The finite generation of $M$, to conclude $M$ is Noetherian (when $R$ is Noetherian).

Question

We know the following proposition.

Proposition. Let $R$ be a Noetherian/Artinian ring and $M$ an $R$-module. If $M$ is finitely generated, then the $R$-module $M$ is Noetherian module.

I was wondering if the condition of "finitely generation" of $M$ is necessary. And for this purpose if tried to find an example.

Examples.

1. We consider the $\Bbb Z$-module $\Bbb Q$, which is the abelian group $(\Bbb Q,+).$ The ring $\Bbb Z$ is a Noetherian ring, the $\Bbb Z$-module $\Bbb Q$ is not finitely generated (since the abelian group $(\Bbb Q,+)$ is not finitely generated) and of course $\Bbb Z$-module $\Bbb Q$ is not a Noetherian module (since itself is again a submodule, not f.g.). Is this idea in the right way?

2. But $\Bbb Z$ is not Artinian, to consider an example for this case. So, what if we take a field $K$ and construct the $K\!$-module $K[X]$? What can we conclude for this case? Is $K[X]$ Noetherian/Artinian as a $K$-module and why?

3. Any other ideas?

Thank you.

Answer 1

No, $K[X]$ is not artinian, because a commutative artinian ring has Krull dimension $0$.

Actually, for commutative rings, artinian $\iff$ noetherian and of Krull dimension $0$.

However, you may consider its field of fractions $K(X)$, which is artinian, being a field.

Answer 2

For a commutative ring, we have that an artinian ring is noetherian, this by the Hopkins-Levitzki theorem.

https://en.wikipedia.org/wiki/Hopkins%E2%80%93Levitzki_theorem

If $A$ is a noetherian ring then $A[x]$ is a noetherian ring, but $A[x]$ is not noetherian as an $A$-module.

https://en.wikipedia.org/wiki/Hilbert%27s_basis_theorem

In general for a non zero commutative ring $A$, $A[x]$ is not a noetherian $A$-module. Consider the ideals $I_n=\sum_{k=0}^nAx^k$. So we have a strict ascending chain of $A$-submodules in $A[x]$. So you have the ascending chain $I_0\subset I_1\subset i_2\subset\dots$, so have found a non statioanary ascending chain.Showing that $A[x]$ is not noetherian as $A$-module.

To see that is no artinian as an a $A$-module, we build a descending chain given by $J_n=\sum_{k=n}^\infty Ax^k$. So you have a strict descending chain of $A$-modules. Showing that $A[x]$ is not artinian as $A$-module.

Answer 3

Any Noetherian module is finitely generated, so the condition is clearly necessary for a module being Noetherian.

Artinian modules can be infinitely generated. The simplest example is the Prüfer group.

Your $K[X]$ cannot be Artinian, because vector spaces over a field are Artinian if and only if they're Noetherian, that is, finite dimensional.