Bourbaki

Lemma 2

Tag 05BT, permanent. Cite it as [Bourbaki, Tag 05BT].

Algebra, chapter I, § 5. Groups operating on a set

p≠0p \neq 0 and σp=(−1)ν(σ)\sigma p = (-1)^{\nu(\sigma)} for σ∈Sn\sigma \in \mathfrak{S}_n.

p(1,2,…,n)=∏i<j(j−i)≠0p(1, 2, \ldots, n) = \prod_{i < j} (j - i) \neq 0 and hence p≠0p \neq 0. On the other hand, if σ∈Sn\sigma \in \mathfrak{S}_n, then

σp(z1,…,zn)=p(zσ(1),…,zσ(n))=∏i<j(zσ(j)−zσ(i)). \sigma p(z_1, \ldots, z_n) = p(z_{\sigma(1)}, \ldots, z_{\sigma(n)}) = \prod_{i < j} (z_{\sigma(j)} - z_{\sigma(i)}).

Let C be the set of ordered pairs (i,j)(i, j) such that 1≤i≤n, 1≤j≤n, i<j1 \leq i \leq n,\ 1 \leq j \leq n,\ i < j. A permutation θ\theta is defined on C by setting θ(i,j)=(σ(i),σ(j))\theta(i, j) = (\sigma(i), \sigma(j)) if (i,j)(i, j) is not an inversion, θ(i,j)=(σ(j),σ(i))\theta(i, j) = (\sigma(j), \sigma(i)) if (i,j)(i, j) is an inversion. This implies σp=(−1)ν(σ)p\sigma p = (-1)^{\nu(\sigma)} p.

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

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Cites

Nothing.

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