Am I correct in describing the group algebra $R[G]$ as $R \otimes_{Z} G$? (As a tensor product of $Z$-algebras.)
There is clearly a map $R \times G$ to $R[G]$, just by sending $(r,g)$ to $rg$, and this map is Z bilinear. Factoring through the tensor product, we see that $1 \otimes g$ maps onto a basis of $R[G]$, so this map is onto. It is also injective, since if $\Sigma r_i g_i = 0$, then each $r_i = 0$ (as $R[G]$ is free over the $g$), so that $\Sigma r_i \otimes g_i = 0$.
This seems like a perfectly natural way to describe the group algebra (and useful, based on the few problems I just solved using it), but I don't recall ever seeing it (though I am admittedly very much a novice at representation theory), and so I am concerned that I might be making a mistake somewhere.
Edit - this can't possibly be right, since $Q[Z_2] \cong Q \times Q$, but $Q \otimes_Z Z_2 = 0$.
$G$ isn't a $\mathbb{Z}$-algebra. It's not even a $\mathbb{Z}$-module when $G$ is nonabelian. What's true is that $R[G] \cong R \otimes_{\mathbb{Z}} \mathbb{Z}[G]$ and that $\mathbb{Z}[G]$ is isomorphic, as an abelian group, to the direct sum of $|G|$ copies of $\mathbb{Z}$.