4
$\begingroup$

This is a follow up of the question Example of a finitely generated faithful torsion module over a commutative ring on MathSE.

Let $M$ be a finitely generated module over a commutative ring $R$ with the property that $\operatorname{Ann}_R x\ne 0$ for all $x \in M$. When $\operatorname{Ann}_R M\ne0$?

The simplest case is $R$ an integral domain. But what about $R$ (local) artinian, or noetherian? (In the counterexample I gave to the linked question $R$ is a commutative ring which is not noetherian.)

$\endgroup$
0
4
$\begingroup$

Over a noetherian ring:

1) the answer is no in general, even for local artinian rings: Let $R$ be the ring given as a $K$-linear space with basis $(1,t_1,t_2)$ and $t_1t_2=0$, it is local artinian if $K$ is a field. Let $M$ be the free $K$-module with basis $(e_1,e_2,f)$; make it an $R$-module by setting $t_ie_i=f$, $t_ie_j=0$ for $i\neq j$, $t_if=0$.

It is faithful because if $\alpha+\alpha_1t_1+\alpha_2t_2$ is in the annihilator, evaluation on $f$ yields $\alpha=0$, then on $e_i$ yields $\alpha_i=0$.

But every element has a nontrivial annihilator: an element $a_1e_1+a_2e_2+bf$ is killed by $a_2t_1-a_1t_2$ which is nonzero if $(a_1,a_2)\neq (0,0)$, and $bf$ is killed by $t_1$.

There are many more examples using large abelian subspaces of nilpotent matrices, for instance, using matrices in the upper right block (say over a field) of $2n\times 2n$-matrices, we obtain a $K$-algebra of dimension $n^2+1$ and a faithful $2n$-dimensional module (with dimension in the sense of linear algebra), hence it's to small to contain free submodules of rank 1 as soon as $n\ge 2$.

2)(edit!) the answer is yes if $R$ is an arbitrary domain.

Indeed, let $M$ be a finitely generated $R$-module such that each element has a nonzero annihilator. Using generators, we can write $M$ as a quotient of $\bigoplus_{j=1}^d R/I_j$ for some nonzero ideals $I_j$ (the annihilators of the $d$ generators). Since $R$ is a domain, $I=I_1\dots I_d\neq 0$; since $I$ annihilates $M$ we deduce that $M$ has a nonzero annihilator.

$\endgroup$
3
  • $\begingroup$ I've made it clear in my question that all domains satisfy the requirement. $\endgroup$
    – user26857
    Mar 20 '15 at 12:27
  • $\begingroup$ You asked several questions. I did't claim I answered all of them and since I give a negative answer in the noetherian case, it was natural to add the easy argument that it works in the case of noetherian domains. $\endgroup$
    – YCor
    Mar 20 '15 at 13:48
  • $\begingroup$ ... but finally it works in the general case of domains, so I edited. $\endgroup$
    – YCor
    Mar 20 '15 at 13:59
2
$\begingroup$

As I proved in the answer to this question, the following holds:

Proposition: For an Artinian ring $A$, the following are equivalent:

  1. $A$ is Gorenstein.
  2. Any finitely generated faithful module $M$ over $A$ contains an element with $(0)$ annihilator.

So in the context of your question, $Ann_A(M)\neq (0)$ if $A$ is not Gorenstein, and in fact this property characterizes Artinian Gorenstein rings. Note that YCor's example is $A=k[x,y]/(x,y)^2$, the smallest Artinian non-Gorenstein ring.

$\endgroup$

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.