Remark

Tag 0663, permanent. Cite it as [Bourbaki, Tag 0663].

Algebra, chapter III, § 9. Norms and traces

Suppose that the homomorphism $\lambda \mapsto \lambda . 1$ of K into A is injective and let K be identified with its image in A; suppose that A admits a finite basis $(e_i)_{1 \leq i \leq n}$ as a K-module. Let s be an automorphism of A such that $s(K) = K$ . Let a be an element of A; then, by transporting the structure

\begin{align*} \operatorname{Tr}_{A/K}(s(a)) &= s(\operatorname{Tr}_{A/K}(a)) \\ N_{A/K}(s(a)) &= s(N_{A/K}(a)). \end{align*}

(27) (28) Consider also a derivation D of A ( $§ 10$ , no. 2) such that $D(K) \subset K$ and write $D(e_i) = \sum_{j=1}^n e_j \mu_{ji}$ where $\mu_{ji} \in K$ ; write

ae_i = \sum_{j=1}^n e_j \lambda_{ji} \quad \text{with} \quad \lambda_{ji} \in K.

Then

D(a)e_i + aD(e_i) = D(ae_i) = \sum_{j=1}^n (D(e_j)\lambda_{ji} + e_j D(\lambda_{ji})).

It follows that

D(a)e_i = \sum_{j=1}^n e_j v_{ji}

with

v_{ji} = D(\lambda_{ji}) + \sum_{k=1}^n (\mu_{jk}\lambda_{ki} - \lambda_{jk}\mu_{ki})

. As

\sum_{i,k} (\mu_{ik}\lambda_{ki} - \lambda_{ik}\mu_{ki}) = 0

, therefore

\operatorname{Tr}_{A/\mathbf{K}}(D(a)) = \sum_{i=1}^n D(\lambda_{ii})

, in other words

\operatorname{Tr}_{A/\mathbf{K}}(D(a)) = D(\operatorname{Tr}_{A/\mathbf{K}}(a)). *

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

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Cites

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