I am studying the book "Rings and Categories of modules" written by Frank W. Anderson and Kent R. Fuller. On page 121, the authors offer a proposition.
9.18 Proposition. If every proper submodule of $M$ is contained in a maximal submodule of $M$, then $Rad M$ is the unique largest superfluous submodule of $M$.
We know the proposition is not necessary. I want to add a condition (suppose condition a) such that the proposition is necessary. I post my effort.
Proposition. If condition a and $Rad M \ll M$, then every proper submodule of $M$ is contained in a maximal module of $M$.
Proof. Let $N$ be a proper submodule of $M$ and $f:M \to M/RadM$ be the natural epimorphism. Then $f(N)\neq M/RadM$ follows from $Rad M \ll M$. Indeed $Rad M \ll M$ $\Leftrightarrow$ for any $L<M$ $\Rightarrow$ $L +Rad M \neq M $,i.e. $L\neq M/RadM$. Then $f(N)$ is contained in a maximal submodule $P\subset M/RadM$. Thus $N$ is a submodule of the maximal submodule $f^{-1}(P)\subset M$.
It seems I don't use the condition a. Who can point out my mistakes? Any help will be welcomed.
A module for which every proper submodule is contained in a maximal submodule has been called a coatomic module, by some authors. I'll use that as an abbreviation.
The second sentence does not follow from the first. A proper submodule might not be contained in any maximal submodule. (Indeed, that is a corollary of examples of modules which do not have any maximal submodules.) So, you are begging the question by assuming the conclusion of this theorem holds for the quotient $M/Rad(M)$.
So, the condition (a) that would fit the bill precisely would be: every proper submodule of $M/Rad (M)$ is contained in a maximal submodule.
So you can say: if $Rad(M)$ is a superfluous submodule and $M/Rad(M)$ is coatomic, then $M$ is coatomic.
Obviously Noetherian modules have that property (owing to the maximal condition on submodules.) Also as obviously a module with a unique maximum submodule would do the job (even if it isn't Noetherian.)