Let $M$ be a non-zero finitely generated module over a commutative ring (with unity) $R$ ; then how to show that $M$ has a maximal submodule ?
I was trying by induction on the minimal no. of generators of $M$ , denote it by $\mu(M)$ . If $\mu(M)=1$, let $M=Rm$ ; then $M\cong R/ann_R(m)$, and since $ann_R(m)$ is a proper ideal, it is contained in some maximal ideal $J$, and then the image of $J/ann_R(m)$ in $M$ is a maximal submodule of $M$ . Now let the claim be true for all $M$ with $\mu (M)<n$ . We want to prove the claim for $M$ with $\mu (M)=n$ . Let $M=Rm_1+...+Rm_n$ where $m_1,...,m_n \in M$ . Then the submodule $N=Rm_1+...+Rm_{n-1}$ satisfies $\mu (N) < n$, so $N$ has a maximal submodule say $L$ . But I am unable to produce a maximal submodule of $M$ in this way . Does this approach at all work , or is there another (simpler) way ?
Pleas help . Thanks in advance
Hint: If $M$ has a proper submodule $K$ such that $M/K$ has a maximal submodule, then $M$ has a maximal submodule as well (why?). Can you find a choice of $K$ such that you can prove $M/K$ has a maximal submodule?