Let $A \subset B$ be a faithfully flat extension of integral domains. If $B$ is integrally closed then I have to show that $A$ is also integrally closed.
Assuming $L,K$ be the field of fractions of the domains $A,B$, respectively, and take $\tilde{A}$ to be the integral closure of $A$ in $K$. Since $B$ is integrally closed and $A \subset B$ we've $\tilde{A} \subset B$.So we finally have a tower of domains as $A \subset \tilde{A} \subset B$ with $A \subset B$ faithfully flat and $A \subset \tilde{A}$ integral. If using these we can show that $A \subset \tilde{A}$ is a flat extension then we are done. But I can't show that. I need some help to complete it. Thanks.
Best Answer
Let $x \in A^\sim$. Then since $x$ in $K$, $x = a/b$ for some $a,b$ in $A$. Consider the ideal $I = (bA:_A a)$. If $I = A$, then we are done.
Since $A \subset B$ is flat, $IB = (bB :_B a)$, and the RHS is $B$ as $a/b \in B$. Furthermore, since $A \subset B$ is faithfully flat, this implies that $I = IB \cap A = B \cap A = A$.