Let $R \subseteq S$ be two commutative rings satisfying the following conditions:
(1) $R$ and $S$ are $\mathbb{C}$-algebras.
(2) $R$ and $S$ are integral domains. Denote their fields of fractions by $Q(R)$ and $Q(S)$.
(3) $S=R[w]$, for some $w \in S$, namely, $R \subseteq S$ is simple.
(4) $\dim(R)=\dim(S) < \infty$, where $\dim$ is the Krull dimension.
(5) $R$ and $S$ are local rings, with $m_RS=m_S$, where $m_R$ is the maximal ideal of $R$, $m_S$ is the maximal ideal of $S$, and $m_RS$ is the ideal generated by $m_R$ in $S$, namely, $m_R$ extended to $S$.
(6) $R \subseteq S$ is flat.
(7) $R \subseteq S$ is algebraic, not integral.
Claim: $R=S$.
Question: Could one prove or find a counterexample to the above claim?
Remark: Without condition (7), the following is a counterexample to the claim: $R=\mathbb{C}[x(x-1),y]_{(x(x-1),y)}$, $S=\mathbb{C}[x,y]_{(x,y)}$, $m_R=x(x-1)R+yR$, $m_S=xS+yS$.
Motivation: Theorem 2 in Richman's paper deals with $R \subseteq S$ having the same fields of fractions. If one insists, I could have assumed flatness and other conditions on non-local domains $R \subseteq S$, $m$ maximal in $R$, $n$ maximal in $S$ with $m=R \cap n$, $mR_m S_n=nS_n$, wishing to conclude that $R_m=S_n$.
The following two questions are relevant: a and b. Also asked in MO.
Thank you very much!
