Let $M$ be a finitely generated module over a noetherian ring $R$ and suppose that $\dim R≤1$. And let be a one to one homomorphism $f:M\to M$. It is true that $\operatorname{Coker}f$ has finite length as $R$-module?

Thanks you all.


A finitely generated module over a noetherian ring is of finite length iff its support is contained into the maximal spectrum. (See here, Proposition 1.6.9.)

Now let $\mathfrak p\in\operatorname{Supp}(\operatorname{Coker}f)$. If $\mathfrak p$ is not maximal, then it is minimal since $\dim R\le 1$. But $(\operatorname{Coker}f)_{\mathfrak p}=\operatorname{Coker}f_{\mathfrak p}$, so the question can be viewed over $R_{\mathfrak p}$. But this is an artinian ring, and then $M_{\mathfrak p}$ is an artinian module, so $f_{\mathfrak p}$ is surjective (see here), that is, $\operatorname{Coker}f_{\mathfrak p}=0$, a contradiction.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.