Why this map is birational?

Mathematics Asked on January 7, 2022

Let $Y$ be a connected normal Noetherian scheme, $f: Xto Y$ is an etale morphism of finite type. We assume $X$ is also connected. I have proved that in this case, $X$ is also normal. Denote the functional field of $Y$ and $X$ by $K(Y)$ and $K(X)$. Then $K(X)$ is a finite separable extension of $K(Y)$. Let $widetilde{Y}$ be the normalization of $Y$ in $K(X)$. Since $X$ is normal and f is dominant, by universal property of normalization, we could get a map $g:Xto widetilde{Y}$. My question is

Why g is birational?

In general, $g$ is not birational. So I guess, in this case, $g$ is birational because $f$ is etale. But I do not know how to prove it. Could you tell me how to prove it or where I could find a proof?

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