Does a finitely generated faithful module over an Artinian ring contain a regular element?

Question

In the text

Nicholson -- Introduction to Abstract Algebra, 4th Ed (2012)

the claim of exercise $8(b)$ of exercise set $11.1$ is:

If $R$ is a left artinian ring with $1\ne 0$, and $M$ is a finitely generated left $R$-module such that $\text{ann}(M)=0$, then $M$ has a submodule isomorphic to $R$.

But in my answer to

$\;\;\;\;\;$ How to show $\operatorname{ann}(M) = \operatorname{ann}(X)$.

I gave a counterexample to the above claim.

I wonder if the claim could be repaired by assuming as an additional hypothesis that $R$ is commutative.

Question:

If $R$ is a commutative artinian ring with $1\ne 0$, and $M$ is a finitely generated $R$-module such that $\text{ann}(M)=0$, must $M$ have a submodule isomorphic to $R$?

Two special cases: The answer is "yes" if

• $R$ is a field.$\\[4pt]$
• $R$ is finite.

That's as far as I've got.

Answer 1

Wow, that is an unfortunate mistake you found.

For semiperfect rings, there is a notion of the basic module of the ring, which is a faithful module that captures a lot about the ring. This is a theorem:

If $R$ is a semiperfect ring, then right basic module is a summand of any generator of Mod-$R$.

For a commutative Artinian ring, the basic module is just $R$ itself. But the missing ingredient, as you see, is that a faithful module need not be a generator of Mod-$R$. A ring for which every faithful f.g. module is a generator of Mod-$R$ is called finitely pseudo-Frobenius. So the best you can say, I think is

If $R$ is a commutative, semiperfect, finitely pseudo-Frobenius ring, then $R$ is a summand of every f.g. faithful module.

The nicest class for which this is all true is commutative quasi-Frobenius rings.