Let $R \subseteq S$ be two commutative $\mathbb{C}$-algebras such that:

**(1)** $R$ and $S$ are integral domains.

**(2)** $Q(R)=Q(S)$, namely, their fields of fractions are equal.

**(3)** $S=R[w]$, for some $w \in S$.

**(4)** $S$ is separable over $R$, namely, $S$ is a projective $S \otimes_R S$-module via $f: S \otimes_R S
\to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$.

Should such $S$ be flat over $R$? I guess no, so please it would be nice to see a counterexample.

Is there a fifth condition that would guarantee flatness of $R \subseteq S$?

Perhaps adding a fifth condition **(5)** $R$ is a UFD (or at least integrally closed) would guarantee
flatness of $R \subseteq S$? (I am not sure).

The above is (almost) question 3 of this question. Also asked here without comments.

Thank you very much!