Hello?
I have a simple question.


Is $\mathbb{Z}\_p$ flat $\mathbb{Z}\_pG$-module for a finite $p$-group $G$?
Here, $p$ is prime and $\mathbb{Z}\_p$ means the integers localized at $(p)$.
If not, is it false even for a finite abelian $p$-group $G$?


Please let me know.



Is $\mathbb{Z}_p$ flat $\mathbb{Z}_pG$-module for a finite $p$-group $G$?
*(This is answering a comment to the main question)*


$\newcommand\ZZ{\mathbb Z}$
If $G$ is cyclic of order $p$, then there is a resolution of $\ZZ$ looking like $$\cdots\to\ZZ G\xrightarrow{d\_{\mathrm{odd}}} \ZZ G\xrightarrow{d\_{\mathrm{even}}}\cdots\to\ZZ G\xrightarrow{d\_{\mathrm{even}}} \ZZ G\xrightarrow{d\_{\mathrm{odd}}}\ZZ G\xrightarrow{\varepsilon}\ZZ$$ in which $\varepsilon$ is the usual augmentation, the odd differentials $d\_{\mathrm{odd}}$ are given by multiplication by $g-1$, and the even ones $d\_{\mathrm{even}}$ are given by multiplication by $1+g+\cdots+g^{p-1}$.


If we tensor over $\ZZ$ this complex with the flat $\ZZ$-module $\ZZ\_{(p)}$, we get a new exact complex $$\cdots\to\ZZ\_{(p)} G\xrightarrow{d\_{\mathrm{odd}}} \ZZ\_{(p)} G\xrightarrow{d\_{\mathrm{even}}}\cdots\to\ZZ\_{(p)} G\xrightarrow{d\_{\mathrm{even}}} \ZZ\_{(p)} G\xrightarrow{d\_{\mathrm{odd}}}\ZZ\_{(p)} G\xrightarrow{\varepsilon}\ZZ\_{(p)}$$ which is clearly a free $\ZZ\_{(p)}G$-resolution of $\ZZ\_{(p)}$. Notice that this is *not* the complex you mentioned in your comment.


Drop the rightmost term, tensor it now with $\ZZ\_{(p)}$ over $\ZZ\_{(p)}G$, and we end up with a complex which looks like
$$\cdots\to\ZZ\_{(p)}\xrightarrow{0} \ZZ\_{(p)}\xrightarrow{p}\cdots\to\ZZ\_{(p)} \xrightarrow{p} \ZZ\_{(p)} \xrightarrow{0}\ZZ\_{(p)}$$ In particular, $Tor\\_1^{\ZZ\_{(p)}G}(\ZZ\_{(p)},\ZZ\_{(p)})$ is $\ZZ\\_{(p)}/p\ZZ\\_{(p)}=\ZZ/p\ZZ$.


In fact, it is not hard (using for example a change-of-rings argument for the localization map $\ZZ G\to\ZZ\\_{(p)}G$) that for a general group one has $$Tor\\_1^{\ZZ\\_{(p)}G}(\ZZ\\_{(p)},\ZZ\\_{(p)})=Tor\\_1^{\ZZ G}(\ZZ\\_{(p)},\ZZ\\_{(p)})=\bigr(Tor\\_1^{\ZZ G}(\ZZ,\ZZ)\bigr)\\_{(p)}=(G\\_\mathrm{ab})\\_{(p)},$$ and the last group is not zero if, say, $G$ is a finite $p$-group.


