Lemma 42.68.27. Let $A$ be a Noetherian local ring. Let $M$ be a finite $A$-module of dimension $1$. Assume $\varphi , \psi : M \to M$ are two injective $A$-module maps, and assume $\varphi (\psi (M)) = \psi (\varphi (M))$, for example if $\varphi $ and $\psi $ commute. Then $\text{length}_ R(M/\varphi \psi M) < \infty $ and $(M/\varphi \psi M, \varphi , \psi )$ is an exact $(2, 1)$-periodic complex.
Proof. Let $\mathfrak q$ be a minimal prime of the support of $M$. Then $M_{\mathfrak q}$ is a finite length $A_{\mathfrak q}$-module, see Algebra, Lemma 10.62.3. Hence both $\varphi $ and $\psi $ induce isomorphisms $M_{\mathfrak q} \to M_{\mathfrak q}$. Thus the support of $M/\varphi \psi M$ is $\{ \mathfrak m_ A\} $ and hence it has finite length (see lemma cited above). Finally, the kernel of $\varphi $ on $M/\varphi \psi M$ is clearly $\psi M/\varphi \psi M$, and hence the kernel of $\varphi $ is the image of $\psi $ on $M/\varphi \psi M$. Similarly the other way since $M/\varphi \psi M = M/\psi \varphi M$ by assumption. $\square$
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