Let $K$ be a field of characteristic $0$, $X_K$ a smooth projective curve over $K$. Denote by $\bar{K}$ the algebraic closure of $K$. The base change morphism $X_{\bar{K}} \to X_K$, induces via the pull-back map, a morphism of the Picard groups, $\mbox{Pic}(X_K) \to \mbox{Pic}(X_{\bar{K}})$. Is this map injective? If not true in general, is there a known condition on $K$ under which this holds true?
2 Answers
This map is injective. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\mathrm{Gal}(\bar{K}/K), H^q(X_{\bar{K}},\mathbb{G}_m))$ converging towards $H^{p+q}(X_{K},\mathbb{G}_m)$. This gives a first terms exact sequence $$0\rightarrow H^1(\mathrm{Gal}(\bar{K}/K), \bar{K}^*)\rightarrow \mathrm{Pic}(X_K)\rightarrow \mathrm{Pic}(X_{\bar{K}})\ ,$$but $\ H^1(\mathrm{Gal}(\bar{K}/K), \bar{K}^*)$ is zero by Hilbert theorem 90.
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$\begingroup$ We seem to have given the same answer at the same time... nevermind! $\endgroup$ Commented Sep 5, 2014 at 19:41
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$\begingroup$ @abx, Loughran: Thank you very much for your answers. $\endgroup$– KaliCommented Sep 6, 2014 at 10:12
Yes this is true. It can be proved using the Hochschild-Serre spectral sequence plus Hilbert's theorem 90 (it is true more generally for any geometrically connected projective variety $X$).
The Hochschild-Serre spectral sequence yields the exact sequence:
$$0 \to H^1(K, \bar{K}[X_\bar{K}]^*) \to \mbox{Pic} X \to \mbox{Pic} X_\bar{K}.$$
See for instance Lemma 6.3 of
- Sansuc - Groupe de Brauer et arithmétiques des groupes algébriques linéaires sur un corps de nombres
where this sequence is extended further to the right.
As $X$ is projective we have $\bar{K}[X_\bar{K}]^* = \bar{K}^*$, hence the first cohomology group vanishes by Hilbert's Theorem 90, thus proving the required injectivity.
Note that the result is false in the affine case in general; for example over $\mathbb{Q}$ there exist algebraic tori with non-trivial Picard groups (again see Sansuc).