and $s' \in S$ with $s'r = r's$. Then we can consider the *Ore localization $S^{-1}k G$*. Recall that every element in $S^{-1}kG$ is of the form $s^{-1}\cdot r$ for $r \in k G$ and $s \in S$ and $s_0^{-1}\cdot r_0 =
s_1^{-1}\cdot r_1$ holds if and only there exists $u_0,u_1 \in R$ satisfying $u_0 r_0 = u_1r_1$ and $u_0s_0 = u_1s_1 $. Addition is given on representatives by $s_0^{-1}r_0 + s_1^{-1}r_1 = t^{-1}(c_0r_0 + c_1r_1)$ for $t = c_0s_0 =
c_1s_1$. Multiplication is given on representatives by $s_0^{-1}r_0 \cdot
s_1^{-1}r_1 = (ts_0)^{-1}cr_1$, where $cs_1 = t r_0$. The zero element is $e^{-1}\cdot 0$ and the unit element is $e^{-1}\cdot e$. The Ore localization $S^{-1}kG$ is a skew field and the canonical map $k G \to S^{-1}kG$ sending $r$ to $e^{-1} \cdot r$ is injective. The functor $S^{-1}kG \otimes_{kG} -$ is exact.

\[rem:The\_Ore\_condition\_for\_group\_rings\] If a torsionfree amenable group $G$ satisfies the Kaplansky Conjecture, i.e., $kG$ contains no zero-divisor, then for $S = kG-\{0\}$ the Ore localization $S^{-1}kG$ exists and is a skew field [@Lueck(2002)]\*[Example 8.16 on page 324]{}. Every torsionfree elementary amenable group satisfies the assumptions above for all fields $k$ . If the group $G$ contains the free group 