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Let $V$ be a free $R$-module of finite rank (or, to start, even a finite-dimensional $K$-vector space), $N,P \in \mathrm{End}(V)$ such that $P,N$ commute, $P$ is idempotent, and $N$ is nilpotent. Consider the endomorphism $\alpha=1+(P+N)(t-1)$ of the $R[t]$-module $V[t]$.

Question. Why is the cokernel of $\alpha$ isomorphic to the image of $P$?

This is used in Rosenberg's proof of the Bass-Heller-Swan Theorem, without proof. Actually one only wants that the cokernel of $\alpha$ is finitely generated as an $R$-module. I know how to deal with the case $N=0$, but it doesn't seem to generalize. This should just be linear algebra, but I don't know how to construct a map $\mathrm{coker}(\alpha) \to \mathrm{im}(P)$ at all.

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    You can imagine elements of $V[t]$ as polynomials in $t$ with coefficients in $V$. This is a $R[t]$-module in the obvious way. So it is possible to handle this without knowledge of tensor products.2012-05-24

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It can be showed that cokernel of $\alpha$ is a finitely generated $R$-module.

Let us compute the image of $\alpha$ as following:

$\alpha=(1-P-N)+(P+N)t$, $\alpha P=P(-N+(1+N)t)$, $\alpha (1-P)=(1-P)(1-N)+(1-P)Nt=(1-P)(1-N+Nt)$.

Replacing $N(1+N)^{-1}$ by $N$, then the image equals $\alpha PV[t]+\alpha (1-P)V[t]=(t-N)PV[t]+(1-P)V[t]$.

Suppose $N^k=0$, then $(t-N)V[t]\supset (t-N)(t^{k-1}+t^{k-2}N+\ldots +N^{k-1})V[t]=t^kV[t]$, hence the image contains $t^kV[t]$, and the cokernel is a quotient of $V+Vt+\ldots+Vt^{k-1}$ which is a finitely generated $R$-module.

However, we have not proved the original question yet.

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    Actually it's not hard to show with your method of proof the stronger statement that the cokernel is isomorphic to the image of P.2012-05-24