Are submodules of free $\mathbb{Z}[G]$-modules flat? if not what conditions on $G$ makes it true? $G$ is an infinite group. 


If $\mathbb{Z}[G]$ is a Prüfer domain then this is true. Can a group ring $\mathbb{Z}[G]$ be a Prüfer domain?


In my case $G=GL\_n(A)$ for $A=k[x\_1,\cdots, x\_n]/I$ for some field $k$ and prime ideal $I$. 



Flatness of submodules of free modules
The augmentation ideal of $\mathbb Z[G]$ is flat if and only if $H\_i(G, \mathbb Z) = 0$ for all $i > 1$. I think this should rule out most cases of $GL\_n(A)$, since these groups generally have interesting and nontrivial homology.


