 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/20h/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

B (x, y) = Tr_{K / Q} (x y) .

It is bilinear because the field trace is linear. If

x \neq = 0

, choose

y = x^{- 1}

. Then

B (x, y) = Tr_{K / Q} (1) = [K : Q] \neq = 0.

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.

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