Let n∈NI. If ∣n∣<r, then cr+([n])=0. If ∣n∣=r, then
cr+([n])=φ∑xφ(1)⊗xφ(2)⊗⋯⊗xφ(r)(7)
where φ belongs to the set of maps from {1,2,...,r} to I which take each value i∈Ini times.
By Prop. 1,
cr(x(n))=∑x(p1)⊗⋯⊗x(pr)
where the summation extends over the set of sequences (p1,...,pr) of r elements of M such that p1+⋯+pr=n. In view of Chap. II, §1, no. 3, Prop. 6, the map cr+, extended by linearity to a map from U(g) to Tr(U+(g)), vanishes on Ur−1(g). It follows that, for r≥∣n∣,
cr+([n])=cr+(x(n))=∑π(x(p1))⊗⋯⊗π(x(pr))(8)
For r>∣n∣, the relation p1+⋯+pr=n implies that at least one of the pi is zero, so cr+([n])=0. For r=∣n∣, the only non-zero terms of the third member of (8) are those for which ∣p1∣=⋯=∣pr∣=1, hence (7).
We return to the proof of Th. 1. We retain the notations of (*) and prove, by descending induction on ∣n∣, that d divides an, which is clear when ∣n∣ is sufficiently large. If d divides an for ∣n∣>r then, putting
u′=u−∣n∣>r∑(an/d)[n]∈U
we have
du′=∣n∣≤r∑an[n](9)
For any map φ from {1,2,...,r} to I, put
eφ=xφ(1)⊗⋯⊗xφ(r)
and aφ=an where n= (Cardφ−1(i))i∈I. By Lemma 1, (9) implies that
dcr+(u′)=φ∈Ir∑aφeφ(10)
so cr+(u′)∈Tr(U+)∩QTr(G). But the submodule G of U+ is a direct factor (Algebra, Chap. VII, §4, no. 3, Cor. of Th. 1), so the submodule Tr(G) is a direct factor of Tr(U+), and hence cr+(u′)∈Tr(G). On the other hand, the xi form a basis of G by hypothesis, so the eφ form a basis of Tr(G). Then (10) proves that d divides the aφ, that is, the an for ∣n∣=r. This proves (*).
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.
Nothing.