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. $\square$

There are also:

• 3 comment(s) on Section 59.67: Galois cohomology

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).