flatness and isomorphism Given: domains $A$ and $B$ such that $A\subseteq B$ and such that their fraction fields are equal. Also $A$ is a unique factorization domain and if $a \in A$ is a unit in $B$, it is a unit in $A$. I want to show that if $A \subseteq B$ is flat then $B=A$. I think I need to use the fact that every element in $B$ can be written as $\frac{a_1}{a_2}$ with $a_1,a_2 \in A$. Does anyone have an idea?
The assumption $A \cap B^* = A^*$ implies that $\mathfrak{p}B \neq B$ for all prime ideals $\mathfrak{p} \subseteq A$, i.e. that $B$ is faithfully flat over $A$. Now use SE/482908.
The assumption $A \cap B^* = A^*$ implies that $\mathfrak{p}B \neq B$ for all prime ideals $\mathfrak{p} \subseteq A$, i.e. that $B$ is faithfully flat over $A$. Now use SE/482908.