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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

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    For an example of a non-finitely generated $R$ module with no maximal submodules, take $R=\mathbb{Z}$ and $M=\mathbb{Q}$.2012-02-19
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    Do your rings have unity?2012-02-19
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    Dear Arturo Magidin the ring is unital ring.2012-02-19
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    Dear msnaber: Its not harder to prove that any proper sub-module is contained in a maximal one. Do you see how to reduce this statement to the case $M=R$? (Here $R$ is viewed as a left $R$-module.)2012-02-19

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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.

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    Dear Alex Becker, if $R=\mathbb{Z}$ and $m=(3)$, then if we take $M=R[x]/(3)[x] \cong \mathbb{F}_{3}[x]$, which is not a field, then $m[x]$ is not a maximal ideal. This is contradict with your argument.2012-02-21
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    @msnaber Wait, is $M$ generated by one element over $R$, as a module? If $x$ is the generator as the module, how do you get $x^2$?2012-02-21
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    Let $R$ be a Noetherian ring, then $R[x]$ is a Noetherian too, that means $R[x]$ is finitely generated. Could you please point out for me, what is the generating element of $R[x]$? Is $R[x]=\lbrace a_{0}+a_{1}x+\cdots+a_{n}x^{n}| a_{i}\in R \rbrace$ ? I beg your pardon for my stupid question.2012-02-21
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    It is true that $R[x]=\{a_0+a_1x+\cdots+a_nx^n|a_i\in R\}$. Since $R[x]$ is Noetherian, every $R[x]$-submodule of $R[x]$ is finitely generated, including $R[x]$ itself, but $R[x]$ is not finitely generated *as an $R$-module*, much less generated by 1 element. Indeed, its generators as an $R$-module are $\{1,x,x^2,\ldots\}$.2012-02-21
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    @Alex Becker: May I please know what $N\left(\frac{a}{b}\right)$ stands for?2014-01-30
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    @Alex Becker: May I please know what N(a b ) stands for? I guess it stands for N+Z(a/b) in which Z(a/b) is the cyclic group generated by a/b.2014-02-03
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    @AlexBecker does this also work in non-commutative ring, save we only deal with left ideal left submodule?2015-05-01
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    @AlexBecker why is $I(x_1)$ a maximal submodule of $M$?2015-05-01
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    What if $R$ is a simple ring?2017-10-25
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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.