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Let $P$ be a cyclic group of order $p^n$. Let $M$ be a $\mathbb{Z}_p[P]$-module and $M_p :=\mathbb{Q}_p\otimes M$ be the associated $\mathbb{Q}_p[P]$-module. When will $M_p$ have a direct summand isomorphic to $\mathbb{Q}_p[P]$?

We know $\mathbb{Q}[P]\simeq \bigoplus_{d\mid p^n}\mathbb{Q}(\zeta_d)$. If $M_p\simeq \bigoplus_{d\mid p^n}M_{p,d}$ where $M_{p,d}$ is a vector space over $\mathbb{Q}(\zeta_d)\otimes \mathbb{Q}_p$ and $m_d$ is its dimension, then $M_p\simeq \bigoplus_d (\mathbb{Q}(\zeta_d)\otimes \mathbb{Q}_p)^{m_d}$ as $\mathbb{Q}_p[P]$-modules. When does $M_p$ have a free rank one direct summand?

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    $\begingroup$ Do you want a condition on $M$ or on $M_p$? I think, if $M$ is projective as $\mathbb{Z}_p[P]$ modules, then $M_p$ is free. On the other hand, if you only care about $M_p$, then the existence of $M$ does not help you. Any finite dimensional $\mathbb{Q}_p$ representation of $P$ contains $P$-stable $\mathbb{Z}_p$ lattice. $\endgroup$
    – Steffen Kionke
    Commented Jul 24, 2018 at 9:41
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    $\begingroup$ One (necessary and sufficient) condition is that the character of $M_p$ contains every linear character of $P$ with multiplicity $>0$. In your notation, if $m_d>0$ for all $d \mid p^n$. This is so since $\mathbb{Q}_p[P]$ is semisimple, and in fact $\mathbb{Q}_p[P]\cong \bigoplus_{d\mid p^n} \mathbb{Q}_p(\zeta_d)$. Is this the kind of condition you're looking for? $\endgroup$
    – Frieder Ladisch
    Commented Jul 24, 2018 at 10:38
  • $\begingroup$ Thanks. These are both good starting points! I am/ was hoping to find (weak) conditions on say the $\dim_{\mathbb{Q}_p}M_p$ to say something about $m_d$? I wasn't sure if it is some standard result that I am unaware of. $\endgroup$
    – debanjana
    Commented Jul 24, 2018 at 11:48

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