Bourbaki

Proposition 4

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

Algebra, V, p. 51, § 8. Norms and traces

Let p be the characteristic exponent of K and E an extension of finite degree of K. Let σ1,…,σn\sigma_1, \ldots, \sigma_n be the distinct K-homomorphisms of E into an algebraic closure Ω\Omega of K, and let pe=[E:K]ip^e = [E:K]_i. For each x∈Ex \in E we have in Ω\Omega

TrE/K(x)=pe⋅∑i=1nσi(x),NE/K(x)=(∏i=1nσi(x))pe. \text{Tr}_{E/K}(x) = p^e \cdot \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \left( \prod_{i=1}^n \sigma_i(x) \right)^{p^e} .

Let E, be the relative separable closure of K in E; then EsE_s is a separable extension of degree nn of K and σ1,…,σn\sigma_1, \ldots, \sigma_n induce distinct K-homomorphisms of E, into R; further, E is ap-radical extension of E, of degreepe^e (VV, p. 44, Prop. 13 and p. 46). So Prop. 4 follows from the Formulae (6), (7), (13), (14) and (12).

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

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Cites

Nothing.

Transcribed from the printed edition, 2003, Springer. Not translated.