Grothendieck group of Abelian categories , with "coefficients" in a ring For an Abelian category $\mathcal C$ the Grothendieck group $G_0(\mathcal C)$ is defined as $$\dfrac{\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]}{\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0  \text{ is exact}   \rangle }$$ (where $[X]$ denotes the isomorphism class of $X$ ) . Now when one says Grothendieck group with coefficients in $\mathbb Q$, one usually means $G_0(\mathcal C)\otimes_{\mathbb Z} \mathbb Q$. 
My question is, in general, given a commutative Noetherian ring $R$ which is torsion-free as a $\mathbb Z$-module (hence the natural map $\mathbb Z\to R$ is flat), what is the difference between $G_0(\mathcal C)\otimes_{\mathbb Z} R $ and $$\dfrac{\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}R[X]}{\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0 \text{ is exact}  \rangle } ?$$ 
 A: Since $R$ is torsion free it is flat so the functor $R \otimes -$ is exact. Then the exact sequence 
$$
0 \to B \to A \to A / B \to 0
$$
becomes 
$$0 \to B \otimes R \to A \otimes R \to A/B \otimes R \to 0.$$
after tensoring with $R$. 
This gives an isomorphism $A/B \otimes R \simeq A \otimes R / B \otimes R$.
A: There is no need for torsion-free-ness.
$R\otimes_{\mathbb Z}-$ is always right exact, so if $A\to B\to C\to 0$ is exact, so is $R\otimes_{\mathbb Z}A\to R\otimes_{\mathbb Z}B\to R\otimes_{\mathbb Z}C\to  0$, so that $R\otimes_{\mathbb Z}B/R\otimes_{\mathbb Z}A = R\otimes_{\mathbb Z}C= R\otimes_{\mathbb Z}(B/A)$; the only "problem" being that $R\otimes_{\mathbb Z}A$ is not per se a submodule of $R\otimes_{\mathbb Z}B$, but what is meant is simply the quotient by the image. 
Now in your situation clearly $R\otimes_{\mathbb Z}\bigoplus_X \mathbb{Z}\cdot [X] \simeq \bigoplus_X R\cdot [X]$ and the image of $R\otimes_{\mathbb Z}\langle [A]-[B]+[C], ...\rangle$ contains all the $[A]-[B]+[C]$, and is also clearly the smallest submodule containing them, so the image is precisely $\langle [A]-[B]+[C], ...\rangle$ where here $\langle ... \rangle$ means "sub-$R$-module generated by" instead of "subgroup generated by"
So no matter $R$, these two are always isomorphic. 
