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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

Arsenaler
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4 Answers4

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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.

Alex Becker
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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 08:09
  • @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 09:35
  • 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
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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 ${1,x,x^2,\ldots}$. – Alex Becker Feb 21 '12 at 16:51
  • @Alex Becker: May I please know what $N\left(\frac{a}{b}\right)$ stands for? –  Jan 30 '14 at 02:09
  • @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. –  Feb 03 '14 at 10:03
  • @AlexBecker does this also work in non-commutative ring, save we only deal with left ideal left submodule? – user10024395 May 01 '15 at 02:20
  • @AlexBecker why is $I(x_1)$ a maximal submodule of $M$? – user10024395 May 01 '15 at 02:31
  • What if $R$ is a simple ring? – 6666 Oct 25 '17 at 21:30
  • related answer: https://math.stackexchange.com/questions/234995/mathbbq-has-no-maximal-subgroups – Zhefeng Shen Nov 04 '24 at 09:29
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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.

rschwieb
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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 $M$.

Nguyen Dang Son
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Here is my attempt: Let M be finitely generated and let n be the least possible cardinality for a generating set (so that M can't be generated by a subset with cardinality less than n). The case where $n=1$ was answered by Alex Becker so let's assume that $n>1$.

Let $\{x_1,x_2,...,x_n \}$ be a generating set of size n. Let $N = \{ x_1,x_2,...x_{n-1} \}$. Then $N \neq M$. Let $\mathbb{S}$ be the collection of all submodules of $M$ that contain $N$ and don't contain $x_n$. Then, $\mathbb{S}$ is partially ordered by inclusion and each linearly ordered subset $D \subset \mathbb{S}$ has an upper bound $\cup D$ (it is in $\mathbb{S}$ as $\forall A \in D \ (x_n \notin A)$ and so $x_n \notin \cup D$). So by Zorn's lemma $\mathbb{S}$ has a maximal element $N'$. $N'$ is a maximal submodule as if $N' \subset_{\neq} N''$ then $x_n \in N''$ by maximality of $N'$ and so $\{x_1,x_2,...,x_n \} \subset N''$ and hence $M \subset N''$.

Mohamed AlFardan
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