I'm searching an example of a finite ring extension $A\subseteq B$ and an Artinian $B$-module $M$ such that $M$ isn't Artinian as an $A$-module. Note that if such an $M$ is also finitely generated over $B$, then by this post it is also Artinian over $A$.
My attempt was to mimic the example $\mathbb{Z}[1/p]/\mathbb{Z}$ of an non-Noetherian, Artinian $\mathbb{Z}$-module over $k[x]$ in order to allow for finite subrings. For example, let $k$ be a field of characteristic $0$. One can show that $k[x,x^{-1}]/k[x]$ is Artinian over $k[x]$ but not Noetherian. So if we can prove that it is not Artinian over e.g. $k[x^2]$ or $k[(x+1)^2]$ then we would be done, but I have no idea if this even works.
Edit: $k[x,x^{-1}]/k[x]$ is Artinian over $k[x^2]$ as it is isomorphic to $(k[x^2,x^{-2}]/k[x^2])^{\oplus 2}$. But maybe it isn't Artinian over $k[(x+1)^2]$?
Edit 2: I think $k[(x+1)^2]$ doesn't work either: an equivalent question is whether $k[x,1/(x-1)]/k[x]$ is Artinian over $k[x^2]$. Now let $M$ be a $k[x^2]$-submodule of $k[x,1/(x-1)]$ containing $k[x]$. Suppose $M$ contains an element in which $1/(x-1)$ appears; as $k[x]\subseteq M$ we may suppose that it is of the form $p(1/(x-1))$ for some non-zero $p=p_1x+\cdots p_n x^n\in k[x]$. If $\deg p\geq 2$, by considering $x^2p(1/(x-1))$, we see that $M$ must contain another element of the form $q(1/(x-1))$ with $q=q_1x+\cdots q_n x^n\in k[x]$ such that $q_n=p_n$ and $q_{n-1}=p_{n-1}+2p_n$. Hence $p-q$ is a polynomial of degree $n-1$. By repeating the procedure, we see that if $M\supsetneq k[x]$, then it must contain $1/(x-1)$. As it then contains also $x/(x-1)$, it must in fact contain $(x-1)^{-1}\cdot k[x]$. We may then proceed inductively to conclude that the non-trivial $k[x^2]$-submodules of $k[x,1/(x-1)]$ are precisely of the form $(x-1)^{-n}k[x]$, and it is then straightforward to see that it must be Artinian. In fact, in this case the $k[x^2]$-submodules of $k[x,1/(x-1)]$ are precisely the $k[x]$-submodules. So now I don't know where to look for an example.