0
$\begingroup$

Let $R$ be a commutative ring. It is Noetherian precisely when all its ideals are finitely generated. However it suffices to check only the prime ideals. That is $R$ is Noetherian if and only if every prime ideal is finitely generated. See the answer to this question, for a proof.

A module $M$ over $R$ is Noetherian if and only if all its submodules are finitely generated. This naturally extends the notion of a Noetherian ring: If we regard $R$ as a module over itself in the natural way, we see that $R$ is Noetherian as a ring, if and only if it is Noetherian as a module.

There is also a notion of a prime submodule: $M$ is prime if and only if for all $a\in R, m\in M$:$$am=0\implies m=0 \,\,{\rm or}\,\, a\in{\rm Ann(M)}.$$ Equivalently $M$ is prime precisely when every non-zero submodule $N\subseteq M$ satisfies Ann$(N)=$Ann$(M)$. See this wiki article for the definition.

Given the terminology it would be natural to assume that we can characterise a Noetherian module in the same way as a Noetherian ring. That is, in order to verify that a module is Noetherian, we would only need to check that the prime submodules are Noetherian.

Is this true?

One can attempt to mimic the proof for the case of rings. If $M$ is not Noetherian we can use Zorn's lemma to construct a submodule maximal among non finitely generated submodules (if a nested union of submodules is finitely generated, then all the generators will belong to some term in the union). However there is no obvious way to show that this submodule is prime.

$\endgroup$
2
  • $\begingroup$ Under the axiom of choice, we can equivalently take a Noetherian module to be a module such that every sub-module is finitely generated. $\endgroup$
    – JPhy
    Aug 9, 2020 at 10:15
  • $\begingroup$ yes of course, but can we characterize Noetherian modules by prime submodules like Noetherian rings ? $\endgroup$
    – Adam Ben
    Aug 9, 2020 at 10:31

1 Answer 1

1
$\begingroup$

Let $R=\mathbb{R}[t_2,t_3,t_4,\cdots]/(t_2^2,t_3^3, t_4^4,\cdots)$. Clearly $R$ is not Noetherian, hence it is not Noetherian as a module over itself.

We claim that no ideal of $R$ is prime (when regarded as a submodule of $R$). Thus vacuously we have that all prime submodules of $R$ are finitely generated. Thus your claim is false.

The key point here is that prime ideals do not need to be prime, when regarded as a submodule. Humans have just chosen to give the two concepts the same name.

Proof of claim:

Let $I\subseteq R$ be a non-zero ideal. Let $\alpha\in I$. We may write $\alpha$ as a polynomial expression in the $t_i$. Pick $r$ such that $t_r$ does not occur in this expression, so $t_r\alpha\neq 0$ and $t_r^{r-1}\alpha\neq 0$. Then $$\{0\}\neq t_r I\subseteq I, $$and $$t_r^{r-1}\notin{\rm Ann}(I),\qquad{\rm but}\qquad t_r^{r-1}\in{\rm Ann}( t_rI).$$ We conclude that $I$ is not a prime submodule of $R$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .