Let $R$ be a commutative ring with unity. Let $M$ be an $R$-module. Let $\mathcal F $ be the collection of all non-cyclic submodules of $M$ and suppose $\mathcal F$ is non-empty. Then by Zorn's lemma, $\mathcal F$ has a maximal element say $N$. Now let $P=Ann _R (M/N)=\{x \in R : xM \subseteq N\}$ .
My question is : If $P\ne R$, then is $P$ a prime ideal of $R$ ?
I was going by contradiction. Let $ab \in P$ but $a,b \notin P$ . Then $N \subsetneq N+aM, N+bM$ . So that $N+aM$ and $N+bM$ are cyclic submodules of $M$. If I can show that this implies $N$ is cyclic, then we are done. Let $N+aM=R(n_1+am_1)$ for $n_1 \in N , m_1 \in M$ . Let $L :=\{m \in M : am \in N\}$. Then $N \subseteq L$ and since $ab M \subseteq N$, so $bM \subseteq L$ . So $N+bM \subseteq L$ , so $L$ is cyclic let $L=Rl$ for some $l \in L$ .I can show that $N=Rn_1+aL$ , but unfortunately that's not enough.
Please help .
By [1, Corollary 5.6], $N$ is a prime submodule of $M$ when $M$ is cyclic. Hence, $Ann_R(M/N)=(N:_RM)$ is a prime ideal of $R$. Now in general cass, if $N=M$, then $Ann_R(M/N)=R$ and so the problem is not true in this cass and if $N\not= M$, then $M$ if cyclic by the maximality of $N$ and the problem is true by above argument.
[1] R. Nekooei and E. Rostami, A Prime Submodule Principle,Algebra Colloq. 21, 697 (2014). https://doi.org/10.1142/S1005386714000649