Bourbaki

Proposition 1

Tag 00QP, permanent. Cite it as [Bourbaki, Tag 00QP].

Algebra, VIII, p. 464, Appendix 4. Trace of an Endomorphism of Finite Rank

Let E be a free A-module. Let (ei)i∈I(e_i)_{i\in I} be a basis of E, and let (ei∗)i∈I(e^*_i)_{i\in I} be its dual basis. Let u∈u\in EndAf(E)^f_A(E). The family (⟨u(ei),ei∗⟩)i∈I(\langle u(e_i), e^*_i\rangle )_{i\in I} has finite support, and its sum is equal to Tr(uu).

It suffices to treat the case when uu is of the form θE(x∗⊗x)\theta_E(x^*\otimes x) with x∈Ex\in E and x∗∈E∗x^*\in E^*. The family (⟨x,ei∗⟩)i∈I(\langle x, e^*_i\rangle )_{i\in I} then has finite support, and we have x=∑i∈I⟨x,ei∗⟩eix=\sum_{i\in I}\langle x, e^*_i\rangle e_i. Consequently, the family (⟨x,ei∗⟩⟨ei,x∗⟩)i∈I(\langle x, e^*_i\rangle \langle e_i, x^*\rangle )_{i\in I} also has finite support, and we have ⟨x,x∗⟩=∑i∈I⟨x,ei∗⟩⟨ei,x∗⟩\langle x, x^*\rangle =\sum_{i\in I}\langle x, e^*_i\rangle \langle e_i, x^*\rangle. Now, we have ⟨u(ei),ei∗⟩=⟨x,ei∗⟩⟨ei,x∗⟩\langle u(e_i), e^*_i\rangle =\langle x, e^*_i\rangle \langle e_i, x^*\rangle for every i∈Ii\in I. This proves the proposition.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

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Cites

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.