# A necessary condition for $Rad M$ to be the unique largest superfluous submodule of $M$.

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 semantic nitpick: You don't apply the adjectives "necessary/sufficient" to propositions, you apply them to conditions (i.e. hypotheses and conclusions mentioned in propositions.) What you really mean to say is: "We know that the condition that $M$ is coatomic is not necessary for $Rad(M)$ to be the unique largest superfluous submodule." Equivalently you could say that "the condition that $Rad(M)$ is the unique largest superfluous submodule is not sufficient to prove $M$ is coatomic." Saying that the proposition isn't sufficient or isn't necessary is ambiguous. – rschwieb Apr 6 '17 at 13:23
• Thank you very much for your advice. – Daisy Apr 7 '17 at 1:40

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.

Then $f(N)\neq M/RadM$ follows from $Rad M \ll M$. [...] Then $f(N)$ is contained in a maximal submodule $P\subset M/RadM$.

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.)

• Thank you very much. Very good answer. – Daisy Apr 7 '17 at 1:56