0
$\begingroup$

I read Prof. Goto's Notes in Commutative Algebra and I met a definition of primary module, like this:

"Suppose that $R$ is a Noetherian ring and $Q\in\operatorname{Spec}(R)$. Let $L\leq M$ be a $R$-submodule of $M$. Then we say that $L$ is a $Q$-primary submodule of $M$ if $n\operatorname{Ass}_R(M/L)=\{Q\}$."

Also, the definition of $\operatorname{Ass}$ in these notes is:

"$\operatorname{Ass}_R M=$ $\{Q\in \operatorname{Spec}(R)\;\mid $ there exists an exact sequence $0\to R /Q \to M$ of $R$-modules$\}$"

I encountered a problem like this:

"Suppose $R$ is a Noetherian commutative ring and let $M$ be a finitely generated $R$-module. Prove that if $\operatorname{Ass}_R M=\{Q\}$, then $I=(0:M)$ is a $Q$-primary ideal of $R$."

This problem is quite immediate, in fact you can try to prove it. Note that "$\operatorname{Ass}_R M\subseteq V(I)$".

I wonder if the converse is true:

If $I=(0:M)$ is $Q$-primary, then $\operatorname{Ass}_R M=\{Q\}$.

If it's wrong, can you show me a counter example? Thank you.

$\endgroup$
  • $\begingroup$ That notes are written in Japanese? $\endgroup$ – user26857 Mar 19 '17 at 9:28
  • $\begingroup$ No. English. But for some reasons he doesn't upload it online $\endgroup$ – chí trung châu Mar 19 '17 at 10:56

Your Answer

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

Browse other questions tagged or ask your own question.