Is $\mathbb{Z}_p$ flat $\mathbb{Z}_pG$-module for a finite $p$-group $G$? 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.
 A: The question can be considered and answered in greater generality: 

Let $R$ be a (not necessarily commutative) ring with unit and let $G$ be a finite group. Then the trivial $RG$-module $R$ is flat iff $|G|$ is invertible in $R$.

Proof: By a result of Benson, $R$ is flat iff it's projective (see Theorem 1.2), which is equivalent to the splitting of the augementation $\epsilon: RG \to R$ (over $RG$). 
Assume $i: R \to RG$ is a spliting of $\epsilon$ and $i(1) = \alpha$. From $g \cdot i(1) = i(g\cdot 1) = i(1)\;\; (g \in G)$ it follows that $\alpha = r \cdot N_G$ for some $r \in R$ and the norm element $N_G = \sum_{g \in G}g$. Appyling $\epsilon$ yields $1=r\cdot |G|$, so $|G|$ is a unit in $R$. 
Conversely, if $|G|$ is a unit in $R$, than $i:R \to RG, 1 \to |G|^{-1}\cdot N_G$ is easily seen to be a splitting of $\epsilon$. 
A: Consider the sequence of Z_p G-modules  0 -> I -> Z_p G -> Z_p -> 0
where I is the augmentation ideal.  Tensor this over Z_p G with Z_p = Z_p G/ I and you get
0 -> I / I^2 - > Z_p -> Z_p -> 0
which is no longer exact at the left; thus the module is not flat.  Indeed, Tor_1(Z_p, Z_p) is exactly I/I^2, if I haven't miscomputed.
A: (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.
