I am struggling at finding a proof for some fact I saw in a paper. I call a variety any integral separated scheme of finite type over a field $k$ of char 0, and a cover of varieties is simply a finite surjective morphism (not necessarily etale). For a cover of normal varieties $\pi: Y\to X$, I know $Aut_X(Y)$ is isomorphic to $Aut_{k(X)}(k(Y))^{op}$ and then we call $\pi$ Galois if $\#Aut_X(Y)=deg(\pi)$ or equivalently if $k(Y)/k(X)$ is Galois. This is the same definition I found in Milne's notes on Etale cohomology for "generically Galois".
The paper now tells that for a Galois cover between normal varieties $\pi:Y\to X$, we have a bijection between subgroups of $Aut_{k(X)}(k(Y))$ and subcovers $Y\to Z \to X$ give by $$H\mapsto [Y\to Y/H\to X] $$ where $Y/H$ is the normalization of $X$ in $k(Y)^H$.
However there is no clear proof of this fact, the paper only mentioning that it is true by Galois theory on fields and a version of Zariski's main theorem for normal varieties si I tried to show it myself.
From a subcover $Y\to Z \to X$, I denote $H$ the subgroup of $Aut_{k(X)}(k(Y))$ such that $k(Z)=k(Y)^H$. Then I basically want to show $Z$ and $Y/H$ are isomorphic. I know they are birational since they have the same function field so let $f:Z\to Y/H$ be a birational map.
I assume the version of Zariski's main theorem the paper is considering is similar to this statement in Liu's book. 
I don't understand neither why $f$ should be quasi-finite nor why having $f$ being an open immersion concludes $f$ is an isomorphism.
Another attempt of myself is taking $U$ an affine open subset of $X$ and then checking $V1,V2,W$ its inverse images in $Z,Y/H$ and $Y$.
That should give me extensions of rings 
and then by normality, I assume $B_1$ and $B_2$ should both be the integral closure of $A$ in $k(Y)^H=k(Z)\subset k(Y)$ so I could conclude $Z=Y/H$. The thing is I do not know how to generalize to non-affine schemes and I don't think Zariski's main theorem is involved in this way.
Thanks for your attention, any hint accepted!
OK so I finally made it through. Actually given a subcover $Y\to Z\to X$ with $Z$ a normal variety, there is a version of the universal property of normalization for relative normalization (found in EGA II 6.3.9) giving a unique factorization $$Z\xrightarrow{f} Y/H \xrightarrow{g} X$$
Now by unicity in the universal property, the composition $Y\to Z\to Y/H$ must be equal to the map $Y\to Y/H$ which is finite and surjective. Hence $f$ is surjective.
Then the morphism $Z\to X$ is finite so by https://stacks.math.columbia.edu/tag/035D $f$ is finite. Finally $f$ is birational and finite so by the version of Zariski's main theorem I talked about, $f$ is an open immersion and it is surjective so $f$ is an isomorphism.