 Flat normalization is an isomorphism

If the normalization of an integral scheme is flat, then it is an isomorphism. On an affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ be the integral closure. Integrality gives surjectivity on spectra, so flatness makes $B$ faithfully flat. For $b=u/v\in B$ with $u,v\in A$ and $v\ne 0$, faithful flatness gives $vB\cap A=vA$. Thus $u=va$ for some $a\in A$, and $b=a$. Consequently $B=A$. This proof does not need finiteness of the normalization.