Solution
= Solution
The <trace form of a number field> is
$$
B(x,y)=\operatorname{Tr}_{K/\mathbb Q}(xy).
$$
It is bilinear because the <field trace> is linear. If $x\ne0$, choose $y=x^{-1}$. Then
$$
B(x,y)=\operatorname{Tr}_{K/\mathbb Q}(1)
=[K:\mathbb Q]\ne0.
$$
\b[Thus no nonzero $x$ lies in the radical, and $B$ is a <nondegenerate bilinear form>. This is the <nondegeneracy of the trace form of a number field>.]