Let A,A′ be two K-algebras, M an A-module and M′ an A′-module. Suppose that M and M′ are free K-modules of respective dimensions n and n′ and consider M⊗KM′ as an (A⊗KA′)-module (§4, no. 3, Example 2). Then, for a∈A and a′∈A′,
TrM⊗M′(a⊗a′)=TrM(a)TrM′(a′)
(16)
NM⊗M′(a⊗a′)=(NM(a))n′(NM′(a′))n.
(17)
Formula (16) follows from II, § 4, no. 4, formula (26) and formula (17) from § 8, no. 6, formula (33).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.