— Let f in A[X] be of degree ≤m.
(i) If ρ:A→B is a ring homomorphism, we have dism(ρf)=ρ(dism(f)).
(ii) Let λ,α1,...,ar be elements of A. If f=X(X−α1)...(X−ar); then we have
dism(f)=λ2m−2i<j∏(αi−αj)2.
(iii) Let a0 be the coefficient of Xm in f; then we have
resm,m−1(f,Df)=resm−1,m(Df,f)=(−1)m(m−1)/2a0 dism(f).
Assertion (i) is clear.
Since Δ~ is homogeneous of degree 2m−2, we have
dism(λf)=λ2m−2dism(f)
for every polynomial f∈A[X] of degree ≤m. Assertion (ii) now follows from formulae (46) and (55).
When f is monic of degree m, we have a0=1 and Assertion (iii) is reduced to formula (47). Bearing in mind (55) and the relation
resm,n(λf,μg)=λnμmresm,n(f,g),
(IV, p. 77), we can pass from there to the case where a0 is invertible in A. 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.
Nothing.