If $f: A\to B$ is faithfully flat and $B$ is an Artinian ring then $A$ is also Artinian.

Question

Let $f : A → B$ be a homomorphism of rings. The map $f$ is called faithfully flat if $B$ is flat $A$-module ($B$ is $A$-module w.r.t. multiplication defined by $ab := f(a)b$) and if for any $A$-module $M, M ⊗_AB = 0$ implies that $M = 0$.

If $f$ is faithfully flat show that if $B$ is an Artinian ring then $A$ is also Artinian.

I think we need to use the exact sequence $0 \to M' \to M \to M'' \to 0$, $M$ is Artinian iff $M',M''$ is Artinian. But I am unable to construct the exact sequence. Can anyone give me a hint? Thank you in advance!

Answer 1

If $I_1\supset I_2\supset\cdots$ is a descending chain of ideals in $A$, then $I_1B\supset I_2B\supset\cdots$ is a descending chain of ideals in $B$, so it is stationary. But for any ideal $I$ of $A$ we have $IB\cap A=I$ (see here), so the original chain is also stationary.

Answer 2

Let $I_1, I_2...$ be a descending chain of ideals of $A$, denote by $J_n$ the ideal of $B$ generated by $f(I_n)$, since $B$ is artinian, there exists $l$ such that $J_l=J_{l+m}$ for every $m$. We have $I_l/I_{l+m}\otimes_AB=J_l/J_{l+m}=0$ since $f$ is faithfully flat, $I_l=I_{l+m}$.