Consider the ring $R = \Bbb Q[x]$ and the multiplicatively closed set $S = \{1, x, x^2, \ldots\}$.
Let $R_x = S^{-1}R$. Consider the module $M = R_x/R$. Determine whether $R_x$ and $M$ are Noetherian/Artinian $R$-modules.
Edit: Note that $R$ is an $R$-submodule of $R_x$ and the quotient is of $R$-modules.
Attempt.
Note that $R_x$ is an $R$-algebra (and hence, $R$-module) via the map $p(x) \mapsto \frac{p(x)}{1}$. Since $R$ is a domain, this map is injective and thus, we may regard $R$ as an $R$-submodule of $R_x$.
Now, $R$ is not an Artinian $R$-module and hence, neither is $R_x$.
On the other hand, we have the chain
$$\langle \overline{x^{-1}} \rangle \subsetneq \langle \overline{x^{-2}} \rangle \subsetneq \langle \overline{x^{-3}} \rangle \subsetneq \cdots$$
of $R$-submodules in $R_x/R$ and hence, $R_x/R$ is not a Noetherian $R$-module. In turn, this implies that $R_x$ is not Noetherian.
Thus, only the question of showing whether or not $M$ is an Artinian $R$-module remains, which I am unable to do.
By Bezout, if $\frac{p(x)}{x^k} \in R_x$ with $p(0) \neq 0$, there is a polynomial $q$ such that $x^k|pq-1$. In particular, in $R_x/R$, $q(x)\frac{p(x)}{x^k}=\frac{1}{x^k}$. It follows that the lattice of sub-$R$-modules of $R_x/R$ is isomorphic to $\mathbb{N} \cup \{\infty\}$ so it doesn’t have any strictly decreasing infinite sequence. So $R_x/R$ is Artinian over $R$.