= Solution
Every Brauer class has a <central division algebra> representative $D$, and every <central simple algebra> is a matrix algebra over such a $D$. Put $d=\deg D$; its <reduced norm> is a homogeneous degree-$d$ polynomial on the $d^2$-dimensional <vector space> $D$.
If $d>1$, then $d^2>d$, so the $C_1$ property supplies a nonzero $x\in D$ with $\operatorname{Nrd}(x)=0$. But every nonzero element of a <division algebra> is invertible, and its <reduced norm> cannot be zero. Thus $d=1$, and a <central division algebra> of degree one is $K$ itself. Every <central simple algebra> over $K$ is therefore a matrix algebra over $K$, giving
$$
\boxed{\operatorname{Br}(K)=0}.
$$
This proves the <trivial Brauer group of a C1 field>. The essential ingredient is the anisotropy of a <division algebra>'s <reduced norm>, not the norm polynomial of an arbitrary matrix algebra, which certainly can vanish on nonzero singular matrices.
Back to article page
