Bourbaki

Proposition 13

Tag 063L, permanent. Cite it as [Bourbaki, Tag 063L].

Algebra, chapter III, § 7. Exterior algebras

Let MM be an AA-module, NN a direct factor submodule of MM which is free of dimension pp and {u}\{u\} a basis of Λp(N)\Lambda^p(N). For an element x∈Mx \in M to belong to NN, it is necessary and sufficient that u∧x=0u \wedge x = 0.

Let PP be a submodule of MM complementary to NN and let y∈N, z∈Py \in N,\ z \in P be such that x=y+zx = y + z. Then u∧x=u∧zu \wedge x = u \wedge z. As Λp(N)\Lambda^p(N) is free of dimension 1, the mapping ϕ:P→P⊗Λp(N)\phi: P \to P \otimes \Lambda^p(N) such that ϕ(p)=p⊗u\phi(p) = p \otimes u is bijective (II, § 3, no. 4, Proposition 4). On the other hand (no. 7, Proposition 10), the composition of the canonical homomorphisms

ψ:P⊗Λp(N)→Λ(P)⊗Λ(N)→Λ(M) \psi: P \otimes \Lambda^p(N) \to \Lambda(P) \otimes \Lambda(N) \to \Lambda(M)

is injective. The mapping ψ∘ϕ\psi \circ \phi is therefore injective, whence the proposition.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 1998, Springer. Not translated.