Maximal submodule in a finitely generated module over a ring 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
 A: 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.
A: There's actually a useful way to rephrase what it means to be finitely generated that is helpful:

A module $M$ is finitely generated iff for every ascending chain $M_0\subseteq M_1\subseteq\ldots$ of submodules such that $\cup_{i=1}^\infty M_i=M$, we have $M_j=M$ for some $j$.

Note that this says that any ascending chain of proper submodules in a f.g. module has a union which is proper. Here we can apply Zorn's Lemma, as advised above.
The above characterization is useful, and not hard to prove. Moreover it is easily dualized for finitely cogenerated modules.  I'm not saying it's any easier than the other solutions offered, but doing it this way provides an additional perspective.
A: Although 8 years past for the question, I still tried to answer with another proof.
Assume $M=Rx_1 + Rx_2+...+Rx_k$, then we have a morphism $f$ from the free module $R^k$ $\to$ M onto. By Zorn's lemma, we can get a maximal submodule in $R^k$ that contains the submodule Ker$f$ and its image is the maximal submodule of $M$.
