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Is there a variety $X$ over $\mathbb{Q}$ and a line bundle $L$ over $X$ (other than the trivial line bundle $\mathcal{O}_X$ ) such that $L_v$ is the trivial line bundle over $X_v=X\times_{\mathbb{Q}}\mathbb{Q}_v$ for every place $v$ of $\mathbb{Q}$ ?

(Answer known. There is a pun on "locally trivial" in the title.)

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  • $\begingroup$ Forgive me for asking, but if the answer is known, could you show it to us? $\endgroup$
    – Hailong Dao
    Commented Jan 2, 2010 at 16:18
  • $\begingroup$ I thought people would like to think about it. $\endgroup$
    – Chandan Singh Dalawat
    Commented Jan 3, 2010 at 3:06
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    $\begingroup$ The moral of this one seems to me "don't let on that you know the answer" :-/ Maybe it's time you answered your own question? $\endgroup$
    – Kevin Buzzard
    Commented Jan 6, 2010 at 16:00
  • $\begingroup$ Sorry for having kept everyone waiting ! I had to be away three days... $\endgroup$
    – Chandan Singh Dalawat
    Commented Jan 7, 2010 at 9:32
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    $\begingroup$ Others may disagree, but I think it's against the spirit of things here to ask a question to which you already know the answer (however nice the question is). It might seem like an abuse of people's willingness to help. $\endgroup$
    – Tom Leinster
    Commented Jan 7, 2010 at 14:11

1 Answer 1

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The following example was provided to me by Colliot-Thélène some years ago : Let $X$ be the complement in $\mathbb{P}_{1,\mathbb{Q}}$ of the three closed points defined by $x^2=13$, $x^2=17$, $x^2=221$. Then $\operatorname{Pic}(X)=\mathbb{Z}/2\mathbb{Z}$ but $\operatorname{Pic}(X_v)=0$ for every place $v$ of $\mathbb{Q}$.

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