Let M be a finite length module over a simple noetherian domain. If M/rad(M) is cyclic does it follow that M is cyclic.
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4$\begingroup$ Yes. If $M/\mathrm{rad}(M)$ is cyclic, then there is $m\in M$ such that $mR+\mathrm{rad}(M)=M$. By a variant of Nakayama's lemma, this implies that $mR=M$. Indeed, if that were not the case, there would be a maximal submodule $N$ containing $mR$ (because $M$ is of finite length). Since $N$ contains $\mathrm{rad}(M)$, we would get $mR+\mathrm{rad}(M)\subseteq N\neq M$ contradicting our ealier conclusion. $\endgroup$Uriya First– Uriya First2025-07-24 08:10:32 +00:00Commented Jul 24 at 8:10
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