Let $R$ be domain(not necessary local) with maximal ideal $\mathfrak{p}$ and $d \in R, d \neq 0$. $(R/(d))_{\mathfrak{p}} = 0 \iff (d) \not\subset \mathfrak{p}$(?). And if $ (d) \subset \mathfrak{p}$ what can we say about $\ell(R_{\mathfrak{p}}/(d))$?
Let $\varphi$ be endomorphism of $R^n$ and $d=\det\varphi$.
I want to prove that $$\ell(\mathrm{coker}\varphi) < \infty, \ell(\ker\varphi) < \infty \iff \ell(R/(d)) < \infty$$ using localizations.
If $d \not\in \mathfrak{p} \implies \ell(\ker\varphi_{\mathfrak{p}}) = 0,\ell(\mathrm{coker}\varphi_{\mathfrak{p}}) = 0, \ell(R_\mathfrak{p}/(d)) = 0$. But what can I say about finiteness of length, if $d \in \mathfrak{p}$?
As you said correctly we may assume that $R$ is local. Let $M=R^n$.
From right to left: we may assume that $d\neq 0$ since otherwise we have that $R$ is artinian hence zerodimensional which means a field since you assumed that $R$ is a domain. For fields it is clear. Let $\varphi^*$ the adjoint mapping of $\varphi$. Then $\varphi\varphi^*=\varphi^*\varphi=dI_n$. Since $R$ is a domain we then have $\mathrm{ker}(\varphi)\subseteq\mathrm{ker}(dI_n)=0$ and thus $\varphi$ is injective. On the other hand $\mathrm{coker}(\varphi)\subseteq M/dM$. So $\mathrm{coker}(\varphi)$ has finite length if $M/dM$ has finite length. But the later module is finitely generated over the ring $R/(d)$ which has finite length and therefore also has finite length.
From left to right: Since $\mathrm{coker}(\varphi)\subset M/dM$ the projection $M/dM\to\mathrm{coker}(\varphi)$ induced by $M\to\mathrm{coker}(\varphi)$ must be an isomorphism. Therefore $M/dM$ is of finite length and finitely generated over $R/(d)$. Since $R/(d)$ is a submodule of $M/dM$ it follows that $R/(d)$ is of finite length.
Is this an exercise from a book? From which one?