Let $A\subseteq B$ be a finite ring extension. Let $M$ be a Noetherian/Artinian $B$-module. Show $M$ is a Noetherian/Artinian $A$-module.
Let $M_0\subset M_1\subset...$ be an ascending chain of $A$-submodules. We have $BM_0\subset BM_1\subset...$, which has to stablize and $BM_i=BM_{i+1}$. But how to conclude $M_i=M_{i+1}$ from here? I think we could use Nakayama, but it is not obvious how to use it. This question is mentioned in this answer but no reference is given. Could anyone give a hint how to use the finiteness condition? Thanks.
For the Noetherian part, let me assemble the argument from the comments by @Aphelli and @metaverse to make it available directly in this answer. All the credit goes to them!
Let $M$ be a Noetherian $B$-module; let us prove by induction on the number of generators $n$ of $M$ that $M$ is also $A$-Noetherian. The claim is trivial if $M=0$, i.e. if $n=0$. Let us assume the statement for modules generated by $n-1$ elements, where $n\geq 1$, and let $m_1,\ldots,m_n$ be generators of $M$. Let $N=\sum_{1\leq i<n}Bm_i$, then we have a SES $$ 0\rightarrow N\rightarrow M\rightarrow M/N\rightarrow 0. $$ By induction hypothesis, $N$ is $A$-noetherian, so we are left to prove that $M/N$ is $A$-Noetherian. As $M/N$ is generated by $m_n+N$, it is isomorphic to $B/I$ for some ideal $I\subseteq B$. As $M$ is $B$-Noetherian, $B/I$ is as well, and hence $B/I$ is a Noetherian ring. Notice that we have an injection of rings $A/I^c\hookrightarrow B/I$ where $I^c=I\cap A$, which is still a finite extension. By the Eakin-Nagata theorem, $A/I^c$ is hence Noetherian. As $B/I$ is finitely generated over $A/I^c$, it is a Noetherian $A/I^c$-module. Hence it is also $A$-Noetherian, so by induction we are done.
Let us now prove the case where $M$ is finitely generated Artinian. By this post, this is equivalent to assuming that $M$ has finite length.
So let $M$ be a finite length $B$-module. Via a filtration of $M$ we can reduce to prove that $B/\mathfrak{m}$ is a finite length $A$-module, where $\mathfrak{m}\subseteq B$ is maximal. Let $\mathfrak{m}^c=\mathfrak{m}\cap A$, then $A/\mathfrak{m}^c\subseteq B/\mathfrak{m}$ is still finite, and $A/\mathfrak{m}^c$ is a domain, while $B/\mathfrak{m}$ is a field. It is elementary to show that then $A/\mathfrak{m}^c$ must be a field to. In particular, $B/\mathfrak{m}$ is a finite dimensional $A/\mathfrak{m}^c$-vector space, and in particular an $A$-module of finite length. Hence $M$ is of finite length over $A$ as well.
I'm not sure though that any Artinian $B$-module is also an Artinian $A$-module, and neither is this claimed in the post you're referencing. But I don't have a counterexample right now; I'll think a bit.