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 but got no answer.


2 Answers 2


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 $G^{\operatorname{ab}}$.

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 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 $\mu_r \subseteq k$. $\square$


Let me give an answer when $\mathrm{char} \, k=0$.

In this situation the double cover $f \colon X \to Y$ is defined by a line bundle $\mathcal{L}$ such that $\mathcal{L}^{\otimes 2} = \mathcal{O}_Y$ and the trace map provides a splitting $$f _* \mathcal{O}_X = \mathcal{O}_Y \oplus \mathcal{L}^{-1}.$$ Assume now that $\mathcal{M}$ belongs to the kernel of $$f^* \colon \mathrm{Pic} \, Y \to \mathrm{Pic} \, X.$$ Then $f^* \mathcal{M} = \mathcal{O}_X$ so, by applying the functor $f_*$ and the projection formula, we obtain $$f_* \mathcal{O}_X = f_* f^* \mathcal{M} = \mathcal{M} \otimes f_* \mathcal{O}_X = \mathcal{M} \oplus (\mathcal{M} \otimes \mathcal{L}^{-1}).$$ By the Krull-Schmidt theorem the decomposition of a vector bundle in a direct sum of indecomposable ones is unique up to permutation of the summands, so we get either $\mathcal{M} = \mathcal{O}_Y$ or $\mathcal{M}=\mathcal{L}$.

Summing up, $\ker \, f^*$ is the subgroup of order $2$ generated by $\mathcal{L}$.

Exactly the same argument shows that if the étale cover is cyclic of degree $n$, then $\ker f^*$ is cyclic of the same order.

  • $\begingroup$ Which part of this argument relies on $\mathrm{char}(k)=0$? $\endgroup$ Nov 8, 2021 at 2:47
  • 1
    $\begingroup$ I think that if char$(𝑘)$ divides the degree of the covering then the trace map does not provide in general a splitting of $f_*\mathcal{O}_X$. $\endgroup$ Nov 13, 2021 at 9:34

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