CUBO A Mathematical Journal
Vol.20, Nβo2, (67–93). June 2018

http: // dx. doi. org/ 10. 4067/ S0719-06462018000200067

Some remarks on the non-real roots of polynomials

Shuichi Otake
1
and Tony Shaska

2

1Department of Applied Mathematics,

Waseda University,

Japan.
2Department of Mathematics and Statistics ,

Oakland University,

Rochester, MI, 48309.

shuichi.otake.8655@gmail.com, shaska@oakland.edu

ABSTRACT

Let f ∈ R(t)[x] be given by f(t,x) = xn + t · g(x) and β1 < · · · < βm the distinct real
roots of the discriminant ∆(f,x)(t) of f(t,x) with respect to x. Let γ be the number of

real roots of g(x) =
∑s

k=0 ts−kx
s−k. For any ξ > |βm|, if n−s is odd then the number

of real roots of f(ξ,x) is γ + 1, and if n − s is even then the number of real roots of

f(ξ,x) is γ, γ + 2 if ts > 0 or ts < 0 respectively. A special case of the above result

is constructing a family of degree n ≥ 3 irreducible polynomials over Q with many
non-real roots and automorphism group Sn.

RESUMEN

Sea f ∈ R(t)[x] dada por f(t,x) = xn + t · g(x) y β1 < · · · < βm las diferentes ráıces
reales del discriminante ∆(f,x)(t) de f(t,x) con respecto de x. Sea γ el número de ráıces

reales de g(x) =
∑s

k=0 ts−kx
s−k. Para todo ξ > |βm|, si n − s es impar entonces el

número de ráıces reales de f(ξ,x) es γ + 1, y si n − s es par entonces el número de

ráıces reales de f(ξ,x) es γ, γ+2 si ts > 0 o ts < 0, respectivamente. Un caso especial

del resultado anterior es construyendo una familia de polinomios irreducibles sobre Q

de grado n ≥ 3 con muchas ráıces no-reales y grupo de automorfismos Sn

http://dx.doi.org/10.4067/S0719-06462018000200067


68 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

Keywords and Phrases: Polynomials, non-real roots, discriminant, Bezoutian, Galois groups.

2010 AMS Mathematics Subject Classification: 12D10, 12F10, 26C10.



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Some remarks on the non-real roots of polynomials 69

1 Introduction

Let f(x) ∈ Q[x] be an irreducible polynomial of degree n ≥ 2 and Gal (f) its Galois group over Q.
Let us assume that over R, f(x) is factored as

f(x) = a

r∏

j=1

(x − αj)

s∏

i=1

(x2 + aix + bi),

where a2i < 4bi, for all i = 1, . . . ,s. The pair (r,s) is called the signature of f(x). Obviously

degf = 2s + r. If s = 0 then f(x) is called totally real and if r = 0 it is called totally complex.

Equivalently the above terminology can be defined for binary forms f(x,z). By a reordering of the

roots we may assume that if f(x) has 2s non-real roots then

α := (1,2)(3,4) · · · (2s − 1,2s) ∈ Gal(f).

In [4] it is proved that if degf = p, for a prime p, and s satisfies

s(s logs + 2 logs + 3) ≤ p

then Gal(f) = Ap,Sp. Moreover, a list of all possible groups for various values of r is given for

p ≤ 29; see [4, Thm. 2]. There are some follow up papers to [4].

In [1] the author proves that if p ≥ 4s + 1, then the Galois group is either Sp or Ap. This
improves the bound given in [4]. The author also studies when polynomials with non-real roots are

solvable by radicals, which are consequences of Table 2 and Theorem 2 in [4]. In [13] the author

uses Bezoutians of a polynomial and its derivative to construct polynomials with real coefficients

where the number of real roots can be counted explicitly. Thereby, irreducible polynomials in Q[x]

of prime degree p are constructed for which the Galois group is either Sp or Ap.

In this paper we study a family of polynomials with non-real roots whose degree is not nec-

essarily prime. Given a polynomial g(x) =
∑s

i=0 tix
i and with γ number of non-real roots we

construct a polynomial f(t,x) = xn + tg(x) which has γ,γ + 1,γ + 2 non-real roots for certain

values of t ∈ R; see Theorem 3.2. The values of t ∈ R are given in terms of the Bezoutian matrix
of polynomials or equivalently the discriminant of f(t,x) with respect to x. This is the focus of

Section 3 in the paper.

While most of the efforts have been focusing on the case of irreducible polynomials over Q

which have real roots, the case of polynomials with no real roots is equally interesting. How should

an irreducible polynomial over Q with all non-real roots must look like? What can be said about

the Galois group of such totally complex polynomials? In [5] is developed a reduction theory for

such polynomials via the hyperbolic center of mass. A special case of Theorem 3.2 provides a class

of totally complex polynomials.

Notation For any polynomial f(x) we denote by ∆(f,x) its discriminant with respect to x. If f is a

univariate polynomial then ∆f is used and the leading coefficient is denoted by led(f). Throughout

this paper the ground field is a field of characteristic zero.



70 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

2 Preliminaries

Let f1(x), f2(x) be polynomials over a field F of characteristic zero and, let n be an integer which

is greater than or equal to max{degf1,degf2}. Then, we put

Bn(f1,f2) : =
f1(x)f2(y) − f1(y)f2(x)

x − y
=

n∑

i,j=1

αijx
n−iyn−j ∈ F[x,y],

Mn(f1,f2) : = (αij)1≤i,j≤n.

The matrix Mn(f1,f2) is called the Bezoutian of f1 and f2. Clearly, Bn(f1,f1) = 0 and hence

Mn(f1,f1) is the zero matrix. The following properties hold true; see [6, Theorem 8.25] for details.

Proposition 1. The following are true:

(1) Mn(f1,f2) is an n × n symmetric matrix over F.

(2) Bn(f1,f2) is linear in f1 and f2, separately.

(3) Bn(f1,f2) = −Bn(f2,f1).

When f2 = f
′
1, the formal derivative of f1 (with respect to the indeterminate x), we often

write Bn(f1) := Bn(f1,f
′
1). From now on, for any degree n ≥ 2 polynomial f(x) ∈ R[x] we will

denote by Mn(f) := Mn(f,f
′) as above. The matrix Mn(f) is called the Bezoutian matrix of f.

Remark 2.1. It is often the case that the matrix M′n(f1,f2) = (α
′
ij)1≤i,j≤n defined by the gener-

ating function

B′n(f1,f2) : =
f1(x)f2(y) − f1(y)f2(x)

x − y
=

n∑

i,j=1

α′ijx
i−1yj−1 ∈ F[x,y]

is called the Bezoutian of f1 and f2. But no difference can be seen between these two definitions

as far as we consider the corresponding quadratic forms

n∑

i,j=1

αijxixj and

n∑

i,j=1

α′ijxixj.

In fact, these two quadratic forms are equivalent over the prime field Q (⊂ F) since we have
M′n(f1,f2) =

tJnMn(f1,f2)Jn, where

Jn =













0 1
1

...

1 0













is an n × n anti-identity matrix. This implies that above two quadratic forms are equivalent over
Q or more precisely, over the ring of rational integers Z.



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Some remarks on the non-real roots of polynomials 71

Let f(x) ∈ R[x] be a degree n ≥ 2 polynomial which is given by

f(x) = a0 + a1x + · · · + anxn

Then over R this polynomial is factored as

f(x) = a

r∏

j=1

(x − αj)

s∏

i=1

(x2 + aix + bi)

for some α1, . . . ,αr ∈ R and ai,bi,a ∈ R, where a2i < 4bi, for all i = 1, · · · ,s.

Throughout this paper, for a univariate polynomial f, its discriminant will be denoted by

∆f. For any two polynomials f1(x), f2(x) the resultant with respect to x will be denoted by

Res(f1,f2,x). We notice the following elementary fact, its proof is elementary and we skip the

details.

Remark 2.2. For any polynomial f(x), the determinant of the Bezoutian is the same as the

discriminant up to a multiplication by a constant. More precisely,

∆f =
1

led(f)2
detMn(f),

where led(f) is the leading coefficient of f(x).

If f(x) ∈ Q[x] is irreducible and its degree is a prime number, say degf = p, then there is
enough known for the Galois group of polynomials with some non-real roots; see [4], [1], [13] for

details. If the number of non-real roots is ”small” enough with respect to the prime degree degf = p

of the polynomial, then the Galois group is Ap or Sp. Furthermore, using the classification of finite

simple groups one can provide a complete list of possible Galois groups for every polynomial of

prime degree p which has non-real roots; see [4] for details.

On the other extreme are the polynomials which have all roots non-real. We called them

above, totally complex polynomials. We have the following:

Lemma 2.1. The followings are equivalent:

i) f(x) ∈ R[x] is totally complex

ii) f(x) can be written as

f(x) = a

n∏

i=1

fi

where fi = x
2 + aix + bi, for i = 1, . . . ,n and ai,bi,a ∈ R, where a2i < 4bi, for all i = 1, . . . ,n.

Moreover, the determinant of the Bezoutian Mn(f) is given by

∆f =
1

led(f)2
detMn(f) =

n∏

i=1

∆fi ·
n∏

i,j,i6=j

(Res(fi,fj,x))
2



72 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

where led(f) is the leading coefficient of f(x).

ii) the index of inertia of Bezoutian M(f) is 0

iii) if ∆f 6= 0 then the equivalence class of M(f) in the Witt ring W(R) is 0.

Proof. The equivalence between i), ii), and iii) can be found in [6].

It is not clear when such polynomials are irreducible over Q. If that’s the case, what is the

Galois group Gal (f)? Clearly the group generated by the involution (1,2)(3,4) · · · (2n − 1,2n) is
embedded in Gal (f). Is Gal (f) larger in general?

3 On the number of real roots of polynomials

For any degree n ≥ 2 polynomial f(x) ∈ R[x] and any symmetric matrix M := Mn(f) with real
entries, let Nf be the number of distinct real roots of f and σ(M) be the index of inertia of M,

respectively. The next result plays a fundamental role throughout this section ([6, Theorem 9.2]).

Proposition 2. For any real polynomial f ∈ R[x], the number Nf of its distinct real roots is the
index of inertia of the Bezoutian matrix Mn(f). In other words,

Nf = σ(Mn(f)) .

Let us cite one more result which says that the roots of a polynomial depend continuously on

its coefficients ([11, Theorem 1.4], [16, Theorem 1.3.1]).

Proposition 3. Let be given a polynomial

f(x) =

n∑

l=0

alx
l ∈ C[x],

with distinct roots α1, . . . ,αk of multiplicities m1, . . . ,mk respectively. Then, for any given a

positive

ε < min
1≤i<j≤k

{
|αi − αj|

2

}
,

there exists a real number δ > 0 such that any monic polynomial g(x) =
∑n

l=0 blx
l ∈ C[x] whose

coefficients satisfy

|bl − al| < δ,

for l = 0, · · · ,n − 1, has exactly mj roots in the disk

D(αj;ε) = {z ∈ C | |z − αj| < ǫ} (j = 1, · · · ,k).



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Some remarks on the non-real roots of polynomials 73

Let n, s be positive integers such that n > s and let

g(t0, · · · ,ts;x) =
s∑

k=0

ts−kx
s−k,

f(n)(t0, · · · ,ts,t;x) = xn + t · g(t0, · · · ,ts;x)
(3.1)

be polynomials in x over E1 = R(t0, · · · ,ts), E2 = R(t0, · · · ,ts,t), respectively. Here, E1 (resp.,
E2) is a rational function field with s+1 (resp.,(s+2)) variables t0, · · · ,ts (resp.,(t0, · · · ,ts,t)).
To ease notation, let us put

g(x) = g(t0, · · · ,ts;x), f(t;x) = f(n)(t0, · · · ,ts,t;x)

and for any real vector v = (v0, · · · ,vs) ∈ Rs+1, we put

gv(x) = g(v0, · · · ,vs;x), fv(t;x) = f(n)(v0, · · · ,vs,t;x). (3.2)

By using Proposition 2, we can prove the next theorem ([13, Main Theorem 1.3]).

Theorem 3.1. Let r = (r0, · · · ,rs) ∈ Rs+1 be a vector such that Ngr = s. Let us consider
fr(t;x) = f

(n)(r0, · · · ,rs,t;x) as a polynomial over R(t) in x and put

Pr(t) = detMn(fr(t;x)) = detMn(fr(t;x),f
′
r(t;x)),

where f′r(t;x) is a derivative of fr(t;x) with respect to x. Then, for any real number ξ > αr =

max{α ∈ R | Pr(α) = 0}, we have

Nfr(ξ;x) =






s + 1 if n − s : odd

s if n − s : even, rs > 0

s + 2 if n − s : even, rs < 0.

By this theorem and a theorem of Oz Ben-Shimol [1, Theorem 2.6], we can obtain an algorithm

to construct prime degree p polynomials with given number of real roots, and whose Galois groups

are isomorphic to the symmetric group Sp or the alternating group Ap ([13, Corollary 1.6]).

In this section, we extend this theorem as follows;

Theorem 3.2. Let r = (r0, · · · ,rs) ∈ Rs+1 be a vector such that gr(x) is a degree s separable
polynomial satisfying Ngr(x) = γ (0 ≤ γ ≤ s). Let us consider fr(t;x) = f(n)(r0, · · · ,rs,t;x) as a
polynomial over R(t) in x and put

Pr(t) = detMn(fr(t;x)) = detMn(fr(t;x),f
′
r(t;x)),

where f′r(t;x) is a derivative of fr(t;x) with respect to x. Then, for any real number ξ > αr =

max{α ∈ R | Pr(α) = 0}, we have

Nfr(ξ;x) =






γ + 1 if n − s : odd

γ if n − s : even, rs > 0

γ + 2 if n − s : even, rs < 0.

(3.3)



74 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

The above theorem can be restated as follows:

Corolary 1. Let f ∈ R(t)[x] be given by

f(t,x) = xn + t ·
s∑

k=0

ts−kx
s−k

and β1 < · · · < βm the distinct real roots of the degree s polynomial

P(t) :=
1

tn−1
∆(f,x)(t).

For any ξ > |βm|, the number of real roots of f(ξ,x) is

Nf(ξ,x) =






γ + 1 if n − s : odd

γ if n − s : even, ts > 0

γ + 2 if n − s : even, ts < 0.

where γ is the number or real roots of g(x) =
f(x)−xn

t
∈ R[x].

The rest of the section is concerned with proving Thm. 3.2.

3.1 The Bezoutian of f(t;x)

First, let us put

A(t0, · · · ,ts,t) = (aij(t0, · · · ,ts,t))1≤i,j≤n = Mn(f(t;x)) ∈ Symn(E2),
B(t0, · · · ,ts) = (bij(t0, · · · ,ts))1≤i,j≤s = Ms(g(x)) ∈ Syms(E1).

For ease of notation, we also write

A(t0, · · · ,ts,t) = A(t) = (aij(t))1≤i,j≤n, B(t0, · · · ,ts) = B = (bij)1≤i,j≤s

and we put B(t) = (bij(t))1≤i,j≤s = t
2B. Then, by Proposition 1, we have

A(t) = Mn(x
n
+ tg(x),nxn−1 + tg′(x))

= nMn(x
n,xn−1) − ntMn(x

n−1,g(x)) + tMn(x
n,g′(x)) + t2Mn(g(x),g

′(x))

= nMn(x
n,xn−1) − nt

s∑

k=0

ts−kMn(x
n−1,xs−k)

+ t

s−1∑

k=0

(s − k)ts−kMn(x
n,xs−k−1) + t2Mn(g(x),g

′(x)).

Lemma 3.1. Let λ,µ,ν be integers such that λ ≥ µ > ν ≥ 0. Then Mλ(xµ,xν) = (mij)1≤i,j≤λ,
where

mij =

{
1 i + j = 2λ − (µ + ν) + 1 (λ − µ + 1 ≤ i, j ≤ λ − ν),
0 otherwise.



CUBO
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Some remarks on the non-real roots of polynomials 75

Proof. By definition, we have

Bλ(x
µ,xν) =

xµyν − xνyµ

x − y

=

µ−ν∑

k=1

xµ−kyν+k−1 =

µ−ν∑

k=1

xλ−(λ−µ+k)yλ−(λ−ν−k+1),

which implies

mij =

{
1 (i, j) = (λ − µ + k,λ − ν − k + 1) (1 ≤ k ≤ µ − ν)
0 otherwise

=

{
1 i + j = 2λ − (µ + ν) + 1 (λ − µ + 1 ≤ i, j ≤ λ − ν),
0 otherwise.

This completes the proof.

Here, let us divide A(t) into two parts Â(t) and Ã(t), where

Â(t) = (âij(t))1≤i,j≤n = nMn(x
n,xn−1) − nt

s∑

k=0

ts−kMn(x
n−1,xs−k)

+ t

s−1∑

k=0

(s − k)ts−kMn(x
n,xs−k−1),

Ã(t) = (ãij(t))1≤i,j≤n = t
2Mn(g(x),g

′(x))

and put lk = n − s + k + 2 (= 2n − (n + s − k − 1) + 1). Then, by lemma 3.1, we have

{
â11(t) = n

â1,lk−1(t) = âlk−1,1(t) = (s − k)ts−kt (0 ≤ k ≤ s − 1).

Moreover, when i + j = lk, we have

âij(t) = −ntts−k + t(s − k)ts−k = −(lk − 2)ts−kt (2 ≤ i, j ≤ lk − 2, 0 ≤ k ≤ s). (3.4)

Remark 3.3. Note that, if s = n − 1, we have

−nt

s∑

k=0

ts−kMn(x
n−1,xs−k) = −nt

s∑

k=1

ts−kMn(x
n−1,xs−k),

Thus, when i + j = lk, equation (3.4) should be modified by

âij(t) = −ntts−k + t(s − k)ts−k = −(lk − 2)ts−kt (2 ≤ i, j ≤ lk − 2, 1 ≤ k ≤ s).

We avoid this minor defect by considering that there is no entries satisfying 2 ≤ i, j ≤ l0 −2 when
s = n − 1 since l0 − 2 = n − s = 1.



76 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

Proposition 4. Put lk = n − s + k + 2. Then

âij(t) =






n (i, j) = (1,1)

(s − k)ts−kt (i, j) = (1,lk − 1) or (lk − 1,1) (0 ≤ k ≤ s − 1)
−(lk − 2)ts−kt i + j = lk, 2 ≤ i, j ≤ lk − 2, (0 ≤ k ≤ s)
0 otherwise.

ãij(t) =

{
bi−(n−s),j−(n−s)t

2 n − s + 1 ≤ i, j ≤ n
0 otherwise.

Proof. The statement for âij(t) has just been proved. For ãij(t), it is enough to see that we can
denote

Ms(g(x)) =

s∑

ℓ=0

s∑

m=1

mtℓtmMs(x
ℓ,xm−1),

Mn(g(x)) =

s∑

ℓ=0

s∑

m=1

mtℓtmMn(x
ℓ,xm−1),

that is, we can obtain Mn(g(x)) from Ms(g(x)) by just replacing s with n for all Ms(x
ℓ,xm), which,

by Lemma 3.1, means that s × s matrix Ms(g(x)) occupies the part {b†ij | n − s + 1 ≤ i, j ≤ n} of
the matrix Mn(g(x)) = (b

†
ij)1≤i,j≤n.

By Proposition 4, we can express the matrix A(t) as follows;

A(t) =











































n 0 .. . 0 stst (s − 1)ts−1t . . . t1t

0 −(n − s)tst −(n − s + 1)ts−1t . . . −(n − 1)t1t −nt0t

.

.

.

...

...

...

...

0

0 −(n − s)tst

...

...

0 0

stst −(n − s + 1)ts−1t

(s − 1)ts−1t
.
.
.

...

...

C(t).
.
. −(n − 1)t1t

...

0

t1t −nt0t 0 0











































. (3.5)

Here, C(t) = (cij(t))1≤i,j≤s = C(t0, · · · ,ts,t) = (cij(t0, · · · ,ts,t))1≤i,j≤s is an s × s symmetric
matrix whose entries are of the form

cij(t0, · · · ,ts,t) = bijt2 + λijt
= bij(t0, · · · ,ts)t2 + λij(t0, · · · ,ts)t (λij = λij(t0, · · · ,ts) ∈ E1).

Next, let A(t)1 = (aij(t)1)1≤i,j≤n = A(t0, · · · ,ts,t)1 = (aij(t0, · · · ,ts,t)1)1≤i,j≤n be the n × n
symmetric matrix obtained from A(t) by multiplying the first row and the first column by 1/

√
n

and then sweeping out the entries of the first row and the first column by the (1,1) entry 1. Here,
let Qm(k;c) = (qij)1≤i,j≤m and Rm(k,l;c) = (rij)1≤i,j≤m be m × m elementary matrices such
that



CUBO
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Some remarks on the non-real roots of polynomials 77

Qm(k;c)=































1

...

1

c

1

...

1































, Rm(k,l;c)=

































1

...

1 c

...

1

...

1

































,

where qkk = c and rkl = c. Moreover, for any m×m matrices M1, M2, · · · , Ml, put
∏l

k=1 Mk =

M1M2 · · ·Ml. Then, we have A(t)1 = tS(t)1A(t)S(t)1, where

S(t)1 = Qn(1;1/
√
n)

s−1∏

k=0

Rn(1,lk − 1; −a1,lk−1(t)/
√
n).

The matrix A(t)1 can be expressed as follows;

A(t)1 =











































1 0 . . . 0 0 0 . . . 0

0 0 . . . −(n − s)tst −(n − s + 1)ts−1t . . . −(n − 1)t1t −nt0t

.

.

.
.
.
.

... ... ... ...

0

0 −(n − s)tst

... ...

0 0

0 −(n − s + 1)ts−1t

0
.
.
.

... ...

C(t)1..
. −(n − 1)t1t

...

0

0 −nt0t 0 0











































. (3.6)

Here, C(t)1 = (cij(t)1)1≤i,j≤s = C(t0, · · · ,ts,t)1 = (cij(t0, · · · ,ts,t)1)1≤i,j≤s is an s×s symmetric
matrix whose entries are of the form

cij(t0, · · · ,ts,t)1 = b̄ij(t0, · · · ,ts)t2 + λij(t0, · · · ,ts)t (b̄ij(t0, · · · ,ts) ∈ E1),

where

b̄ij(t0, · · · ,ts) = bij(t0, · · · ,ts) −
(s − i + 1)(s − j + 1)

n
ts−i+1ts−j+1 (3.7)

for any i, j (1 ≤ i, j ≤ s). We put b̄ij(t0, · · · ,ts) = b̄ij and B̄ = (b̄ij)1≤i,j≤s.

3.2 Some results for the Bezoutian of f
r
(t;x)

Let r = (r0, · · · ,rs) ∈ Rs+1 be a vector as in Theorem 3.2. We put

Ar(t) = (a
(r)
ij (t))1≤i,j≤n = A(r0, · · · ,rs,t) ∈ Symn(R(t)),

Br = (b
(r)
ij )1≤i,j≤s = B(r0, · · · ,rs) ∈ Syms(R)



78 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

and Br(t) = t
2Br. Let us also put Ar(t)1 = A(r0, · · · ,rs,t)1. By equation (3.6), the matrix Ar(t)1

can be expressed as follows;

Ar(t)1 =











































1 0 . . . 0 0 0 . . . 0

0 0 . . . −(n − s)rst −(n − s + 1)rs−1t . . . −(n − 1)r1t −nr0t

.

.

.
.
.
.

... ... ... ...

0

0 −(n − s)rst

... ...

0 0

0 −(n − s + 1)rs−1t

0
.
.
.

... ...

Cr(t)1..
. −(n − 1)r1t

...

0

0 −nr0t 0 0











































.

Here, Cr(t)1 = (c
(r)
ij (t)1)1≤i,j≤s = C(r0, · · · ,rs,t)1 and

c
(r)
ij (t)1 = b̄ij(r0, · · · ,rs)t

2 + λij(r0, · · · ,rs)t (b̄ij(r0, · · · ,rs),λij(r0, · · · ,rs) ∈ R).

Note that, by equation (3.7), we have

b̄ij(r0, · · · ,rs) = b
(r)
ij −

(s − i + 1)(s − j + 1)

n
rs−i+1rs−j+1 (1 ≤ i, j ≤ s).

To ease notation, we put b̄ij(r0, · · · ,rs) = b̄
(r)
ij and B̄r = (b̄

(r)
ij )1≤i,j≤s.

In particular, since

Ms(gr) = Ms

(

rsx
s,

s−1∑

k=0

(s − k)rs−kx
s−k−1

)

+ Ms

(

s∑

k=1

rs−kx
s−k,g′r

)

=

s−1∑

k=0

(s − k)rsrs−kMs(x
s,xs−k−1) + Ms

(

s∑

k=1

rs−kx
s−k,g′r

)

,

we have

b
(r)
1,k+1 = b

(r)
k+1,1 = (s − k)rsrs−k (0 ≤ k ≤ s − 1) (3.8)

by Lemma 3.1 and hence

b̄
(r)
1j = (s − j + 1)rsrs−j+1 −

s(s − j + 1)

n
rsrs−j+1 (3.9)

= (s − j + 1)
(

1 −
s

n

)

rsrs−j+1 (1 ≤ j ≤ s).

Lemma 3.2. Put B̄r(t) = t
2B̄r. Then, Br(ξ) and B̄r(ξ) are equivalent over R for any real number

ξ and we have σ(B̄r(ξ)) = Ngr for any non-zero real number ξ.

Proof. Let us denote by B∗r = (b
(r,∗)
ij )1≤i,j≤s (B̄

∗
r = (b̄

(r,∗)
ij )1≤i,j≤s) the matrix obtained from

Br (B̄r) by multiplying the first row and the first column by 1
/

±
√

b
(r)
11

(

1
/

±
√

b̄
(r)
11

)

(the sign



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 79

before

√

b
(r)
11

(

√

b̄
(r)
11

)

are the same as the sign of rs; see the definition of d
(

d̄
)

below) and then

sweeping out the entries of the first row and the first column by the (1,1) entry 1. Since b11 = sr
2
s

(> 0) and b̄11 = s(1 − s/n)r
2
s (> 0) by (3.8) and (3.9), we have

B∗r =
tTBrT, B̄

∗
r =

tT̄B̄rT̄, (3.10)

where

T = Qs(1;1/d)

s∏

k=2

Rs(1,k; −b
(r)
1k /d) (d =

√
s · rs),

T̄ = Qs(1;1/d̄)

s∏

k=2

Rs(1,k; −b̄
(r)
1k /d̄) (d̄ =

√

s(1 − s/n) · rs).

Note that in [13, Lemma 3.3], we have proved b
(r,∗)
ij = b̄

(r,∗)
ij (1 ≤ i, j ≤ s) and hence t

2B∗r = t
2B̄∗r,

which, by (3.10), implies that symmetric matrices Br(ξ) and B̄r(ξ) are equivalent over R for any
real number ξ. Then, since Ngr = σ(Br) = σ(Br(ξ)) for any ξ ∈ R \ {0}, the latter half of the
statement have also been proved.

3.3 Nonvanishingness of some coefficients

In this subsection, we prove the next lemma.

Lemma 3.3. Let

Φ(x) = Φ(t0, · · · ,ts;x) =
s∑

k=0

hs−k(t0, · · · ,ts)xs−k ∈ E1[x] (3.11)

be the characteristic polynomial of B̄. Then, hs−k(t0, · · · ,ts) is a non-zero polynomial in E1 for
any k (1 ≤ k ≤ s).

Proof. Lemma 3.3 is clear for s = 1, since we have

B = M1(t1x + t0) =
[

t21

]

and hence, by equation (3.7),

B̄ =

[

t21 −
1

n
t21

]

=

[

n − 1

n
t21

]

.

Next, suppose s ≥ 2. Then, by equation (3.7) and the definition of the Bezoutian, we have
hs−k(t0, · · · ,ts) ∈ R[t0, · · · ,ts] for any k (1 ≤ k ≤ s). Thus, we have only to prove that
hs−k(t0, · · · ,ts) 6= 0 for any k (1 ≤ k ≤ s), which is clear from the next Lemma 3.4.

Lemma 3.4. Suppose s ≥ 2 and put u0 = us = 1, u1 = t1 and uk = 0 (2 ≤ k ≤ s − 1).
Then, hs−k(u0, · · · ,us) is a non-constant polynomial in R(t1) for any k (1 ≤ k ≤ s), i.e.,
hs−k(u0, · · · ,us) ∈ R[t1] \ R (1 ≤ k ≤ s).



80 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

To prove lemma 3.4, let us put u = (u0, · · · ,us) and

gu(x) = g(u0, · · · ,us;x) = xs + t1x + 1 ∈ R(t1)[x],
fu(t;x) = x

n
+ tgu(x) ∈ R(t1,t)[x] (n > s),

Au(t) = (a
(u)
ij (t))1≤i,j≤n = A(u0, · · · ,us,t) ∈ Symn(R(t1,t)),

Bu = (b
(u)
ij )1≤i,j≤s = B(t0, · · · ,us) ∈ Syms(R(t1)), Bu(t) = t

2Bu.

Then, by equation (3.5), we have

Au(t) =











































n 0 .. . 0 st 0 . . . t1t

0 −(n − s)t 0 . . . −(n − 1)t1t −nt

.

.

.

... ... ... ...

0

0 −(n − s)t

... ...
0 0

st 0

0
.
.
.

... ...

Cu(t).
.
. −(n − 1)t1t

...

0

t1t −nt 0 0











































,

where Cu(t) = (c
(u)
ij (t))1≤i,j≤s = C(u0, · · · ,us,t) and

c
(u)
ij (t) = bij(u0, · · · ,us)t

2 + λij(u0, · · · ,us)t (λij(u0, · · · ,us) ∈ R(t1)).

Moreover, by equation (3.6), we also have

Au(t)1 =











































1 0 . . . 0 0 0 . . . 0

0 0 . . . −(n − s)t 0 . . . −(n − 1)t1t −nt

.

.

.
.
.
.

... ... ... ...

0

0 −(n − s)t

... ...

0 0

0 0

0
.
.
.

... ...

Cu(t)1..
. −(n − 1)t1t

...

0

0 −nt 0 0











































.

Here, Cu(t)1 = (c
(u)
ij (t)1)1≤i,j≤s = C(u0, · · · ,us,t)1 and

c
(u)
ij (t)1 = b̄ij(u0, · · · ,us)t

2
+ λij(u0, · · · ,us)t (b̄ij(u0, · · · ,us) ∈ R).

Note that, by equation (3.7), we have

b̄
(u)
ij =






b
(u)
11 − (s

2/n) (i, j) = (1,1)

b
(u)
1s − (s/n)t1 (i, j) = (1,s) or (s,1)

b
(u)
ss − (1/n)t

2
1 (i, j) = (s,s)

b
(u)
ij otherwise.

(3.12)



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 81

Let us put B̄u = (b̄
(u)
ij )1≤i,j≤s and B̄u(t) = t

2B̄u. Then, since

Ms(gu) = Ms(x
s + t1x + 1,sx

s−1 + t1)

= sMs(x
s,xs−1) + t1Ms(x

s,1) − st1Ms(x
s−1,x) − sMs(x

s−1,1)

+ t21Ms(x,1) + t1Ms(1,1),

we have

(a) if s = 2,

Bu =

[

2 t1

t1 t
2
1 − 2

]

,

(b) if s ≥ 3,

b
(u)
ij =






s (i, j) = (1,1)

t1 (i, j) = (1,s) or (s,1)

(1 − s)t1 i + j = s + 1, 2 ≤ i, j ≤ s − 1
−s i + j = s + 2

t21 (i, j) = (s,s),

0 otherwise,

which, by equation (3.12), implies

(a′) if s = 2,

B̄u =

[

2(n − 2)/n (n − 2)t1/n

(n − 2)t1/n (n − 1)t
2
1/n − 2

]

,

(b′) if s ≥ 3,

b̄
(u)
ij =






s(n − s)/n (i, j) = (1,1)

(n − s)t1/n (i, j) = (1,s) or (s,1)

(1 − s)t1 i + j = s + 1, 2 ≤ i, j ≤ s − 1
−s i + j = s + 2

(n − 1)t21/n (i, j) = (s,s),

0 otherwise.

Therefore, if s ≥ 3, the matrix B̄u = (b̄
(u)
ij )1≤i,j≤s has the expression of the form

































s(n − s)/n 0 0 0 · · · 0 (n − s)t1/n
0 0 0 · · · 0 (1 − s)t1 −s
0 0

...

(1 − s)t1 −s 0

0
...

... ... ... ... ...
... 0 (1 − s)t1

... ...

0

0 (1 − s)t1 −s

...

0

(n − s)t1/n −s 0 · · · 0 0 (n − 1)t21/n

































.



82 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

Here, let us denote by

Φu(x) =

s∑

k=0

h
(u)
s−kx

s−k = Φ(u0, · · · ,us;x)
(

=

s∑

k=0

hs−k(u0, · · · ,us)xs−k
)

the characteristic polynomial of B̄u. Note that since we have h
(u)
s−k ∈ R[t1] by the proof of Lemma

3.3, we have only to prove h
(u)
s−k is non-constant for any k (1 ≤ k ≤ s).

By the above expression of B̄u, we have

(a′′) if s = 2,

Φu(x) = x
2 −

(n − 1)t21 − 4

n
x +

(n − 2)t21 − 4n + 8

n
,

(b′′) if s ≥ 3,

Φu(x) =







































































































































































































∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

x − s(n − s)/n −(n − s)t1/n

x (s − 1)t1 s

. . .

...

s

. . .

...

...

x + (s − 1)t1 s

...

s x

... ... . . .

(s − 1)t1 s x

−(n − s)t1/n s x − (n − 1)t
2
1
/n

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

(s is odd),

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

x − s(n − s)/n −(n − s)t1/n

x (s − 1)t1 s

. . .

...

s

x (s − 1)t1

...

(s − 1)t1 x + s

...

... . . .

(s − 1)t1 s x

−(n − s)t1/n s x − (n − 1)t
2
1
/n

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

(s is even).

Example 3.1. (1) Put s = 7 and n = 10. Then, we have

gu(x) = x
7
+ t1x + 1, fu(t;x) = x

10
+ t(x7 + t1x + 1),



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 83

Φu(x) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

x − 21/10 0 0 0 0 0 −3t1/10

0 x 0 0 0 6t1 7

0 0 x 0 6t1 7 0

0 0 0 x + 6t1 7 0 0

0 0 6t1 7 x 0 0

0 6t1 7 0 0 x 0

−3t1/10 7 0 0 0 0 x − 9t
2
1
/10

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=x7 +

(

−
9

10
t21 + 6t1 −

21

10

)

x6 +

(

−
27

5
t31 −

351

5
t21 −

63

5
t1 − 147

)

x5

+

(

324

5
t41 −

2106

5
t31 +

1197

5
t21 − 588t1 +

3087

10

)

x4

+

(

1944

5
t51 +

5832

5
t41 +

5859

5
t31 +

16758

5
t21 +

6174

5
t1 + 7203

)

x3

+

(

−
5832

5
t61 +

34992

5
t51 −

21546

5
t41 +

50274

5
t31 −

95697

10
t21 + 14406t1 −

151263

10

)

x2

+

(

−
34992

5
t71 +

11664

5
t61 −

81648

5
t51 +

15876

5
t41 −

111132

5
t31 +

21609

5
t21 −

151263

5
t1

− 117649

)

x +
69984

5
t71 +

2470629

10
.

(2) Put s = 8 and n = 12. Then, we have

gu(x) = x
8 + t1x + 1, fu(t;x) = x

12 + t(x8 + t1x + 1)

and

Φu(x) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

x − 8/3 0 0 0 0 0 0 −t1/3

0 x 0 0 0 0 7t1 8

0 0 x 0 0 7t1 8 0

0 0 0 x 7t1 8 0 0

0 0 0 7t1 x + 8 0 0 0

0 0 7t1 8 0 x 0 0

0 7t1 8 0 0 0 x 0

−t1/3 8 0 0 0 0 0 x − 11t
2
1
/12

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=x8 +

(

−
11

12
t21 +

16

3

)

x7 +

(

−152t21 −
640

3

)

x6 +

(

539

4
t41 − 256t

2
1 − 1024

)

x5

+

(

22736

3
t41 +

45824

3
t21 + 16384

)

x4 +

(

−
26411

4
t61 −

22736

3
t41 +

31744

3
t21 + 65536

)

x3

+

(

−
355348

3
t61 − 213248t

4
1 −

1064960

3
t21 − 524288

)

x2 +

(

1294139

12
t81 +

1075648

3
t61

+
1404928

3
t41 +

1835008

3
t21 −

4194304

3

)

x −
823543

3
t81 +

16777216

3
.

Proof of Lemma 3.4. To prove Lemma 3.4, it is enough to prove degh
(u)
s−k ≥ 1 for any k (1 ≤ k ≤ s).

This is clear for s = 2 by (a′′) and we suppose s ≥ 3 hereafter. To prove degh(u)s−k ≥ 1 (1 ≤ k ≤ s),



84 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

let us compute the leading term of h
(u)
s−k (∈ R[t1]). Then, since h

(u)
s−k is the coefficient of the term

h
(u)
s−kx

s−k of the characteristic polynomial Φu(x), we need to maximize the degree in t1 when we
take ‘s − k’ x and the remaining k elements from R[t1].

(a) Suppose s is odd. Let us divide the case into three other sub-cases.

(a1) Suppose k is odd and 1 ≤ k ≤ s − 2.
In this case, the degree of the leading term of h

(u)
s−k is k + 1. In fact, it is obtained by taking

(a11) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(a12) ‘k − 1’ (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1).

First, suppose we take the (s,s) entry x − (n − 1)t21/n from the s-th row. Then we must
take the (1,1) entry from the first row. Next, let us proceed to the (s − 1)-th row. If we take the
(s −1,s− 1) entry x from the (s−1)-th row, then we must also take x from the second row, while
if we take (s−1)t1 from the (s−1)-th row, then we must also take (s−1)t1 from the second row.
The situation is the same for the (s − 2)-th row, the (s − 3)-th row ... and so on, which implies
that (s − 1)t1 must occur in pair.

Hence, the leading term of h
(u)
s−k is

−
n − 1

n
t21 ·

(

(s − 3)/2

(k − 1)/2

)

{(−1) · (s − 1)2t21}
(k−1)/2

((

n

0

)

= 1 (n ≥ 0)
)

and the degree of this term is k + 1 (≥ 2).
(a2) Suppose k is odd and k = s.

If k = s, h
(u)
s−k = h

(u)
0 is the constant term of Φu(x). In this case, the degree of the leading term

of h
(u)
0 is s. In fact, it is obtained by taking

(a21) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(a22) If s ≥ 5 (⇔ (s,k) 6= (3,3)), ‘(s−3)/2’ pairs of (s−1)t1 from entries of the form (i,s+1−i)
(2 ≤ i ≤ (s − 1)/2, (s + 3)/2 ≤ i ≤ s − 1),

(a23) (s − 1)t1 from the ((s + 1)/2,(s + 1)/2) entry x + (s − 1)t1,

(a24) −s(n − s)/n from the (1,1) entry x − s(n − s)/n

or by taking

(a25) all anti-diagonal entries.



CUBO
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Some remarks on the non-real roots of polynomials 85

Therefore, the leading term of h
(u)
0 is

−
n − 1

n
t21 · {(−1) · (s − 1)

2t21}
(s−3)/2 · (s − 1)t1 ·

(

−
s(n − s)

n

)

+ (−1) ·
(

−
n − s

n
t1

)2

· {(−1) · (s − 1)2t21}
(s−3)/2 · (s − 1)t1

=
(n − s)(s − 1)

n
· (−1)(s−3)/2(s − 1)s−2ts1

= (−1)(s−3)/2
(n − s)(s − 1)s−1

n
ts1

for any s (s ≥ 3) and the degree of this term is s.
(a3) Suppose k is even.

In this case, we have 2 ≤ k ≤ s − 1 and the degree of the leading term of h(u)s−k is k + 1. In fact, it
is obtained by taking

(a31) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(a32) If s ≥ 5 (⇔ (s,k) 6= (3,2)), ‘(k−2)/2’ pairs of (s−1)t1 from entries of the form (i,s+1− i)
(2 ≤ i ≤ (s − 1)/2, (s + 3)/2 ≤ i ≤ s − 1),

(a33) (s − 1)t1 from the ((s + 1)/2,(s + 1)/2) entry x + (s − 1)t1.

Therefore, the leading term of h
(u)
s−k is

−
n − 1

n
t21 ·

(

(s − 3)/2

(k − 2)/2

)

{(−1) · (s − 1)2t21}
(k−2)/2 · (s − 1)t1

for any s (s ≥ 3) and the degree of this term is k + 1 (≥ 3).
(b) Suppose s is even (s ≥ 4). We also divide this case into three other sub-cases.
(b1) Suppose k is odd.

In this case, we have 1 ≤ k ≤ s − 1 and the degree of the leading term of h(u)s−k is k + 1. In fact, it
is obtained by taking

(b11) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(b12) ‘(k − 1)/2’ pairs of (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1).

Therefore, the leading term of h
(u)
s−k is

−
n − 1

n
t21 ·

(

(s − 2)/2

(k − 1)/2

)

{(−1) · (s − 1)2t21}
(k−1)/2

and the degree of this term is k + 1 (≥ 2).
(b2) Suppose k is even and 2 ≤ k ≤ s − 2.
In this case, the degree of the leading term of h

(u)
s−k is k. In fact, it is obtained by taking



86 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

(b21) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(b22) ‘(k − 2)/2’ pairs of (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1),

(b23) −s(n − s)/n from the (1,1) entry x − s(n − s)/n

or by taking

(b24) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(b25) I f s ≥ 6 (⇔ (s,k) 6= (4.2)), ‘(k−2)/2’ pairs of (s−1)t1 from entries of the form (i,s+1−i)
(2 ≤ i ≤ (s − 2)/2, (s + 4)/2 ≤ i ≤ s − 1),

(b26) s from the ((s + 2)/2,(s + 2)/2) entry x + s

or by taking

(b27) ‘k/2’ pairs of (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1)

or by taking

(b28) One pair of −(n − s)t1/n from the (1,s) and the (s,1) entry,

(b29) ‘(k − 2)/2’ pairs of (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1).

Here, note that if we take the (s,1) entry −(n − s)t1/n from the s-th row, we must also take the
(1,s) entry −(n − s)t1/n from the first row.

Therefore, the leading term of h
(u)
s−k is

−
n − 1

n
t21 ·

(

(s − 2)/2

(k − 2)/2

)

{(−1) · (s − 1)2t21}
(k−2)/2

·

(

−
s(n − s)

n

)

−
n − 1

n
t21 ·

(

(s − 4)/2

(k − 2)/2

)

{(−1) · (s − 1)2t21}
(k−2)/2

· s +
(

(s − 2)/2

k/2

)

{(−1) · (s − 1)2t21}
k/2

+

(

(−1) ·
{−(n − s)}2

n2
t21

)

·

((s − 2)/2

(k − 2)/2

)

{(−1) · (s − 1)2t21}
(k−2)/2

=

(

s(n − s)(n − 1)

n2

((s − 2)/2

(k − 2)/2

)

−
s(n − 1)

n

((s − 4)/2

(k − 2)/2

)

− (s − 1)2
((s − 2)/2

k/2

)

−
(n − s)2

n2

((s − 2)/2

(k − 2)/2

)

)

{(−1) · (s − 1)2t21}
(k−2)/2t21.

for any s (s ≥ 4). Then, since
((s − 4)/2

(k − 2)/2

)

=
s − k

s − 2

((s − 2)/2

(k − 2)/2

)

,
((s − 2)/2

k/2

)

=
s − k

k

((s − 2)/2

(k − 2)/2

)

,

we have



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 87

s(n − s)(n − 1)

n2

((s − 2)/2

(k − 2)/2

)

−
s(n − 1)

n

((s − 4)/2

(k − 2)/2

)

(3.13)

− (s − 1)2
((s − 2)/2

k/2

)

−
(n − s)2

n2

((s − 2)/2

(k − 2)/2

)

=

(

s(n − s)(n − 1)

n2
−
s(s − k)(n − 1)

n(s − 2)
−

(s − 1)2(s − k)

k
−

(n − s)2

n2

)

((s − 2)/2

(k − 2)/2

)

=
s{
(

k(k + s2 − 4s + 2) − s3 + 4s2 − 5s + 2
)

n − k(k + s2 − 4s + 2)}

nk(s − 2)

((s − 2)/2

(k − 2)/2

)

.

Hence, if the above value becomes zero, we have

(

k(k + s2 − 4s + 2) − s3 + 4s2 − 5s + 2
)

n − k(k + s2 − 4s + 2) = 0,

which implies

k(k + s2 − 4s + 2) = 0, −s3 + 4s2 − 5s + 2 = 0 (3.14)

or

n =
k(k + s2 − 4s + 2)

k(k + s2 − 4s + 2) − s3 + 4s2 − 5s + 2
. (3.15)

Here, (3.14) is impossible since −s3 + 4s2 − 5s + 2 = −(s − 1)2(s − 2) and s ≥ 4. Also, (3.15) is
impossible since, for any s ≥ 4 and 2 ≤ k ≤ s − 2, we have

k(k + s2 − 4s + 2) ≥ 2(2 + s2 − 4s + 2) ≥ 2(s − 2)2 > 0

and

k(k + s2 − 4s + 2) − s3 + 4s2 − 5s + 2

≤ (s − 2){(s − 2) + s2 − 4s + 2} − s3 + 4s2 − 5s + 2
= −s2 + s + 2

= −(s + 1)(s − 2) < 0,

which implies n < 0, a contradiction. Thus, the above value (3.13) is non-zero and the degree of

the leading term of h
(u)
s−k is k.

(b3) Suppose k is even and k = s.

If k = s, h
(u)
s−k = h

(u)
0 is the constant term of Φu(x). In this case, the degree of the leading term

of h
(u)
0 is s. In fact, it is obtained by taking

(b31) −(n − 1)t21/n from the (s,s) entry x − (n − 1)t
2
1/n,

(b32) ‘(s − 2)/2’ pairs of (s − 1)t1 from entries of the form (i,s + 1 − i) (2 ≤ i ≤ s − 1),

(b33) −s(n − s)/n from the (1,1) entry x − s(n − s)/n

or by taking



88 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

(b34) all anti-diagonal entries.

Therefore, the leading term of h
(u)
0 is

−
n − 1

n
t21 · {(−1) · (s − 1)

2t21}
(s−2)/2 ·

(

−
s(n − s)

n

)

+ (−1) ·
(

−
n − s

n
t1

)2

· {(−1) · (s − 1)2t21}
(s−2)/2

= (−1)(s−2)/2
(n − s)(s − 1)s−1

n
ts1

and the degree of this term is s (s ≥ 4).

Lemma 3.5. Let v = (v0, · · · ,vs) ∈ Rs+1 be a real vector and n (> s) be an integer. Put

Pv(t) = detMn(fv(t;x)) = detMn(f
(n)

(v0, · · · ,vs,t;x))

and αv = max{α ∈ R | Pv(α) = 0}. If there exists a real number ρ0 (> αv) such that Nfv(ξ;x) = γ0
for any ξ > ρ0, we have Nfv(ξ;x) = γ0 for any ξ > αv.

Proof. Put Av(t) = Mn(fv(t;x)). Then, by Proposition 2, we have γ0 = σ(Av(ξ)) for any ξ > ρ0.
Let us also put

R = {ρ ∈ R | ρ > αv, σ(Av(ξ)) = γ0 for any ξ > ρ}.

Since R is a nonempty set (ρ0 ∈ R) having a lower bound αv, R has the infimum ρv; ρv = inf R.
Then, it is enough to prove ρv = αv. Here, suppose to the contrary that ρv > αv and we denote
by

Ωv(t;x) =

n∑

k=0

ωk(t)x
k ∈ R(t)[x]

the characteristic polynomial of Av(t). Note that ωk(t) ∈ R[t] (0 ≤ k ≤ n) and for any ξ >
αv, Ωv(ξ;x) has n non-zero real roots (counted with multiplicity) since Av(ξ) is symmetric and
detAv(ξ) 6= 0. Then, by Proposition 3, there exists a positive real number δ such that ρv −δ > αv
and for any ξ ∈ [ρv − δ,ρv + δ], Ωv(ξ;x) has the same number of positive and hence negative
real roots with Ωv(ρv;x). On the other hand, since ρv = inf R, there exist real numbers ξ+
(ρv < ξ+ < ρv + δ) and ξ− (ρv − δ < ξ− < ρv) such that σ(Av(ξ+)) 6= σ(Av(ξ−)), which implies
Ωv(ξ+;x) and Ωv(ξ−;x) have different number of positive and hence negative real roots. This is
a contradiction and we have ρv = αv.

3.4 Proof of Theorem 3.2

Let r = (r0, · · · ,rs) ∈ Rs+1 be the vector as in Theorem 3.2 and put

n0 =

{
(n − s + 1)/2, n − s − 1 : even

(n − s + 2)/2, n − s − 1 : odd.



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 89

When n − s ≥ 2, we inductively define the matrix Ar(t)k = (a
(r)
ij (t)k)1≤i,j≤n (2 ≤ k ≤ n − s)

as the matrix obtained from Ar(t)k−1 by sweeping out the entries of the k-th row (k-th column)
by the (k,l0 − k) entry −(n − s)rst ((l0 − k,k) entry −(n − s)rst). That is, we define Ar(t)k =
tSr(t)kAr(t)k−1Sr(t)k, where

Sr(t)k =






n∏

m=l0−k+1

Rn

(

l0 − k,m; −
a
(r)
km(t)k−1

−(n − s)rst

)

(2 ≤ k ≤ n0)

Rn

(

l0 − k,k; −
a
(r)
kk (t)k−1

−2(n − s)rst

)

n∏

m=k+1

Rn

(

l0 − k,m; −
a
(r)
km(t)k−1

−(n − s)rst

)

(n0 < k ≤ n − s).

Then, if n − s ≥ 1, we can express the matrix Ar(t)n−s as follows;

Ar(t)n−s =



































1 0 . . . 0

0 0 . . . −(n − s)rst
...

...

...

0 O
0 −(n − s)rst 0 0

O Cr(t)n−s



































.

Note that a
(r)
km(t)k−1 and a

(r)
kk (t)k−1 appearing in Sr(t)k are degree 1 monomials in t and hence

the numbers −a
(r)
km(t)k−1/(−(n− s)rst), −a

(r)
kk (t)k−1/(−2(n − s)rst) appearing in Sr(t)k are just

real numbers. Therefore, the entries of the s×s symmetric matrix Cr(t)n−s = (c
(r)
ij (t)n−s)1≤i,j≤s

(n − s ≥ 1) are of the form

c
(r)
ij (t)n−s = b̄

(r)
ij t

2
+ λ̄

(r)
ij t (λ̄

(r)
ij ∈ R). (3.16)

Moreover, since the matrix

Dr(t)n−s =















1 0 . . . 0

0 0 . . . −(n − s)rst
...

...

...

0

0 −(n − s)rst 0 0

















90 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

is equivalent to the matrix

D̄r(t)n−s =





























































































































































1

0 −(n − s)rst

−(n − s)rst 0

..
.

0 −(n − s)rst

−(n − s)rst 0























(n − s : odd)


























1

−(n − s)rst

0 −(n − s)rst

−(n − s)rst 0

..
.

0 −(n − s)rst

−(n − s)rst 0



























(n − s : even)

over R, we have

σ(Dr(ξ)n−s) = σ(D̄r(ξ)n−s) =






1 n − s : odd

0 n − s : even, rs > 0

2 n − s : even, rs < 0

(3.17)

for any real number ξ > αr (≥ 0). Here, note that since Pr(0) = 0, we have αr ≥ 0.
Next, let Φr(t;x), Ψr(t;x) be characteristic polynomials of B̄r(t), Cr(t)n−s, respectively. Then,

by equations (3.11) and (3.16), we have

Φr(t;x) = x
s + h

(r)
s−1t

2xs−1 + · · · + h(r)1 t
2s−2x + h

(r)
0 t

2s

(

h
(r)
s−k = hs−k(r0, · · · ,rs) ∈ R (1 ≤ k ≤ s)

)

,

Ψr(t;x) = x
s +

(

h
(r)
s−1t

2 + ψs−1(t)
)

xs−1 + · · ·

+
(

h
(r)
1 t

2s−2 + ψ1(t)
)

x +
(

h
(r)
0 t

2s + ψ0(t)
)

(ψ0(t), · · · ,ψs−1(t) ∈ R[t],degψs−k(t) < 2k (1 ≤ k ≤ s)) .

Here, let us divide the proof into next two cases.

(i) The case h
(r)
0 h

(r)
1 · · ·h

(r)
s−1 6= 0.

In this case, we have

Ψr(t;x) = x
s + h

(r)
s−1t

2

(

1 +
ψs−1(t)

h
(r)
s−1t

2

)

xs−1 + · · ·

+ h
(r)
1 t

2s−2

(

1 +
ψ1(t)

h
(r)
1 t

2s−2

)

x + h
(r)
0 t

2s

(

1 +
ψ0(t)

h
(r)
0 t

2s

)



CUBO
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Some remarks on the non-real roots of polynomials 91

and 1+ψs−k(t)
/

h
(r)
s−kt

2k → 1 (t → ∞) for any k (1 ≤ k ≤ s). Moreover, since h(r)0 h
(r)
1 · · ·h

(r)
s−1 6= 0,

we have h
(r)
0 6= 0, which implies that for any non-zero real number ξ, Φr(ξ;x) have s non-zero real

roots (counted with multiplicity). Thus, there exists a real number ρ0 (> αr) such that for any
real number ξ > ρ0, Ψr(ξ;x) have the same number of positive (hence also negative) real roots
with Φr(ξ;x) by Proposition 3, which implies σ(Cr(ξ)n−s) = σ(B̄r(ξ)) and hence σ(Cr(ξ)n−s) =
Ngr = γ (ξ > ρ0) by Lemma 3.2. Then, by the equation (3.17), we have

σ(Ar(ξ)n−s) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0

for any ξ > ρ0, which implies

Nfr(ξ;x) = σ(Ar(ξ)) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0

for any ξ > ρ0 since Ar(ξ) and Ar(ξ)n−s are equivalent over R. Hence, by Lemma 3.5, we have

Nfr(ξ;x) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0

for any ξ > αr.

(ii) General case.

Let ε0 be a positive real number and for any vector v ∈ Rs+1, set

α′v = max{|α| | α ∈ C,Pv(α) = 0}.

Clearly, we have α′v ≥ αv for any v ∈ Rs+1. Here, let us put ρ′0 = α′r + ε0. Then, by Lemma 3.5,
it is enough to prove the next claim.

Claim 1. For any real number ξ > ρ′0, we have

Nfr(ξ;x) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0.

Proof. By the assumption that gr(x) is a separable polynomial of degree s and the fact that the
non-real roots must occur in pair with its complex conjugate, there exists a real number δ0 such
that for any vector v = (v0, · · · ,vs) ∈ Rs+1 satisfying |r − v|0 = max0≤k≤s{|rk − vk|} < δ0, gv(x)
is also a degree s separable polynomial satisfying Ngv = Ngr = γ by Proposition 3.

(S1) If a vector v ∈ Rs+1 satisfies |r − v|0 < δ0, then gv(x) is also a degree s
separable polynomial satisfying Ngv = Ngr = γ.



92 Shuichi Otake and Tony Shaska CUBO
20, 2 (2018)

Next, we put

P(t) =
∑

k≥0

xk(t0, · · · ,ts)tk = detA(t) (A(t) = A(t0, · · · ,ts,t))

and let us consider P(t) as a polynomial over E1 = R(t0, · · · ,ts) in t. Then, since xk(t0, · · · ,ts) ∈
R[t0, · · · ,ts] for any k ≥ 0, there exists a real number δ1 > 0 such that for any vector v ∈ Rs+1
satisfying |r − v|0 < δ1, we have |α

′
r − α

′
v| < ε0 by Proposition 3;

(S2) If a vector v ∈ Rs+1 satisfies |r − v|0 < δ1, we have |α′r − α
′
v| < ε0.

Here, let ξ be any real number such that ξ > ρ′0 = α
′
r + ε0 and let

Ω(t0, · · · ,ts,ξ;x) =
n∑

k=0

yk(t0, · · · ,ts)xk ∈ E1[x]

be the characteristic polynomial of the Bezoutian

A(t0, · · · ,ts,ξ;x) = Mn(f(n)(t0, · · · ,ts,ξ;x),f(n)(t0, · · · ,ts,ξ;x)′).

Here, f(n)(t0, · · · ,ts,ξ;x)′ is the derivative of

f(n)(t0, · · · ,ts,ξ;x) =
n∑

k=0

zk(t0, · · · ,ts)xk ∈ E1[x]

with respect to x. Then, since zk(t0, · · · ,ts) ∈ R[t0, · · · ,ts] (0 ≤ k ≤ n), we also have yk(t0, · · · ,ts) ∈
R[t0, · · · ,ts] (0 ≤ k ≤ n). Moreover, since ξ > ρ′0 > αr, we have detAr(ξ) = detA(r0, · · · ,rs,ξ) 6=
0.

By these arguments, we can also deduce that there exists a positive real number δ2 such that
for any vector v ∈ Rs+1 satisfying |r − v|0 < δ2, the characteristic polynomial Ωv(ξ;x) have the
same number of positive and hence negative real roots with Ωr(ξ;x) (counted with multiplicity),
which implies Nfr(ξ;x) = σ(Ar(ξ)) = σ(Av(ξ)) = Nfv(ξ;x).

(S3) If a vector v ∈ Rs+1 satisfies |r − v|0 < δ2, we have Nfr(ξ;x) = Nfv(ξ;x).

Put δ = min{δ0,δ1,δ2} > 0. Then, there exists a vector w = (w0, · · · ,ws) ∈ Rs+1 such that

(a) |r − w|0 < δ, (b) h
(w)
0 h

(w)
1 · · ·h

(w)
s−1 6= 0.

Here, we put h
(w)
s−k = hs−k(w0, · · · ,ws) for any k (1 ≤ k ≤ s). In fact, since hs−k(t0, · · · ,ts) is a

non-zero polynomial for any k (1 ≤ k ≤ s) by Lemma 3.3, the product
∏s

k=1 hs−k(t0, · · · ,ts) is
also non-zero, which implies that there exists a vector w ∈ Rs+1 satisfying (a) and (b).

Let w ∈ Rs+1 be the vector as above. Then, since |r − w|0 < δ ≤ δ0, gw(x) is a degree s
separable polynomial satisfying Ngw = γ by (S1) and also, by (S2), we have αw ≤ α′w < α′r +ε0 =
ρ′0 < ξ. Thus, by (b) and the case (i), we have

Nfw(ξ;x) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0,



CUBO
20, 2 (2018)

Some remarks on the non-real roots of polynomials 93

which, by (S3), implies

Nfr(ξ;x) =






γ + 1 n − s : odd

γ n − s : even, rs > 0

γ + 2 n − s : even, rs < 0.

Since ξ is any real number such that ξ > ρ′0, this completes the proof of Claim and hence the proof
of Theorem 3.2.

Proposition 5. Let g(x) =
∑s

i=0 aix
i be a polynomial in R[x] such that ∆g 6= 0 and

f(t,x) = xn + t · g(x) (3.18)

If g(x) is totally complex, (n − s) is even, and as > 0 then f(β,x) is totally complex for all
β > max{α |∆(f,x)(α) = 0}.

Proof. We have to show that f(β,x) has no real roots. Since g(x) is totally complex we have that
γ = 0. Nf(β,x) = γ as β > max{α |∆(f,x)(α) = 0} and as > 0, so Nf(β,x) = γ = 0. Hence, f(β,x)
is totally complex.

Let K := Q(t,a0, . . . ,as) be the field of transcendental degree s + 1 and g(x) =
∑s

i=0 aix
i.

Then we have the following.

Corolary 2. Let K := Q(t,a0, . . . ,as) be the field of transcendental degree s+1, g(x) =
∑s

i=0 aix
i

and

f(t,x) = xn + t · g(x)

For any value of (λ0, . . . ,λs) ∈ Zs+1, if g(λ0, . . . ,λs,x) ∈ Z[x] is irreducible and satisfies the
conditions of the Eisenstein criteria, then f(x) is irreducible, over Q.

We also note:

Remark 3.4. It can be verified computationally by Maple that if n ≤ 9 and 1 ≤ s < n then the
Galois group Gal K(f,x) is isomorphic to Sn.

Remark 3.5. Polynomials in Eq. (3.18) for s = 1 and t = 1 has been treated by Y. Zarhin in
[18] while studying Mori trinomials. It is shown there that the Galois group of f(x) over Q is
isomorphic to Sn; see [18, Cor. 3.5] for details.

In general, if we let K := Q(t,a0, . . . ,as) be the field of transcendental degree s + 1, for
1 ≤ s < n, then we expect that Gal K(f) ∼= Sn for all n ≥ 1. If true, this would generalize Zarhin’s
result to a more general class of polynomials.

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	Introduction
	Preliminaries
	On the number of real roots of polynomials
	The Bezoutian of f(t; x)
	Some results for the Bezoutian of fr(t ; x)
	Nonvanishingness of some coefficients
	Proof of Theorem ??