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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)