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.
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?
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?
Any other ideas?