5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.