2
$\begingroup$

Let $M$ be an $R$-module of finite length and $N$ a maximal submodule of $M.$

Is there an element $m$ in $M$ such that $m(N:M)=0$?

It is a generalization of this result:

In a Notherian ring $R,$ all minimal prime ideals have non-zero annihilator.

$\endgroup$
2
  • 2
    $\begingroup$ What do you mean by m(N:M)=0? $\endgroup$
    – Fan Zheng
    Nov 3, 2015 at 15:19
  • $\begingroup$ @FanZheng: Presumably $(N:M)\subset R$ is the annihilator of $M/N$ and the OP is treating $R$ as acting on the right.. $\endgroup$ Nov 3, 2015 at 19:48

1 Answer 1

4
$\begingroup$

Put $P=(N:M)$.

Because $N$ is a maximal submodule, $P$ is a maximal ideal.

Because $M$ has finite length, there is some minimal $k$ such that $MP^k=MP^{k+1}$. By Nakayama, there exists $s\in 1+P$ such that $MsP^k=0$.

Put $T=MsP^{k-1}$. If $T=0$, then $MP^{k-1}\subset MP^k$, contradicting minimality of $k$. So $T\neq 0$. But by the preceding paragraph, $TP=0$.

So if $k\neq 0$, then any nonzero $m\in T$ will do. But if $k=0$, then $Ms=0$. Therefore $s\in ann(M)\subseteq (N:M)=P$ and hence $1\in P$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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