— Let V be a vector space over N, (uσ)σ, r a family of mappings of V into itself satisfying (1) to (4) and V0 the set of x∈V such that uσ(x)=x for all σ∈Γ.
a) V0 is a vector sub-K-space of V and the K-linear mapping φ of N⊗KV0 into V which maps λ⊗x to λx is injective.
b) If Γ is finite, then φ is bijective and V0 is a K-structure on V.
It is clear that V0 is a vector sub-K-space of V.
The formula uσ∘φ=φ∘(σ⊗IdV0) shows that the kernel W of φ is stable under the mappings σ⊗IdV0; by the Cor. to Prop. 6 there exists therefore a subspace W0 of V0 such that W=N⊗KW0. If x belongs to W0 we then have x=φ(1⊗x)=0, hence W0=0 and so W=0. This proves a).
Suppose now that Γ is finite; we have to show that φ is surjective, or equivalently that V0 generates the vector N-space V. Thus let f be an N-linear form on V whose restriction to V0 is zero. Let x∈V; for every λ∈N the element yλ=∑σ∈Γuσ(λx) of V clearly belongs to V0, whence f(yλ)=0, that is,
σ∈Γ∑f(uσ(x))σ(A)=0. By Dedekind's theorem (V, p. 27, Cor. 2) we thus have
f(uσ(x))=0 for each
σ∈Γ; in particular, taking
σ=ε we find
f(x)=0, which means that
f=0. This proves b).
Let M be a vector space over N; for each σ∈Γ let Mσ be the vector space over N with the same underlying additive group as M, with the external law (A,x)↦σ(λ)x. Write V=∏σ∈ΓMσ; the underlying additive group of V is that of all mappings of Γ into M, with the external law defined by
(λ.h)(σ)=σ(λ)h(σ)(λ∈N,h∈V,σ∈Γ). (The product
σ(λ)h(σ) is calculated in the vector space
M.) Further, we define on
N⊗KM a vector space structure over
N by the formula
λ(i∑μi⊗xi)=i∑λμi⊗xi Finally we denote by
ψ the K-linear mapping of
N⊗KM into
V characterized by the relation
ψ(λ⊗x)(σ)=σ(λ).x for
A∈N,x∈M and
σ∈Γ. It is clear that
ψ is N-linear
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.