Bourbaki

Proposition 5

Tag 062Z, permanent. Cite it as [Bourbaki, Tag 062Z].

Algebra, chapter III, § 7. Exterior algebras

(i) Let (xi)1≤i≤n(x_i)_{1 \leq i \leq n} be a finite sequence of elements of the module MM; for every permutation σ\sigma in the symmetric group SnS_n,

xσ(1)∧xσ(2)∧⋯∧xσ(n)=εσ⋅x1∧x2∧⋯∧xn x_{\sigma(1)} \wedge x_{\sigma(2)} \wedge \cdots \wedge x_{\sigma(n)} = \varepsilon_\sigma \cdot x_1 \wedge x_2 \wedge \cdots \wedge x_n

where εσ\varepsilon_\sigma denotes the signature of the permutation σ\sigma.

(ii) If there exist two distinct indices i,ji, j such that xi=xjx_i = x_j, the product

x1∧x2∧⋯∧xn x_1 \wedge x_2 \wedge \cdots \wedge x_n

is zero.

(i) First of all, since x∧x=0x \wedge x = 0 for all x∈Mx \in M by definition of the ideal g′\mathfrak{g}', also, for x,yx, y in MM,

x∧y+y∧x=(x+y)∧(x+y)−x∧x−y∧y=0. x \wedge y + y \wedge x = (x + y) \wedge (x + y) - x \wedge x - y \wedge y = 0.

This establishes (6) in the case n=2n = 2. The general case then follows from § 4, no. 6, Lemma 3.

(ii) Under the hypothesis of (ii), there exists a permutation σ∈Sn\sigma \in S_n such that σ(1)=i\sigma(1) = i and σ(2)=j\sigma(2) = j; then the left hand side of (6) is zero for this permutation and hence so is the right hand side.

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.