# Picard groups of abelian étale covers

Let $X$ be a scheme (you can assume that $X$ is proper and smooth over an algebraically closed field) and $T$ is a finite subgroup of $\text{Pic } X$ (of order prime to the characteristic). Does there exist a finite étale cover $f:Y\to X$ (possibly Galois with Galois group $T$) for which the map $f^* :\text{Pic }X\to\text{Pic }Y$ identifies $\text{Pic } Y$ with the quotient $(\text{Pic }X)/T$?

Of course (except for the claim about $f$ being Galois) it suffices to treat the case when $T$ is cyclic (of order $m$ say). Then we take a generator $L$ of $T$ and $$Y = \text{Spec} {}_X \bigoplus_{i=0}^{m-1} L^i$$ (the standard construction showing that $H^1(X, \mu_m)$ classifies such covers). I think I can show that the kernel of $\text{Pic }X\to\text{Pic }Y$ is $T$, but I don't know how to prove that it is surjective...

It also reminds me of the Hilbert class field: I don't know much algebraic number theory, but I think that the construction is similar: $K$ is a number field, $X = \text{Spec }\mathcal{O}_K$, $T = \text{Pic }X$ is the class group, then $Y = \text{Spec }\mathcal{O}_H$ where $H$ is the Hilbert class field of $K$, its maximal abelian unramified extension.

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The answer is "no," even for curves. Say $X$ is a genus 2 curve over $\mathbf{C}$, and $f:Y \to X$ is any finite etale degree $d$ morphism of degree $> 1$. Then $Y$ is a smooth projective curve of genus $d+1 > 2$. In particular, $\mathrm{Pic}(Y)$ has dimension $>2$, so it cannot be a quotient of $\mathrm{Pic}(X)$.

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OOPS! Yes, you're right, thank you! But what if we assume that $X$ and $Y$ have trivial $\text{Pic}^0$, does it help? Is there a good description of the cokernel of $f^*: \text{Pic }X\to \text{Pic }Y$ (or just the next group in some long exact sequence extending $f^*$)? –  Piotr Achinger Aug 20 '12 at 0:52
For finite Galois covers f:Y ---> X with group G, the best tool I know is the Hochschild-Serre spectral sequence for etale cohomology with coefficients in G_m (i.e., H^p(G,H^q(Y,G_m)) ===> H^{p+q}(X,G_m)) and the fact that H^1(X,G_m) = Pic(X). This gives an LES involving the map Pic(X) ---> Pic(Y)^G. It doesn't suggest a good statement when Pic^0 is 0, but does lead to counterexamples to hopes of the form Pic(Y) = Pic(X)/T; we just need to find such Y and G with Pic(Y)^G strictly smaller than Pic(Y). –  anon Aug 20 '12 at 1:42