# An example where $Pic(X) = H^0(k,Pic(\overline{X}))$?

Let $X$ be a geometrically integral smooth projective variety over a number field $k$. Then if $X$ is everywhere locally soluble, we have $Pic(X) = H^0(k,Pic (\overline{X}))$, where $\overline{X}=X \times_k \overline{k}$. This can be proved using the Hochschild-Serre spectral sequence and the fundamental exact sequence of class field theory.

My question is about the converse: If the canonical injection $Pic(X) \hookrightarrow H^0(k,Pic (\overline{X}))$ is an isomorphism, then is $X$ everywhere locally soluble?

An example of a class of varieties for which this holds is for Brauer-Severi varieties, however my intuition tells me that this is not the case in general. Does anybody have a good counterexample?

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Note that the title of your question asks for an example of nonequality but the body asks for an example of equality. I found this momentarily confusing when I first read your question. –  Pete L. Clark Dec 10 '10 at 17:43
woops yes of course thanks for pointing this out... –  Daniel Loughran Dec 10 '10 at 22:17

No, it is possible for $X$ not to have points everywhere locally and still have every $k$-rational divisor class be represented by a $k$-rational divisor (this is the "Picard equality" you want).

For instance, the Picard equality holds if $X$ has points everywhere locally except at a single place. Many examples of genus one curves satisfying this property -- over any number field $k$ -- are given in this paper of mine.

There should be no lack of other examples as well. In some sense the simplest family is as follows: it is enough to find hyperelliptic quartic curves $C: y^2 = P_4(x)$ which do not have points everywhere locally and for which the Jacobian elliptic curve $E$ has $E(k)$ finite of odd order. The point here is that then the "elliptic Kummer sequence"

$0 \rightarrow E(k)/2E(k) \rightarrow H^1(k,E[2]) \rightarrow H^1(k,E)[2] \rightarrow 0$

gives an isomorphism $H^1(k,E[2]) \rightarrow H^1(k,E)[2]$. This means that for each torsor $C \in H^1(k,E)[2]$ there is a unique rational divisor class of degree $2$, so $C$ comes from a hyperelliptic quartic curve iff this divisor class is represented by a rational divisor.

Actually, a simpler argument can be given if we assume a little more, that $E(k) = \{0\}$. (By a recent theorem of Mazur and Rubin, such curves exist over every number field $k$.) Then the degree zero part of $H^0(\operatorname{Pic} C) = E(k) = 0$, so the equality of $H^0(\operatorname{Pic} C) = \operatorname{Pic}(C)$ for torsors is equivalent to having period equals index, so if you can find a hyperelliptic quartic curve $C$ with Jacobian having trivial Mordell-Weil group and failing to have points over more than one completion, this will give an example of what you want.

In practice, it should be easy to write down such curves $C$ over $\mathbb{Q}$, say.

Note that the situation is different for genus one curves over a $p$-adic field $k$: by a theorem of Roquette-Lichtenbaum, the order of $C$ in $H^1(k,\operatorname{Jac} C)$ is equal to the order of $H^0(\operatorname{Pic} C)/ \operatorname{Pic}(C)$.

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Thanks for the useful answer! –  Daniel Loughran Dec 10 '10 at 22:30