For every extension K′ of K and every element (ti)i∈I of K′I, we have
(48) PcrdA(K′)/K′(∑i∈Iti⊗ei)=P((ti)i∈I,X).
Let φ:K[T]→K′ be the unique K-algebra homomorphism that sends Ti to ti for every i∈I; it defines on K′ the structure of a K[T]-algebra. The K′-algebra A(K[T])(K′) can be identified with A(K′) (transitivity of extension of scalars), where the element 1⊗(∑Ti⊗ei) is identified with the element ∑ti⊗ei of A(K′). We denote by φ:K[T][X]→K′[X] the K-algebra homomorphism deduced from φ. By formula (21) of III, §9, No. 3, p. 544, the characteristic polynomial of ∑ti⊗ei with respect to the K′-algebra A(K′) is the image by φ of the characteristic polynomial of ∑Ti⊗ei with respect to the K[T]-algebra A(K[T]), that is, of Pn. In other words, we have
(49) PcA(K′)/K′(∑i∈Iti⊗ei;X)=P((ti)i∈I,X)n;
Lemma 5 then follows from Lemma 2 of VIII, p. 339.
Consider the specific case K′=K of Lemma 5. We have
(50) PcrdA/K(∑i∈Itiei;X)=P((ti)i∈I,X)
for every element (ti)i∈I of KI. Since the polynomial P in K[T,X] is homogeneous of degree n, it can be expanded uniquely as
P(T,X)=r=0∑n(−1)rBr(T)Xn−r(51)
where Br is a homogeneous polynomial of degree r in K[T]. By formulas (46), (50), and (51), we have
br(i∈I∑tiei)=Br((ti)i∈I)
for every element (ti)i∈I of KI. Proposition 6 follows.
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.