectively over $\mathbb{Q}$. $\endgroup$
    – Alex B.
    Commented Apr 29, 2016 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
    Commented May 6, 2016 at 23:01

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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 $\mathbb{Q}$.

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