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Let $f: X\rightarrow Y$ be proper etale morphism between varieties over the field of complex numbers. Does there exists a finite group $G$ such that $Y$ is the categorical quotient of $X$ under the free action of $G$? Or in other words is every etale cover a principal bundle?

We can consider the group scheme $Aut_Y(X)$ on $Y$, whose $T$ valued points are $Aut_{T}(X\times_Y T)$ i.e., the group of automorphisms of $X\times_Y T$ which commute with the projection to $T$. It is clear the $X$ is the categorical quotient of $X$ by $Aut_Y(X)$. But it is not clear why $Aut_Y(X)$ should be the group $Y\times G$, for some finite group $G$.

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    $\begingroup$ Not in general: the function field of $X$ might not be a Galois extension of the function field of $Y$. But there always exists a cover of the type you want which dominates $X$, i.e. a $G$-torsor $Z\to Y$ and a subgroup $H\subseteq G$ such that $X=Z/H$. The problem is that $H$ might not be normal in $G$. The question is probably better for math.SE. $\endgroup$ – Piotr Achinger Jan 21 at 9:36
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Not every etale covering is a principal bundle under a group $G$. This is easiest to see using the Galois correspondence for etale coverings: the category of finite etale coverings of $X$ is equivalent to the category of finite continuous $\pi_1^{et}(X,x)$-sets, for a chosen basepoint $x\in X$. Passed across this formalism, you are asking whether for every finite continuous $\pi_1^{et}(X,x)$-set $S$ (corresponding to the covering $Y\to X$), there is a free action of a finite group $G$ on $S$, commuting with the $\pi_1^{et}(X,x)$-action, whose quotient is the single point (corresponding to the trivial covering $X\to X$). In other words, the action of $G$ on $S$ should be freely transitive.

But in general there are plenty of examples where there is no $\pi_1^{et}(X,x)$-equivariant freely transitive action of a finite group $G$ on $S$. For example, suppose that $H\leq\pi_1^{et}(X,x)$ is a non-normal open subgroup. Then the coset space $S:=\pi_1^{et}(X,x)/H$ with the left-regular action of $\pi_1^{et}(X,x)$ provides a counterexample. Indeed, since $H$ is non-normal, there is an element $g\in\pi_1^{et}(X,x)$ such that $gHg^{-1}\neq H$. But the stabiliser of the identity coset $H\in S$ is $H$, whereas the stabiliser of the coset $gH\in S$ is $gHg^{-1}$. Hence there is no $\pi_1^{et}(X,x)$-equivariant permutation of $S$ taking $H$ to $gH$. So there's no chance of a $\pi_1^{et}(X,x)$-equivariant freely transitive action on $S$.

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