The Second Discriminant of a Univariate Polynomial

arXiv:1609.00840v1 [] 3 Sep 2016

Dongming Wanga,c and Jing Yangb

Abstract
We define the second discriminant D2 of a univariate polynomial f of degree greater than 2
as the product of the linear forms 2 rk − ri − rj for all triples of roots ri , rk , rj of f with i < j
and j 6= k, k 6= i. D2 vanishes if and only if f has at least one root which is equal to the
average of two other roots. We show that D2 can be expressed as the resultant of f and a
determinant formed with the derivatives of f , establishing a new relation between the roots
and the coefficients of f . We prove several notable properties and present an application of
D2 .
Keywords: Determinant, discriminant, polynomial ideal, resultant, root configuration.

1. Introduction
The discriminant of a univariate polynomial f = f (x) may be defined as a function of
the coefficients of f in x, whose vanishing is a necessary and sufficient condition for f to
have multiple roots for x. The term discriminant was used early by Sylvester in [8] and it
will be referred to as the first discriminant hereinafter. The first discriminant of f contains
information about the nature of the roots1 of f and has played a fundamental role in the
study of polynomial equations. It has many remarkable properties [3, 4] and has been used
in diverse areas ranging from algebraic geometry and Galois theory to bifurcation analysis
and number theory.
To define the first discriminant D1 of f , one considers the simple form ri − rj for any
pair of roots ri , rj of f with i 6= j and takes the product of all such forms as D1 , which can
be expressed as the resultant of f and its derivative. In this paper, we define the second
discriminant D2 of f (of degree greater than 2) as the product of the linear forms 2 rk −ri −rj
for all triples of roots ri , rk , rj of f with i < j and j 6= k, k 6= i.
More concretely, let
f = xn + an−1 xn−1 + · · · + a1 x + a0

(1)

be any univariate polynomial of degree n ≥ 3 in x with real or complex coefficients. Let
r1 , . . . , rn be the n roots of f for x over C, the field of complex numbers. By a symmetric
triple of roots, we mean a triple (ri , rk , rj ) of roots of f with i < j and j 6= k, k 6= i such
a School

of Mathematics and Systems Science, Beihang University, Beijing 100191, China.
of Information Science and Engineering, Guangxi University for Nationalities,
Nanning 530006, China. E-mail: yangjing0930@gmail.com
c Centre National de la Recherche Scientifique, 75794 Paris, France. E-mail: Dongming.Wang@lip6.fr
1 For example, if the discriminant of a cubic polynomial f with real coefficients is positive, then f has no
complex root [5].
b College

that rk = (ri + rj )/2. Then, obviously, D2 = 0 if and only if f has a symmetric triple of
roots. We will show that D2 can be expressed as the resultant of f and a determinant formed
with the derivatives of f , and thus as a polynomial in a0 , . . . , an−1 with rational coefficients.
Several other properties of D2 will also be proved, highlighting the geometric interest of the
symmetric triples of roots. The second discriminant D2 complements the well-known first
discriminant D1 of f in depicting the structural properties such as distribution, position,
and configuration of the roots of f .
In the following section, the second discriminant D2 for an arbitrary univariate polynomial f of degree n is defined formally in terms of the roots of f ; some simple properties of
D2 are then proved. In Sections 3 and 4, we show that D2 as a polynomial in the coefficients
of f is irreducible of total degree 3 (n − 1)(n − 2)/2. In Sections 5 and 6, we elaborate D2
with resultants and ideals from the perspective of modern algebra, which leads to different
ways for the construction of D2 . In Section 7, we provide exact formulas for the degrees of
some determinant polynomials involved in the construction of D2 . Finally, an application of
D2 to the classification of root configurations is presented and the paper is concluded with
some remarks in Section 8.
2. Symmetric Triples of Roots and the Second Discriminant
Let f ∈ C[x] be as in (1) with deg(f, x) = n ≥ 3 and r1 , . . . , rn be the n roots of f over
C as above. Consider any two roots ri and rj . We call (ri + rj )/2 the average of ri and rj .
For any triple r = (ri , rk , rj ), where
f (ri ) = f (rj ) = f (rk ) = 0 and i < j, j 6= k, k 6= i,
if rk is the average of ri and rj , i.e., rk = (ri + rj )/2, then r is called a symmetric triple of
roots of f . We are interested in the condition under which f has symmetric triples of roots.
Recall that the first discriminant of f may be defined as
Y
Y
(ri − rj ).
D1 =
(ri − rj )2 = ±
1≤i<j≤n

1 ≤ i, j ≤ n
i 6= j

D1 = 0 if and only if f has a multiple root. To obtain the condition under which f has a
symmetric triple of roots, we define the second discriminant D2 of f as follows:
Y
(2 rk − ri − rj ),
(2)
D2 =
1 ≤ i, j, k ≤ n
i < j, j 6= k, k 6= i

a symmetric polynomial of total degree n(n − 1)(n − 2)/2 in r1 , . . . , rn . For the sake of
simplicity, we shall write i < j 6= k for the range of i, j, k determined by 1 < i, j, k ≤ n and
i < j, j 6= k, k 6= i.
Remark 1. D1 = 0 does not imply D2 = 0, and vice versa.
Proposition 1. (a) If D1 = 0 and D2 6= 0, then any root of f has multiplicity not greater
than 2.
(b) If D1 6= 0 and D2 = 0, then there exist pairwise distinct ri , rj , rk with i < j 6= k such
that 2 rk − ri − rj = 0.

2

Proof. (a) Suppose that f has a root with multiplicity m > 2 under the condition D1 = 0
and D2 6= 0, e.g., ri = rj = rk (i < j 6= k). This is then a special case of
rk = (ri + rj )/2,
so D2 = 0, which leads to contradiction.
(b) D1 6= 0 implies that ri , rj , rk are pairwise distinct for any i < j 6= k and D2 = 0
implies the existence of ri , rj , rk with i < j 6= k such that 2 rk − ri − rj = 0.
Theorem 1. D2 = 0 if and only if f has a symmetric triple of roots.
Proof. (=⇒) D2 = 0 implies that there exist ri , rj , rk such that rk = (ri + rj )/2. Thus
(ri , rk , rj ) is a symmetric triple of roots which we seek for.
(⇐=) Suppose that (ri , rk , rj ) is a symmetric triple of roots that f has. Then rk =
(ri + rj )/2. It follows that
Y
D2 =
(2 rk − ri − rj ) = 0.
i<j6=k

The second discriminant D2 defined above is a polynomial in the roots r1 , . . . , rn of f .
This polynomial is symmetric with respect to the roots, so D2 can be expressed as another
polynomial in the coefficients a0 , . . . , an−1 of f . We will provide explicit formulas and simple
algorithmic approaches for the construction of the polynomial in ai , together with several
properties about D2 .
3. Expression of the Second Discriminant
In this section, we show that the second discriminant D2 of f can be expressed as a
polynomial in a0 , . . . , an−1 , the coefficients of f . The expression of D2 we have discovered
as the resultant of f and the determinant of a shifting matrix formed with the derivatives
f (1) , . . . , f (n) of f , given in the following theorem, appears pretty amazing. It is puzzling
how and why the derivatives of f get occurred in H so structurally. We will answer this
question in Lemma 6 by linking H to the resultant of two other polynomials derived from f .
As usual, denote by det(M ) the determinant of any square matrix M and by res(f, g, x)
the Sylvester resultant of any two polynomials f and g with respect to x.
Theorem 2. The second discriminant D2 of f is equal to the resultant of f and a determinant H formed with the derivatives of f with respect to x. More precisely,
D2 = res(f, H, x),
where H is the (n − 2)th leading principal minor of the following matrix
 (2)

f (4)
f (6)
f (2l)
f
·
·
·
·
·
·
·
·
·
·
·
·
4!
6!
(2l)!
 2!

 f (1) f (3) f (5) · · · · · · f (2l−1) · · · · · · 
 1!

3!
5!
(2l−1)!


(2)
(4)
(2l−2)


f
f
f
·
·
·
·
·
·
·
·
·
·
·
·
 0

2!
4!
(2l−2)!


(3)
(2l−3)
(1)


f
f
f
M = 0

·
·
·
·
·
·
·
·
·
·
·
·
1!
3!
(2l−3)!




(2)
(2l−3)
f
f
 0

0
·
·
·
·
·
·
·
·
·
·
·
·
2!
(2l−3)!




..
..
..
 ..

..
..
 .
..
.. 
.
.
.
..
..
..
.
..
..
..
.
.
.
and f (ı) denotes the ıth derivative of f .

3

(3)

Note that
res(f, H, x) =

n
Y

H(rk ),

k=1

where r1 , . . . , rn are the n roots of f as before. To prove Theorem 2, we only need to show
that for each k, H(rk ) is the product of 2 rk − ri − rj for all i, j with i < j 6= k. The proof
will be divided into two parts. In the first part, it is shown that for any i, j with i < j 6= k,
2 rk − ri − rj is a divisor of H(rk ) (see Lemmas 1 and 2). The second part is devoted to
(n−1)(n−2)
2
proving that the leading term of H(rk ) with respect to rk is (2 rk )
(see Lemma 3).
Lemma 1. If rk = (ri + rj )/2 for i < j 6= k, then H(rk ) = 0.
Proof. It suffices to show that the lemma holds for k = 1, i = 2, and j = 3. Denote by Ωγl
the set of all γ-tuples obtained from (l, . . . , n) by deleting n − γ components, where l is a
positive integer not greater than n. Let bı = x − rı for ı = 1, . . . , n. By calculus, it is easy
to verify that
( P
bı1 · · · bın− ,  = 0, . . . , n − 1;
(ı1 ,...,ın− )∈Ωn−
f ()
1
=
(4)
!
1,
 = n.
Let cı = r1 − rı for ı = 2, . . . , n and suppose that r1 = (r2 + r3 )/2. Then c2 + c3 = 0.
Substituting x = r1 into (4) and observing that any term bı1 · · · bın− involving x−r1 vanishes
at x = r1 , we have
f ()
!

x=r1

X

=

bı1 · · · bın−

(ı1 ,...,ın− )∈Ωn−
1

x=r1

= tn− + c2 c3 tn−−2 + c2 tn−−1 + c3 tn−−1
= c2 c3 tn−−2 + (c2 + c3 )tn−−1 + tn−
= −c22 tn−−2 + tn− ,
where
tn− =









P

cı1 · · · cın−

if 3 ≤  ≤ n − 1;

(ı1 ,...,ın− )∈Ωn−
4

1

if  = n;

0

if  ≤ 2 or  ≥ n + 1.

4

Substitution of tn− into H(r1 ) yields

H(r1 ) =

−c22 tn−4
−c22 tn−3

−c22 tn−6 + tn−4
−c22 tn−5 + tn−3

−c22 tn−8 + tn−6
−c22 tn−7 + tn−5

······
−c22 tn−2−2 + tn−2
· · · · · · −c22 tn−2−1 + tn−2+1

0
0

−c22 tn−4
−c22 tn−3

−c22 tn−6 + tn−4
−c22 tn−5 + tn−3

······
······

−c22 tn−2 + tn−2+2
−c22 tn−2+1 + tn−2+3

0
..
...
.

0
..
...
.

−c22 tn−4 + tn−2
..
...
.

······
..
..
..

−c22 tn−2+2 + tn−2+4
..
...
.

0

0

0

······

0

0
0

0
0

0
0

······
······

0
0

0

0

0

······

0

······

0

······

0

0

0

······
······
······

0
0
0

······
······
······

0
0
0

0
0
0

0
0
0

······
..
..
..

0
..
...
.

······
..
..
..

0
..
...
.

0
..
...
.

0
..
...
.

· · · · · · −c22 t2−6 + t2−4
· · · · · · −c22 t2−5 + t2−3

· · · · · · −c22 + t2
· · · · · · −c22 t1 + t3

1
t1

0
0

· · · · · · −c22 t2−4 + t2−2
· · · · · · −c22 t2−3 + t2−1

· · · · · · −c22 t2 + t4
· · · · · · −c22 t3 + t5

−c22 + t2
−c22 t1 + t3

1
t1

5

.

For each ı = n − 2, . . . , 2, add the ıth column multiplied by c22 to the (ı − 1)th column of
H(r1 ) iteratively. It follows that

H(r1 ) =

0 tn−4
0 tn−3
0
0
0
0
..
..
...
...
.
.
0
0
0
0

0
0
0
0

tn−6
tn−5
tn−4
tn−3
..
...
.
0
0
0
0

······
tn−2
· · · · · · tn−2+1
· · · · · · tn−2+2
· · · · · · tn−2+3
..
..
..
...
..
.
..
······
..
.
······
...
······
...
······
.

······
······
······
······
..
..
..

0
0
0
0
..
...
.

0
0
0
0
..
...
.

······ 1
· · · · · · t1
· · · · · · t2
· · · · · · t3

0
0
1
t1

= 0.

Lemma 2. For any i < j 6= k, the linear form 2 rk − ri − rj divides H(rk ).
Proof. Let rk = (ri + rj )/2, where i and j are arbitrary but fixed. By Lemma 1, H(rk ) = 0.
It follows that
[rk − (ri + rj )/2] | H(rk ),

or (2 rk − ri − rj ) | H(rk )

over Q (the field of rational numbers).
Note that u | v stands for “u divides v” as usual. Let cı = r1 − rı for ı = 2, . . . , n. It is
easy to verify that
f ()
= t∗n− ,
(5)
!
x=r1

where
t∗n−

=





P

cı1 · · · cın−

if 1 ≤  ≤ n − 1;

(ı1 ,...,ın− )∈Ωn−
2

Let M (r1 ) be the matrix obtained from
 ∗
tn−2 t∗n−4
 t∗n−1 t∗n−3

 0
t∗n−2

M (r1 ) =  0
t∗n−1

 .
..
 ..
..
..
..
.

1

if  = n.

M in (2) by replacing x with r1 . Then

t∗n−6 · · · · · ·
t∗n−2
······
t∗n−5 · · · · · · t∗n−2+1 · · · · · · 

t∗n−4 · · · · · · t∗n−2+2 · · · · · · 

.
t∗n−3 · · · · · · t∗n−2+3 · · · · · · 


..
..
..
..

..
..
..
..
..
..
..
..

n− n−
Since Cn−1
r1 is the leading coefficient of t∗n− with respect to r1 , t∗n− can be written as
n− n−
t∗n− = Cn−1
r1 + O(r1n− ),

where O(r1n− ) denotes terms of degree less than n −  in r1 . Now let Mn (r1 ) be the (n − 2)th
leading principal minor of the matrix obtained from M (r1 ) by replacing each entry t∗n− with
n− n−
Cn−1
r1 . Then
Mn (r1 ) = m1 (r1 ), . . . , mı (r1 ), . . . , mn−2 (r1 ) ,
6

where

mı (r1 ) =

 
T
n−2ı+ n−2ı+
n−2ı n−2ı


C
r
,
.
.
.
,
C
r
,
.
.
.
,
0,
.
.
.
.
.
.
.
.
.
.
.
.
,
0

n−1 1
n−1
1

{z
}
|



max(n−2ı−2,0) terms









n−2


if ı ≤
and 0 ≤  ≤ min(n − 3, 2 ı − 1);


2





T



0

0, . . . . . . . . . , 0, Cn−1
r10 , . . . , Cn−1
r1 , . . .


|
{z
}




2ı−n terms








n−2



and 0 ≤  ≤ 2 n − 2 ı − 3.
if ı >
2

Therefore, Mn (r1 ) has the following form:
n−2 n−2
Cn−1
r1

n−4 n−4
Cn−1
r1

n−6 n−6
Cn−1
r1

n−1 n−1
Cn−1
r1

0

n−3 n−3
Cn−1
r1
n−2 n−2
Cn−1 r1
n−1 n−1
Cn−1
r1

0
..
...
.

0
..
...
.

n−5 n−5
Cn−1
r1
n−4 n−4
Cn−1 r1
n−3 n−3
Cn−1
r1
n−2 n−2
Cn−1 r1

0
Mn (r1 ) =

..
...
.

······

n−2 n−2
Cn−1
r1

······

······

n−2+1 n−2+1
Cn−1
r1
n−2+2 n−2+2
Cn−1 r1
n−2+3 n−2+3
Cn−1
r1
n−2+4 n−2+4
Cn−1 r1

······

······
······
······
..
..
..

..
...
.

······
······
..

..

.

..

Apparently, the above expression for Mn (r1 ) remains valid when r1 is substituted by rk for
any k > 1.
Lemma 3. Mn (rk ) = (2 rk )

(n−1)(n−2)
2

for k = 1, . . . , n.

Proof. We prove the lemma for k = 1. The proof applies for any k 6= 1. Substitution of
ı−1
ı
Cnı = Cn−1
+ Cn−1
into mı (rk ) yields


n−2ı+
n−2ı+−1 n−2ı+
n−2ı
n−2ı−1 n−2ı

+ Cn−2
)rk
,
 (Cn−2 + Cn−2 )rk , . . . , (Cn−2


T



.
.
.
,
0,
.
.
.
.
.
.
.
.
.
.
.
.
,
0


|
{z
}




max(n−2ı−2,0) terms








n−2

and 0 ≤  ≤ min(n − 3, 2 ı − 1);
if ı ≤
mı (rk ) =
2



T


 0, . . . . . . . . . , 0, C 0 r0 , . . . , (C  + C −1 )r , . . .


n−2
n−2
n−2
k
k

{z
}
 |


2ı−n
terms









n−2


and 0 ≤  ≤ 2 n − 2 ı − 3,
if ı >
2


where Cn−2
= 0 for  < 0 and  > n − 2, and ⌈γ⌉ denotes the smallest integer that is not less
than the rational number γ. For any positive integer l, denote by col the lth column and by

7

rol the lth row of this matrix. Then
coı +coı+1 ·r 2

Mn (rk ) =========k=
ı=n−3,...,1

n−2
P

ı=0
n−2
P

ı
Cn−2
rkn−2
ı
Cn−2
rkn−1

ı=0
n−2
P

ı=0
n−2
P
ı=0

n−2
P
ı=0

=

n−2
X

ı
Cn−2
rkn

ı
Cn−2
rkn+1

..
...
.

ı
Cn−2
rkn−2

0

n−4
P

ı=0
n−3
P

ı=0
n−2
P

ı=0
n−2
P
ı=0

n−4
P
ı=0

ı
Cn−2
rkn−4

······

ı
Cn−2
rkn−3

······

ı
Cn−2
rkn−2

······

ı
Cn−2
rkn−1

······

n−2
P

ı=1
n−2+1
P

ı=1
n−2+2
P

ı=1
n−2+3
P
ı=1

..
...
.

..

..

..

n−6
P

ı
Cn−2
rkn−4

······

ı
Cn−2
rkn−2+1

······

ı
Cn−2
rkn−2+2

······

ı
Cn−2
rkn−2+3

······

..
...
.

..

..

roı −roı−1 ·rk

==========
ı=2,...,n−2

..

n−2
P

n−5 n−5
Cn−2
rk

······

n−2 n−2
rk
Cn−2
ı=0
n−2+1 n−2+1
Cn−2
rk

······

ı=0

n−3 n−3
Cn−2
rk

ı
Cn−2
rkn−2

n−6 n−6
Cn−2
rk

······

······

0

n−2 n−2
Cn−2
rk

n−4 n−4
Cn−2
rk

n−2+2 n−2+2
· · · · · · Cn−2
rk

······

0

0

n−3 n−3
Cn−2
rk

n−2+3 n−2+3
· · · · · · Cn−2
rk

······

0
..
...
.

0
..
...
.

n−2 n−2
Cn−2
rk
..
...
.

n−2+4 n−2+4
· · · · · · Cn−2
rk
..
..
..
...
..
.

······
..
..
..

ı
Cn−2
rkn−2 · Mn−1 (rk ) = (2 rk )n−2 · Mn−1 (rk ).

ı=0

Since M3 (rk ) = 2 rk , it is easy to verify that
Mn (rk ) = (2 rk )

(n−1)(n−2)
2

by induction.
Proof of Theorem 2. By Lemma 2,
hYn

i<j6=k

(2 rk − ri − rj )

i

H(rk ).

Hence there exists a polynomial P = P (r1 , . . . , rn ) such that
n
Y

H(rk ) = P

(2 rk − ri − rj ).

i<j6=k

Observe that both of the leading terms of
is equal to (2 rk )

(n−1)(n−2)
2

Qn

i<j6=k

(2 rk − ri − rj ) and H with respect to rk

. This implies that P = 1, so that
H(rk ) =

n
Y

(2 rk − ri − rj ).

i<j6=k

8

Therefore,
res(f, H, x) =

n
Y

H(rk ) =

Y

(2 rk − ri − rj ) = D2 .

i<j6=k

k=1

4. Irreducibility and Degree of the Second Discriminant
Using Theorem 2, one can easily verify that
(1) for n = 3, D2 = −2 a32 + 9 a1 a2 − 27 a0 ;
(2) for n = 4, D2 is an irreducible polynomial of total degree 9, and more explicitly:
D2 = 216 a0a83 − 72 a1 a2 a73 + 16 a32 a63 − 2304 a0a2 a63 + 72 a21 a63 + 672 a1a22 a53 − 144 a42a43
+ 5310 a0a1 a53 + 7446 a0a22 a43 − 2346 a21a2 a43 − 1278 a1a32 a33 + 324 a52 a23
− 9675 a20a43 −28950 a0a1 a2 a33 − 6804 a0a32 a23 + 1658 a31a33 + 9423 a21a22 a23 −1296 a1a42 a3
+51600 a20a2 a23 +31890 a0a21 a23 +19440 a0a1 a22 a3 +1296 a0a42 −17262 a31a2 a3 +972 a21a32
− 120000 a20a1 a3 − 28800 a20a22 − 5040 a0a21 a2 + 9261 a41 + 160000 a30;
(3) for n = 5, D2 is an irreducible polynomial of total degree 18, consisting of 521 terms, in
a0 , . . . , a4 .
In what follows, we prove that D2 is an irreducible polynomialPof total degree 3 (n −
1)(n − 2)/2 in a0 , . . . , an−1 for any n ≥ 3. For this purpose, let si = rk1 · · · rki be the sum
of all the possible, distinct products of i elements taken from r1 , . . . , rn for i = 1, . . . , n. The
sum si of products is called the elementary symmetric polynomial of degree i in r1 , . . . , rn .
It is easy to show that the Vieta formula an−i = (−1)i si holds for i = 1, . . . , n.
Proposition 2. D2 (a0 , . . . , an−1 ) ∈ Q[a0 , . . . , an−1 ] is irreducible over Q.
Proof. Let P ∈ Q[a0 , . . . , an−1 ] be a nonconstant irreducible polynomial and suppose that
P | D2 . We show that D2 | P .
P
Substituting Vieta’s formula an−i = (−1)i rk1 · · · rki into P and D2 , we obtain two
symmetric polynomials P̄ and D̄2 in Q[r1 , . . . , rn ], respectively. Then P̄ = 0 is equivalent
to P = 0, and so is D̄2 to D2 . Since P is nonconstant, so is P̄ . As P | D2 , P̄ | D̄2 ; so P̄
contains at least one irreducible factor of D̄2 , say 2 r1 − r2 − r3 . Therefore, every 2 rk − ri − rj
is a factor of P̄ because P̄ is symmetric with respect to r1 , . . . , rn . It follows that D̄2 | P̄ .
Hence D̄2 and P̄ differ only by a nonzero constant factor, and so do D2 and P . Therefore,
D2 | P and thus D2 is irreducible over Q.
For simplicity, we write a for (a0 , a1 , . . . , an−1 ) and deg(F, a) for the total degree of F
in a from now on.
Proposition 3. deg(D2 , a) = 3 (n − 1)(n − 2)/2.
Proof. Set B0 = D2 . For i = 1, . . . , n, let Ci be the homogeneous part of Bi−1 of the highest
total degree in a+ = (aP
0 , . . . , an−1 , r1 ) and let Bi be obtained from Ci by substituting Vieta’s
formula an−i = (−1)i rk1 · · · rki = Un−i r1 + Vn−i , where Un−i 6= 0 and deg(Un−i , r1 ) =
deg(Vn−i , r1 ) = 0. Then
Ni
i
Ci = Sn−i aN
+ Tn−i = Bi ,
n−i + Tn−i = Sn−i (Un−i r1 + Vn−i )
Ni Ni
r1 = Ci+1 ,
Sn−i Un−i

9

where Ni = deg(Ci , an−i ), Sn−i is the leading coefficient of Ci with respect to an−i , and
Sn−i Un−i 6= 0. Therefore, the total degrees of Ci , Bi , Ci+1 in a+ remain the same for
i = 1, . . . , n, so deg(C1 , a+ ) = deg(Cn , a+ ). Note that Cn is the leading term of D2 ,
expressed in terms of the roots r1 , . . . , rn as in (2), with respect to r1 and deg(Cn , a+ ) =
deg(Cn , r1 ) = 3 (n − 1)(n − 2)/2. Thus deg(D2 , a) = deg(C1 , a+ ) = 3 (n − 1)(n − 2)/2 and
the proposition is proved.
5. The Second Discriminant with Resultants
The following three polynomials will play a significant role in this and later sections:


x+y
− f (x)
f
f (y) − f (x)
2
f1 (x, y) =
, f2 (x, y) =
,
y−x
y−x
2


x+y
+ f (x)
f (y) − 2 f
2
f3 (x, y) =
.
(y − x)2
2
The rational functions on the right-hand side of the above equalities can all be simplified to
polynomials in x and y.
Proposition 4. res(f, res(f1 , f2 , y), x) = 0 if and only if D1 D2 = 0.
Proof. (⇐=) Let
R1 (x) = res(f1 , f2 , y),

R2 = res(f, R1 , x).

We want to show that, if D1 D2 = 0, then there exist ri and rj such that
f (ri ) = f (rj ) = 0,

f1 (ri , rj ) = 0,

f2 (ri , rj ) = 0.

(6)

For this purpose, first suppose that D1 = 0. Then there exist ri = rj , i 6= j, such that
f (ri ) = f (rj ) = 0 and f ′ (ri ) = f ′ (rj ) = 0 (where ′ is the derivation operator). Note that
f1 (x, y) =

n−1
X
k=0

f (k+1) (x)
(y − x)k ,
(k + 1)!

f2 (x, y) =

n−1
X
k=0

f (k+1) (x)
(k + 1)!



y−x
2

k

.

Substitution of x = ri and y = rj into the above expressions shows that (6) holds in this
case.
Now supposethat 
D2 = 0 and D1 6= 0. Then there exist ri 6= rj such that f (ri ) =
f (rj ) = 0 and f

ri +rj
2

f1 (ri , rj ) =

= 0. It follows that

f (rj ) − f (ri )
= 0,
rj − ri

f
f2 (ri , rj ) =




ri + rj
− f (ri )
2
= 0.
rj − ri
2

Thus (6) holds as well.
In any case, f1 (ri , y) and f2 (ri , y) have a common zero rj for y. Therefore,
R1 (ri ) = res(f1 (ri , y), f2 (ri , y), y) = 0.
10

Hence f (x) and R1 (x) have a common root ri for x. This implies that R2 = 0.
(=⇒) res(f, res(f1 , f2 , y), x) = 0 implies that there exist ri and rj , i < j, such that
f (ri ) = 0,

f1 (ri , rj ) = f2 (ri , rj ) = 0.

Thus f (rj ) = f1 (ri , rj )(rj − ri ) + f (ri ) = 0.
If rj = ri , then f (x) has a multiple root and thus D1 = 0. Otherwise,
 

.
ri + rj
f2 (ri , rj ) = 2 f
− f (ri )
(rj − ri ) = 0
2


r +r
implies that f i 2 j = 0, so f has three roots, which form a symmetric triple. Therefore
D2 = 0.
Using similar ideas, we can prove the following proposition, which shows how to construct
D2 via resultant computation twice.
Proposition 5. res(f, res(f1 , f3 , y), x) = 0 if and only if D2 = 0.
Proof. (⇐=) Let
F (x) = res(f1 , f3 , y),

E = res(f, F, x).

We show that, if D2 = 0, then there exist ri and rj such that
f (ri ) = f (rj ) = 0,

f1 (ri , rj ) = 0,

f3 (ri , rj ) = 0.

(7)

First suppose that there exist ri = rj = rk , i < j 6= k, such that f (ri ) = f (rj ) = f (rk ) =
0. Then f ′ (ri ) = f ′′ (ri ) = 0. Note that
f1 (x, y) =

n−1
X
k=0

f (k+1) (x)
(y − x)k ,
(k + 1)!
n−1
P

n−1
P f (k+1) (x)
f (k+1) (x)
(y − x)k −
(k + 1)!
k=0 (k + 1)!
y−x
2

f1 − f2
f3 (x, y) = y − x = k=0
2

n−2
X
1 f (k+2) (x)
k
(y − x) .
=
2− k
2
(k + 2)!



y−x
2

k

k=0

Substitution of x = ri and y = rj into the above expressions shows that (7) holds in this
case.
Suppose otherwise that there exist ri 6= rj such that f (ri ) = f (rj ) = 0 and f ((ri + rj )/2) =
0. Then it follows from D2 = 0 that


ri + rj
f (rj ) − 2 f
+ f (ri )
f (rj ) − f (ri )
2
= 0, f3 (ri , rj ) =
f1 (ri , rj ) =
= 0,
rj − ri
(rj − ri )2
2
so (7) holds as well.
In any case, f1 (ri , y) and f3 (ri , y) have a common zero rj for y. Therefore,
F (ri ) = res(f1 (ri , y), f3 (ri , y), y) = 0.
11

Hence f (x) and F (x) have a common root ri for x. This implies that E = 0.
(=⇒) res(f, res(f1 , f3 , y), x) = 0 implies that there exist ri and rj , i < j, such that
f (ri ) = 0,

f1 (ri , rj ) = f3 (ri , rj ) = 0.

Moreover, f (ri ) = 0 and f1 (ri , rj ) = 0 imply that f (rj ) = 0.
Consider first the case when ri = rj . In this case, D1 = 0 and thus f ′ (ri ) = 0. The
following calculation shows that f ′′ (ri ) = 0:


x + ri
f′
− f ′ (ri )
2
f ′′ (ri ) = lim
x−r
x→ri

i

2





x + ri
x + ri
− f (ri )
f (x) − f
f
2
2
−
x − ri
x − ri
2
2
= lim
x − ri
x→ri

2
x + ri
f (x) + f (ri ) − 2 f
2
= 2 lim
x→ri
(x − ri )2
2
= 2 lim f3 (x, ri ) = 2 f3 (ri , rj ) = 0.
x→ri

Therefore, there exists an rk such that k 6= i, k 6= j and rk = ri = rj , which implies that
2 rk − ri − rj = 0. Hence D2 = 0.
Now consider the case when rj 6= ri . In this case,
 


(rj − ri )2
ri + rj
=0
f3 (ri , rj ) = f (ri ) + f (rj ) − 2 f
2
2
implies that f ((ri + rj )/2) = 0, so x = (ri + rj )/2 is a root of f . Therefore D2 = 0.
Since D2 is irreducible over Q, there exist a positive integer q and a nonzero constant
c ∈ Q such that
D2q = c · res(f, res(f1 , f3 , y), x).
In what follows, we prove that q = 2. For simplicity, we write F for res(f1 , f3 , y) and E for
res(f, res(f1 , f3 , y), x).
Theorem 3. E = c D22 , where c is a nonzero rational number.
The proof of this theorem requires Lemmas 4 and 11, of which the latter shows that
deg(E, a) ≤ 3 (n − 1)(n − 2) + 2 (n − 2).
Lemma 4. For any k, j with 1 < k 6= j, r1 − 2 rk + rj divides F (r1 ).
Proof. It suffices to show that F (r1 ) = 0 when r1 = 2 rk − rj for any fixed k, j satisfying
1 < k 6= j.
According to the theory of resultants [7, pp. 228f], there exist polynomials A1 (x, y) and
A3 (x, y) such that
F (x) = A1 (x, y)f1 (x, y) + A3 (x, y)f3 (x, y).
12

Suppose that rj 6= r1 . Since f (r1 ) = f (rk ) = f (rj ) = 0, substitution of x = r1 and y = rj
into f1 and f3 yields
f (rj ) − f (r1 )
= 0,
rj − r1


r1 + rj
+ f (r1 )
f (rj ) − 2 f
f (rj ) − 2 f (rk ) + f (r1 )
2
f3 (r1 , rj ) =
=
= 0.
(rj − r1 )2
(rj − r1 )2
2
2

f1 (r1 , rj ) =

Suppose otherwise that rj = r1 . Then rk = (r1 + rj )/2 = r1 , which implies that x = r1 is a
root of f with multiplicity greater than 2. Thus f (r1 ) = f ′ (r1 ) = f ′′ (r1 ) = 0. It follows that
n−1
X

f (k+1) (r1 )
(rj − r1 )k = f ′ (r1 ) = 0,
(k + 1)!
k=0

n−2
X
1 f (k+2) (r1 )
f ′′ (r1 )
f3 (r1 , rj ) =
2− k
(rj − r1 )k =
= 0.
2
(k + 2)!
2!

f1 (r1 , rj ) =

k=0

Hence, in both cases we have f1 (r1 , rj ) = f3 (r1 , rj ) = 0. Therefore
F (r1 ) = A1 (r1 , rj )f1 (r1 , rj ) + A3 (r1 , rj )f3 (r1 , rj ) = 0,
so r1 − 2 rk + rj divides F (r1 ).
Proof of Theorem 3. By Lemma 4, (r1 − 2 rk + rj ) | F (r1 ) for arbitrarily chosen k, j with
1 < k 6= j. Hence
Y
(r1 − 2 rk + rj ) | F (r1 ).
1<k6=j

It follows from the theory of resultants [4, p. 398] that
E=

n
Y

i=1

F (ri ) =

Y

(ri − 2 rk + rj )2 · K = D22 · K

i<j6=k

for some polynomial K in r1 , . . . , rn . By Proposition 5, there exist a nonzero constant c and
an integer q ≥ 2 such that E = c D2q .
On the other hand, by Lemma 11, deg(E, a) ≤ 3 (n−1)(n−2)+2 (n−2); by Proposition 3,
deg(D2 , a) = 3 (n − 1)(n − 2)/2. Under these constraints, the only possibility for E = c D2q
to hold is that K is a constant and q = 2.
6. The Second Discriminant with Ideals
In searching for explicit representations of D2 in terms of the coefficients of f , we have
discovered the amazingly structured matrix M formed with the derivatives of f shown in (3).
In what follows, we establish an inherent connection between the (n − 2)th leading principal
minor H of M and res(f1 , f3 , y), which reveals the hidden mystery for the structure of M .
Let hf1 , . . . , fm i denote the ideal generated by f1 , . . . , fm in a ring of polynomials. The
polynomials f1 , . . . , fm are called the generators of the ideal.

13

Lemma 5.




x+y
f (x), f (y), f
, w(x − y) − 1
2




x+y
− f (x), w(x − y) − 1
= f (x), f (y) − f (x), f
2
= hf (x), f1 (x, y), f2 (x, y), w(x − y) − 1i

= hf (x), f1 (x, y), f3 (x, y), w(x − y) − 1i .
Proof. Let the four ideals in the above identity be denoted successively by I1 , . . . , I4 . It is
obvious that I1 = I2 . We only need to show that I2 = I3 and I3 = I4 .


x+y
y−x
(1) Since f (y) − f (x) = f1 · (y − x) and f
− f (x) = f2 ·
, we have I2 ⊂ I3 .
2
2
On the other hand,
f1 = −w [f (y) − f (x)] − [w(x − y) − 1]
 
x+y
f2 = −2 w f
2



f (y) − f (x)
,
y−x


f
− f (x) − 2 [w(x − y) − 1]




x+y
− f (x)
2
,
y−x

so I3 ⊂ I2 .
x−y
· f3 . I4 ⊂ I3 can be easily deduced from f3 =
2
−2 wf1 + 2 wf2 − f3 [w(x − y) − 1].

(2) I3 ⊂ I4 follows from f2 = f1 +

Therefore I1 = I2 = I3 = I4 .
Lemma 6. Let
g1 (x, y) = f1 (x − y, x + y),

g3 (x, y) = f3 (x − y, x + y),

G = res(g1 , g3 , y).

Then G = H 2 , where H is as in Theorem 2.
Proof. For any rational number γ, denote by ⌊γ⌋ the biggest integer that is not greater than
γ. Taking Taylor expansion for g1 and g3 at x, we have
f (x + y) − f (x − y)
2y
f (x + y) − f (x) f (x − y) − f (x)
−
=
2y
2y
n
(k)
X
 f (x) k−1

1
y
1 − (−1)k
=
2
k!

g1 (x, y) =

k=1

=

⌊ n−1
2 ⌋

X f (2 k+1) (x)
y2 k
(2 k + 1)!
k=0

14

and
f (x + y) + f (x − y) − 2 f (x)
2 y2


f (x + y) − f (x) f (x − y) − f (x)
1
·
+
=
2y
y
y
n
(k)
X
 f (x) k−2

1
=
y
1 + (−1)k
2
k!

g3 (x, y) =

k=2

=

⌊n
2 ⌋−1

X f (2 k+2) (x)
y2 k .
(2 k + 2)!
k=0

Let g1∗ and g3∗ be obtained from g1 and g3 by replacing y 2 with z. Then
f (n−1)
(n−1)!

res(g1∗ , g3∗ , z)

0
..
..
..

=

f (n)
n!

0
..
..
..

f (n−3)
(n−3)!
f (n−1)
(n−1)!

..
..
..

f (n−2)
(n−2)!
f (n)
n!

..
..
..

f (n−5)
(n−5)!
f (n−3)
(n−3)!

..
..
..

f (n−4)
(n−4)!
f (n−2)
(n−2)!

..
..
..

Therefore,

······ 


· · · · · ·  jnk
..

..

..


2



−1

······ 



······  n− 1
..

..

..





= ± H.

2

G = res(g1 , g3 , y) = [res(g1∗ (x, z), g3∗ (x, z), z)]2 = H 2 .
Corollary 1. D2 ∈ hf (x), g1 (x, y), g3 (x, y)i ∩ Q[a0 , . . . , an−1 ].
Proof. Let K = Q[a0 , . . . , an−1 ]. By Lemma 6,
res(f, G, x) = res(f, H 2 , x) = D22 .
According to the theory of resultants, there exist A1 (x, z), A2 (x, z) ∈ K[x, z] such that
H = A1 (x, z)g1∗ (x, z) + A2 (x, z)g3∗ (x, z).
Similarly, there exist B1 (x), B2 (x) ∈ K[x] such that
D2 = res(f, H, x) = B1 (x)f (x) + B2 (x)H
= B1 (x)f (x) + B2 (x)[A1 (x, z)g1∗ (x, z) + A2 (x, t)g3∗ (x, z)].
Substituting z = y 2 , one gets
D2 = B1 (x)f (x) + A1 (x, y 2 )B2 (x)g1 (x, y) + A2 (x, y 2 )B2 (x)g3 (x, y)
∈ hf, g1 , g3 i ∩ K.
The corollary is proved.
Theorem 4. D2 ∈ hf (x), f1 (x, y), f3 (x, y)i ∩ Q[a0 , . . . , an−1 ].
15

Proof. Replace x and y in f (x), f1 (x−y, x+y), f3 (x−y, x+y) by (Y + X)/2 and (Y − X)/2,
respectively. Since D2 ∈ hf (x), f1 (x − y, x + y), f3 (x − y, x + y)i and D2 does not involve x
and y,
 


X +Y
D2 ∈ f
, f1 (X, Y ), f3 (X, Y ) .
2
Furthermore, from


X +Y
1
1
f
= (Y − X)2 f3 (X, Y ) + (Y − X)f1 (X, Y ) + f (X, Y ),
2
2
2
one can deduce
hf (X), f1 (X, Y ), f3 (X, Y )i =



 
X +Y
, f1 (X, Y ), f3 (X, Y ) .
f
2

Therefore,
D2 ∈ hf (X), f1 (X, Y ), f3 (X, Y )i.
Substitution of X = x and Y = y back to the above expression, we have
D2 ∈ hf (x), f1 (x, y), f3 (x, y)i.
The proof is complete.
Corollary 2.
D2 ∈





x+y
, w(x − y) − 1 ∩ Q[a0 , . . . , an−1 ].
f (x), f (y), f
2

Proof. It follows from Lemma 5 and Theorem 4.
Proposition 6.
hD2 i =





x+y
, w(x − y) − 1 ∩ Q[a0 , . . . , an−1 ].
f (x), f (y), f
2

Proof. Let I1 and I4 be as in the proof of Lemma 5, which implies that
I1 ∩ K = I4 ∩ K,
where K = Q[a0 , . . . , an−1 ]. We proceed to show that hD2 i = I4 ∩ K.
Since E = res(f, res(f1 , f3 , y), x), E ∈ I4 ∩ K. Let (ā0 , . . . , ān−1 , x̄, ȳ, w̄) be any zero of
I4 and h be any polynomial in I4 ∩ K. Then E(ā0 , . . . , ān−1 ) = h(ā0 , . . . , ān−1 ) = 0. By
Theorem 4, D2 (ā0 , . . . , ān−1 ) = 0, so D2 and h have a nonconstant common divisor. As D2
is irreducible over Q, D2 | h.
On the other hand, by Corollary 2




x+y
D2 ∈ f (x), f (y), f
, w(x − y) − 1 = I1 = I4 .
2
Since D2 | h for any h ∈ I4 ∩ K, the intersection I4 ∩ K is a principal ideal generated by D2 .
Therefore,
hD2 i = I4 ∩ K = I1 ∩ K.

16

Proposition 7. Let si be the elementary symmetric polynomial of degree i in r1 , . . . , rn and
vi = an−i − (−1)i si for i = 1, . . . , n. Then
D Y
E
(2 rk − ri − rj ), v1 , . . . , vn ∩ Q[a0 , . . . , an−1 ]
i<j6=k

= h2 r1 − r2 − r3 , v1 , . . . , vn i ∩ Q[a0 , . . . , an−1 ] = hD2 i .

Proof. Let K = Q[a0 , . . . , an−1 ] as before and
*
+
Y
J1 =
(2 rk − ri − rj ), v1 , . . . , vn ∩ K,
i<j6=k

J2 = h2 r1 − r2 − r3 , v1 , . . . , vn i ∩ K.

Proof of J1 = hD2 i. Note first that each vi (1 ≤ i ≤ n) is a polynomial monic
and linear in an−i . Dividing D2 by vn , . . . , v1 with respect to a0 , . . . , an−1 respectively,
one can obtain a remainder R in r1 , . . . , rn . Then there exist polynomials A1 , . . . , An ∈
Q[a0 , . . . , an−1 , r1 , . . . , rn ] such that
D2 = A1 v1 + · · · + An vn + R.
Substituting an−i = (−1)i si into the above formula and by Theorem 2, we have
Y
R=
(2 rk − ri − rj ).
i<j6=k

Therefore, D2 can be written as a linear combination of polynomials in J1 . This implies
that D2 ∈ J1 and thus hD2 i ⊂ J1 .
To show that J1 ⊂ hD2 i, let h be any polynomial in J1 . Then the greatest common
divisor gcd(h, D2 ) of h and D2 is contained in the ideal J1 . As D2 is irreducible over Q,
gcd(h, D2 ) is either a nonzero constant, or equal to D2 . If gcd(h, D2 ) is a nonzero constant,
then
Q J1 is equal to the unit ideal, which is not possible because for any r1 , . . . , rn satisfying
i<j6=k (2 rk − ri − rj ) = 0, there always exist a0 , . . . , an−1 such that v1 = · · · = vn = 0, i.e.,
J1 always has zeros. Therefore, gcd(h, D2 ) = D2 and D2 | h. It follows that h ∈ hD2 i.
Proof of J1 = J2 . Since J1 ⊂ J2 holds obviously, we only need to show that J2 ⊂ J1 .
Observe that
\
J1 ⊃
h2 rk − ri − rj , v1 , . . . , vn i ∩ K.
i<j6=k

Since J1 = hD2 i is a prime ideal, there exist i < j 6= k such that
J1 ⊃ Jijk = h2 rk − ri − rj , v1 , . . . , vn i ∩ K
and Jijk is prime. Note that v1 , . . . , vn are symmetric in r1 , . . . , rn . Hence the primality
of Jijk implies the primality of Jıκ for all ı <  6= κ. Therefore, all the Jıκ are identical.
Hence
J1 ⊃ J2 = h2 r1 − r2 − r3 , v1 , . . . , vn i ∩ K.
As shown by Propositions 6 and 7, there are several ideals with different generators whose
intersections with K are equal to hD2 i. The generator D2 of the principal ideal hD2 i, which
is an elimination ideal of I1 = · · · = I4 or J1 = J2 , can be obtained by computing the
reduced lexicographical Gröbner basis of any of the ideals Iı and J (see [2, Lemma 6.8]).
17

7. Degrees of Some Determinant Polynomials
The two determinant polynomials H and F = res(f1 , f3 , y), defined in Theorem 2 and
Proposition 5 respectively, can be used for the construction of the second discriminant D2 .
In what follows, we provide some simple formulas for the exact degrees of H and F in x,
which may be used for complexity analysis of D2 .
Lemma 7. deg(H, x) ≤ (n − 1)(n − 2)/2.
Proof. Let g1 , g3 and g1∗ , g3∗ be as Lemma 6 and its proof. Then
G = res(g1 , g3 , y) = [res(g1∗ (x, z), g3∗ (x, z), z)]2 = H 2 ,
where H is as in Theorem 2. Now consider
∆(a, x, y, α) =

g1 (x, y)

g3 (x, y)

g1 (x, α)

g3 (x, α)



(y − α)

and let ν = (x, y, α) and ñ = 2 ⌊ n−1
2 ⌋. It is easy to see that ∆ is of degree 2 n − 4 in ν and
deg(∆, α) = deg(∆, y) = ñ − 1, deg(∆, a) ≤ 2. Let ∆ be written as
X
δijk xi y j αk .
∆=
0 ≤ i ≤ 2n − 4
0 ≤ j, k ≤ ñ − 1

Then for every term xi y j αk occurring in ∆, i+j+k ≤ deg(∆, ν) = 2 n−4, so i ≤ 2 n−4−j−k.
Denote by
!
2X
n−4
i
δijk x
B = (bj+1,k+1 ) =
i=0

the ñ × ñ Bézout matrix of g1 and g3 with respect to y. It follows that deg(bjk , x) ≤
2 n−2−j −k. Let (k1 , . . . , kñ ) denote an arbitrary permutation of (1, . . . , ñ) and Ḡ = det(B).
Then
2 deg(H, x) = deg(H 2 , x) = deg(G, x) = deg(Ḡ, x)
≤
≤

max

(k1 ,...,kñ )

max

(k1 ,...,kñ )

deg(b1k1 · · · bñ
ñ
X

kñ−1 , x) =

max

(k1 ,...,kñ )

ñ
X

deg(bjkj , x)

j=1

[2 n − 2 − j − kj ] = (2 n − 2) · ñ −

ñ
X
j=1

j=1

j−

min

(k1 ,...,kñ )

ñ
X

kj

j=1

= (2 n − 2) · ñ − (ñ + 1) · ñ = (n − 1)(n − 2).
Therefore, deg(H, x) ≤ (n − 1)(n − 2)/2.
Lemma 7 provides an upper bound for deg(H, x). In what follows, we show that the
bound can be achieved for a particular polynomial. Thus the degree of H constructed from
the generic form of f is equal to the bound.
Lemma 8. For a0 = · · · = an−1 = 0, deg(H, x) = (n − 1)(n − 2)/2.

18

Proof. When a0 = · · · = an−1 = 0, H becomes the (n − 2)th leading principal minor of the
following matrix

 2 n−2
Cn x
Cn4 xn−4 Cn6 xn−6 · · · · · ·

 1 n−1
Cn3 xn−3 Cn5 xn−5 · · · · · · 
 Cn x



0
Cn2 xn−2 Cn4 xn−4 · · · · · · 


.

0
Cn1 xn−1 Cn3 xn−3 · · · · · · 





..
..
..
..

.
.. 
...
...
...
.
.
.
.
Simple calculation shows that
Cn2 xn−2
Cn1 xn−1

Cn4 xn−4
Cn3 xn−3

Cn6 xn−6
Cn5 xn−5

······
······

0
0
..
...
.

Cn2 xn−2
Cn1 xn−1
..
...
.

Cn4 xn−4
Cn3 xn−3
..
...
.

······
······
..
..
..

co1 ÷xn−2
co2 ÷xn−4 ,con−2 ×xn−4

co3 ÷xn−3 ,con−3 ×xn−3
···

=================
ro2 ÷x
...
ron−2 ÷xn−3

=

Cn2

Cn4

Cn6

······

Cn1

Cn3
Cn2
Cn1

Cn5
Cn4
Cn3

······
······

..
..
..

..
..
..

······
..
..
..

Cn2
Cn1

Cn4
Cn3

Cn6
Cn5

······
······

0

Cn2

Cn4

······

0
..
..
..

Cn1

Cn3

······
..
..
..

0
0
..
..
..

..
..
..

..
..
..

x(n−2)+1+2+···+(n−3)

x

(n−1)(n−2)
2

(n−1)(n−2)
.
2
= cn x
.

In what follows, we prove that cn 6= 0.
Let
n
2⌊ n ⌋
U = Cn2 z 2 + Cn4 z 4 + · · · + Cn 2 z 2⌊ 2 ⌋ ,

2⌊ n+1 ⌋−1 2⌊ n+1 ⌋−1
z 2
,
V = Cn1 z + Cn3 z 3 + · · · + Cn 2

and Ū and V̄ be obtained from U/z 2 and V /z, respectively, by replacing z 2 with t. Then
cn = ± res(Ū , V̄ , t). If cn = 0, then U/z 2 and V /z have at least one common zero, say z̄,
where z̄ 6= 0. Note that
U + V = (z + 1)n − Cn0 .
Substituting z = z̄ into the above equation, we have (z̄ +1)n −1 = 0. Similarly, (U −V )|z=z̄ =
(z̄ − 1)n − 1 = 0. Therefore, there exist two unit roots u1 , u2 such that z̄ + 1 = u1 and
z̄ − 1 = u2 , which leads to u1 − u2 = 2. In other words, u1 and u2 have the same imaginary
part and the difference of their real parts is 2. This can happen only when u1 = 1 and
19

u2 = −1. Therefore, z̄ = 0, which leads to contradiction since z̄ is nonzero. Hence the
conclusion holds.
The following theorem follows from Lemmas 7 and 8.
Theorem 5. deg(H, x) = (n − 1)(n − 2)/2.
Similarly, we have the following theorem.
Theorem 6. deg(F, x) = (n − 1)(n − 2).
This theorem is established by proving the following two lemmas.
Lemma 9. deg(F, x) ≤ (n − 1)(n − 2).
Proof. Let
∆(a, x, y, α) =

f1 (x, y)
f1 (x, α)

f3 (x, y)
f3 (x, α)


(y − α)

and ν = (x, y, α). It is easy to see that ∆ is of degree 2 n − 4 in ν and deg(∆, α) =
deg(∆, y) = n − 2, deg(∆, a) ≤ 2. Let ∆ be written as
X
δijk xi y j αk .
∆=
0 ≤ i ≤ 2n − 4
0 ≤ j, k ≤ n − 2

Then for every term xi y j αk occurring in ∆, i+j+k ≤ deg(∆, ν) = 2 n−4, so i ≤ 2 n−4−j−k.
Denote by
!
2X
n−4
i
δijk x
B = (bj+1,k+1 ) =
i=0

the (n−1)×(n−1) Bézout matrix of f1 and f3 with respect to y. It follows that deg(bjk , x) ≤
2 n − 2 − j − k. Let (k1 , . . . , kn−1 ) denote an arbitrary permutation of (1, . . . , n − 1) and
F̄ = det(B). According to the theory of resultants [1], F = res(f1 , f3 , y) = ± F̄ . Therefore,
deg(F, x) = deg(F̄ , x) ≤

≤

max

(k1 ,...,kn−1 )

max

(k1 ,...,kn−1 )

deg(b1k1 · · · bn−1,kn−1 , x) =

max

(k1 ,...,kn−1 )

n−1
X

(2 n − 2 − j − kj ) = (2 n − 2)(n − 1) −

j=1

n−1
X
j=1

= 2 (n − 1)2 − (n − 1)n = (n − 1)(n − 2).

j−

n−1
X

deg(bjkj , x)

j=1

min

(k1 ,...,kn−1 )

n−1
X

kj

j=1

Lemma 10. For a0 = · · · = an−1 = 0, deg(F, x) = (n − 1)(n − 2).
Proof. When a0 = · · · = an−1 = 0, f = xn . We first prove
 that x = 0 is equivalent to F = 0.
(=⇒) If x = 0, then f1 = y n−1 and f3 = 2 − 1/2n−2 y n−2 . In this case, f1 and f3 have
a common zero and thus F = 0.
(⇐=) Let F = 0; then f1 and f3 have at least one common zero for y, say ȳ. Then
n

ȳ + x
n
+ xn
ȳ − 2
ȳ n − xn
2
f1 (x, ȳ) =
= 0, f3 (x, ȳ) =
= 0.
ȳ − x
(ȳ − x)2
2
20

Suppose that x 6= 0 and let t̄ = ȳ/x. Then the above equalities imply that

n
1 1
t̄n = 1,
+ t̄
= 1.
2 2
Therefore, there exist two unit roots u1 and u2 such that t̄ = u1 and (1 + t̄)/2 = u2 , which
implies that u2 = (1 + u1 )/2. This can happen only when u1 = u2 = 1; so ȳ = x. Thus
f1 (x, ȳ) =

y n − xn
= xn−1 + xn−2 ȳ + · · · + xȳ n−2 + xȳ n−1 = nxn−1 = 0,
y−x

which implies that x = 0. This contradicts the assumption that x 6= 0. Therefore, x = 0.
Since x = 0 and F = 0 are equivalent, there exist a nonzero constant c and an integer
N ≥ 1 such that F = c xN . It remains to show that N = (n − 1)(n − 2).
Let

f1 (x, y) f3 (x, y)
∆(x, y, α) =
(y − α)
f1 (x, α) f3 (x, α)
and ν = (x, y, α). It is easy to see that ∆ is homogeneous of degree 2 n − 4 in ν and
deg(∆, α) = deg(∆, y) = n − 2. Let ∆ be written as
X
δijk xi y j αk .
∆=
0 ≤ i ≤ 2n − 4
0 ≤ j, k ≤ n − 2

Then for every term xi y j αk occurring in ∆, i+j+k = deg(∆, ν) = 2 n−4, so i = 2 n−4−j−k.
Denote by
!
2X
n−4
i
δijk x
B = (bj+1,k+1 ) =
i=0

the (n − 1) × (n − 1) Bézout matrix of f1 and f3 with respect to y and let F̄ = det(B).
According to the theory of resultants [1], F = res(f1 , f3 , y) = ± F̄ , so deg(F̄ , x) ≥ 1.
Note that for any entry bjk in B, either bjk = 0 or deg(bjk , x) = 2 n − 2 − j − k. Let
(k1 , . . . , kn−1 ) denote an arbitrary permutation of (1, . . . , n−1). Then either b1k1 · · · bn−1,kn−1 =
0, or
deg(b1k1 · · · bn−1,kn−1 , x) =

n−1
X

deg(bjkj , x) =

n−1
X

(2 n − 2 − j − kj )

j=1

j=1

= (2 n − 2)(n − 1) −

n−1
X
j=1

j−

n−1
X

kj

j=1

= 2 (n − 1)2 − (n − 1)n = (n − 1)(n − 2).
Note that F̄ 6= 0, so deg(F̄ , x) = (n − 1)(n − 2). It follows that deg(F, x) = (n − 1)(n −
2).
The following lemma has been used for the proof of Theorem 3.
Lemma 11. deg(E, a) ≤ 3 (n − 1)(n − 2) + 2 (n − 2).

21

Proof. Let N = deg(F, x); then N ≤ (n − 1)(n − 2) according to Lemma 9. Moreover, from
the proof of Lemma 9 we know that deg(F, a) ≤ 2 (n − 2). Since E is a determinant formed
with n rows of f -coefficients and N rows of F -coefficients, the degree of each f -coefficient
is at most 1, and the degree of each F -coefficient is at most 2 (n − 2), the degree of E is at
most N · 1 + n · 2 (n − 2) ≤ 3 (n − 1)(n − 2) + 2 (n − 2). The proof is complete.
From Proposition 3 and Theorem 3 the following corollary follows.
Corollary 3. deg(E, a) = 3 (n − 1)(n − 2).
The result of this corollary allows us to reduce the upper bound 3 (n− 1)(n− 2)+ 2 (n− 2)
of deg(E, a) to 3 (n − 1)(n − 2), the exact degree of E in a, which is also the degree of D22
in a.
Remark 2. The determinant polynomials F and H are both irreducible over Q[a]. The
irreducibility of H is obvious because D2 = res(f, H, x) is irreducible and that of F can
be proved by using the symmetry of F (r1 ) with respect to r2 , . . . , rn .2 Hence F and H do
not have any common divisor. On the other hand, G is obtained from f1 and f3 via linear
transformation and resultant computation and F is connected to H via G by the relations
res(f, F, x) = res(f, G, x) = [res(f, H, x)]2 ,
deg(F, x) = deg(G, x) = 2 deg(H, x),
and G = H 2 . However, it is unclear whether there is any direct connection between F
and G. Note that F and thus D2 are constructed from f , f1 , and f3 naturally; yet the
occurrence of the sequences of odd derivatives and even derivatives of f with respect to x in
the determinant expressions of H and G remains uninterpretable. Meaningful interpretations
of the occurrence might be figured out by exploring direct connections between F and G.
8. Application and Remarks
In this section, we illustrate the usefulness of the second discriminant by an application (to
the classification of root configurations for the cubic polynomial) and discuss the possibility
of introducing discriminants of higher order.
The form ri − rj in D1 can be viewed as the vector from rj to ri , considered as two points
in the complex plane. Similarly, the form 2 rk −ri −rj in D2 can be viewed as twice the vector
from the middle point of ri and rj to rk . The signs of D1 and D2 carry information about
the distribution, position, and relative configuration of the roots r1 , . . . , rn of f . Therefore,
D1 and D2 can be used to explore such structural properties of the roots of f without exactly
computing them out.
For the cubic polynomial f = x3 + a2 x2 + a1 x + a0 , we have the following Lagrange
formula with radicals for its three roots:
r1 =

−a2 + ω 1 c1 + ω 2 c2
,
3

r2 =

−a2 + ω 0 c1 + ω 2 c2
,
3

r3 =

−a2 + ω 2 c1 + ω 1 c2
,
3

2 Let F (a, x) = F (a, x)F (a, x) with deg(F , x) 6= 0. In this equality, substitution of x by r and
1
2
1
1
elimination of each ai by using Vieta’s formula yield F̄ (r1 , . . . , rn ) = F̄1 (r1 , . . . , rn )F̄2 (r1 , . . . , rn ), where
F̄ , F̄1 ,Q
and F̄2 are all symmetric with respect to r2 , . . . , rn . From the proof of Theorem 3, one sees that
F̄ = c k6=j (r1 − 2 rk + rj ) for some constant c. Thus F̄1 has at least one divisor r1 − 2 rk + rj for some
Q
j 6= k. The symmetry of F̄1 with respect to r2 , . . . , rn implies that 1<k6=j (r1 − 2 rk + rj ) is also a divisor
of F̄1 . Therefore, F̄1 differs from F̄ only by a nonzero constant, and so does F1 from F . It follows that F2
is a constant. This proves the irreducibility of F .

22

where ω = e

2π
3 i

= − 21 +
c1 =

√

3
2 i

and

q
p
3
(D2 + 2 −3 D1 )/2,

c2 =

q
p
3
(D2 − 2 −3 D1 )/2.

Using the above formula, one can classify the roots of f into 9 types of configurations
according to the signs of D1 and D2 as shown in Table 1 (cf. [9]).

Table 1. Types of configurations for the roots r1 , r2 , r3 of the cubic polynomial f , where
Re(r1 ) ≥ Re(r2 ) ≥ Re(r3 ) and the red points of small, middle, and large sizes stand respectively for single, double, and triple roots of f .
D2 < 0

D2 = 0

D2 > 0

r1 , r2 , r3 ∈ R,
r1 − r2 < r2 − r3

r1 , r2 , r3 ∈ R,
r1 − r2 = r2 − r3

r1 , r2 , r3 ∈ R,
r1 − r2 > r2 − r3

r1 , r2 , r3 ∈ R
r1 = r2 > r3

r1 , r2 , r3 ∈ R
r1 = r2 = r3

r1 , r2 , r3 ∈ R
r1 > r2 = r3

r1 , r2 ∈ C, r3 ∈ R
Re(r1 ) = Re(r2 ) > r3

r1 , r3 ∈ C, r2 ∈ R
Re(r1 ) = Re(r2 ) = r3

r2 , r3 ∈ C, r1 ∈ R
r1 > Re(r2 ) = Re(r3 )

D1 > 0

D1 = 0

D1 < 0

The second discriminant can also be used in the root formula with radicals and to classify the types of configurations of the four roots for the general quartic polynomial. The
classification in this case is somewhat involved and will be presented in a forthcoming paper
[6].
The second discriminant of a univariate polynomial f , a concept we have introduced, is
defined as the product of all possible linear forms 2 rk − ri − rj in the roots ri , rk , rj of f with
i < j 6= k, so its vanishing is a necessary and sufficient condition for f to have a symmetric
triple of roots, i.e., a triple (ri , rk , rj ) of roots of f such that rk = (ri + rj )/2. We have shown
that the second discriminant of f can be expressed as the resultant of f and a determinant
formed with the derivatives of f and it possesses several notable properties3 and can be used
to analyze the structure of the roots of f .
3 Our

experiments also show that, when a0 , . . . , an−1 take integer values, D2 6≡ 2 mod 4 for n > 3.

23

We may naturally consider the product of linear forms in d roots of f for any n ≥ d ≥ 4.
The product should be symmetric with respect to the n roots of f and the linear form should
be chosen such that its vanishing constrains the d general roots of f to form a degenerate
configuration which is geometrically interesting. Then one can try to establish conditions
for f to have d roots forming the degenerate configuration.
For n ≥ d = 4, linear forms of interest in four roots ri , rj , rk , rl of f could be taken of
the following type
ri + rj − rk − rl , or 3 rl − ri − rj − rk .
(8)
The former is twice the difference between the average of the two roots ri and rj and that
of the two roots rk and rl , while the latter is three times the difference from the root rl
to the average of the three roots ri , rj , rk . When the roots are considered as points in the
complex plane, the average of two or three roots may be interpreted as the middle point or
the centroid of the two or three points, respectively. Using the first linear form in (8), one
may define
Y
(ri + rj − rk − rl ).
D3 =
i 6= j 6= k 6= l
i < j, k < l, i < k

For n = 4, D3 can be expressed as a polynomial in the coefficients of f and this polynomial
has been used in the root formula of f with radicals. How to express D3 as a polynomial in
the coefficients of f for arbitrary n > 4 and what properties D3 may have are questions that
remain for further investigation. Similar questions may be asked for D3 defined by using the
other linear form, and for D4 , D5 , . . . , when they are properly defined.
It should be pointed out that the ideas and methodologies used in the study of D2
provide a new approach to explore the properties of D1 . It may be generalized to investigate
D3 , D4 , . . . and to discover other mysteries about the roots of f .
References
[1] Bézout, É.: Théorie générale des équations algébriques. Ph.D. thesis, Pierres, Paris
(1779).
[2] Buchberger, B.: Gröbner bases: an algorithmic method in polynomial ideal theory. Multidimensional Systems Theory, pp. 184–232. Reidel, Dordrecht (1985).
[3] Cohen, H.: Resultants and discriminants. A Course in Computational Algebraic Number
Theory, pp. 119–123. Springer, New York (1993).
[4] Gelfand, I. M., Kapranov, M. M., Zelevinsky, A. V.: Discriminants, Resultants and Multidimensional Determinants. Birkhäuser, Boston (1994).
[5] Gow, R.: Some properties and uses of the discriminant of a polynomial. Irish Mathematical Society Bulletin, 24, 12–19 (1990).
[6] Hong, H., Wang, D., Yang, J., Zhao, T.: Solution formulas for quartic equations without
or with constraints. In preparation (2016).
[7] Mishra, B.: Algorithmic Algebra. Springer, New York (1993).
[8] Sylvester, J. J.: On a remarkable discovery in the theory of canonical forms and of hyperdeterminants. Philosophical Magazine, 4(2), 391–410 (1851).
[9] Zhao, T., Wang, D., Hong, H.: Solution formulas for cubic equations without or with
constraints. Journal of Symbolic Computation, 46(8), 904–918 (2011).
24

