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 pullback 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 HochschildSerre 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.

$\begingroup$ We seem to have given the same answer at the same time... nevermind! $\endgroup$ Sep 5, 2014 at 19:41


$\begingroup$ @abx, Loughran: Thank you very much for your answers. $\endgroup$– KaliSep 6, 2014 at 10:12
Yes this is true. It can be proved using the HochschildSerre spectral sequence plus Hilbert's theorem 90 (it is true more generally for any geometrically connected projective variety $X$).
The HochschildSerre 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 nontrivial Picard groups (again see Sansuc).