Examples of modules with no maximal proper submodules.

Question

An exercise in chapter 0 in Noncommutative Algebra by Benson Farb and R. Keith Dennis indirectly asks the reader to come up with an example of an $R$-module with no maximal proper submodules.

I'm interested in more examples of $R$-modules (or left $R$-modules) with this property that either aren't $\mathbb{Z}$-modules or are different from the examples given below. My immediate goal in doing this is just getting better intuition about modules.

I tried to come up with one for a bit and failed, so I searched for examples on the site and found $(\mathbb{Q}, +)$ as a $\mathbb{Z}$-module, as discussed in these two questions:

• About non existence of maximal subgroup in additive group $\mathbb{Q}$
• $(\mathbb{Q},+)$ has no maximal subgroups

A little while after reading those answers and thinking about the associated proofs, I came up with two examples.

1. The trivial $R$-module $0$ has no proper submodules at all and hence no maximal proper submodules.
2. The dyadic rationals $(\mathbb{D}, 2)$ where $\mathbb{D}$ are rationals of the form $\frac{a}{2^b}$.

For proof of (2), I can cite this fact. I can also look at a proper subgroup $A$, consider an element $\frac{c}{2^d}$ in $ \mathbb{Q} \setminus A$ and show that $A \oplus \left(\frac{c}{2^{d+1}}\right) > A \oplus \left(\frac{c}{2^d}\right) > A$.

Answer 1

Observe that such a module can only exist if $R$ is not a field, because every vector space contains a maximal sub-vector space (as it has a basis). One immediate generalization to any commutative ring $R$ is the following:

Let $R$ be a commutative ring which is not a field, and let $M$ be a divisible $R$-module, i.e. for all $m\in M$ and $r\in R\setminus\{0\}$ there exists $m'\in M$ with $m=rm'$ (or in other words $rM=M$ for all non-zero $r$, i.e. multiplication by $r$ is surjective). Then $M$ has no maximal submodules.

Indeed, suppose by contradiction that $N\subsetneq M$ is maximal, and take $m\in M\setminus N$. As $M/N$ is simple, we have $M/N\cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}\subseteq R$. In particular, $\operatorname{Ann}_R(M/N)=\mathfrak{m}$. Now as $R$ is not a field we can take $r\in\mathfrak{m}\setminus\{0\}$, and by divisibility there exists $m'\in M$ with $rm'=m$. But then $m+N=rm'+N=r(m'+N)=0_{M/N}$ as $r\in \operatorname{Ann}_R(M/N)$, and thus $m\in N$, contradiction.

I think one may immediately generalize this to the non-commutative case (assuming that $(0)$ isn't a maximal left ideal).