EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5996 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Irreducible Polynomials Over a Finite Field Amara Chandoul1,∗, Abdallah Assiry2 1{Department of Mathematics, Higher Institute of Informatics and Multimedia of Sfax (ISIMS), Sfax University, Sfax, Tunisia 2 Departement of Mathematics, College of Science, Umm Al-Qura University. Mecca 21955, Saudi Arabia Abstract. The irreducibility of a polynomial over a finite field refers to whether the polynomial, with coefficients in that field, cannot be factored into nontrivial polynomials. It is surprising to discover that there exist very efficient but still little-known divisibility criteria. In this paper, we give some irreducibility criterions of a given polynomial with coefficients in Fq[X], were Fq is a finite field. The arguments can be extended to discuss our results, including potential applications or future research directions. 2020 Mathematics Subject Classifications: 11A05, 11C08, 11T06, 11T55, 12E05 Key Words and Phrases: Polynomial, irreducible polynomial, finite field, divisibility, divisibility criteria 1. Introduction Finite fields, also known as Galois fields, are fundamental structures in mathematics with far-reaching applications in coding theory, cryptography, combinatorics, and compu- tational algebra. At the heart of many of these applications lies the study of irreducible polynomials over finite fields. These polynomials serve as the building blocks for con- structing finite field extensions, enabling the representation and manipulation of elements in higher-dimensional spaces. The theory of irreducible polynomials over finite fields is both rich and elegant, blending algebraic rigor with practical utility. From the enumeration of irreducible polynomials to the development of efficient algorithms for their construction and testing, this area of research has witnessed significant advancements over the past century. Moreover, the study of specific families of irreducible polynomials—such as binomials, trinomials, and cyclotomic polynomials—has led to deep insights into their structural properties and their role in theoretical and applied contexts. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5996 Email addresses: amarachandoul@yahoo.fr (Amara Chandoul), aaassiry@uqu.edu.sa (A. Assiry) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 2 of 9 Some of application of irreducibility is to explore its relationship with special elements, as Salem formal power series, which are algebraic elements (roots of monic integer polyno- mials) with a special property: they have exactly one conjugate outside the unit circle and all other conjugates on the unit circle. the connection to irreducible polynomials arises because Salem numbers are defined by their minimal polynomials, which are irreducible [1]. 2. Preliminaries Let Fq represent the finite field containing q elements, with q being a power of a prime. The ring of polynomials whose coefficients lie in Fq is denoted by Fq[X], while Fq(X) signifies the field of rational functions over Fq. The field of formal Laurent series in X−1 over Fq is denoted by Fq((X −1)) and is defined as: Fq((X −1)) = { ∞∑ n=n0 anX −n ∣∣∣ an ∈ Fq for some integer n0 } . For an element w = ∑+∞ n=n0 anX −n ∈ Fq((X −1)), we define its integer part [w] as: [w] = {∑0 n=n0 anX −n if n0 ≤ 0, 0 if n0 > 0. The fractional part of w is denoted by {w} and is given by: {w} = w − [w] = +∞∑ n=1 anX −n. A non-Archimedean absolute value | · | defined on a field Fq((X −1))∗ by |w| = e−n0 , where n0 is the smallest index such that an0 ̸= 0. If w = 0, we set |w| = 0. This absolute value makes Fq((X −1)) a complete and locally compact metric space. Let Fq((X −1)) denote the algebraic closure of Fq((X −1)). The absolute value | · | extends uniquely to Fq((X −1)), and we use the same notation for this extended absolute value. We define a non-constant polynomial P over F as irreducible if P = QH with Q and H in F, which implies that either Q or H is a constant. Otherwise, it is called to be reducible. Irreducible polynomials are commonly studied in fields such as number theory, combi- natorics, and algebraic geometry. They also play a significant role in practical domains, including coding theory, cryptography, complexity theory, and computer science[2]. A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 3 of 9 Constructing and Characterizing irreducible polynomials over finite fields Fq is one of the main challenges in the theory of finite fields that has just lately received attention. During the last fifty years, constructions of irreducible polynomials over finite fields have been extensively studied. However, it is recognized that there is no general criteria for determining whether such a polynomial is reducible or irreducible. Despite this, numerous tests referred to as irreducibility criteria have been established to offer valuable insights into specific classes of polynomials. In [3], Lipka derived conditions for the irreducibility of integer polynomials of the form f(X) = anX n + · · ·+ a1X + a0p k, where p is a prime number and p ∤ a0. For instance, he demonstrated that such a polynomial is irreducible over Q for all but finitely many positive integers k. For older results, one can see [4, 5]. In the first part of this paper we prove an irreducibility criterion for lacunary polynomials with coefficients in Fq[X], which is similar to the first result of Lipka. In [6], Ben Nasr and kthiri proved, using the upper Newton polygon, that Theorem 1. Let Λ(Y ) = Y d + λd−1Y d−1 + · · · + λ0 be a polynomial with λi ∈ Fq[X], λ0 ̸= 0, and deg λd−2 > deg λi for all i ̸= d− 2. Suppose further that deg λd−2 is odd and satisfies deg λd−2 ≥ 2 deg λd−1. Then, Λ(Y ) is irreducible over Fq[X]. Recall that Newton polygons can be used to determine the behavior of roots in poly- nomials over a field. Let P (Y ) = AsY s +As−1Y s−1 +As−3Y s−3 + · · ·+A1Y +A0 be a polynomial over Fq[X], and assume, for simplicity, that AsA0 ̸= 0. To each term Ai of P (Y ), we assign the point in the following manner: • If Ai ̸= 0 take the point : (i,degAi) • If Ai = 0 disregard the nonexistent point : (i,∞) Then, the points we plot for the polynomial P (Y ) are (0,degA0), · · · (s,degAs) We begin by connecting the points with line segments. The process starts at the point (0, degA0), which is connected to a point, denoted as B, that yields the least feasible slope for the first line segment. Subsequently, B is connected to the next point to its right, ensuring the smallest possible slope for the second line segment, and so forth. To geometrically visualize this construction, imagine a string fixed at one end to the point (0,degA0) and being pulled counterclockwise around the points by the other end. As the string wraps around each point, it forms a “bend” at that location. The process concludes once all finite points have been wrapped in this manner. A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 4 of 9 There should be a note here for points of the form (i,deg 0). Because deg 0 = ∞, these points have no interfer on our construction because it does not yield small slopes while constructing line segments. It is better to completely ignore such points together. The final structure is referred to as the upper Newton polygon of P (Y ). Observe that the upper Newton polygon consists of a sequence of line segments with strictly decreasing and distinct slopes. The condition AsA0 ̸= 0 ensures that the polygon begins at a finite point and terminates at another finite point. For example, the slope of a line segment in the Newton polygon of P (Y ) connecting the point (r, degAr) to the point (r +m,degAr+m) is computed as: Slope = k = degAr+m − degAr m . Denote by KP the set of slopes. This result of Ben Nasr and kthiri was a continuation of an idea we started in [7]. We established a well-known criterion for irreducibility, which can be formulated as follows: Theorem 2. Let Λ(Y ) = Y d + λd−1Y d−1 + · · · + λ0 be a polynomial with coefficients λi ∈ Fq[X], where λ0 ̸= 0. If deg λd−1 > deg λi for all i ̸= d− 1, then Λ is irreducible over Fq[X]. However, the authors did not refer to our findings, maybe because they were unaware of it or because they were focused on studying Pisot numbers. Then their results on irreducibility was an unintended incidental result. In a second part of this paper, we generalize these two results and provide a new criterion of polynomials’s irreducibility over Fq[X]. 3. Results Now, we are prepared to give the main results. Theorem 3. Let Fq (q = pn, n ≥ 2 and p be a prime) be a finite field of characteristic p. Consider the polynomial P (Y ) = BnAsY s +BmAs−1Y s−1 +As−2Y s−2 + · · ·+A1Y +A0, defined over Fq[X], such that AsAs−1A0 ̸= 0. Here, B is an irreducible factor of both As and As−1, but B does not divide AsAs−1. If the inequality n > ms+ (s− 1)(degAs −m degB) + max (max0≤i≤s−2 degAi, deg(B mAs−1)) degB holds, then the polynomial P is irreducible over Fq[X]. Proof. Assume that P can be expressed as the product P (Y ) = Q(Y )H(Y ), where Q and H are in Fq[X][Y ]. let A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 5 of 9 Q(Y)= BdQjY j +Qj−1Y j−1 +Qj−2Y j−2 + · · ·+Q1Y +Q0 and H(Y)= Bn−dHkY k +Hk−1Y k−1 +Hk−2Y k−2 + · · ·+H1Y +H0 where j+k = s, QjHk = As, Q0H0 = A0, As−1 = BdQjHk−1+Bm−dHkQj−1. Assume that d ≤ n− d. Consider the factorization of the polynomials P and Q in the algebraic closure of the field of formal Laurent series Fq((X−1)). We can express P and Q as: P (Y ) = As n∏ i=1 (Y − ωi) and Q(Y ) = Qj j∏ i=1 (Y − ωi), where ωi are in Fq((X−1)) for all i = 1, . . . , n. Consider now the non-Archimedean absolute value defined over a field where each element ωi belongs to the algebraic closure of Fq((X −1)) for all i = 1, . . . , n, and set a real number α ≥ 0 such that |As| > eαmax |Ai| i ̸=s Then, applying Viète’s theorem, we obtain |ω1 · · ·ωs| = |ω1| · · · |ωs| = |A0| |As| < |A0| eαmax |Ai| i ̸=s < 1 eα , thus, we must have, for any j := 1, · · · , n, |ωj | < 1 eα/s . Consequently, we find |ω1 · · ·ωj | < 1 ejα/s . On the other hand, we have |ω1 · · ·ωj | = ∣∣∣∣Q0 Qj ∣∣∣∣ = ∣∣∣∣ Q0 Bdqj ∣∣∣∣ ≥ 1 |Bm| |as| . To reach a contraduction, it is still necessary to chose α such that 1 |Bm| |as| ≥ 1 ejα/s . It can be sufficient to choose α such that |Bm| |as| ≤ eα/s. Or, equivalently α is greater than or equal to the quantity smdeg(B) + s ( deg(As)− n deg(B) ) . A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 6 of 9 A possible expression for α could be smdegB + s(degAs − n degB). However, this results in a contradiction when n > ms+ (s− 1)(degAs −mdegB) + max( degAi 0≤i≤s−2 , BmAs−1) degB . What was to be proved. Theorem 4. Let Fq (q = pn, n ≥ 2 and p be a prime) be a finite field of characteristic p and let P (Y ) = BnAsY s +BmAs−2Y s−2 +As−3Y s−3 + · · ·+A1Y +A0 be a polynomial over Fq[X], such that AsAs−2A0 ̸= 0, B is an irreducible polynomial over Fq and B ∤ AsAs−2. If n−m is odd, and n > ms+ (s− 1)(degAs −mdegB) + max( degAi 0≤i≤s−3 , BmAs−2) degB , then P is irreducible over Fq[X]. Proof. Suppose that P (Y ) = Q(Y )H(Y ), where Q,H ∈ Fq[X][Y ]. let Q(Y)= QjY j +Qj−1Y j−1 +Qj−2Y j−2 + · · ·+Q1Y +Q0 and H(Y)= HkY k +Hk−1Y k−1 +Hk−2Y k−2 + · · ·+H1Y +H0 where j+ k = s, QjHk = As, Q0H0 = A0 and BmAs−1 = QjHk−1 +HkQj−1. Assume that n−d ≥ d. Then, using precisely the same justifications as in the proof of Theorem 3. The remainder of the proof is similar to that of Theorem 4, using max( degAi 0≤i≤s−3 , BmAs−2) instead of max( degAi 0≤i≤s−2 , BmAs−1) Theorem 5. Let P (Y ) = Y n + An−1Y n−1 + · · · + An−k+1Y n−k+1 + An−kY n−k + · · · + A1Y +A0 with A0 ̸= 0 and |An−k| > |Ai| i ̸=n−k be a polynomial over Fq[X] of degree ≥ k + 1. If degAn−k > k maxdegAn−i 1≤i≤k−1 and k ̸ | degAn−k, then, P has no root in Fq((X −1)) with modulus strictly greater than 1. Moreover P is irreducible over Fq[X −1]. To prove this theorem, We use the idea of a Newton polygon and the following propo- sition : Proposition 1. (Weiss, pp 73-75) Let P (Y ) = AnY n +An−1Y n−1 + · · ·+A1Y +A0 with AnA0 ̸= 0 be a polynomial over Fq[X]. Let KP the set of slopes of its Newton polygon. Then for all k = degAr+m − degAr m ∈ KP A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 7 of 9 • (i) P has m roots α1, · · · , αm such that | α1 |= · · · =| αm |= e−k • (ii) the polynomial Pk(Y ) = m∏ i=1 (Y − αi) and P (Y ) = ∏ k∈KP Pk(Y ) Proposition 2. Let P (Y ) = Y n+An−1Y n−1+· · ·+A1Y +A0 with A0 ̸= 0 be a polynomial over Fq[X] of degree ≥ k + 1. Then, P has exactly k roots with modulus strictly greater than 1 and all the remaining roots lie inside of the unit disc if and only if |An−k| > |Ai| for all i ̸= n− k. Proof. ”=⇒” Let ω1, · · · , ωn be the roots of P . Assuming that |ω1| ≥ · · · ≥ |ωk| > |ωn−k+1| ≥ · · · ≥ |ωn| Using the Viète theorem, we get |An−l| = | ∑ i1<··· |Ai| for all i ̸= n − k. It is easy to remark, using the Viète theorem, that P has at least one root with modulus strictly greater than 1. Assume now, that P has l ̸= k roots ω1, · · · , ωl such that |ω1| ≥ · · · ≥ |ωl| > |ωn−l+1| ≥ · · · ≥ |ωn|. Therefore, we get two cases: Case 1 : l < k, then from the above we can conclude that |An−l| > |Ai| for all i ̸= n− l, which contraduct our assume. Case 2 : l > k, we have |An−l| = | ∑ i1<··· | ∑ i1<··· k maxdegAn−i 1≤i≤k−1 , therefore, the upper Newton polygon of P contains the line segment joining (n, 0) to (n− 1, degA− n− 1). The slope of this line segment is −degAn−k k . Applaying the proposition 1, item (i), P has n− (n− k) = k Pisot elements with same absolute value | ω1 |= · · · =| ωk |= q degAn−k k A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 8 of 9 So that, ω1, · · · , ωk /∈ Fq((X −1)). Now, assume that P decomposes as P (Y ) = Q(Y )H(Y ), such that Q,H are defined as follow: Q(Y)= Ys +Qs−1Y s−1 +Qs−2Y s−2 + · · ·+Q1Y +Q0 and H(Y)= Yt +Ht−1Y t−1 +Ht−2Y t−2 + · · ·+H1Y +H0 where Q(Y ), H(Y ) ∈ Fq[X][Y ] \ Fq. Case 1 : If one of the polynomials Q or H, say Q, vanishes all the k roots with ab- solute values greater than 1. which implies that everyone of the roots of H have an absolute values less than 1. which, given | H0 |≥ 1, is not possible. Case 2 : If Q have l roots with absolute values greater than 1 and H have l roots with absolute values greater than 1, such that l +m = k. Then, degQs−l > degQi i ̸=s−l , and degHt−m > degQi i ̸=t−m . Or degAn−k > k maxdegAn−i 1≤i≤k−1 , then degQs−l = degHt−m. In addition degAn−k = deg ∑ i+j=k Qs−iHt−j = degQs−l + degHt−m = 2degQs−l. Absurd. Then, we deduce that P is irreducible over Fq[X]. Acknowledgements The authors extend their appreciation to Umm Al-Qura University, Saudi Arabia for funding this research work through grant number: 25UQU4270201GSSR01. Funding This research work was funded by Umm Al-Qura University, Saudi Arabia under grant number : 25UQU4270201GSSR01. References [1] Oussama Dammak and Saber Mansour. On salem formal power series. European Journal of Pure and Applied Mathematics, 15(3):1321–1330, 2022. [2] Ahmed Cherchem, Soufyane Bouguebrine, and Hamza Boughambouz. On the con- struction of irreducible and primitive polynomials from fqm [x] to fq [x]. Finite Fields and Their Applications, 78:101971, 2022. [3] Stephan Lipka. Über die irreduzibilität von polynomen. Mathematische Annalen, 118(1):235–245, 1941. A. Chandoul, A. Assiry / Eur. J. Pure Appl. Math, 18 (2) (2025), 5996 9 of 9 [4] Ravindranathan Thangadurai. 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