EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 3, 2022, 1321-1330
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On Salem formal power series

Oussama Dammak1,2, Saber Mansour1,3,∗

1 Department of Mathematics, Umm Al-Qura University, College of First Common year,
P.O. Box 14035, Holly Makkah 21955, Saudi Arabia
2 University of Gabes, Science Faculty of Gabes, Omar Ibn AlKhattab-Zrig Road, 6029
Gabes, Tunisia
3 Sfax University, Department of Mathematics, Faculty of Sciences of Sfax, Soukra Road,
3018 Sfax, BP 802, Tunisia

Abstract. In this paper, we will take a look at Salem elements dealing with formal power series
on Fq, where Fq is a finite field . Our main result, presents a criteria for an element to be the
smallest Salem element (SSE) via an order extending a given order in Fq. Moreover, we provide
the CFE of the (SSE) for each n.

2020 Mathematics Subject Classifications: 11Kxx, 11K16, 11T06

Key Words and Phrases: Finite fields, Formal Power Series, Pisot elements, Salem elements,
continued fractions.

1. Introduction

Let α1 be an algebraic integer of degree n with Galois conjugates α2, α3, · · · , αn. If{
|α1| > 1

and |αi| < 1, ∀2 ≤ i ≤ n,

α1 is called to be a Pisot number. If
|α1| > 1

and |αi| = 1, for 2 ≤ i ≤ n,
and |αj | ≤ 1, for 2 ≤ j ≤ n, j ̸= i,

α1 is called to be a Salem number.
The set of the so called, Pisot numbers, is usually denoted by S, it is denoted by T ,

the set of Salem numbers. Thus, Pisot numbers are commonly referred to as S−numbers,

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4463

Email addresses: omdammak@uqu.edu.sa (O. Dammak), samansour@uqu.edu.sa (S. Mansour)

https://www.ejpam.com 1321 © 2022 EJPAM All rights reserved.



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1322

while Salem numbers are referred to as T−numbers. The sets S and T appears in a va-
riety of algebraic number theory problems, Diophantine approximation, Fourier analysis,
distribution, the so-called β-expansions, etc. The set S was defined approximately simul-
taneously and separately by C. Pisot [9] and Vihayaraghavan [18, 19]. R. Salem [11] a few
years later, defined the set T . However, some of the first research in this approach and
related to uniform distributions were published earlier by Thue [17].

In 1919, Hardy showed that if α is an algebraic integer such that αn → 0( mod 1) as
n → ∞, then α is a Pisot number.

Later, this concept, was investigated by Salem, who proved that the only algebraic
numbers that have this property of being badly distributed modulo 1 are S-numbers [12].
In brief, this occurs because for all algebraic number α and n ∈ N∗, the sum of the nth
powers of α and its conjugates is an integer.

when α is an S−number, the nth powers of the conjugates of α tends to 0 as n tends
to ∞, because they all have modulus strictly less than 1. This fundamental property of
S-numbers raises the following significant unanswered question about the characterization
of the set S. Asuume α > 1 is a real number with αn → 0( mod 1) as n → ∞.

Can we then conclude, under no other assumptions, that θ is an algebraic integer in
the set S ? This is possibly one of the oldest unsolved problems involving S-numbers, as
it appears in [12].

The sets S and T have been widely investigated, and a large number of results are
known about them. Here are a handful of the more notable results. Because they exclu-
sively include algebraic numbers, both S and T are clearly countable sets. Furthermore, S
contains infinitely many limit points, because it is possible to consider each integer a ≥ 2
as a limit of a sequence of elements in S. Then, S contains an infinity of limit points.

The derived set of S is denoted by S′ and contains all of S’s limits points. Equally,
the derived set of T , is the set of all T limits points, and it is denoted by T ′.

While the product of two algebraic integers is also an algebraic integer, Pisot numbers

are not. For instance, We have just shown, that both 2 and
1 +

√
5

2
are Pisot; however,

their product 1 +
√
5 with conjugate equals to 1−

√
5 clearly is not a Pisot number.

If the two Pisot numbers are equal, their product is a Pisot number as well.
The sets S and T are tightly connected and contain fascinating linkages, as one might

expect given their comparable definitions. In [10] it was proved, by Salem, that S′ ⊂ S
is closed. So, it must have a smallest element because it is bounded below. According to
Siegel [14], the smallest known Pisot number is equal to the largest root of x3 = x + 1,
which is around 1.3247179.

The smallest Pisot number of degree n ≥ 3 was identified by Dufresnoy and Pisot [5].
They arrived to the following theorem :

Theorem 1. Let an be the smallest Pisot number of degree n ≥ 3, then the following
assertions holds
i) Pn(z) = zn − zn−1 − zn−2 + z2 − 1 is the minimal polynomial of an

ii) the sequence (an)n≥1 is increasing and eventually converges to
1 +

√
5

2
≈ 1.61803, a



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1323

root of x2 = x+ 1.

It’s worth noting that similar claims concerning the set T have yet to be discovered.
There are, indeed, two of the most well-known unanswered questions about S and T -
numbers. The first question concerns T ’s limit point. The set S is known to be contained
in T ′, or, to put it another way, any point of S is a limit point of T, on both sides. Salem
[11] was the first to demonstrate this astounding fact by creating polynomial sequences
that provided the needed T−numbers. Although it is known that S ⊆ T ′, There’s no way
of knowing if the set T has any limit points other than the ones identified in S; This is an
inquiry that is addressed in [12], but it is yet unsolved.

Small T−numbers are the subject of the second open question. We can’t assume the
existence of a lowest Salem number because T isn’t closed. While we can’t assume that
T contains a smallest element, there is a possibility. It is conjectured that 1.1762808..., a
root of the 10th degree polynomial

x10 + x9 − x7 − x6 − x5 − x4 − x3 + x+ 1,

discovered by Lehmer [7] in 1933, is the smallest Salem number.
There hasn’t been a smaller Salem number in almost eighty years. All Salem numbers

< 1.3 of degree 20 over [3], are included in the list of 39 Salem numbers given in [2]; it will
enough for the applications below. There are currently around 47 Salem numbers < 1.3
are known and the list is believed to be comprehensive up to degree 44..

Although this has not been proven, it is widely assumed that Lehmer’s Salem number
1.1762808... is an isolated point of T.

It was stated and proven some basic results about Salem numbers in [15], and then a
survey of the literature about them was conducted. Chris Smyth’s intention was to sup-
plement rather than duplicate other general treatises on these numbers. This is especially
true of Bertin and her coauthors’ work [1], as well as Ghate and Hironaka’s application-
rich Salem number survey [6]. He did, however, cite some findings from Salem’s classic
monograph [13].

Moreover, the concept of the Mahler measure of a matrix arose from the investigation
of Salem numbers, which appeared as Mahler measures of graphs. It was demonstrated
that certain limit points of these Salem numbers are Pisot numbers, providing yet another
example of a more general result. it was focused on an interlacing construction for Salem
numbers, which evolved historically from the graph construction but is far more general
(For more details one can see [8]).

Chandoul et al. [4], established that the minimal polynomial of the so-called, smallest
Pisot element (SPE) of degree n in in Fq((X

−1)) is P (Y ) = Y n − aXY n−1 − an, where a
is the least element of the finite field Fq \ {0} (as a finite total ordered set).

It was shown that the sequence of SPEs in the case of degree n is decreasing. Moreover
it converges to aX.

But why is finding the smallest element of a set so important?
The answer is summed up in the possibility of discovering a total order and a point

that reduces all the properties of the set which is the smallest according to this order.



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1324

The smallest Salem element (SSE) of a given degree n in Fq((X − 1)) is presented in
this work.

The following is how the paper is structured: In Section 2, we define the lexicographic
order on in Fq((X

−1)) and provide some early definitions. We offer the (SSE) of degree n
in in Fq((X

−1)) in Section 3. Section 4 investigates the (SSE)’s CFE over in in Fq((X
−1)).

2. Formal power series

Let Fq be a field with q elements of characteristic p, Fq[X] the set of polynomials of
coefficients in Fq and Fq(X) its field of fractions. The set Fq((X

−1)) of formal power series
over Fq is defined as follows

Fq((X
−1)) = {

+∞∑
j=s

ajX
−j : aj ∈ Fq, as ̸= 0 with s ∈ Z}.

Let ω =

+∞∑
j=s

ajX
−j ∈ Fq((X

−1)). The polynomial part of ω is denoted by [ω] and its

fractional part is denoted by {ω}. We remark that ω = [ω] + {ω}. As in Sprindz̃uk [16] a
non archimedean absolute value on Fq((X

−1)) is definied by | ω |= e−s. Clearly, we have,

| P |= edegP , for all P ∈ Fq[X], and, | P
Q

|= edegP − degQ, for all Q ∈ Fq[X], such that

Q ̸= 0.

It is well known that Fq((X
−1)) is complete. In terms of the metric provided by this

absolute value, Fq((X
−1)) is locally compact.

An algebraic closure of Fq((X
−1)) is denoted by Fq((X

−1)). It is worth noting that
the absolute value has a distinct extension to Fq((X

−1)). We will use the same symbol | · |
for the two absolute values, slightly abusing the notations.

As Fq is a finite total order set, we denote by ⪯ a totally order on Fq.
Now, we extend ⪯ to the field of formal power series as follows :

Let w =

+∞∑
i=m

wiX
−i and v =

+∞∑
i=k

viX
−i with wmvk ̸= 0, then, w ⪯ v if and only if

m > k or w = v or m = k and there exists j ≥ m, such that wi = vi, for i < j and
wj ⪯ vj .

Let w ∈q ((X−1)), with |w| > 1 is called to be a Salem element if it is algebraic over

q[X] whose conjugates wi in Fq((X
−1)) have modulus | wi |≤ 1, with at least one case of

equality.

Theorem 2. Let w such that |w| > 1, be an element of q((X
−1)), These two affirmations

are equivalent:
1) w is a Salem element.



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1325

2) the minimal polynomial P of w can be written as P (Y ) = Y s + As−1Y
s−1 + · · · + A0,

with Ai ∈q [X] \ {0}, Ai ∈q [X] for i = 0, . . . , s− 1, and |As−1| = max
i ̸=s−1

|Ai|.

Now, we will go over the main results.

3. Main result

Theorem 3. Let S(n) = {Salem elements of degree n}, n ≥ 2 and a is the least element
of Fq\{0}, then wn = inf S(n) is a Salem element of minimal polynomial

Pn(Y ) = Y n − aXY n−1 − aY + aX − a

In addition, the sequence (wn)n≥1 of wn = inf S(n) is decreasing one and converges to aX.

The following lemmas are needed to prove this theorem.

Lemma 1. Let P (Y ) = AdY
d + · · · + A0 with Ai ∈q [X], Ad ̸= 0 and |An−1| > |Ai|, for

all i ̸= n− 1. Then P has exactly one root w ∈q ((X
−1)) such that |w| > 1. Furthermore

[w] = −
[
An−1

An

]
.

Lemma 2. Let H(Y ) = Y d −AY d−1 −B, A,B ∈ Fq[X] \ {0}, degA ≥ degB. Then H
is irreducible over Fq[X].

Proof. According to Lemma 1, H has exactly one root w such that | w |> 1 and
[w] = A. Let wi be the other roots of H, 2 ≤ i ≤ d and w = w1.

Because H is a monic polynomial, then, we have
d∑

k=1

wk
i ∈ Fq[X], for every k ∈ N,

which implies lim
m→+∞

{wm} = 0.

Let P (Y ) = Y n +An−1Y
n−1 + · · ·+A0 be the minimal polynomial of w, it is obvious

that An−1 = −A, since [w] = A. From Theorem 2, the polynomial P satisfies degAn−1 ≥
max
i ̸=n−1

degAi.

Let now H(Y ) = P (Y )Q(Y ), with Q(Y ) = Y m+Bm−1Y
m−1+ · · ·+B0. Suppose that

m ≥ 1, then
Bm−1 +An−1 = −A and A0B0 = −B (1)

∑
i+ j = s
0 ≤ i ≤ n
0 ≤ j ≤ m

AiBj = 0 ; s ∈ {1, 2, · · · , d− 2}. (2)

Since An−1 = −A, then from (1) Bm−1 = 0. Let i0 ∈ {0, 1, · · · ,m} such that degBi0 =
max
0≤i≤m

degBi.



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1326

If Bi0 ̸= 0, then deg(An−1Bi0) > deg(AiBj), (i, j) ̸= (n− 1, i0). Consequently

deg


∑

i+ j = n+ i0 − 1
0 ≤ i ≤ n
0 ≤ j ≤ m

AiBj


= deg(An−1Bi0),

there is a contradiction with (2).
Finally, we arrive to H(Y ) = Y mP (Y ), if m ≥ 1, which leads to a contradiction (due

to the fact that B ̸= 0) and arising H(Y ) = P (Y ).

Proof. of Theorem 3. Consider the polynomial Pn(Y ) = Y n−aXY n−1−aY +aX−a,
such that a is the least element of Fq\{0},. It follows from lemma 2 that P is irreducible.
Moreover, P has exactly one root wn satisfying | wn |> 1 and [wn] = aX, all of whose
other conjugates wi satisfy | wi |≤ 1 with at least one case of equality. Because P is a

monic polynomial, wn is a Salem element of degree n. Additionally, if wn = aX +
1

h
, then

h is an algebraic formal power series satisfying

anhn − (aX)n−1hn−1 −
n−2∑
k=0

(
n− 1
k

)
(aX)khk = 0

and using lemma 1, we have

[h] =
Xn−1

a
. (3)

Now, we consider an other Salem element vn ̸= wn of degree n such that [vn] = aX,
then from Theorem 2 the minimal polynomial of vn can be written as F (Y ) = Y n −

aXY n−1−
n−2∑
i=0

AiY
i where degAi ≤ 1, with at least one case of equality. Let vn = aX+

1

g
,

then

F

(
aX +

1

g

)
=

(
aX +

1

g

)n

−
(
aX +

1

g

)n−1

−
n−2∑
j=0

Aj

(
aX +

1

g

)j

=
n∑

j=0

Aj

j∑
k=0

(
j
k

)
(aX)j−kg−k

; An = 1, An−1 = aX

=
n∑

k=0

 n∑
j=n−k

Aj

(
j

n− k

)
(aX)j+k−n

 gk = 0.



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1327

Let

Bk =
n∑

j=n−k

Aj

(
j

n− k

)
(aX)j+k−n. (4)

Then
n∑

k=0

Bkg
k = 0.

We have, by using lemma 1 and the equation (4)

[g] =


(aX)n−1 −

n−2∑
i=1

iAi(aX)i−1

n−2∑
i=0

Ai(aX)i

 ,

• if (A1, . . . , An−2) ̸= (0, . . . , 0), we have deg[g] < n− 1 = deg[h] then
1

h
⪯ 1

g
,

• if (A1, . . . , An−2) = (0, . . . , 0), we have [g] = an−1

A0
Xn−1 and from (3), we obtain

1
h = aX−(n−1) + · · · ⪯ A0

an−1X
−(n−1) + · · · = 1

g (A0 ̸= an if not vn is not a Salem ele-
ment.

Hence, we get in the two cases
1

h
⪯ 1

g
, which implies that wn ⪯ fn, and consequently

wn is the (SSE) of degree n.

Since wn = aX +
1

h
, then from (3) |wn−aX| = |1

h
| = | a

Xn−1
| = e−(n−1), consequently

lim
n→+∞

wn = aX.

4. CFE of the SSE

Let J = {f ∈q ((X
−1))/ | f |< 1} and let T : J → J be the map given by

T (ω) :=
1

ω
−
[
1

ω

]
, ω ̸= 0, T (0) = 0,

recall that the map T generate the continued fraction expansion of ω of the form

ω = A0 +
1

A1 +
1

A2 +
1

A3 +
.. . +

1

An +
.. .

, (5)



O. Dammak, S. Mansour / Eur. J. Pure Appl. Math, 15 (3) (2022), 1321-1330 1328

where An =
[

1
Tn−1(w)

]
. When used as a shorthand for (5) we can write

ω = [A0;A1, A2, . . .].

Let ω the (SSE) of degree qn + 1, with n ∈ N, then ω = [A0, A1, . . . , As, . . .], where
A0 = aX, A1 = Xqn−1

A2s = aXXq(2s)n−q(2s−1)n
and A2s+1 = Xqn−1Xq(2s+1)n−q(2s)n .

Proof. Let P (Y ) = Y qn+1−aXY qn−aX the minimal polynomial of (SPE) w = wqn+1.

Let z0 = ω, A0 = [z0] = aX, U0 = 1, V0 = −aX, R0 = 0, T0 = −aX and zs+1 =
1

zs − [zs]
,

then from lemma 1 and lemma 2 we know that zs satisfies the equation

Usz
qn+1
s + Vsz

qn

s +Rszs + Ts = 0

with deg Vs > max(degUs, degRs, deg Ts) for all s ≥ 1 and

Us+1 = UsA
qn+1
s + VsA

qn
s +RsAs + Ts,

Vs+1 = UsA
qn
s ,

Rs+1 = Vs +AsUs,
Ts+1 = Us,

As+1 = −
[
Vs+1

Us+1

]
Now one shows, using a simple recurrence on s , that

U2s = 1, U2s+1 = −aX,

V2s = −(aX)Xqn−1Xq(2s−1)n−q(2s−2)nq
n

,

V2s+1 = a(XXq(2s)n−q(2s−1)n
)q

n
,

Rs = 0, T2s = −aX, T2s+1 = 1,

A2s = aXXq(2s)n−q(2s−1)n

and A2s+1 = Xqn−1Xq(2s+1)n−q(2s)n .

Example 1. The minimal polynomial of the (SSE) w of degree 2 over 2((X
−1)) is P (Y ) =

Y 2 −XY −X, so w =
∞∑

i=−1

wiX
−i is defined by

w−1 = w0 = w1 = 1,
w2n = 0
w2n+1 = wn

, for all n ≥ 0.

The continued fraction of w is w = [X,X3, X11, X48, · · · ].



REFERENCES 1329

Acknowledgements

The authors would like to thank the Deanship of Scientific Research at Umm Al-Qura
University for supporting this work grant code: 22UQU4340610DSR02.

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