MAHLER'S MEASURE AND ITS RELATED TWO OPEN PROBLEMS

by

Yun Wang
B. Sc., University of Northern British Columbia, 2022

THESIS SUBMITTED IN PARTIAL FULFILMENT OF
THE REQUIREMENTS OF THE DEGREE OF
MASTER OF SCIENCE
IN
MATHEMATICS

UNIVERSITY OF NORTHERN BRITISH COLUMBIA
April 2024

© Yun Wang, 2024

Abstract

√
The main achievement of the thesis is the proof that 1 + 17 is not a Mahler measure
of an algebraic number. This answers a question of A. Schinzel posted in [6] in 2004.
We also show that, theoretically, there exists an algorithm to reduce the shortness of
a polynomial without changing its Mahler measure, a problem considered in [5] by
J. McKee and C. Smyth. However the number of computations required makes this
algorithm infeasible.

ii

Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ii

Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

iii

1 Introduction

1

√

2 Is 1 + 17 a Mahler measure of an algebraic number?

2

3 Does there exists an algorithm to reduce the shortness of a polynomial without
changing its Mahler measure?

19

A Complementary facts

23

iii

Chapter 1
Introduction
My graduate thesis is about Mahler measure. In this paper, I plan to solve two relevant
open problems. We start by the key denition.

Denition 1 Let P(z) = a0 zd + a1 zd−1 + . . . + ad ∈ Z[z] with a0 ̸= 0, and let its zeros in
C be α1 , α2 , . . . , αd . The Mahler measure, M(P), is dened to be the product of |a0 | and
all |αi | for which |αi |> 1, where 1 ≤ i ≤ d .

M(P) = |a0 |∏|αi |>1 |αi |
The Mahler measure of an algebraic number α is denoted by M(α).

1

Chapter 2
√
Is 1 + 17 a Mahler measure of an algebraic number?
This question was asked by A.Schinzel [6] and quoted by A.Dubickas [2] ,J.McKee and
C.Smyth [5], P.A.Fili, L.Potmeyer, and M.Zhang [3], among others.
The Mahler measure of an algebraic number is dened as the Mahler measure of its
minimal polynomial over Z.

Lemma 1 Let OK be the ring of algebraic integers of a number eld K. If
n

f (x) = a ∏(x − αi ) ∈ OK [x]
i=1

then aα1 . . . αs is an algebraic integer for 1 ≤ s ≤ n.
The following proof will make this lemma more clear.

Proof 1 f (x) = a ∏ni=1 (x − αi ) = axn − aσ1 xn−1 + · · · + (−1)n aσn . It's trivial to see that
σ1 = α 1 + α 2 + · · · + α n ;

σ2 = α1 α2 + α1 α3 + · · · + α1 αn + α2 + · · · + α3 αn−1 αn ;

2

Chapter 2

3
...
σn = α1 α2 . . . αn .

The numbers a, σ1 , σ2 , σn are all algebraic numbers because f (x) ∈ OK [x].
In order to prove that aα1 . . . αs is an algebraic integer, it suces to show that it is a
root of a nonzero monic polynomial in OK [x]. It is not dicult to check that

F(x) = ∏ (x − aαρ(1) . . . αρ(s) )
ρ∈Sn

is such polynomial. Here Sn is the permutation group on the set of n elements. Indeed,the coecients of F are sums of symmetric functions in α1 , . . . , αn multiplied by
powers of a. Each monomial in these functions is of the form ak αni11 . . . αnimm with some
positive integer m and k ≥ max{n1 , . . . , nm }. By the Fundamental Theorem of Symmetric Functions, the coecients of F are polynomials in elementary symmetric functions

σ1 , . . . , σn of the roots of f . Further, by examining the standard procedure for the conversion of a symmetric function into a polynomial in elementary symmetric functions,
as outlined, for example in [4] we conclude that the monomials occurring in these polymn
1
nomials are of the form ak σm
1 . . . σn , where k, m1 , . . . , mn are nonnegative integers, and

k ≥ m1 +· · ·+mn , hence they are algebraic integers because f ∈ OK [x] and, consequently,
aσi , i = 1 . . . n are algebraic integers.

By using this lemma, we have the following corollary.

Corollary 1 If f ∈ Z[x] is a nonzero polynomial,then M( f ), the Mahler's measure of f ,
is an algebraic integer.

√

Proposition 1 Let K = Q( d), where d > 1 is a square-free integer then every element
of OK has the form

Chapter 2

4

⎧
√
⎪
⎪
⎨β = b + c d,

if d ̸≡ 1

mod 4

√
⎪
⎪
⎩β = b + c 1+ d ,
2

if d ≡ 1

mod 4.

,

where b and c are any rational integers.

Proposition 2 Let β > 1 be an irrational algebraic integer in a real quadratic eld
√
Q( d) and let β′ be its algebraic conjugate. If |β′ |≤ 1, then β is a Mahler's measure
of a monic irreducible polynomial in Z[x].

Proof 2 M( f ) = β for f (x) = (x − β)(x − β′ ). We can see f ∈ Z[x] because its coecients
are symmetric functions of algebraic numbers.
The case of |β′ |> 1 is more interesting. In this direction we have the following
theorem.

√

Theorem 1 Let β > 1 be an irrational algebraic integer in a real quadratic eld Q( d)
and suppose that |β′ |> 1. Then β is not a Mahler's measure for any irreducible, monic
polynomial in Z[x].

Proof 3 For a contradiction, suppose that f ∈ Z[x] is a monic irreducible polynomial
and M( f ) = β. Let
n

f (x) = ∏(x − αi ).
i=1

Suppose that |αi |> 1 for 1 ≤ i ≤ s and |αi |≤ 1 for s + 1 ≤ i ≤ n. By the denition of
Mahler measure, we have
s

β = M( f ) = ∏|αi |, 1 ≤ i ≤ s.
i=1

Next,we claim that

β = εα1 . . . αs where ε ∈ {−1, +1}.

Chapter 2

5

For this note that if |αi |> 1 and αi ∈ C \ R then |ᾱi | is also greater than 1, so it
occurs in the product α1 . . . αs . Hence this product is a real number and, consequently

β = ±α1 . . . αs .
Further, we must have s < n, since otherwise M( f ) will be equal to the constant
term in f (x) which is a rational integer.
For convenience we will denote the conjugates αs+1 , . . . , αn by α′1 , . . . , α′r , so that

n = s + r.
√
Let L = Q(α1 , . . . , αn ) and K = Q( d), so

Q ⊂ K ⊂ L.
Then L/K and L/Q are Galois extensions. A nite extension L/K is called a Galois
extension if |Aut(L/K)|= [L : K]. Let G = Gal(L/Q) and H = Gal(L/K). Let D = |G| be
the order of G. Then |H|= D2 , as [K : Q] = 2. In particular, D is even. Then we have

σ(β) = β for every σ ∈ H
while

σ(β) = β′ for every σ ∈ G \ H.
The rst statement implies that

Every σ ∈ H is a permutation of the set S = {α1 , . . . , αs }
and a permutation of the set R = {α′1 , . . . , α′r }

(2.1)

In order to see this, notice that |α1 . . . αs | is strictly larger than absolute value of any

Chapter 2

6

other product of s conjugates from the set {α1 , . . . , αn } because |αi |> 1 for i = 1 . . . s.
Since |σ(β)|= |β|= |∏si=1 σ(α1 ) . . . σ(αs )|, all σ(αi ) for 1 ≤ i ≤ s must belong to S. Fur-

/
ther, σ is one-to-one from S to S, so a permutation of S. This implies that σ(R) ∩ S = 0,
so that σ is also a permutation of R.

Now consider

L = ∏ σ(α1 . . . αs ) = ∏ σ(α1 ) . . . σ(αs ).
σ∈G

σ∈G

The Galois group G acts transitively on the set {α1 , . . . , αn } because f is irreducible.
Clearly L is a product of conjugates and because of transitivity, each conjugate occurs
in L with the same frequency. Hence
sD

sD

L = (α1 . . . αn ) n = ((−1)n an ) n ,
where D = |G| is the order of G, and an = f (0) is the constant term of f .
Observe that G is a disjoint union of two cosets G = H

⋃︁

σ(H), where σ is any auto-

morphism from G \ H.
By α1 . . . αs = εβ and σ(β) = β for σ ∈ H and σ(β) = β′ for σ ∈ G \ H we get

∏ β′.

L = ∏ σ(εβ) = ∏ εβ
σ∈H

σ∈G

σ∈G\H

With our notation for β we have ββ′ = N(β) =

⎧
⎪
⎪
⎨b2 − c2 d,

if d ̸≡ 1 mod 4

⎪
2
2
⎪
⎩ (2b+c) −c d ,

if d ≡ 1 mod 4

4

Hence

D

D

D

sD

L = (β) 2 (β′ ) 2 = N(β) 2 = ((−1)n an ) n

.

Chapter 2

7

We get
2s

|an | n = |N(β)|= |ββ′ |> |β| because |β′ |> 1.
However |an |= |α1 . . . αs ||αs+1 . . . αn |≤ β. Thus

n

β ≥ |an |> β 2s
and we conclude that 2s > n, so 2s > s + r, s > r.
We shall show that the last inequality, s > r, together with (2.1) contradicts the irreducibility of f .
For this let
s

r

i=1

i=1

f1 (x) = ∏(x − αi ) and f2 (x) = ∏(x − α′i ).
The coecients of these polynomials are symmetric functions of {α1 , . . . , αs } and

{α′1 , . . . , α′r }, respectively. Hence by (2.1) they are preserved by any σ from H, also
they are algebraic integers. By Galois theory we conclude that the coecients of both
polynomials are algebraic integers in the eld K. Now, let σ be any automorphism in

G \ H, then fi (x)σ( fi (x)) for i = 1 and i = 2 are in Z[x], because σ(K) = K, and its
restriction to K is the nonidentity automorphism. Further f (x) = f1 (x) f2 (x). We get

f 2 (x) = f (x)σ( f (x)) = ( f1 (x)σ( f1 (x)))( f2 (x)σ( f2 (x))).
The degree of integer polynomial f2 (x)σ( f2 (x)) is 2r < n. However f 2 (x) as a product of
two irreducible factors of degree n cannot have a factor of degree 2r < n. This completes
the proof of Theorem 1.

The main theorem

Chapter 2

8

In [6] Schinzel studied the conditions under which a quadratic algebraic integer is
a Mahler measure of an algebraic number. Let M = {M(α)|α ∈ Q̄}, where Q̄ is the
algebraic closure of Q̄.He proves there two theorems:

Theorem 2 A primitive real quadratic integer β is in M if and only if there exists a
rational integer a such that β > a > |β′ | and a | ββ′ , where β′ is the conjugate of β. If the
condition is satised then β = M(β/a) and a = N(a, β), where N denotes the absolute
norm.
Here `primitive' means that β has no rational integer factor, other than ±1. Let J be
an ideal of OK . The absolute norm of J is the number of residue class of in OK , that is

N(J ) = |OK /J |.For quadratic integers that are not primitive he considers the numbers
pβ, where p is a rational prime and β a primitive algebraic integer, and proves

Theorem 3 Let K be a quadratic eld with discriminant ∆ > 0, β, β′ be primitive conjugate integers of K and p a prime. If
1.

pβ ∈ M ,
then either there exists an integer r such that
2.

pβ > r > p β′ | and r | ββ′

p∤r

or
3.

β ∈ M and p splits in K
Conversely, (2) implies (1), while (3) implies (1) provided either
4.

⎫
⎧
(︄
√ )︄2 ⎬
⎨
1+ ∆
β > max −4β′ ,
⎭
⎩
4

Chapter 2

9

or
5.

p>

√
∆.

√
√
In Schinzel's notation 1 + 17 = 2β, where β = 1+2 17 is primitive and p = 2. With
√
K = Q( 17) we have ∆ = 17. Thus condition (2) fails, condition (3)is satised but

without (4) or (5). This fact motivates Schinzel's question:
√
Is 1 + 17 a Mahler measure of an algebraic number?

Let α be an algebraic number and suppose that M(α) = β. Then, by denition of

M(α), we would have M(α) = M( f ), where f ∈ Z[x] is the minimal polynomial of α. So
√
f is irreducible in Z[x]. However we shall prove that it is impossible for β = 1 + 17.
More specically I shall prove the following theorem.

√

Theorem 4 Let f ∈ Z[x] be irreducible over Q[x]. If M( f ) = 1 + 17 then 2 divides the
content of f and

f (x) = 4x2s ± 2xs − 4.

Note: In algebra, the content of a nonzero polynomial with integer coecients is the
greatest common divisor of its coecients.This theorem implies that f is not a minimal
√
polynomial of an algebraic number and consequently β = 1 + 17 is not a measure of
an algebraic number. For example, we can check directly that

M(4x2 − 2x − 4) = M(4x2 + 2x − 4) = β.
The content of both polynomials is 2, c(4x2 − 2x − 4) = c(4x2 + 2x − 4) = 2.

Proof 4 Suppose that f ∈ Z[x] is irreducible over Q[x] and M( f ) = β. The rst step
consists of showing that

Chapter 2

10

Claim 1
f (x) = 4xn + a1 xn−1 + · · · + an−1 x − 4.

Proof of claim 1 For this we are following the steps of Theorem 1.
Let f (x) = a ∏ni=1 (x − αi ), where a is a positive integer and suppose that |αi |> 1 for

i = 1 . . . s while |αi |≤ 1 for i = s + 1, . . . , n.

Again we use the notation α′i = αs+i for i = 1 . . . r, where r = n − s, S = {α1 , . . . , αs }
√
and R = {α′1 , . . . , α′r }, L = Q(α1 , . . . , αn ) and K = Q( 17) = Q(β), G = Gal(L/Q), and

H = Gal(L/K). Let D = |G|. Then again |H|= D/2, every σ from H is a permutation
of S and R. Now

M( f ) = εaα1 . . . αs , where ε ∈ {−1, +1}.
This time we dene

L = ∏ σ(aα1 . . . αs )
σ∈G

and conclude that
sD

L = aD (α1 . . . αn ) n = aD (−1)sD

(︂ a )︂ sD
n

a

n

,

where an = f (0).
On the other hand aα1 . . . αs = εβ, so that

L = ∏ σ(β)
σ∈H

∏ σ(β) = (ββ′)D/2 = (−16)D/2,
σ∈G\H

√
√
as ββ′ = (1 + 17)(1 − 17) = −16.
By comparing both expressions of |L| we get

a

D(n−s)
n

sD

|an | n = 4D

Chapter 2

11

which simplies to

(a

D(n−s)
n

sD

n

n

|an | n ) D = (4D ) D and so ar |an |s = 4n .

Further

√
|an |= |aα1 . . . αn |≤ |aα1 . . . αs |= β = 1 + 17.
Since the previous equation shows that |an | is a power of 2 we conclude that

|an |≤ 4.
Now consider the polynomial

g(x) = sign(an )xn f (x−1 ).
If f (x) = axn + a1 xn−1 + · · · + an−1 x + an then g(x) = η(an xn + an−1 xn−1 + · · · + a1 x + a),
where η = sign(an ). We know that g is irreducible over Q[x] and that M(g) = M( f )
because of reciprocality. However, the leading coecient of g is |an |, while its constant
coecient is ηa. By applying the same argument to g as we applied to f we conclude
that a ≤ 4.
The inequalities |an |≤ 4 and a ≤ 4 together with ar |an |s = 4n show that |an |= a = 4.
Hence r = s, so n = 2s is even. Further we notice that every σ ∈ G \ H maps S onto R
and vice versa. Indeed, we have

|an |= a|α1 . . . αs ||α′1 . . . α′r | and so 4 = 4|α1 . . . αs ||α′1 . . . α′r |.
and since |α1 . . . αs |= β4 we deduce that |α′1 . . . α′r |= β4 . But β4 = | β4′ |= |σ( β4 )|. Hence

|σ(α′1 . . . α′r )|= |σ(α′1 ) . . . σ(α′r )|= |α1 . . . αs |.

Chapter 2

12

Hence the last term has strictly the largest value among absolute values of any choice
of s conjugates, hence must have σ(R) = S. By applying σ to both sides of this equality
√
√
and noticing that σ2 ∈ H, because σ2 ( 17) = 17, we also nd out that σ(S) = R.
Thus we have

an = (−1)n aα1 . . . αn = (4α1 . . . αs )(α′1 . . . α′r ) = (εβ)(εσ(β/4)) = −4,
where σ is any automorphism in G \ H.
We have proved that f and g have the following forms:
n−1

n−1

f (x) = 4xn + ∑ ai xn−i − 4,

g(x) = 4xn − ∑ an−i xn−i − 4.

i=1

i=1

This concludes the proof of claim 1.

In the step 2, we separate f , g into 4 new polynomials and introduce some arithmetical facts.
We denote the roots of g by γ1 , . . . , γs and γ′1 , . . . , γ′s , where γi = (α′i )−1 , and γ′i =
√
(αi )−1 , for i = 1 . . . s. In what follows we use the fact that K = Q( 17) has class number
1. This implies that every irreducible element in OK is a prime element and the greatest
common divisor is dened. Consequently the content of a polynomial is dened and
we denote it by c( f ). It is determined uniquely up to a unit factor.
We need to establish some arithmetical facts about OK
We have

√
• u = 4 + 17 is the fundamental unit. That means that group of unit of OK is of
the form U = {±un : n ∈ Z},
√

√

• π1 = −3+2 17 and π2 = −3−2 17 are primes,

Chapter 2

13

• π1 π2 = −2,
√

• 1+2 17 = uπ21 ,
√

• 1−2 17 = −u−1 π22 .
Next we dene four polynomials:
s

s

f̆ (x) = 4 ∏(x − α′i )

f̂ (x) = 4 ∏(x − αi ),
i=1

i=1

and
s

ĝ(x) = 4 ∏(x − γi ),
i=1

s

ğ(x) = 4 ∏(x − γ′i ).
i=1

Then by Lemma 1, all four polynomials are in OK [x], and

• 4α1 . . . αs = εβ,

4α′1 . . . α′s = εβ′ ,

• 4γ1 . . . γs = −εβ,

4γ′1 . . . γ′s = −εβ′ ,

• 4 f (x) = f̂ (x) f̆ (x) and 4g(x) = ĝ(x)ğ(x),
• f̂ (0) = (−1)s εβ = (−1)s εu2π21 ,
• f̆ (0) = (−1)s εβ′ = −(−1)s εu−1 2π22 ,
• ĝ(0) = −(−1)s εβ = −(−1)s εu2π21 ,
• ğ(0) = −(−1)s εβ′ = (−1)s εu−1 2π22 .
We can see that all zeros of f̂ lie strictly outside the unit circle while the zeros of

f̆ lie inside or on the unit circle. We shall show that, in fact, the zeros of f̆ must
lie strictly inside the unit circle. For this, suppose that a zero of f̆ , α lies on the
unit circle. Then α ̸= 1 because 4 f = f̂ f̆ , by our assumption is irreducible over Q.
Suppose then that |α|= 1 and α ∈ C \ R. Then ᾱ = α−1 is another zero of f̆ . Then
by transitivity of action of G on the zeros of f , {σ(α−1 )|σ ∈ G} = {α−1
i : i = 1 . . . n}

Chapter 2

14

must be the set of the zeros of f , thus showing that f is reciprocal, so that we have

f (x) = 4 ∏ni=1 (x − αi ) = 4 ∏ni=1 (x − α−1
i )
n

x f (x

−1

n
n

) = 4x ∏(x
i=1

−1

n

n

n

(x−α−1
i ) = −4

−αi ) = 4 ∏(1−xαi ) = 4 ∏(−αi ) ∏
i=1

i=1

i=1

n

∏(x−α−1
i ) = − f (x)
i=1

, because 4 ∏ni=1 (−αi ) = an = −4. Now substituting x = 1 gives − f (1) = f (1). Hence

f (1) = 0 which contradicts the irreducibility of f . We have

c(4 f ) = 4c( f ) = c( f̂ )c( f̆ ).

This implies that 4|c( f̂ )c( f̆ ). For any σ ∈ G \ H we have σ( f̂ ) = f̆ and σ(ĝ) = ğ. Thus,

2|c( f̂ ) if and only if 2|c( f̆ ).Further, 4 = π21 π22 and π1 π2 = −2, so if 2 ∤ c( f̂ ) then either
π1 ∤ c( f̂ ) or π2 ∤ f̂ . So We have two possibilities
1. 2|c( f̂ ) and 2|c( f̆ )
or
2. π21 |c( f̂ ) and π22 |c( f̆ ).
Note that we cannot have π21 |c( f̆ ) and π22 |c( f̂ ) because π2 does not divide the constant
term of f̂ . We claim that if the second possibility occurs then 2|c(ĝ), so also 2|c(ğ).
We have

ĝ(x) =

4 s −1
4εu
x f̆ (x ) =
xs f̆ (x−1 ).
2
s
−(−1) 2π2
f̆ (0)

Hence c(ĝ) = c(± 2u
)c(xs f̆ (x−1 )) = c(± 2u
)c( f̆ ), we deduce that 2|c(ĝ) because π22 |c( f̆ ).
π2
π2
2

2

Similarly, we show that 2|c(ğ). However M( f ) = M(g), and if we prove that g(x) =

4x2s ± 2xs − 4 then it would imply that f (x) = 4x2s ± 2xs − 4 as well. Therefore if the
second case occurs we can work with polynomial g instead of f , so without loss of
generality we can assume that the rst case occurs. We thus conclude that

Chapter 2

15

f̂ 1 (x) =

s−1
1
f̂ (x) = 2xs + ∑ Ai xs−i + (−1)s εuπ21
2
i=1

and
s−1
1
ε
s
f̆ 1 (x) = f̆ (x) = 2x + ∑ Ãi xs−i − (−1)s π22
2
u
i=1

are in OK [x], and f = f̂ 1 (x) f̆ 1 (x) Here Ãi are algebraic conjugates of Ai , i = 1 . . . s.
In the nal step, we'll show that all coecients Ai is equal to 0 with Schur lemma.

The following is Schur [7] lemma, employed in the Schur-Cohn algorithm to determine the distribution of roots of a complex polynomial relative to the unit circle. The
version below is presented in Wikipedia [8].

Lemma 2 Let p be a complex polynomial of degree n ≥ 1 and let p∗ be dened by
p∗ (z) = zn p(z̄−1 ). Dene T p by T p = p(0)p − p∗ (0)p∗ , and let δ = T p(0).

1. If δ ̸= 0 then p, T p, and p∗ share zeros on the unit circle.
2. If δ > 0 then p and T p have the same number of zeros inside the unit circle.
3. If δ < 0 then p∗ and T p have the same number of zeros inside the unit circle.
If δ < 0 the T p and p∗ have the same number of roots inside the unit circle.
We apply this lemma to
s−1

p(x) = xs f̂ 1 (x−1 ) = (−1)s εuπ21 xs + ∑ Ai xi + 2
i=1

and to p ∗ (x) = fˆ1 (x). Here p has all its roots inside the unit circle. We get
s−1

T p(x) = ∑ (2Ai − (−1)s εuπ21 As−i )xi + 4 − ε2 u2 π41 .
i=1

Chapter 2

16
δ = 4 − ε2 u2 π41 ≈ −2.56 < 0

The polynomial p∗ has no roots inside the unit circle, therefore the same is true about

T p. The degree of T p is less than s. Suppose that deg T p = i for some i, 1 ≤ i ≤ s − 1.
Then the leading coecient of T p is 2Ai − (−1)s εuπ21 As−i . Since all roots of T p lie
outside of the unit circle, we must have

|2Ai − (−1)s εuπ21 As−i |< |4 − ε2 u2 π41 |= |T p(0)|.
s−i − (−1)s ε 1 π2 whose
Now we apply the same argument to p = f˘1 = 2xs + ∑s−1
i=1 Ãi x
u 2
i
roots lie inside the unit circle. Then p∗ (x) = −(−1)s ε u1 π22 xs + ∑s−1
i=1 Ãi x + 2.
s−1

T p(x) = ∑ (−(−1)s εu−1 π22 Ãs−i − 2A˜i )xi + ε2 u−2 π42 − 4.
i=1

We nd the corresponding

1
δ = ε2 2 π42 − 4 ≈ −1.56 < 0
u
We conclude as in the previous case that

|

−ε
ε
1
(−1)s π22 Ãs−i − 2Ãi |= |2Ãi + (−1)s π22 Ãs−i |< | 2 π42 − 4|.
u
u
u

From both inequalities we get

1
ε
|2Ai −(−1)s εuπ21 As−i ||2Ãi + (−1)s π22 Ãs−i |= |N(2Ai −(−1)s εuπ21 As−i )|< |4−ε2 u2 π41 || 2 π42 −4|= 4,
u
u
where N is the norm from K to Q. Further

2Ai − (−1)s εuπ21 As−i = −π1 (π2 Ai − (−1)s εuπ1 As−i )

Chapter 2

17

and

ε
ε
2Ãi + (−1)s π22 Ãs−i = −π2 (π1 Ãi − (−1)s π2 Ãs−i ).
u
u
Hence

1
|N(π2 Ai + (−1)s εuπ1 As−i )|= |N(2Ai − (−1)s εuπ21 As−i |< 2.
2
We conclude that π2 Ai − (−1)s εuπ1 As−i is a unit.
However we have

|π2 Ai + (−1)s εuπ1 As−i |< |
and
sε

|π1 Ãi − (−1)

u

π2 Ãs−i |< |

4 − ε2 u2 π41
|< 4.562
π1

1 4
π −4
u2 2

π2

|< 0.4385

The last inequality excludes the possibility π2 Ai + (−1)s εuπ1 As−i = ±1. It remains
the possibility that π2 Ai + (−1)s εuπ1 As−i = ±uk with k ̸= 0. However then π1 Ãi −
√
(−1)s uε π2 Ãs−i = ±u−k , but max(|uk |, |u−k |) ≥ u = 4 + 17 > 4.562, hence this possibility
is also excluded. Finally, we have proved that T p has degree 0, so that

π2 Ai + (−1)s εuπ1 As−i = 0 for i = 1 . . . s − 1.
This implies that π1 and π2 divide each Ai for i = 1 . . . s − 1, so also divide each Ãi . Thus

2|Ai , so also 2|A˜i . Hence Ai = 2Bi with Bi ∈ OK for alli. and we get
π2 Bi + (−1)s εuπ1 Bs−i = 0 for i = 1 . . . s − 1.
We can repeat the same argument again and conclude that 2|Bi for all i. After several
repetitions we get

2k |Ai for every positive integer k and all i.

Chapter 2

18

Hence all coecients Ai are zero. We have

fˆ1 = 2xs + (−1)s εuπ21 and f˘1 = 2xs − (−1)s εu−1 π22 .
Finally,with d = s we get

f (x) =

1
f̂ f̆ = fˆ1 f˘1 = 4x2s ± 2xs − 4.
4

This completes the proof of Theorem 4.

Chapter 3
Does there exists an algorithm to reduce
the shortness of a polynomial without changing its Mahler measure?
J. McKee and C. Smyth in [5] mentioned that so far no algorithm that can reduce the
shortness of a polynomial without changing its Mahler measure is known. To partially
answer this problem, we need to use the theorem of Dobrowolski [1] and its corollary.

Denition 2 Let P(x) = ∑ni=0 ai xn−i ∈ C[x]. The length of P denoted by L(P) is the sum
of absolute values of the coecients of P, that is, L(P) = |a0 |+ · · · + |an |.

Denition 3 Let P(z) ∈ Z[z]. A short polynomial for P is a polynomial of minimum
length of the shape P(z)Q(z), where Q(z) is a product of cyclotomic polynomials.

Denition 4 The shortness of a polynomial P(z) ∈ Z[z] is the length of a short polynomial for P. The shortness of an algebraic integer α is the shortness of its minimal
polynomial.
In the following theorem, fc denotes the product of all cyclotomic factors of f .

19

Chapter 3

20

Theorem 5 (Dobrowolski [1]) Let f ∈ Z[x], f (0) ̸= 0, be a polynomial with k nonzero
coecients. There are positive constants c1 , c2 , depending only on k, and polynomials

f0 , f2 ∈ Z[x] such that if
deg fc ≥ (1 −

1
) deg f
c1

then either

• f (x) = f0 (xl ), where deg f0 ≤ c2
or

• f (x) = (∏i Φqi (xli )) f2 (x), where mini {li } > max{ 2c11 deg f , deg f2 }.
k−2

The size of the constants are: ci ≤ exp(3⌊ 4 ⌋ si k2 log k) with s1 = 0.636 and s2 = 1.06
k−2

for f with reciprocal exponents; ci ≤ exp(3⌊ 2 ⌋ti k2 log k) with t1 = 1.81 and t2 = 2.841
for f that does not have reciprocal exponents.
Note: In the second case in the original paper we have mini {li } ≥ max{ 2c11 deg f , deg f2 }
was not sharp, however the proof implies a sharp inequality.

Corollary 2 Let f (x) = ∑ki=1 ai xni ∈ Z[x], f (0) ̸= 0, be a polynomial with k nonzero coecients. If the second case of Theorem 5 occurs then f2 (x) = ± ∑ki= j ai xni with some

j, 1 < j < k.
Note: In this theorem, we assume that f (x) = ∑ki=1 ai xni with k nonzero coecients, the
exponents n1 , . . . , nk are strictly decreasing; fc denotes the product of all cyclotomic
factors of f , fn denotes the product of all noncyclotomic factors and possibly a constant,
so that f = fc fn . When we say f has reciprocal exponents, the exponents of x in

xdeg f f (x−1 ) are the same as in f (x). Φq denotes the qth cyclotomic polynomial. If
f (x) = f0 (xl ) occurs then also fn (x) is a polynomial in xl . Hence, this case in the
theorem is not interesting, because M( f (x)) = M( f (xl )) for any polynomial f , so if we
are studying Mahler measure we can assume that f (x) ̸= f0 (xl ) with l > 1.
Let fn ∈ Z[x] be a monic and noncyclotomic polynomial, and fc (x) ∈ Z[x] be a product

Chapter

21

of cyclotomic polynomials. Corollary 2 implies that if fc fn has the minimal number of
nonzero terms, we must have

deg fc < (1 −

1
) deg( fc fn )
c1

since otherwise the part of fn fc which contains power of x with exponents less than
some m, would form the polynomial f2 that is a multiple of fn and some cyclotomic
polynomials, and has even smaller number of nonzero terms and which contradicts
that fc fn has minimal number of terms.
Concerning the minimal length of fc fn we notice that if fc fn has k nonzero terms then

L( fc fn ) ≥ k. Thus for the shortest length we need to examine polynomials which have
fewer than L( fn ) nonzero terms. The inequality

deg fc < (1 −

1
) deg( fc fn )
c1

implies that

deg fc < (c1 − 1) deg fn .
This means that if we want to nd a polynomial with smallest number of nonzero
coecients that is a multiple of fc and fn , we can limit the search for polynomials

fc with deg fc < (c1 − 1) deg fn . The same inequality applies for the search of shortest
polynomials, but we have to replace k in c1 by L( fn ).

However, we tried to use the formula in the theorem 5 to calculate c1 and nd a
bound of maximum degree of the product of cyclotomic polynomials required for the
search, but even a very small values of k resulted in a bound exceeding computer's
limit. Hence, the algorithm exists only theoretically and cannot be implemented for

Chapter
calculations.

22

Appendix A
Complementary facts
• All elds considered are subelds of C, so we do not discuss separability.
• A eld L is an extension of a eld K if K ⊆ L, and the operations of K are those
of L restricted to K.

• A splitting eld of a polynomial with coecients in a eld is the smallest eld
extension of that eld which contains the zeros of the polynomial.

• The automorphism group of a eld extension L/K is the group consisting of eld
automorphisms of L that x K, that is they are identities on K.

• If f is a polynomial over F and if E is its splitting eld over F , then G(E/F)
denotes all automorphisms of E that x F and it is called the Galois group of f
over F.

• The symmetric group dened over any set is the group whose elements are all the
bijections from the set to itself, and whose group operation is the composition of
functions.

Theorem 6 Let K be a quadratic eld.Let d = d(K). Let p be a rational prime. Then
• ⟨p⟩ splits ⇔ ( dp ) = 1,
23

Chapter A

24

• ⟨p⟩ ramies ⇔ ( dp ) = 0,
• ⟨p⟩ is inert ⇔ ( dp ) = −1,
where ( dp ) is the Legendre symbol for p > 2 and Kronecker symbol for p = 2.

Denition 5 Let d be a nonsquare integer with d ≡ 0 or 1 mod 4. The Kronecker symbol
(︁ d )︁
2

is dened by

⎧
⎪
⎪
⎪
if d ≡ 0 mod 4,
⎪0,
⎪
(︃ )︃ ⎪
⎨
d
= 1,
if d ≡ 1 mod 8,
⎪
2
⎪
⎪
⎪
⎪
⎪
⎩−1, if d ≡ 5 mod 8

(A.1)

Bibliography
[1]

E. Dobrowolski, Mahler's measure of a polynomial in terms of the number of
its monomials, Acta Arith. 123, 2006, 201231.

[2]

A. Dubickas, Mahler measures in a cubic eld, Czechoslovak Math. J.,
56(131), 2006, 949956.

[3]

P.A. Fili, L. Pottmeyer, and M. Zhang, On the behavior of Mahler's measure
under iteration, Monathsh. Math. 193 (2020), 6186.

[4]

C. R. Hadlock, Field theory and its classical problems, MMA, CM 19, 1978.

[5]

J. McKee and C. Smyth, Around the unit circle, Springer UTX 2021.

[6]

A. Schinzel, On values of the Mahler measure in a quadratic eld (solution
of a problem of Dixon and Dubickas),Acta Arith. 113.4 (2004), 401408.

[7]

I. Schur, Über Potenzreichen, die im Innern des Einheitskreises beschränkt
sind, J. Reine Amgew. Math. 147(1917), 205232.

[8]

Wikipedia contributors, LehmerSchur algorithm, Wikipedia, The Free Encyclopedia, Retrived on Dec. 30, 2023, https://en.wikipedia.org/w/ind
ex.php?title=Lehmer%E2%80%93Schur_algorithm&oldid=1189645486

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