Bourbaki

Proposition 12

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

Algebra, chapter III, § 7. Exterior algebras

Let M be a projective A-module. For elements x1,x2,…,xnx_1, x_2, \ldots, x_n of M to be linearly dependent, it is necessary and sufficient that there exist λ≠0\lambda \neq 0 in A such that

λx1∧x2∧⋯∧xn=0. \lambda x_1 \wedge x_2 \wedge \cdots \wedge x_n = 0.

The condition is necessary (with no hypothesis on M), for if for example λx1\lambda x_1 (with λ≠0\lambda \neq 0) is a linear combination of x2,…,xnx_2, \ldots, x_n, relation (22) holds (no. 3, Proposition 5). We show that the condition is sufficient by induction on nn; for n=1n = 1, it is a trivial consequence of the definition. Suppose therefore that n>1n > 1 and that condition (22) holds for some λ≠0\lambda \neq 0. If

λx2∧x3∧⋯∧xn=0, \lambda x_2 \wedge x_3 \wedge \cdots \wedge x_n = 0,
then the induction hypothesis implies that x2,…,xnx_2, \ldots, x_n are linearly dependent and hence a fortiori x1,x2,…,xnx_1, x_2, \ldots, x_n are. If λx2∧x3∧⋯∧xn≠0\lambda x_2 \wedge x_3 \wedge \cdots \wedge x_n \neq 0, it follows from no. 8, Corollary 3 to Theorem 1 that there exists an alternating (n−1)(n - 1)-linear form ff such that f(λx2∧x3∧⋯∧xn)=μ≠0f(\lambda x_2 \wedge x_3 \wedge \cdots \wedge x_n) = \mu \neq 0. Since
x1∧(λx2)∧⋯∧xn=0, x_1 \wedge (\lambda x_2) \wedge \cdots \wedge x_n = 0,
it follows from no. 8, Corollary 3 to Theorem 1 that μx1\mu x_1 is a linear combination of λx2,x3,…,xn\lambda x_2, x_3, \ldots, x_n; hence x1,x2,…,xnx_1, x_2, \ldots, x_n are linearly dependent.

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

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Cites

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