TITLE: étale covers and torsion line bundles
QUESTION [10 upvotes]: Let $n \geq 2$ be an integer, $X$ a smooth variety over a field $k$ containing $\mu_n$ and $G$ a cyclic group of order $n$ acting on it. Assume that the action is free. Then the morphism $\pi: X \to Y=X/G$ is etale. I'm trying to understand why $\pi_\ast \mathcal{O}_X$ decomposes as 
$$
\pi_\ast \mathcal{O}_X=\mathcal{O}_Y \oplus L \oplus L^2 \oplus \cdots \oplus L^{n-1} 
$$ 
where $n$ is the order of $G$ and $L$ is a line bundle on $Y$ such that $L^n \simeq \mathcal{O}_X$. 
Can anybody help me please?

REPLY [8 votes]: By construction the sheaf $\pi_* \mathcal{O}_X$ is a $G$-equivariant vector bundle of rank $n$ on $Y$ and, since $G$ is a cyclic group, the representation of $\pi_* \mathcal{O}_X$ as a $G$-module splits into direct summands which are all line bundles. 
Now take any isomorphism $G \cong \mathbb{Z}/ n \mathbb{Z}$, and call $L$ the eigensheaf of $\pi_* \mathcal{O}_X$ corresponding to the generator $\bar{1} \in G$. Clearly $L$ is a $n$-torsion line bundle on $X$ and moreover for any $k \in \mathbb Z$ the eigensheaf corresponding to $\bar{k} \in G$ is precisely $L^k$. 
So you obtain the desired splitting.