Let $M$ be a non-zero finitely generated module over a commutative ring (with unity) $R$ ; then how to show that $M$ has a maximal submodule ?

I was trying by induction on the minimal no. of generators of $M$ , denote it by $\mu(M)$ . If $\mu(M)=1$, let $M=Rm$ ; then $M\cong R/ann_R(m)$, and since $ann_R(m)$ is a proper ideal, it is contained in some maximal ideal $J$, and then the image of $J/ann_R(m)$ in $M$ is a maximal submodule of $M$ . Now let the claim be true for all $M$ with $\mu (M)<n$ . We want to prove the claim for $M$ with $\mu (M)=n$ . Let $M=Rm_1+...+Rm_n$ where $m_1,...,m_n \in M$ . Then the submodule $N=Rm_1+...+Rm_{n-1}$ satisfies $\mu (N) < n$, so $N$ has a maximal submodule say $L$ . But I am unable to produce a maximal submodule of $M$ in this way . Does this approach at all work , or is there another (simpler) way ?

Pleas help . Thanks in advance

  • $\begingroup$ What if take $N=M/Rm_n$? $\endgroup$ – user26857 Nov 28 '17 at 19:13
  • $\begingroup$ You can use Zorn's Lemma to "find" a maximal submodule of $M$, just like it is used to show every non-zero ring has a maximal ideal. Finite generation of $M$ over $R$ is not necessary here (although for a constructive proof it is). $\endgroup$ – Matthé van der Lee Dec 24 '17 at 14:46
  • $\begingroup$ @MatthévanderLee: An arbitrary nonzero module need not have a maximal submodule. The Zorn's lemma proof you have in mind doesn't work (without some finite generation hypothesis) because one can't conclude that the union of a chain of proper submodules is again proper. (This issue doesn't arise when showing that a nonzero ring has a maximal ideal --- the union of a chain of proper ideals must be proper because otherwise the identity element would be contained in one of the ideals.) $\endgroup$ – Beren Sanders Nov 23 '19 at 22:11
  • $\begingroup$ @BerenSanders: You are absolutely right, my mistake. $M=\mathbb{Q}$ is a case in point; it does not have any maximal subgroups (=$\mathbb{Z}$-submodules). $\endgroup$ – Matthé van der Lee Nov 25 '19 at 18:56

Hint: If $M$ has a proper submodule $K$ such that $M/K$ has a maximal submodule, then $M$ has a maximal submodule as well (why?). Can you find a choice of $K$ such that you can prove $M/K$ has a maximal submodule?


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