THE DISCRIMINANT OF A COMPOSITION OF TWO POLYNOMIALS
JOHN CULLINAN
I needed this Lemma in two of my papers: once to compute the discriminant of an iterated
rational function, and another to extract the discriminant of f (x) from that of f (g(x)) (where I
already knew the formula for f (g(x))). It never actually made it into the final versions of the
papers, so I’m posting it here. See any source on discriminants and resultants for the necessary
background.
Lemma. Let f (x) and g(x) be polynomials of degrees and δ, respectively, with non-zero discriminant and no common roots. Then the discriminant of the composition f (g(x)) is given by
disc f (g(x)) = (−1)δ(3δ−2δ−1)/2 `(f (x))δ−1 `(g(x))(δ−δ−1) (disc f (x))δ Res(f (g(x)), g 0 (x)),
where ` denotes the leading term of the polynomial.
Proof. The definition of discriminant in terms of the resultant is gives us
δ
disc f (g(x)) = (−1)( 2 ) `(f (g(x))−1 Res(f (g(x)), f 0 (g(x))g 0 (x)).
Applying the identity Res(P, QR) = Res(P, Q) Res(P, R), we focus on the resultant Res(f (g(x)), f 0 (g(x))):
Y
Res(f (g(x)), f 0 (g(x))) = (−1)δ((−1)δ) `(f 0 (g(x)))δ
f (g(θ))
{θ : f 0 (g(θ))=0}
δ

= (−1)δ((−1)δ) `(f 0 (g(x)))δ 
{ρ :
δ((−1)δ)
= (−1)
Y
f (ρ)
f 0 (ρ)=0}
δ
`(f (g(x))) (`(f (x)))− Res(f (x), f 0 (x))
0
δ
= (−1)δ((−1)δ) `(f 0 (g(x)))δ (`(f (x)))−δ `(f (x))δ (disc f (x))δ .
Putting this resultant back into the discriminant formula above and working out the leading terms
yields the desired formula.
Department of Mathematics, Bard College, Annandale-On-Hudson, NY 12504
E-mail address: cullinan@bard.edu