Bourbaki

Proposition 6

Tag 0660, permanent. Cite it as [Bourbaki, Tag 0660].

Algebra, chapter III, § 9. Norms and traces

Let AA be a commutative KK-algebra admitting a finite basis over KK and VV an AA-module admitting a finite basis over AA. Then VV admits a finite basis over KK and for every AA-endomorphism uu of VV, if uKu_K is the mapping uu considered as a KK-endomorphism of VV,

Tr⁡(uK)=Tr⁡A/K(Tr⁡(u)),det⁡(uK)=NA/K(det⁡(u))Pc⁡(uK;X)=NA[X]/K[X](Pc⁡(u;X)). \begin{align*} \operatorname{Tr}(u_K) &= \operatorname{Tr}_{A/K}(\operatorname{Tr}(u)), \qquad \det(u_K) = N_{A/K}(\det(u)) \\ \operatorname{Pc}(u_K; X) &= N_{A[X]/K[X]}(\operatorname{Pc}(u; X)). \end{align*}
(24)

Let (ai)1≤i≤m(a_i)_{1 \leq i \leq m} be a basis of AA over KK and (ej)1≤j≤n(e_j)_{1 \leq j \leq n} a basis of VV over AA; then (aiej)(a_i e_j) is a basis of VV over KK (II, § 1, no. 13, Proposition 25). On the other hand the third of formulae (24) can be deduced from the second applied to the endomorphism X−uˉX - \bar{u} of the A[X]A[X]-module A[X]⊗AVA[X] \otimes_A V (§ 8, no. 10). It will therefore suffice to show the first two formulae in (24). We shall first establish the following lemma:

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

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Transcribed from the printed edition, 1998, Springer. Not translated.