Bourbaki

Definition 2

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

Algebra, chapter IV, § 6. Symmetric functions

Let ff be a monic polynomial of A[X]A[X] of degree mm, and denote by EE the AA-algebra A[X]/(f)A[X]/(f) and by xx the canonical image of XX in EE. We define the discriminant of ff, written dis⁡(f)\operatorname{dis}(f), as the discriminant DE/A(1,x,…,xm−1)D_{E/A}(1, x, \ldots, x^{m-1}) of the basis (1,x,…,xm−1)(1, x, \ldots, x^{m-1}) of the AA-algebra EE.

For every positive integer kk we write ρk=Tr⁡E/A(xk)\rho_k = \operatorname{Tr}_{E/A}(x^k). By III, p. 549, Def. 2 amounts to the formula

dis⁡(f)=det⁡(pi+j)0≤i,j<m \operatorname{dis}(f) = \det(p_{i+j})_{0 \leq i,j < m}

*Examples. — 1) If ff is a monic polynomial of degree 0 or 1, we have dis⁡(f)=1\operatorname{dis}(f) = 1, by (45). 2) Let f(X)=X2+αX+βf(X) = X^2 + \alpha X + \beta be a monic polynomial of degree 2. Newton's relations may be written in the form (IV, p. 75)

p0=2p1+α=0p2+αp1+2β=0, p_0 = 2 \\ p_1 + \alpha = 0 \\ p_2 + \alpha p_1 + 2\beta = 0 ,

whence p1=−ap_1 = -a, p2=a2−2pp_2 = a^2 - 2p. It follows that we have

dis⁡(f)=det⁡(2−α−aα2−2p)=α2−4β \operatorname{dis}(f) = \det \begin{pmatrix} 2 & -\alpha \\ -a & \alpha^2 - 2p \end{pmatrix} = \alpha^2 - 4\beta

Let B be a commutative ring, ρ\rho a homomorphism from A to B and ξ1,…,ξm\xi_1, \ldots, \xi_m elements of B such that

ρf(X)=(X−ξ1)⋯(X−ξm). \rho f(X) = (X - \xi_1) \cdots (X - \xi_m).
Let us write M for the matrix (ρ(pi+j))0≤i,j<m(\rho(p_{i + j}))_{0 \leq i, j < m} and D for the Van der Monde matrix (ξij+1)0≤i,j<m(\xi_i^{j+1})_{0 \leq i, j < m}. By Example 4 of IV, p. 75 we have
ρ(pk)=ξ1k+⋯+ξmk, \rho(p_k) = \xi_1^k + \cdots + \xi_m^k,
whence M=′D.D;M = 'D . D ; by III, p. 532 we have D=∏i>j(ξi−ξj)D = \prod_{i > j} (\xi_i - \xi_j) and det⁡M=(det⁡D)2\det M = (\det D)^2, that is
ρ(dis⁡(f))=∏i<j(ξi−ξj)2 \rho(\operatorname{dis}(f)) = \prod_{i < j} (\xi_i - \xi_j)^2
Moreover (on writing D for the derivation ddX\frac{d}{dX}) we have
D(ρf)(ξi)=(ξi−ξ1)⋯(ξi−ξi−1)(ξi−ξi+1)⋯(ξi−ξm) D(\rho f)(\xi_i) = (\xi_i - \xi_1) \cdots (\xi_i - \xi_{i-1})(\xi_i - \xi_{i+1}) \cdots (\xi_i - \xi_m)
for 1≤i≤m1 \leq i \leq m, whence
ρ(dis⁡(f))=(−1)m(m−1)/2∏i≠j(ξi−ξj)=(−1)m(m−1)/2∏i=1mD(ρf)(ξi). \rho(\operatorname{dis}(f)) = (-1)^{m(m-1)/2} \prod_{i \neq j} (\xi_i - \xi_j) = (-1)^{m(m-1)/2} \prod_{i=1}^m D(\rho f)(\xi_i).
By Cor. 1 of IV, p. 80 applied to the universal decomposition of ff, we finally obtain
res⁡(f,Df)=res⁡(Df,f)=(−1)m(m−1)/2dis⁡(f). \operatorname{res}(f, Df) = \operatorname{res}(Df, f) = (-1)^{m(m-1)/2} \operatorname{dis}(f).

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

Cited by

Cites

Transcribed from the printed edition, 2003, Springer. Not translated.