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

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 $$R$$.

$$\;\;\;\;\;$$ 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.

• Does this help you? math.stackexchange.com/a/3187153/164860. Here $R$ is a commutative Artinian ring, $M$ is a faithful $2$-generated $R$-module, and every element of $M$ has nonzero annihilator. Thus $M$ can't have a submodule isomorphic to $R$. Note that another special case where the answer is "yes" is if $R$ is any reduced Noetherian ring, see e.g. math.stackexchange.com/questions/1269660/… Dec 25 '20 at 23:32
• @Badam Baplan: Thanks for the reference, and yes, the example given in the accepted answer in that link appears to give an answer of "no" to my question. Thanks again. Should I now delete my question? Dec 25 '20 at 23:59
• Hmm.. I think it's a worthwhile question to keep searchable because it addresses an error in a textbook and points to a useful counterexample (I think it is valuable to reinforce good mse posts by linking to them). Perhaps you could write up a short answer to your own question and accept it. Dec 26 '20 at 0:05
• I took the liberty of editing the title to make it more specific / searchable. Dec 26 '20 at 0:07
• @Badam Baplan: I'll leave it for you or someone else to answer. Thanks again. Dec 26 '20 at 0:09

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.