Let $f:R\to S$ be a local finite monomorphism .If $M$ is an Artinian $S$-module, is it an Artinian $R$-module?

Question

$(R,m)$ and $(S,n)$ are local rings (commutative Noetherian with 1).
Let $f:R\to S$ be a local homomorphism/monomorphism ($f(m)\subseteq n$), such that the natural induced homomorphism $R/m\to S/n$ is an isomorphism, and such that $S$ is finite $R$-module via $f$ (So it is integral extension). Let $M$ is an $S$-module (not necessarily finite). So $M$ is also an $R$-module via $f$.

If $M$ is an Artinian $S$-module, is it an Artinian $R$-module?
If $M$ is an Artinian $R$-module, is it an Artinian $S$-module?

The first is true for $M=S$, since in this case S is of finite length.

Thank you.

Answer 1

Since Artinian modules for $S$ and for the completion of $S$ are the same we may assume that $S$ and $R$ are complete.

Let $E_S$ be the injective hull of $S/n$. Then every Artinian $S$-module is a submodule of $E_S^N$ for some $N\ge0$. Thus it suffices to prove that $E_S$ is Artinian as an $R$-module.

Let $I:=f(m)S$. Then the fiber $S/I$ is finite length as an $S$-module, hence a finite dimensional $k:=R/m$-vector space. Now consider the annihilator $A$ of $f(m)$ in $E_S$. It is also the annihilator of $I$ in $E_S$ which is Matlis dual to $S/I$. This implies that $A$ is also a finite dimensional $k$-vector space. Let $d$ be its dimension.

By the universal property of the injective hull, each $R$-homomorphism $k\to E_S$ extends to a homomorphism $E_S\to E_R$. This gives an $R$-homomorphism $\phi:E_S\to E_R^d$.

Claim: $\phi$ is injective. Let $K=\ker\phi$. Then by construction $K\cap A=0$. But every element of $E_S$ is killed by some power of $I$. Hence every element of $K$ is killed by some power of $m$. Thus $K=0$.

Now we are done since $E_R^d$ is clearly $R$-Artinian.

The opposite direction is obvious.