Lemma 11.5.3. The Brauer group of an algebraically closed field is zero.
Proof. Let $k \subset K$ be a finite central skew field extension. For any element $x \in K$ the subring $k[x] \subset K$ is a commutative finite integral $k$-sub algebra, hence a field, see Algebra, Lemma 10.36.19. Since $k$ is algebraically closed we conclude that $k[x] = k$. Since $x$ was arbitrary we conclude $k = K$. $\square$
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