Integral closure $\tilde{A}$ is flat over $A$, then $A$ is integrally closed

One can use the following:

1. If $A\subset B$ is an integral extension and $B$ is flat over $A$, then $B$ is faithfully flat over $A$.

2. If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$. (Matsumura, Commutative Ring Theory, Exercise 7.2)

Remark. This argument shows that the condition "$\tilde{A}$ is a finitely generated $A$-module" is superfluous.

Let $A\subseteq B$ be an extension of integral domains such that $B_{\mathfrak p}=x_{\mathfrak p}A_{\mathfrak p}$ for every prime ideal $\mathfrak p$ of $A$ and some element $x_{\mathfrak p}\in B_{\mathfrak p}$. Then $x_{\mathfrak p}$ is invertible in $B_{\mathfrak p}$ and therefore $B_{\mathfrak p}=A_{\mathfrak p}$. One then gets

$B\subseteq\bigcap\limits_{\mathfrak p} B_{\mathfrak p} =\bigcap\limits_{\mathfrak p} A_{\mathfrak p}=A$.