Doubtful solution to an exercise on faithful flatness in Matsumura's Commutative Algebra

Question

The exercise on page 30. It says that:

Let $A, B$ be integral domain having the same field of fractions, $B supseteq A$. Prove that $B$ is faithfully flat over $A$ only when $B = A$.

My solution seems to be easy, which makes me doubt its correctness. My idea is to prove that for any $x in B$, $x in A$ so that $B subseteq A$. By assumption that $A, B$ have the same field of fraction, any element of $B$ is of the form $a/b$ for $a, b in A$, $b not= 0$. By faithful flatness, the sequence of $A$-modules
$$0 rightarrow A rightarrow A[X]/(bX - a) rightarrow 0$$
is exact if and only if the sequence of $B$-modules
$$0 rightarrow underbrace{A otimes_A B}_{B} rightarrow A[X]/(bX - a) otimes_A B rightarrow 0$$
is exact. We have
begin{align*}
A[X]/(bX - a) otimes_A B &cong A[X]/(bX - a) otimes_{A[X]} A[X] otimes_A B\
&cong A[X]/(bX - a) otimes_{A[X]} B[X]\
&cong B[X]/(bX - a)\
&cong B &text{ because }a/b in B
end{align*}
So the second sequence is exact: $0 rightarrow B rightarrow B rightarrow 0$. This implies the first sequence is exact and that can only occurs when $a/b in A$.

Badam Baplan · Answer

There is fatal flaw towards the end of your proof when you say $B[X]/(bX - a) cong B$ because $a/b in B$.
For example, let $A$ be a domain and $a in A$.  Then $a^2 / a = a in A$ but $A[X]/(aX - a^2)$ is not even a domain, so clearly is not isomorphic to $A$.
A correct solution might look like:
Let $A subseteq B$ and $a/b in B$.  Then $(b :_A a) = (b :_A a)B cap A = (bB :_A aB) cap A = B cap A = A$.  The first equality uses that $I = IB cap A$ for all $A$-ideals under faithfully flat extension.  The second equality is a fundamental property of colon ideals under flat extensions.  So $a = bc$ for some $c in A$, i.e. $a/b in A$.

Doubtful solution to an exercise on faithful flatness in Matsumura's Commutative Algebra

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