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Let $R$ be a ring and $R[\mathbb{Z}]$ be the group ring obtained from ring $R$ and group $\mathbb{Z}=<s>$.

Suppose that $M$ be a $R[\mathbb{Z}]$-module and it is isomorphic to $R^n$ as $R$-module.

We can describe $R[\mathbb{Z}]$-module structure as a matrix $H\colon R^n\to R^n$ which corresponds to a map $s\colon M\to M$ defined by $v\to s\cdot v$.

Now, we can regard $H$ a $n\times n$ square matrix over $R[\mathbb{Z}]$ via the inclusion $R\to R[\mathbb{Z}]$.

Is it true that $M\cong \operatorname{Coker}(sI-H\colon R[\mathbb{Z}]^n\to R[\mathbb{Z}]^n)$? (I think that it should be.)

This is not the homework.

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I think that it might be obtained by tensoring from the sequence $R[\mathbb{Z}]\to R[\mathbb{Z}]\to R$ defined by $x\to (s-1)x$, $y\to \epsilon(y)$, where $\epsilon$ is augmentation map. – Moduletheory Jun 3 '11 at 7:06
It looks right to me, but I'm not an expert in homological algebra. – Jim Belk Jun 3 '11 at 17:03

I think the answer is yes, and can just be checked directly.
If you choose a basis $e_1,\ldots,e_n$ for $M$ as an $R$-module, then you get a nautral surjection $R[\mathbb Z]^n \to M,$ via the action of $R[\mathbb Z]$ on the $e_i$.

If you apply $s$ to the $e_i$, it acts via $H$ (by definition of $H$), and so this map factors through the stated cokernel. On the other hand, if you compute this cokernel, using the relation $sI = H$ to replace every occurence of $s$ by an action of the matrix $H$, then you see that every coset in the cokernel has a coset representative consisting of an element in $R^n$.

In other words, the natural $R$-linear inclusion $R^n \to R[\mathbb Z]^n$ gives an $R$-linear section to the surjection $R[\mathbb Z]^n \to M$, which becomes surjective after passing to the cokernel of $sI - H$. This proves that the cokernel is mapping isomorphically onto $M$.

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