Let $X \to Y$ be a cyclic etale cover of smooth projective geometrically connected curves over some field $k$. Then the map is classified by an element of the cohomology group $H^1\_{et}(Y\_{k\_s}, \mu\_n)$; in other words the data of the covering is equivalent to giving a line bundle on $Y$ together with a trivialization of its $n$-th power. It follows that the induced map on Picard schemes given by pullback is not injective. My question is: what can be said about the kernel? I am especially interested in the special case of a degree $2$ cover; i.e are there are any other elements in the kernel besides the trivial bundle and the bundle classified by the covering. This one is given explicitly by pushing forward the structure sheaf on $X$ and modding out by the structure sheaf on $Y$.


I tried asking this on [MSE](https://math.stackexchange.com/q/2200771/39599) but got no answer.



Line bundles and Cyclic Covers of Curves

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> > **Theorem.** Let $X \to Y$ be an étale Galois cover with group $G$ of proper geometrically integral schemes over any field $k$. Then we have an exact sequence
> >  $$0 \to \operatorname{Hom}(G,k^\times) \to \operatorname{Pic}(Y) \to \operatorname{Pic}(X)^G.$$
> >  In particular, the kernel has size at most $n = |G^{\operatorname{ab}}|$, with equality if and only if $n$ is invertible in $k$ and $\mu\_r \subseteq k$, where $r$ is the exponent of $G^{\operatorname{ab}}$.
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*Proof.* An étale Galois cover with group $G$ is the same thing as a $G$-torsor, and we have a Hochschild–Serre spectral sequence (see e.g. Milne's [*Étale cohomology*](http://www.jmilne.org/math/CourseNotes/LEC.pdf) notes, **Thm 14.9**)
$$E\_2^{pq} = H^p(G,H^q(X,\mathbb G\_m)) \Rightarrow H^{p+q}(Y,\mathbb G\_m).$$
The exact sequence of low degree terms is
$$0 \to H^1(G,\mathbb G\_m(X)) \to H^1(Y,\mathbb G\_m) \to H^1(X,\mathbb G\_m)^G \to H^2(G,\mathbb G\_m(X)) \to \ldots .$$
Since $X$ is proper and geometrically integral, the global sections of $\mathcal O\_X$ are just the constants $k$, hence the global sections of $\mathbb G\_m$ are just $k^\times$. This clearly has the trivial $G$-action, so
$$H^1(G,\mathbb G\_m(X)) = \operatorname{Hom}(G,k^\times).$$
This proves the first statement. The second statement follows since
$$\operatorname{Hom}(G,k^\times) = \operatorname{Hom}(G^{\operatorname{ab}},k^\times),$$
and for any finite abelian group $G$ there is a noncanonical isomorphism $G^{(p')} \cong \operatorname{Hom}(G,\bar k^\times)$, where $(-)^{(p')}$ denotes the prime to $p$-part if $\operatorname{char} k = p > 0$ (and the entire group if $\operatorname{char} k = 0$).


Since the group generated by the images of homomorphisms $G \to \bar k^\times$ is the subgroup $\mu\_r$ for $r$ the exponent of $G$, we conclude that they are all realised over $k$ if and only if $\mu\_r \subseteq k$. $\square$


