I know that $\mathbb{Q}[x]$ is noetherian but not artinian as a $\mathbb{Q}[x]$-module. The question is, is there a proper ideal $M\neq 0$ of $\mathbb{Q}[x]$ such that the quotient $\mathbb{Q}[x]/M$ is an artinian $\mathbb{Q}[x]$-module?
I tried to find an ideal such that the quotient will be artinian, but could not find one. So my guess is that for all proper ideals, the quotient is not artinian. But I failed to come up with a proof for it.
That's very odd, because as it turns out, every quotient of $\mathbb Q[x]$ by a nonzero submodule will be an Artinian $\mathbb Q[x]$ module.
This is because $\mathbb Q[x]$ is a principal ideal domain, and that means every ideal is generated by a polynomial. If $I=(f)$ where $f$ is of degree $n$, then $\mathbb Q[x]/I$ is $n$-dimensional as a $\mathbb Q$ vector space.
Each such submodule of the quotient is in particular a $\mathbb Q$ subspace, so no strictly ascending chain of submodules cannot have more than $n$ links, and so the quotient module is Artinian.
In fact, in Cohen, Commutative rings with restricted minimum condition, Duke Math. J. 17 (1950), 27-42. it is noted that among commutative rings, the non-Artinian ones for which every quotient by a nonzero ideal is Artinian are exactly the one-dimensional Noetherian domains.