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I was reading a chapters homology with local coefficients. Where one of the preliminary sections asks us to compute

$$\mathbb{Z}_{+} \otimes_{\mathbb{Z}[\mathbb{Z}/2]}\mathbb{Z}_{-}$$

Here $\mathbb{Z}_{+}$ is the integers acted upon by $\mathbb{Z}[\mathbb{Z}/2]$ via the trivial representation and $\mathbb{Z}_{-}$ is the integers acted upon the nontrivial representation. More formally, acted upon by the representation $\rho: \mathbb{Z}/2 \rightarrow Aut(\mathbb{Z})$ that sends the generator $1$ of $\mathbb{Z}/2$ to the automorphism $\sigma: \mathbb{Z} \rightarrow \mathbb{Z}$ that sends $m \mapsto -m$.

I haven't fully thought this problem through, however, I notice immediately that since $\mathbb{Z}[\mathbb{Z}/2]$ is commutative, we have that this abelian group $\mathbb{Z}_{+} \otimes_{\mathbb{Z}[\mathbb{Z}/2]}\mathbb{Z}_{-}$ is a representation of $\mathbb{Z}/2$. It looks like it only has two generators and also looks torsion free, therefore I think this maybe the tensor of the two representations $\mathbb{Z}_{+}$ and $\mathbb{Z}_{-}$.

If this is true, I was wondering if one could make a more general assertion, namely given an abelian group $A$ and two representations $(V, \rho)$ and $(W, \sigma)$ over a Euclidean domain $\mathfrak{k}$, is $V \otimes_{\mathfrak{k}[G]} W$ the tensor of the representations?

Also, with homology with local coefficients, we are often tensoring over $\mathbb{Z}[\pi_{1}(X)]$ which often results in just an abelian group since the group ring may not be commutative. Is there a weaker statement that can be made in this case?

I'm sorry for the vagueness of this question, I wanted to get some feedback from people who have thought about this before.

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Consider a group $G$ and a commutative ring $k$, and form the algebra $kG$. There is an algebra augmentation $\varepsilon :kG\to k$ that makes $k$ into what is the trivial $kG$ module, so that $g\lambda =\lambda$ for every $g\in G,\lambda\in k$.

Claim For any $kG$ module $M$, $k\otimes_{kG} M$ is canonically isomorphic to the module of coinvariants $M_G$; the quotient of $M$ by the submodule generated by $gx-x,x\in M,g\in G$.

Proof Consider the map $f:M\to k\otimes_{kG} M$ that sends $m$ to $1\otimes m$. Note that $gm$ maps to $1\otimes gm=g1\otimes m=1\otimes m$. Thus the kernel of this map is at least as big as the submodule generated by $gx-x,x\in M,g\in G$. Now define $g:k\times M\to M_G$ that sends $(\lambda,m)\to \lambda\overline m$. This is bilinear and balanced since $f(\lambda,gm) = f(\lambda,m)=f(\lambda g,m)$, and thus descends to a map $g: k\otimes_{kG} M\to M_G$, which is in fact the inverse of $f$.

Alternatively, the trivial $kG$ module is the quotient $kG/J$ where $J$ is the kernel of the augmentation $\varepsilon$, which is the ideal generated by $g-1,g\in G$. Then by general considerations,

$$kG/J\otimes_{kG} M \simeq M/JM$$

and $JM$ is precisely the submodule spanned by $gx-x,g\in G,x\in M$.

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