Corollary

Tag 06LE, permanent. Cite it as [Bourbaki, Tag 06LE].

— Put $m = [F : E]$ ; then for every $x \in E$ we have

\text{Tr}_{F/K}(x) = m \cdot \text{Tr}_{E/K}(x) \\ \text{N}_{F/K}(x) = \text{N}_{E/K}(x)^m .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.