Bourbaki

Corollary 1

Tag 06FC, permanent. Cite it as [Bourbaki, Tag 06FC].

Algebra, chapter IV, § 6. Symmetric functions

— (i) Let λ,α1,…,αp\lambda, \alpha_1, \ldots, \alpha_p be elements of A and suppose that f(X)=λ(X−α1)…(X−αp)f(X) = \lambda (X - \alpha_1) \ldots (X - \alpha_p). We have

res⁡p,q(f,g)=λqg(α1)…g(αp). \operatorname{res}_{p,q}(f, g) = \lambda^q g(\alpha_1) \ldots g(\alpha_p) .
(ii) Let μ,β1,…,βq\mu, \beta_1, \ldots, \beta_q be elements of A and suppose further that g(X)=μ(X−β1)…(X−βq)g(X) = \mu (X - \beta_1) \ldots (X - \beta_q). Then we have
res⁡p,q(f,g)=λqμp∏1≤i≤p1≤j≤q(αi−βj). \operatorname{res}_{p,q}(f, g) = \lambda^q \mu^p \prod_{\substack{1 \leq i \leq p \\ 1 \leq j \leq q}} (\alpha_i - \beta_j) .
Assertion (i) follows directly from the formulae (28), (32) and (36). Now Assertion (ii) follows from (i).

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, 2003, Springer. Not translated.