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Is $\mathbb{Z}_{p}[G]$ a PID, where $G=(\mathbb{Z}_{p},+)$ is the additive group of the $p$-adics $\mathbb{Z}_{p}$?

I am studying a paper where the authors implicitly use that claim, but it is unclear to me. (I am a little bit embarassed by the fact that I cannot solve this myself.)

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    $\begingroup$ What does the notation $\mathbb{Z}[\mathbb{Z}]$ stand for? $\endgroup$ – Alex Provost Mar 11 '16 at 15:04
  • $\begingroup$ What's $\Bbb Z[\Bbb Z]$? $\endgroup$ – user228113 Mar 11 '16 at 15:04
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    $\begingroup$ Do you mean the group ring whose basis elements are elements of $\mathbb{Z}$? In other words, the ring where the elements are finite sums of the form $\sum n_i e_{m_i}$ with $n_i,m_i\in\mathbb{Z}$ and multiplication defined as $(n_1 e_{m_1})\cdot (n_2 e_{m_2})=(n_1n_2)e_{m_1+m_2}$? $\endgroup$ – Michael Burr Mar 11 '16 at 15:04
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    $\begingroup$ By $\mathbb{Z}_p$ do you mean the $p$-adics or integers mod $p$? $\endgroup$ – Eric Wofsey Mar 11 '16 at 15:15
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    $\begingroup$ @Eric Wofsey: $p$-adics. $\endgroup$ – jdh Mar 11 '16 at 15:16
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This isn't true; in fact, $\mathbb{Z}_p[G]$ is not even Noetherian. For instance, take the augmentation ideal $I$, i.e. the ideal generated by $\{g-1:g\in G\}$. If $I$ were finitely generated, there would be a finite subset $F\subset G$ such that $I$ is generated by the elements $g-1$ for $g\in F$. But if $H\subseteq G$ is the subgroup generated by $F$ and $J$ is the ideal generated by the elements $g-1$ for $g\in F$, it is easy to see that the canonical quotient map $\mathbb{Z}_p[G]\to\mathbb{Z}_p[G/H]$ factors through the quotient $\mathbb{Z}_p[G]\to\mathbb{Z}_p[G]/J$. Thus if $J$ is all of $I$, $H$ must be all of $G$. But $G$ is not finitely generated, so this is impossible.

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