Assume $A \subseteq B=A[b]$ are integral domains, $b \in B$ is algebraic over $A$ (but not necessarily integral over $A$), and $A$ and $B$ have the same field of fractions. (Notice that $b=u/v$ for some $u, v \in A$).
I do not mind to further assume that: $A$ and $B$ are noetherian, $\mathbb{C} \subset A$, $B$ is a UFD (but I do not want to assume that $A$ is a UFD), $B$ is regular (but I do not want to assume that $A$ is regular), $B$ is separable over $A$, and if it helps I agree to assume that $b$ is integral over $A$.
My question: I wish to show that $B$ is flat over $A$. (Notice that if $B$ is flat and integral over $A$, then it is faithfully flat over $A$, and then since $A$ and $B$ have the same field of fractions, it follows that $A=B$).
I have tried to show flatness using the follwong ideas:
(1) Corollary 9 (and Theorem 5) of Wang.
But the problem is that $A$ is not a UFD, so $B$ is not isomorphic to $A[Z]/(vZ-u)$, but to $A[Z]/(h_1,\ldots, h_m)$.
(2) Lemma 1 (with Theorem 1) of Richman.
Any help will be appreciated.
