Let $R$ be a unital ring, $M$ is a finitely generated $R$-module.
My question is to prove that there exist a maximal submodule in $M$. However I have no strategy to prove that except using the idea of Zorn lemma.
Can any body help me to solve this problem?
Also, please give a counter example for the case that if $M$ is not finitely generated.
Thank for reading. I beg your pardon for my poor English
Best Answer
Finitely generated case: Let $M$ be generated by $x_1,\ldots,x_n$ over $R$. If $n=1$, $M$ is generated by one element over $R$, so if $I$ is a maximal ideal of $R$ then $I(x_1)$ is a maximal submodule of $M$. To see that a maximal ideal $I\subset R$ exists, consider the set of ideals of $R$ partially ordered by inclusion. Observe that if $\{J_\alpha\}_{\alpha\in A}$ is a chain of ideals in $R$ then $\bigcup\limits_{\alpha\in A} J_\alpha$ is an ideal, as $1$ is not in the (why?), so $\{J_\alpha\}_{\alpha\in A}$ has an upper bound, thus by Zorn's lemma we have a maximal ideal. I will leave the inductive step to you.
To see that $\mathbb Q$ as a $\mathbb Z$-module has no maximal submodule, observe that any submodule $N$ of $\mathbb Q$ must not contain some $\frac{a}{b}$, thus does not contain $\frac{a}{2b}$, so $N\subset N(\frac{a}{b})\subset \mathbb Q$. Thus no maximal submodule exists.