When every proper submodule of a free module is contained in a maximal submodule

Question

Let $(R, m)$ be a commutative local ring (it is not Noetherian in general) and $F$ be a free $R$-module. Under what conditions every proper submodule of $F$ is contained in a maximal submodule.

Answer 1

I believe that the answer is that every submodule of a free $R$-module is contained in a maximal submodule if and only if $R$ is a perfect ring (for a commutative ring, this is equivalent to dcc on principal ideals).
You can weaken local to the condition that $R$ has no infinite sets of orthogonal idempotents. (Update: I added an update to show that if even a single infinite rank free module has the property then $R$ must be pefect.)

For commutative rings (local is not needed), the answer of Jack Schmidt here shows that every submodule of a free module (he writes projective but just uses free and for local rings, as in the question, there is no difference) is contained in a proper maximal submodule if and only if every non-zero module has a maximal submodule.

For commutative rings with no infinite set of orthogonal idempotents, it is shown in Hamsher, Ross M. “Commutative rings over which every module has a maximal submodule.” Proc. Amer. Math. Soc. 18 (1967) 1133–1137 that every nonzero module contains a maximal submodule if and only if $R$ is perfect. Since a local ring cannot have any idempotents except $0$ and $1$, this covers the local case.

Slight update. For commutative rings $R$ in general, every module has a maximal proper submodule (or equivalently every submodule of a free module is contained in a maximal proper submodule) if and only if the Jacobson radical $J(R)$ is $T$-nilpotent and $R/J(R)$ is von Neumann regular. See for example C. Faith, Rings whose modules have maximal submodules, Publ. Mat. 39 (1) (1995), 201–214.

Update 2. As pointed out by @JeremyRickard, one can interpret the question as putting conditions on $F$ and $R$. If $F$ is finitely generated, then obviously every submodule is contained in a maximal proper submodule by Zorn's lemma for any ring. So the only interesting case is when there is some infinite rank free module. But the proof of Lemma 1 of Hamsher, Ross M. “Commutative rings over which every module has a maximal submodule.” Proc. Amer. Math. Soc. 18 (1967) 1133–1137 shows that if the free module on countably many generators has the property that every proper module is contained in a maximal submodule, then $J(R)$ is $T$-nilpotent and for a local ring, that immediately implies being perfect. Now if the free module on an uncountable generating set has the property that every proper submodule is contained in some maximal submodule, then Jack Schmidt's argument mentioned above shows the same is true for the free module on a countable basis (just map the free module on a bigger basis onto it and pull back the proper submodule) and so again $R$ is perfect. Thus for a commutative local ring $R$ the following are equivalent:

1. $R$ is perfect;
2. There is some free $R$-module of infinite rank where every proper submodule is contained in a maximal submodule
3. Every submodule of a free $R$-module is conatined in a maximal proper submodule.
4. Every non-zero $R$-module has a proper submodule.