Flat algebra over a Dedekind domain

Question

Let $B$ be a flat algebra over a Dedekind domain $A$. Let $f\in B$ be such that for every maximal ideal $\mathfrak m$ of $A$, the image of $f$ in $B/\mathfrak mB$ is not a zero divisor. How can I show that $B/fB$ is flat over $A$? (Liu, Algebraic Geometry and Arithmetic Curves, Exercise 2.12.)

Answer 1

Since the flatness is a local property one can assume that $A$ is local, so $A$ is a DVR. Let $\mathfrak m$ be the maximal ideal of $A$, and $a\in A$ such that $\mathfrak m=aA$.
For DVRs flatness is equivalent with torsion-free.
Then it's enough to prove that $B/fB$ is a torsion-free $A$-module. (Note that this is equivalent to $a$-torsion-free.) Let $b\in B$ such that $ab\in fB$. We want to show that $b\in fB$.
There is $b'\in B$ such that $ab=fb'$. Then $fb'\in\mathfrak mB$, and since $f$ is a non zerodivisor on $B/\mathfrak mB$ we get $b'\in\mathfrak mB$. But $\mathfrak mB=aB$, and hence $b'=ab''$. Plugging this into $ab=fb'$ we obtain $a(b-fb'')=0$. But $B$ is $A$-flat, so it's torsion-free. It follows that $b=fb''\in fB$ and we are done.

Answer 2

First of all you can suppose that $A$ and $B$ are local ring by with $A$ neotherian, by localization, as $A$ is a Dedekind domain. Now, See Bourbaki, Algèbre Commutative, paragraph 5, number 2, theorem 1, and the equivalence of the first and third assertions of the theorem :

More details here. But as you will see, whatever you do/write, you will write something meaning that some Tor of an $A$-module is zero, which is the so called "local criterion of flatness."