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.


    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.

  • $\begingroup$ 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/… $\endgroup$ Dec 25 '20 at 23:32
  • $\begingroup$ @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? $\endgroup$
    – quasi
    Dec 25 '20 at 23:59
  • $\begingroup$ 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. $\endgroup$ Dec 26 '20 at 0:05
  • $\begingroup$ I took the liberty of editing the title to make it more specific / searchable. $\endgroup$ Dec 26 '20 at 0:07
  • 1
    $\begingroup$ @Badam Baplan: I'll leave it for you or someone else to answer. Thanks again. $\endgroup$
    – quasi
    Dec 26 '20 at 0:09

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.


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