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Every (non-zero) finitely generated module over a ring has a maximal proper submodule by a simple application of Zorn's lemma.

I am interested in the following question, with two variants.

Variant 1: For which groups $G$ does every (non-zero) $\mathbb CG$-module have a maximal proper submodule.

Variant 2: For which groups $G$ does every countably generated (non-zero) $\mathbb CG$-module have a maximal proper submodule.

What I know about this:

  1. If $G$ is finite, then the answer to Variant 1 (and hence 2) is yes since every (non-zero) module $M$ over an Artinian ring $R$ has a maximal proper submodule (if $J$ is the radical of $R$, then $JM$ is proper because $J$ is nilpotent and so $M/JM$ is semisimple and hence has a simple quotient, whose kernel pulls back to a maximal submodule of $M$)?
  2. If $G$ is commutative, then by the first theorem of this paper Variant 1 holds for $\mathbb CG$ iff $G$ is locally finite (using complex group rings are semiprimitive and are von Neumann regular iff the group is locally finite)?
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    $\begingroup$ Just a remark (which you probably observed but for everybody): for every $G$ there exists a nonzero $\mathbf{Z}G$-module with no maximal proper submodule: say $\mathbf{Q}$ with the trivial action. Whence the restriction to $\mathbf{C}G$-modules. $\endgroup$
    – YCor
    Commented Aug 18, 2018 at 22:53
  • $\begingroup$ If a module fails to have a maximal submodule then every generating set has the property that removing any finite subset from the generating set still yields a generating set. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 19, 2018 at 16:18
  • $\begingroup$ The latter amounts to saying that a module with no maximal proper submodule has no nonzero finitely generated quotient. (In more general structures, e.g., groups, it rather says that an object with no maximal proper subobject is infinitely generated over every proper subobject.) $\endgroup$
    – YCor
    Commented Aug 19, 2018 at 20:45
  • $\begingroup$ Yes since finitely generated quotients have simple quotients. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 19, 2018 at 22:01
  • $\begingroup$ Since for noetherian commutative rings, finitely cogenerated modules are the same as artinian modules (and are countably generated) — all this is in Lam's book—, every nonzero module admits a nonzero countably generated quotient. In particular, the existence of a nonzero module with no maximal proper submodule already implies that of a countably generated one. $\endgroup$
    – YCor
    Commented Oct 24, 2018 at 22:37

1 Answer 1

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Here is a partial answer about Variant 2: there are countably generated $\mathbb C[\mathbb Z]$-modules with no proper quotient.

First note that $\mathbb C[\mathbb Z]\cong \mathbb C[x,x^{-1}]$ is a PID. The simple modules are just the quotients by maximal ideals, that is by ideals generated by irreducible elements, and any such quotient is a torsion module.

In the answer to Injective indecomposable modules over Laurent polynomial rings, it is shown that there are injective=divisible $\mathbb C[x,x^{-1}]$-modules that have countable dimension over $\mathbb C$. These are obviously countably generated. They do not have simple quotients because any quotient of a divisible module is divisible and any torsion divisible module is trivial. Hence they do not have maximal submodules.

I suspect that a similar argument shows that $\mathbb C[\mathbb Z^n]\cong \mathbb C[x_1^{\pm 1},\cdots,x_n^{\pm 1}]$ also has countably generated divisible modules.

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  • $\begingroup$ Over a PID $A$ it's straightforward that the indecomposable modules the field of fractions and quotient of localizations $A[t^{-1}]/A$, $t$ irreducible: except the fraction field, the other ones are countably generated since $A[t^{-1}]$ is generated by $\{t^{-n}:n\ge 0\}$ as $A$-module. So indecomposable $\mathbf{C}[x,x^{-1}]$-modules are $\mathbf{C}(t)$ and $\mathbf{C}[t,t^{-1},(t-a)^{-1}]/\mathbf{C}[t,t^{-1}]$ for $a\in\mathbf{C}$. Except the first one. This is very standard (but I didn't catch the idea to use this PID group ring to solve your question). $\endgroup$
    – YCor
    Commented Oct 24, 2018 at 20:56
  • $\begingroup$ Divisible torsion $C[Z^2]$-modules are much more tricky to consider. It's unclear to me whether you can embed any simple module into a countably generated one. I think that I read/heard that their injective envelope is known to be uncountably generated. $\endgroup$
    – YCor
    Commented Oct 24, 2018 at 21:02
  • $\begingroup$ We just need at least one divisible is a countably generated module and that any simple is not divisible. @YCor $\endgroup$
    – Benjamin Steinberg
    Commented Oct 24, 2018 at 21:18
  • $\begingroup$ @YCor, maybe sounding like londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/jlms/… works for the trivial module. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 24, 2018 at 21:21
  • $\begingroup$ I only realized this morning the connection of my question to divisible modules @YCor $\endgroup$
    – Benjamin Steinberg
    Commented Oct 24, 2018 at 21:22

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