Let f be a monic polynomial of A[X] of degree m, and denote by E the A-algebra A[X]/(f) and by x the canonical image of X in E. We define the discriminant of f, written dis(f), as the discriminant DE/A(1,x,…,xm−1) of the basis (1,x,…,xm−1) of the A-algebra E.
For every positive integer k we write ρk=TrE/A(xk). By III, p. 549, Def. 2 amounts to the formula
dis(f)=det(pi+j)0≤i,j<m
*Examples. — 1) If f is a monic polynomial of degree 0 or 1, we have dis(f)=1, by (45). 2) Let f(X)=X2+αX+β 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,
whence p1=−a, p2=a2−2p. It follows that we have
dis(f)=det(2−a−αα2−2p)=α2−4β
Let B be a commutative ring, ρ a homomorphism from A to B and ξ1,…,ξm elements of B such that
ρf(X)=(X−ξ1)⋯(X−ξm). Let us write M for the matrix
(ρ(pi+j))0≤i,j<m and D for the Van der Monde matrix
(ξij+1)0≤i,j<m. By Example 4 of IV, p. 75 we have
ρ(pk)=ξ1k+⋯+ξmk, whence
M=′D.D; by III, p. 532 we have
D=∏i>j(ξi−ξj) and
detM=(detD)2, that is
ρ(dis(f))=i<j∏(ξi−ξj)2 Moreover (on writing D for the derivation
dXd) we have
D(ρf)(ξi)=(ξi−ξ1)⋯(ξi−ξi−1)(ξi−ξi+1)⋯(ξi−ξm) for
1≤i≤m, whence
ρ(dis(f))=(−1)m(m−1)/2i=j∏(ξi−ξj)=(−1)m(m−1)/2i=1∏mD(ρf)(ξi). By Cor. 1 of IV, p. 80 applied to the universal decomposition of
f, we finally obtain
res(f,Df)=res(Df,f)=(−1)m(m−1)/2dis(f).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.