I came across this problem in Enochs' and Jenda's Relative Homological Algebra.
Suppose $A$ is a subring of $B$. Show that $B$ is faithfully flat as an $A$-module if and only if $B/A$ is a flat $A$-module.
Unfortunately, I haven't been able to come up with a solution to it. I've tried showing that if $\mathfrak{m}$ is a maximal ideal of $A$, then $B/\mathfrak{m}B$ is non-zero, but I can't seem to find how to use the fact that $B/A$ is flat. I can't find an exact sequence which upon tensoring with $B/A$ involves $B/\mathfrak{m}B$. Perhaps I'm unaware of some relation of quotients of modules and tensors?
Suppose $B/A$ is flat. We have a short exact sequence $0\to A \to B\to B/A\to 0$. Tensoring this with any $A$-module $M$, we get a long exact sequence $$\operatorname{Tor}_1(A,M)\to\operatorname{Tor}_1(B,M)\to\operatorname{Tor}_1(B/A,M)\to A\otimes M\to B\otimes M\to B/A\otimes M\to 0.$$
Since $A$ and $B/A$ are flat, the Tor terms involving them are $0$. It follows that $\operatorname{Tor}_1(B,M)=0$, and $A\otimes M\to B\otimes M$ is injective. If $M$ is nonzero, then $A\otimes M\cong M$ is nonzero, so $B\otimes M$ is nonzero. Since $M$ is arbitrary, this implies $B$ is faithfully flat.
For the converse, apply the same long exact sequence, using the fact that if $B$ is a faithfully flat $A$-algebra then $A\otimes M\to B\otimes M$ is injective for all $M$.