Bourbaki

Proposition 11

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

Algebra, chapter IV, § 6. Symmetric functions

— Let f in A[X]A[X] be of degree ≤m\leq m.

(i) If ρ:A→B\rho : A \to B is a ring homomorphism, we have dis⁡m(ρf)=ρ(dis⁡m(f))\operatorname{dis}_m(\rho_f) = \rho(\operatorname{dis}_m(f)).

(ii) Let λ,α1,...,ar\lambda, \alpha_1, ..., a_r be elements of A. If f=X(X−α1)...(X−ar)f = X(X-\alpha_1) ... (X-a_r); then we have

dis⁡m(f)=λ2m−2∏i<j(αi−αj)2. \operatorname{dis}_m(f) = \lambda^{2m-2} \prod_{i < j} (\alpha_i - \alpha_j)^2 .

(iii) Let a0a_0 be the coefficient of XmX^m in ff; then we have

resm,m−1(f,Df)=resm−1,m(Df,f)=(−1)m(m−1)/2a0 dism(f). \text{res}_{m,m-1}(f, Df) = \text{res}_{m-1,m}(Df, f) = (-1)^{m(m-1)/2} a_0 \text{ dis}_m(f) .

Assertion (i) is clear.

Since Δ~\tilde{\Delta} is homogeneous of degree 2m−22m - 2, we have

dism(λf)=λ2m−2dism(f) \text{dis}_m(\lambda f) = \lambda^{2m-2} \text{dis}_m(f)

for every polynomial f∈A[X]f \in A[X] of degree ≤m\leq m. Assertion (ii) now follows from formulae (46) and (55).

When ff is monic of degree mm, we have a0=1a_0 = 1 and Assertion (iii) is reduced to formula (47). Bearing in mind (55) and the relation

resm,n(λf,μg)=λnμmresm,n(f,g), \text{res}_{m,n}(\lambda f, \mu g) = \lambda^n \mu^m \text{res}_{m,n}(f, g),

(IV, p. 77), we can pass from there to the case where a0a_0 is invertible in AA. Now the general case follows by Prop. 11, (i) and Lemma 5 (IV, p. 79).

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.