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Proof. Let $D$ be a finite dimensional division algebra over $K$ with center $K$. We have seen that

\[ D \otimes _ K K^{sep} \cong \text{Mat}_ d(K^{sep}) \]

uniquely up to inner isomorphism. Hence the determinant $\det : \text{Mat}_ d(K^{sep}) \to K^{sep}$ is Galois invariant and descends to a homogeneous degree $d$ map

\[ \det = N_\text {red} : D \longrightarrow K \]

called the reduced norm. Since $K$ is $C_1$, if $d > 1$, then there exists a nonzero $x \in D$ with $N_\text {red}(x) = 0$. This clearly implies that $x$ is not invertible, which is a contradiction. Hence $\text{Br}(K) = 0$. $\square$

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