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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 $M=\mathbb{Q}$. –  Arturo Magidin Feb 19 '12 at 6:56
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Do your rings have unity? –  Arturo Magidin Feb 19 '12 at 7:09
    
Dear Arturo Magidin the ring is unital ring. –  Arsenaler Feb 19 '12 at 7:14
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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.) –  Pierre-Yves Gaillard Feb 19 '12 at 7:28

1 Answer 1

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. –  Arsenaler Feb 21 '12 at 8:09
    
@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 $x^2$? –  Alex Becker Feb 21 '12 at 9:35
    
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. –  Arsenaler Feb 21 '12 at 16:04
    
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 $\{1,x,x^2,\ldots\}$. –  Alex Becker Feb 21 '12 at 16:51
    
@Alex Becker: May I please know what $N\left(\frac{a}{b}\right)$ stands for? –  user112535 Jan 30 at 2:09

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