Lemma 59.67.12. Let k be an algebraically closed field and K/k a field extension of transcendence degree 1. Then for all q \geq 1, H_{\acute{e}tale}^ q(\mathop{\mathrm{Spec}}(K), \mathbf{G}_ m) = 0.
Proof. Recall that H_{\acute{e}tale}^ q(\mathop{\mathrm{Spec}}(K), \mathbf{G}_ m) = H^ q(\text{Gal}(K^{sep}/K), (K^{sep})^*) by Lemma 59.59.2. Thus by Proposition 59.67.4 it suffices to show that if K'/K is a finite field extension, then \text{Br}(K') = 0. Now observe that K' = \mathop{\mathrm{colim}}\nolimits K'', where K'' runs over the finitely generated subextensions of k contained in K' of transcendence degree 1. Note that \text{Br}(K') = \mathop{\mathrm{colim}}\nolimits \text{Br}(K'') which reduces us to a finitely generated field extension K''/k of transcendence degree 1. Such a field is the function field of a curve over k, hence has trivial Brauer group by Lemma 59.67.11. \square
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