EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5709 ISSN 1307-5543 – ejpam.com Published by New York Business Global Base-β(C) Representations and Generalizations of Irreducibility Criteria for Polynomials over Any Imaginary Quadratic Fields Phitthayathon Phetnun1, Narakorn Rompurk Kanasri1,∗ 1 Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen, 40002, Thailand Abstract. Let K be an imaginary quadratic field with the ring of integers OK . In the authors’ earlier work, the so-called base-β(C) representation for nonzero elements of OK was determined, where C is a complete residue system modulo β. Using such a representation, irreducibility criteria for polynomials in OK [x] were established. In this paper, we provide the explicit shapes of all base- β(C) representations for nonzero elements of OK . Generalizations of such irreducibility criteria for polynomials in OK [x] under a certain condition are also established. 2020 Mathematics Subject Classifications: 11R04, 11R09, 11R11 Key Words and Phrases: Imaginary quadratic field, Ring of integers, Complete residue system, Prime element, Irreducible polynomial 1. Introduction An old and interesting problem in mathematics is determining irreducible polynomials over a field. A polynomial with integer coefficients is said to be irreducible in Z[x] if it is an irreducible element of the polynomial ring Z[x]. A polynomial f(x) of positive degree in Z[x] is primitive if the greatest common divisor of its coefficients is 1. It is well known that a nonconstant polynomial in Z[x] is irreducible in Z[x] if and only if it is both irreducible over Q and primitive in Z[x]. Among many irreducibility criteria for polynomials in Z[x], we are interested in a classical result of A. Cohn, given by Pólya and Szegö [10]: if a prime p is expressed in the decimal representation as p = an10 n + an−110 n−1 + · · ·+ a110 + a0, then the polynomial f(x) = anx n + an−1x n−1 + · · ·+ a1x+ a0 is irreducible in Z[x]. This result was subsequently generalized to any base b by Brillhart et al. [2] and Murty [5]. In ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5709 Email addresses: phitthayathon@kkumail.com (P. Phetnun), naraka@kku.ac.th (N.R. Kanasri) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 2 of 15 1982, Filaseta [3] generalized this result in another way by considering wp instead of p: if w and b are positive integers with w < b and wp = bmbm + bm−1b m−1 + · · ·+ b1b+ b0 is the base-b representation of wp, where p is a prime, then the polynomial f(x) = bmxm+ bm−1x m−1 + · · ·+ b1x+ b0 is irreducible over Q. In another direction, let K = Q( √ m) with a unique squarefree integer m ̸= 1, be a quadratic field and OK denotes the set of algebraic integers that lie in K, called the ring of integers of K. We note that OK is an integral domain and K is its quotient field. Moreover, the set of units in the polynomial ring OK [x] is U(OK), the group of units in OK . The quadratic field K is said to be real if m > 0 and imaginary if m < 0. By letting the following notation σm =  √ m if m ̸≡ 1 (mod 4), 1 + √ m 2 if m ≡ 1 (mod 4), we have OK = {a+ bσm | a, b ∈ Z}. More specifically, we see that the ring of Gaussian integers Z[i] is the ring of integers of Q(i) [1]. It is important to note that every prime element of OK is irreducible; the converse is not generally true but holds if OK is a unique factorization domain. We say that a nonzero polynomial p(x) ∈ OK [x] is irreducible in OK [x] if it is an irreducible element of OK [x], in other words, p(x) is not a unit and p(x) = f(x)g(x) in OK [x] implies f(x) or g(x) is a unit in OK . Polynomials that are not irreducible are said to be reducible. For β = a+ bσm ∈ OK , we denote the norm of β by N(β) =  a2 −mb2 if m ̸≡ 1 (mod 4), a2 + ab+ b2 ( 1−m 4 ) if m ≡ 1 (mod 4). We have seen from [1] that if N(β) = ±p, where p is a rational prime, then β is an irreducible element. In addition, ifK is an imaginary quadratic field, then |β|2 = N(β) ∈ N for all β ∈ OK\{0} and |β| = 1 for all β ∈ U(OK). Let us recall the divisibility and congruence for elements of OK as follows: for α, β ∈ OK with α ̸= 0, we say that α divides β, denoted by α | β, if there exists δ ∈ OK such that β = αδ. For α, β, γ ∈ OK with γ ̸= 0, we say that α is congruent to β modulo γ, denoted by α ≡ β (mod γ), if γ | (α− β). A complete residue system modulo β in OK is defined as in the following definition [9]. Definition A. A complete residue system modulo β in OK , abbreviated by CRS(β), means a set of |N(β)| elements C = {α1, α2, . . . , α|N(β)|} in OK which satisfies the follow- ing: (i) for each α ∈ OK , there is αi ∈ C such that α ≡ αi (mod β); P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 3 of 15 (ii) αi ̸≡ αj (mod β) for all i, j ∈ {1, 2, . . . , |N(β)|} with i ̸= j. For example, the set C = { x+ yi | x = 0, 1, . . . , a2 + b2 d − 1 and y = 0, 1, . . . , d− 1 } (1) is a CRS(β), where β = a+ bi ∈ Z[i] with d = gcd (a, b) [11]. It is clear that the set C′ := {x+ yi | x = 0, 1, . . . ,max{|a|, |b|} − 1 and y = 0, 1, . . . , d− 1} ⊆ C. In 2017, Singthongla et al. [12] established the irreducibility criterion for polynomials in Z[i][x] using the CRS(β) in (1). The result is as follows. Theorem A. Let β ∈ {2± 2i, 1± 3i, 3± i} or β = a+ bi ∈ Z[i] be such that |β| ≥ 2+ √ 2 and a ≥ 1. For a Gaussian prime π, if π = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β), where n ≥ 1, Re(αn) ≥ 1, and α0, α1, . . . , αn−1 ∈ C′ satisfying Re(αn−1) Im(αn) ≥ Re(αn) Im(αn−1), then f(x) is irreducible in Z[i][x]. Afterward, Kanasri et al. [4] generalized Theorem A by considering ωπ (ω ∈ Z[i]\{0}) instead of π as the following theorem. Theorem B. Let β = a + bi ∈ Z[i] be such that |β| ≥ 2|ω| + √ 2 and a ≥ |ω|. For a Gaussian prime π, if ωπ = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β), where n ≥ 1, Re(αn) ≥ 1, and α0, α1, . . . , αn−1 ∈ C′ satisfying Re(αn−1) Im(αn) ≥ Re(αn) Im(αn−1), then f(x) is irreducible over Q(i). For any quadratic field K = Q( √ m), Tadee et al. [13] proved for the case m ̸≡ 1 (mod 4) that the set C = { x+ yσm | x = 0, 1, . . . , |N(β)| d − 1 and y = 0, 1, . . . , d− 1 } (2) is a CRS(β), where β = a + bσm ∈ OK with d = gcd (a, b). Recently, Phetnun et al. [8] verified that (2) is also a CRS(β) in any quadratic field K = Q( √ m) with m ≡ 1 (mod 4). These results extend the CRS(β) for Gaussian integers in (1) to any quadratic field. Moreover, they determined the so-called base-β(C) representation for nonzero elements of OK and extended Theorem A to any imaginary quadratic field using such a representation. It was shown for any m < 0 in [8] that the set C′ := {x+ yσm | x = 0, 1, . . . ,max{|a|, |b|} − 1 and y = 0, 1, . . . , d− 1} ⊆ C. Some results in [8] described above are as follows. P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 4 of 15 Definition B. Let K = Q( √ m) be an imaginary quadratic field. Let β ∈ OK\{0} and C be the CRS(β) as in (2). We say that η ∈ OK\{0} has a base-β(C) representation if η = αnβ n + αn−1β n−1 + · · ·+ α1β + α0, (3) where n ≥ 1, αn ∈ OK\{0}, and αi ∈ C (i = 0, 1, . . . , n−1). If αi ∈ C′ (i = 0, 1, . . . , n−1), then (3) is called a base-β(C′) representation of η. Theorem C. Let K = Q( √ m) be an imaginary quadratic field with m ̸≡ 1 (mod 4). Let β = a+b √ m ∈ OK be such that |β| ≥ 2+ √ 1−m and a ≥ 1+ √ 1−m. For an irreducible element π of OK with |π| ≥ |β|, if π = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β) is a base-β(C′) representation with Re(αn) ≥ 1 satisfying Re(αn−1) Im(αn) ≥ Re(αn) Im(αn−1), then f(x) is irreducible in OK [x]. Theorem D. Let K = Q( √ m) be an imaginary quadratic field with m ≡ 1 (mod 4). Let β = a+ bσm ∈ OK be such that |β| ≥ 2 + √ (9−m)/4, a ≥ 1, and a+ (b/2) ≥ 1. For an irreducible element π of OK with |π| > √ (9−m)/4 (|β| − 1), if π = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β) is a base-β(C′) representation with Re(αn) ≥ 1 satisfying Re(αn−1) Im(αn) ≥ Re(αn) Im(αn−1), then f(x) is irreducible in OK [x]. Recently, Phetnun and Kanasri [7] established further irreducibility criteria for poly- nomials in OK [x], which extended Theorem 3.22 in [12] to any imaginary quadratic field. They also provided elements of β that can be applied to these criteria but not to Theorem C and Theorem D. In the present work, we provide the explicit shapes of all base-β(C) representations for nonzero elements of OK , where K is an imaginary quadratic field. In addition, we generalize Theorem C and Theorem D by considering ωπ instead of π, where ω ∈ OK\{0} and π is a prime element, which in turn extend Theorem B to any imaginary quadratic field. 2. Explicit Shapes of Base-β(C) Representations Let K be an imaginary quadratic field and η, β = a+ bσm be nonzero elements of OK . Let C be the base-β(C) representations as in (2). If η ∈ C, it is clear from Definition B that η cannot be written as a base-β(C) representation. Assume henceforth that η ̸∈ C. By Definition A, there exists a unique α0 ∈ C such that η ≡ α0 (mod β), so η = γ0β + α0 for some γ0 ∈ OK\{0}. If γ0 ∈ C, then the process stops, otherwise, there exists a unique α1 ∈ C such that γ0 ≡ α1 (mod β), yielding γ0 = γ1β + α1 for some γ1 ∈ OK\{0}. It P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 5 of 15 follows that η = γ1β 2 + α1β + α0. Continuing the process, we obtain the sequence (γi)i≥0 of elements of OK such that γi−1 = γiβ + αi (i ≥ 0) (4) with γ−1 = η. If there exists n ∈ N ∪ {0} such that γn ∈ C, then η = γ0β + α0, η = γ1β 2 + α1β + α0, ... η = γnβ n+1 + αnβ n + · · ·+ α1β + α0 (5) are n+ 1 base-β(C) representations of η. The following theorem shows that the equations in (5) are all base-β(C) representations of η for the case γn ∈ C for some n ∈ N ∪ {0}. Theorem 1. Let K = Q( √ m) be an imaginary quadratic field and let β, η ∈ OK\{0} with η ̸∈ C. Let (γi)i≥0 be the sequence described in (4). If γn ∈ C for some n ∈ N ∪ {0}, then all equations in (5) are the base-β(C) representations of η. Proof. By the description mentioned above, all n + 1 equations in (5) are base-β(C) representations of η. Next, let η = δkβ k + δk−1β k−1 + · · ·+ δ1β + δ0 (6) be any base-β(C) representation of η. Then k ∈ N, δi ∈ C (0 ≤ i ≤ k−1), and δk ∈ OK\{0}. Suppose that k ≥ n + 2. By the last equation in (5) together with (6), we obtain that η ≡ α0 (mod β) and η ≡ δ0 (mod β). It follows from Definition A(ii) that α0 = δ0. Then η1 := γnβ n + αnβ n−1 + · · ·+ α2β + α1 = δkβ k−1 + δk−1β k−2 + · · ·+ δ2β + δ1, so α1 ≡ δ1 (mod β). Again, by Definition A(ii), we obtain α1 = δ1 and so η2 := γnβ n−1 + αnβ n−2 + · · ·+ α2 = δkβ k−2 + δk−1β k−3 + · · ·+ δ2. Proceeding in the same manner, we deduce that αi = δi (0 ≤ i ≤ n) and γn = δkβ k−n−1+ · · · + δn+2β + δn+1, implying γn ≡ δn+1 (mod β). Since γn, δn+1 ∈ C, it follows that γn = δn+1. Thus we have δkβ k−n−2 + · · ·+ δn+3β + δn+2 = 0, (7) so δn+2 ≡ 0 (mod β). If k = n+ 2, then (7) implies δk = 0, a contradiction. If k > n+ 2, then δn+2 ∈ C and thus δn+2 = 0 by Definition A(ii). Continuing similarly, we finally obtain δk = 0, which is a contradiction. Hence k ≤ n+ 1, implying the base-β(C) representation of η in (6) is one of that in (5) as desired. The base-β(C) representations for the case γn ∈ C for some n ∈ N ∪ {0} are shown in the following example. P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 6 of 15 Example 1. Let K = Q( √ −1), β = −3 + 2i, and η = −439 − 258i. Then d = 1, N(β) = 13, and thus C = {0, 1, . . . , 12} is a CRS(β). One can compute that η = (63 + 128i)β + 6, 63 + 128i = (7− 38i)β + 8, 7− 38i = (−7 + 8i)β + 2, −7 + 8i = 4β + 5. Thus, γ0 = 63+128i, γ1 = 7− 38i, γ2 = −7+ 8i, γ3 = 4 ∈ C and α0 = 6, α1 = 8, α2 = 2, α3 = 5. By Theorem 1, we conclude that η = γkβ k+1 + αkβ k + · · ·+ α1β + α0 (0 ≤ k ≤ 3) are the base-β(C) representations of η. To provide the explicit shapes of the base-β(C) representations of η for the case γi ̸∈ C for all i ∈ N ∪ {0}, we require the following lemma. Lemma 1. Let K = Q( √ m) be an imaginary quadratic field and let β = a + bσm ∈ OK such that |β| ≥ 2 and d = gcd(a, b). Let S =  √( |N(β)| d − 1 )2 −m(d− 1)2 if m ̸≡ 1 (mod 4),√( |N(β)| d − 1 )2 + ( |N(β)| d − 1 ) (d− 1) + (d− 1)2 ( 1−m 4 ) if m ≡ 1 (mod 4). (8) For η ∈ OK with |η| ≥ S, if η = γ0β+α0, where γ0 ∈ OK\{0} and α0 ∈ C, then |η| ≥ |γ0|. Proof. Suppose to the contrary that |γ0| > |η|. Then |γ0| > |γ0β + α0| ≥ |γ0||β| − |α0| and thus |α0| > |γ0|(|β| − 1). Since α0 ∈ C, we have |α0| ≤ S and hence |γ0| > |η| ≥ S ≥ |α0| > |γ0|(|β| − 1) ≥ |γ0|, which is a contradiction. Theorem 2. Let K = Q( √ m) be an imaginary quadratic field and let β = a+bσm, η ∈ OK be such that |β| ≥ 2, η ̸∈ C, and |η| ≥ S, where S is defined as in (8). Let (γi)i≥0 be the sequence described in (4). If γi ̸∈ C for all i ∈ N ∪ {0}, then there exist nonnegative integers u and v with u < v such that γu = γv and the base-β(C) representations of η are of the form η = γkβ k+1 + ∑ 0≤i≤k αiβ i (0 ≤ k ≤ u), (9) η = γkβ k+1 + ∑ u |γ1β + α1| ≥ |γ1||β| − |α1| and thus |γ1| < 2|γ1| ≤ |γ1||β| < S + |α1| ≤ S + S = 2S ≤ N. Proceeding in the same manner, we obtain |γi| ≤ N for all i ∈ N ∪ {0}. Claim 2. There exist nonnegative integers u and v with u < v such that γu = γv. Proof of Claim 2. By Claim 1, we have |γi| ≤ N for all i ∈ N ∪ {0}. For each i ∈ N ∪ {0}, we denote γi := ai + biσm and A := {(a0, b0), (a1, b1), (a2, b2), . . .}, where ai, bi ∈ Z for all i ∈ N ∪ {0}. We first show that A is a finite set by considering the following two cases: Case 1. m ̸≡ 1 (mod 4). Then a2i −mb2i = N (γi) = |γi|2 ≤ N2 for all i ∈ N ∪ {0}. Thus a2i ≤ N2 and b2i ≤ N2 and so |ai| ≤ N and |bi| ≤ N . It follows that the set A is finite. Case 2. m ≡ 1 (mod 4). Then a2i + aibi + b2i (1 − m)/4 = N (γi) = |γi|2 ≤ N2 for all i ∈ N∪{0}, so (2ai + bi) 2 −mb2i = 4a2i +4aibi + b2i −mb2i ≤ 4N2. It follows that |bi| ≤ 2N and |2ai + bi| ≤ 2N and thus |ai| ≤ |2ai| ≤ |2ai + bi| + |bi| ≤ 2N + 2N = 4N, implying the set A is finite. Consequently, (au, bu) = (av, bv) for some nonnegative integers u and v with u < v. This implies that γu = au + buσm = av + bvσm = γv. Next, we show that the base-β(C) representations of η are the equations in (9), (10), (11), and (12). Recall that η = γ0β + α0 and γi−1 = γiβ + αi (i ≥ 1), (13) where γi ∈ OK\{0} and αi ∈ C for all i ≥ 0. Then we obtain η = γkβ k+1 + αkβ k + αk−1β k−1 + · · ·+ α1β + α0 (0 ≤ k ≤ u), (14) P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 8 of 15 which are base-β(C) representations in (9). Again, using (13) repeatedly, we get η = γkβ k+1 + αkβ k + · · ·+ αu+1β u+1 + αuβ u + · · ·+ α1β + α0 (u < k < v), which are base-β(C) representations in (10). Since γu = γv, we obtain from (13) that γu = γu+1β + αu+1 = γu+2β 2 + αu+2β + αu+1 ... = γuβ v−u + αvβ v−u−1 + αv−1β v−u−2 + · · ·+ αu+2β + αu+1. (15) Substituting (15) into (14) with k = u leads to η = γuβ v+1+αvβ v+αv−1β v−1+ · · ·+αu+1β u+1+αuβ u+αu−1β u−1+ · · ·+α1β+α0. (16) Again, substituting (15) into (16) yields η = γuβ 2v−u+1 + αvβ 2v−u + αv−1β 2v−u−1 + · · ·+ αu+1β v+1 + αvβ v + αv−1β v−1 + · · ·+ αu+1β u+1 + αuβ u + αu−1β u−1 + · · ·+ α1β + α0. (17) Continuing this process, in general, we have η = γuβ (v−u)n+u+1 + αvβ (v−u)n+u + αv−1β (v−u)n+u−1 + · · ·+ αu+1β (v−u)n+2u−v+1 + · · ·+ ∑ u 2 and |η| = √ 525 > √ 294 = S. One can compute that η = (1 + 2 √ −5)β + 11, P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 9 of 15 1 + 2 √ −5 = (−1 + √ −5)β + 15, −1 + √ −5 = (−2 + √ −5)β + (17 + √ −5), −2 + √ −5 = (−2 + √ −5)β + (16 + √ −5). Thus γ0 = 1+2 √ −5, γ1 = −1+ √ −5, γ2 = −2+ √ −5 = γ3, and so γi = γ2 for all i ≥ 2. One can see that γi ̸∈ C for all i ∈ N ∪ {0}. Note that α0 = 11, α1 = 15, α2 = 17 + √ −5, and α3 = 16 + √ −5 = αi for all i ≥ 3. Using Theorem 2 with u = 2 and v = 3, we conclude that the base-β(C) representations of η are η = γkβ k+1 + αkβ k + · · ·+ α1β + α0 (0 ≤ k ≤ 2) and η = γ2β n+3 + α3β n+2 + · · ·+ α3β 3 + α2β 2 + α1β + α0 for all n ∈ N. Example 3. Let K = Q( √ −7), β = 2 + 3σ−7, and η = 3112 − 13810σ−7. Then d = 1, N(β) = 28, and thus C = {0, 1, . . . , 27} is a CRS(β). We have |β| = √ 28 > 2 and |η| = √ 348140024 > 27 = S. One can compute that η = (−2405− 1319σ−7)β + 8, −2405− 1319σ−7 = (−713 + 164σ−7)β + 5, −713 + 164σ−7 = (−97 + 91σ−7)β + 27, −97 + 91σ−7 = (2 + 17σ−7)β + 1, 2 + 17σ−7 = (4 + σ−7)β, 4 + σ−7 = (−3 + 2σ−7)β + 22, −3 + 2σ−7 = (−1 + σ−7)β + 5, −1 + σ−7 = (−3 + 2σ−7)β + 17, −3 + 2σ−7 = (−1 + σ−7)β + 5. Thus γ0 = −2405 − 1319σ−7, γ1 = −713 + 164σ−7, γ2 = −97 + 91σ−7, γ3 = 2 + 17σ−7, γ4 = 4 + σ−7, γ5 = −3 + 2σ−7 = γ7 = γ9 = · · · , and γ6 = −1 + σ−7 = γ8 = γ10 = · · · . One can see that γi ̸∈ C for all i ∈ N ∪ {0}. We also have α0 = 8, α1 = 5, α2 = 27, α3 = 1, α4 = 0, α5 = 22, α6 = 5 = α8 = α10 = · · · , and α7 = 17 = α9 = α11 = · · · . Using Theorem 2 with u = 5 and v = 7, we conclude that the base-β(C) representations of η are η = γkβ k+1 + αkβ k + · · ·+ α1β + α0 (0 ≤ k ≤ 5), η = γ6β 7 + α6β 6 + α5β 5 + α4β 4 + α3β 3 + α2β 2 + α1β + α0, η = γ5β 2n+6 + α7β 2n+5 + α6β 2n+4 + · · ·+ α7β 7 + α6β 6 + α5β 5 + α4β 4 + α3β 3 + α2β 2 + α1β + α0 (n ∈ N), η = γ6β 2n+7 + α6β 2n+6 + α7β 2n+5 + α6β 2n+4 + · · ·+ α7β 7 + α6β 6 + α5β 5 + α4β 4 + α3β 3 + α2β 2 + α1β + α0 (n ∈ N). P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 10 of 15 3. Generalizations of Irreducibility Criteria for Polynomials in OK [x] Let K = Q( √ m) be an imaginary quadratic field. In this section, we extend Theorem B to any imaginary quadratic field, which in turn generalize Theorem C and Theorem D by considering ωπ instead of π, where ω ∈ OK\{0} and π is a prime element. To prove these, we first recall the essential three lemmas. Lemma 2. [12] Let f(x) = αnx n + αn−1x n−1 + · · ·+ α1x+ α0 ∈ C[x] be such that n ≥ 2 and |αi| ≤ M (0 ≤ i ≤ n− 2) for some positive real number M . If f(x) satisfies (i) Re(αn) ≥ 1, Re(αn−1) ≥ 0, and Im(αn−1) ≥ 0; and (ii) Re(αn−1) Im(αn) ≥ Re(αn) Im(αn−1), then any complex zero α of f(x) satisfies either Re(α) < 0 or |α| < ( 1 + √ 1 + 4M ) /2. Lemma 3. [8] Let K = Q( √ m) be an imaginary quadratic field with m ̸≡ 1 (mod 4). Let β = a+ b √ m ∈ OK be such that a ≥ 1 + √ 1−m and M = √ (max{a, |b|} − 1)2 −m(d− 1)2, (19) where d = gcd(a, b). Then √ 1−m(|β| − 1) ≥ M . Lemma 4. [8] Let K = Q( √ m) be an imaginary quadratic field with m ≡ 1 (mod 4). Let β = a+ bσm ∈ OK be such that a ≥ 1 and M = √ (max{a, |b|} − 1)2 + (max{a, |b|} − 1)(d− 1) + (d− 1)2 ( 1−m 4 ) , (20) where d = gcd(a, b). Then √ (9−m)/4(|β| − 1) ≥ M . If f(x) is a nonconstant polynomial in OK [x], we say that f(x) = g(x)h(x) in OK [x] is a proper factorization if both g(x) and h(x) have a smaller degree than f(x). We now proceed to our second main result and start with the case m ̸≡ 1 (mod 4). Theorem 3. Let K = Q( √ m) be an imaginary quadratic field with m ̸≡ 1 (mod 4). Let β = a + b √ m ∈ OK and ω ∈ OK\{0} be such that |β| ≥ 2|ω| + √ 1−m and a ≥ |ω|+ √ 1−m. For a prime element π of OK with |π| ≥ |β|, if ωπ = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β) is a base-β(C′) representation with n ≥ 2 and Re(αn) ≥ 1 satisfying the condition (ii) of Lemma 2, then f(x) has no proper factorization in OK [x]. Moreover, (i) if δ ∤ f(x) for all δ ∈ OK\U(OK), then f(x) is irreducible in OK [x]; (ii) if OK is a unique factorization domain, then f(x) is irreducible over K. P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 11 of 15 Proof. Suppose to the contrary that f(x) has proper factorization in OK [x]. Then f(x) = g(x)h(x) for some nonconstant polynomials g(x) and h(x) in OK [x], so ωπ = g(β)h(β). Since π is a prime element, either π | g(β) and h(β) | ω or π | h(β) and g(β) | ω. This implies that either |ω| ≥ |h(β)| or |ω| ≥ |g(β)|. Without loss of generality, we may assume that |ω| ≥ |g(β)|. Since αi ∈ C′ for all i ∈ {0, 1, . . . , n− 1}, we have |αi| ≤ M for all i ∈ {0, 1, . . . , n− 1}, where M is defined as in (19). We now show that |β| − 1 + √ 1 + 4M 2 ≥ |ω|. (21) Since |β| ≥ 2|ω| + √ 1−m, we have |β|2 − ( 2|ω|+ 1 + √ 1−m ) |β| + 2|ω| + √ 1−m = (|β| − 1) [ |β| − ( 2|ω|+ √ 1−m )] ≥ 0. As |ω|2 + |ω| ≥ 2|ω|, we obtain 4 [ |β|2 − ( 2|ω|+ 1 + √ 1−m ) |β|+ |ω|2 + |ω|+ √ 1−m ] ≥ 0 and so [2|β| − (2|ω|+ 1)]2 = 4|β|2−4(2|ω|+1)|β|+4|ω|2+4|ω|+1 ≥ 1+4 √ 1−m (|β| − 1) . Again, |β| ≥ 2|ω|+ √ 1−m implies 2|β| − (2|ω|+1) > 0 and thus, 2|β| − (2|ω|+1) ≥ √ 1 + 4 √ 1−m (|β| − 1). Hence |β| ≥ 2|ω|+ 1 + √ 1 + 4 √ 1−m (|β| − 1) 2 . As a ≥ |ω|+ √ 1−m, it follows from Lemma 3 that √ 1−m (|β| − 1) ≥ M . Therefore, |β| ≥ 2|ω|+ 1 + √ 1 + 4M 2 = |ω|+ 1 + √ 1 + 4M 2 , which proves (21). Now, we have that deg g(x) ≥ 1, so g(x) can be expressed in the form g(x) = ε ∏ i(x− γi), where ε ∈ OK is the leading coefficient of g(x) and the product is over the set of complex zeros of g(x). It follows from Lemma 2 that any zero γ of g(x) satisfies either Re(γ) < 0 or |γ| < 1 + √ 1 + 4M 2 . (22) In the former case, we have |β − γ| ≥ Re(β − γ) = Re(β) − Re(γ) > Re(β) = a ≥ |ω|+ √ 1−m > |ω|. In the latter case, we obtain by using (21) and (22) that |β − γ| ≥ |β| − |γ| > |β| − 1 + √ 1 + 4M 2 ≥ |ω|. From both cases and |ε| ≥ 1, we deduce that |ω| ≥ |g(β)| = |ε| ∏ i |β − γi| ≥ ∏ i |β − γi| > |ω|, which is a contradiction. We have that f(x) has no proper factorization in OK [x]. Then, (i) is true. If OK is a unique factorization domain, f(x) is irreducible over K. To see this, suppose to the P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 12 of 15 contrary that f(x) = g1(x)h1(x) for some nonconstant polynomials g1(x) and h1(x) in K[x]. Since every element of K is of the form α/r, where α ∈ OK and r ∈ Z\{0}, we may take g2(x) = rg1(x) and h2(x) = sh1(x), where g2(x), h2(x) ∈ OK [x] with deg g2(x) ≥ 1, deg h2(x) ≥ 1 and r, s ∈ Z\{0} so that rsf(x) = g2(x)h2(x). Let π be a prime divisor of rs in OK . Considering the contents of the above polynomials and using Gauss’s lemma for OK , we get that π divides g2(x) or h2(x), yielding rs π f(x) = g3(x)h3(x) for some nonconstant polynomials g3(x), h3(x) ∈ OK [x]. Continuing in the same manner, we finally get f(x) = g(x)h(x), where g(x) and h(x) are positive degree polynomials in OK [x], which is a contradiction. This proves (ii) and we complete the proof. We note that the condition a ≥ |ω|+ √ 1−m in Theorem 3 can be reduced to a ≥ |ω| for the case of Gaussian integers and one can see that Z[i] is a unique factorization domain. This implies that Theorem 3 extends Theorem B to any imaginary quadratic field with m ̸≡ 1 (mod 4). Applying Theorem 3, we can find irreducible polynomials as the following examples. Example 4. Let K = Q( √ −2), β = 9−3 √ −2, ω = 2+ √ −2, and π = −8177−28932 √ −2. Then d = 3 and thus C′ = { x+ y √ −2 | x = 0, 1, . . . , 8 and y = 0, 1, 2 } . One can see that |β| = √ 99 > 2 √ 6 + √ 3 = 2|ω| + √ 1−m and a = 9 > √ 6 + √ 3 = |ω| + √ 1−m. Since N(π) = 1740984577 is a rational prime, we have that π is an irreducible element and so is a prime element because OK is a unique factorization domain. Now, we have |π| > |β| and ωπ = 41510− 66041 √ −2 = (8 + 4 √ −2)β4 + 6β3 + (4 + 2 √ −2)β2 + 6β + (2 + √ −2), which is a base-β(C′) representation of ωπ. Moreover, Re(α4) = 8 > 1 and Re(α3) Im(α4) = (6)(4 √ 2) > (8)(0) = Re(α4) Im(α3). By Theorem 3, we obtain that f(x) = (8+4 √ −2)x4+ 6x3 + (4 + 2 √ −2)x2 + 6x + (2 + √ −2) has no proper factorization in OK [x] and so is irreducible over K. Observe that f(x) is reducible in OK [x] because f(x) = (2 +√ −2) [ 4x4 + (2− √ −2)x3 + 2x2 + (2− √ −2)x+ 1 ] . Example 5. Let K = Q( √ −1), β = 15 − 16i, ω = 3 + 2i, and π = −831565 + 715166i. Then d = 1 and thus C′ = {0, 1, . . . , 15}. Note that |β| = √ 481 > 2 √ 13 + √ 2 = 2|ω| +√ 1−m and a = 15 > √ 13 + √ 2 = |ω| + √ 1−m. Since N(π) = 1202962756781 is a rational prime, we get that π is a Gaussian prime. Now, we have |π| > |β| and ωπ = −3925027 + 482368i = 17β4 + 3β3 + 7β2 + 5β + 13, which is a base-β(C′) representation of ωπ. Moreover, Re(α4) = 17 > 1 and Re(α3) Im(α4) = (3)(0) ≥ (17)(0) = Re(α4) Im(α3). By Theorem 3, we obtain that f(x) = 17x4 + 3x3 + 7x2 + 5x+ 13 has no proper factorization in Z[i][x] and so is irreducible over K. Since 3 and 7 are Gaussian primes, it follows that f(x) is irreducible in Z[i][x]. P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 13 of 15 By taking ω ∈ U(OK) in Theorem 3, we obtain that the polynomial f(x) has no proper factorization in OK [x]. Since π is also an irreducible element, then δ ∤ f(x) for all δ ∈ OK\U(OK), by Lemma 4 in [8]. This shows that f(x) is irreducible in OK [x]. Therefore, Theorem 3 is a generalization of Theorem C by considering ωπ instead of π, where ω ∈ OK\{0} and π is a prime element. For the case m ≡ 1 (mod 4), we now prove the following. Theorem 4. Let K = Q( √ m) be an imaginary quadratic field with m ≡ 1 (mod 4). Let β = a + bσm ∈ OK and ω ∈ OK\{0} be such that |β| ≥ 2|ω| + √ (9−m)/4, a ≥ 1, and a+ (b/2) ≥ |ω|. For a prime element π of OK with |π| > √ (9−m)/4 (|β| − 1), if ωπ = αnβ n + αn−1β n−1 + · · ·+ α1β + α0 =: f(β) is a base-β(C′) representation with n ≥ 2 and Re(αn) ≥ 1 satisfying the condition (ii) of Lemma 2, then f(x) has no proper factorization in OK [x]. Moreover, (i) if δ ∤ f(x) for all δ ∈ OK\U(OK), then f(x) is irreducible in OK [x]; (ii) if OK is a unique factorization domain, then f(x) is irreducible over K. Proof. Suppose to the contrary that f(x) has proper factorization in OK [x], namely f(x) = g(x)h(x) for some nonconstant polynomials g(x) and h(x) in OK [x]. By the same proof as in Theorem 3, we have that either |ω| ≥ |h(β)| or |ω| ≥ |g(β)|. Without loss of generality, we may assume that |ω| ≥ |g(β)|. Since αi ∈ C′ for all i ∈ {0, 1, . . . , n− 1}, we have |αi| ≤ M for all i ∈ {0, 1, . . . , n− 1}, where M is defined as in (20). We now show that |β| − 1 + √ 1 + 4M 2 ≥ |ω|. (23) As |β| ≥ 2|ω|+ √ (9−m)/4, we have |β|2− ( 2|ω|+ 1 + √ (9−m)/4 ) |β|+2|ω|+ √ (9−m)/4 = (|β| − 1) [ |β| − ( 2|ω|+ √ (9−m)/4 )] ≥ 0. Since |ω|2 + |ω| ≥ 2|ω|, it follows that 4 [ |β|2 − ( 2|ω|+ 1 + √ (9−m)/4 ) |β|+ |ω|2 + |ω|+ √ (9−m)/4 ] ≥ 0 and so [2|β| − (2|ω|+ 1)]2 = 4|β|2 − 4(2|ω| + 1)|β| + 4|ω|2 + 4|ω| + 1 ≥ 1 + 4 √ (9−m)/4 (|β| − 1) . Again, |β| ≥ 2|ω| + √ (9−m)/4 shows that 2|β| − (2|ω| + 1) > 0 and thus, 2|β| − (2|ω| + 1) ≥√ 1 + 4 √ (9−m)/4 (|β| − 1). Hence, |β| ≥ 2|ω|+ 1 + √ 1 + 4 √ (9−m)/4 (|β| − 1) 2 . Since a ≥ 1, it follows from Lemma 4 that √ (9−m)/4 (|β| − 1) ≥ M . Thus, |β| ≥ 2|ω|+ 1 + √ 1 + 4M 2 = |ω|+ 1 + √ 1 + 4M 2 , P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 14 of 15 which proves (23). The remaining proof is similar to that of Theorem 3, so we omit it here. We end this paper with the following examples, illustrating the use of Theorem 4. Example 6. Let K = Q( √ −3), β = 7 + σ−3, ω = 2, and π = 54343 + 63615σ−3. Then d = 1 and thus C′ = {0, 1, 2, 3, 4, 5, 6}. One can see that |β| = √ 57 > 4 + √ 3 = 2|ω| + √ (9−m)/4, a = 7 > 1, and a + (b/2) = 7 + (1/2) > 2 = |ω|. Note that OK is a unique factorization domain and π is a prime element because N(π) = 10457059819 is a rational prime. Now, we have |π| > √ (9−m)/4 (|β| − 1) and ωπ = 108686 + 127230σ−3 = 8β5 + 2β4 + 4β3 + 2β2 + 6β + 2, which is a base-β(C′) representation of ωπ. Moreover, Re(α5) = 8 > 1 and Re(α4) Im(α5) = (2)(0) ≥ (8)(0) = Re(α5) Im(α4). By Theorem 4, we obtain that f(x) = 8x5+2x4+4x3+ 2x2+6x+2 has no proper factorization in OK [x] and so is irreducible over K. It is noticed that f(x) is reducible in OK [x] because f(x) = 2(4x5 + x4 + 2x3 + x2 + 3x+ 1). Example 7. Let K = Q( √ −7), β = 11+3σ−7, ω = 2+σ−7, and π = 158481+166844σ−7. Then d = 1 and thus C′ = {0, 1, . . . , 10}. One can see that |β| = √ 172 > 2 √ 8 + 2 = 2|ω|+ √ (9−m)/4, a = 11 > 1, and a+ (b/2) = 11 + (3/2) > √ 8 = |ω|. Note that OK is a unique factorization domain and π is a prime element because N(π) = 107231671997 is a rational prime. Now, we have |π| > √ (9−m)/4 (|β| − 1) and ωπ = −16726 + 659013σ−7 = 31β4 + 4β3 + 3β2 + 9β + 5, which is a base-β(C′) representation of ωπ. Moreover, Re(α4) = 31 > 1 and Re(α3) Im(α4) = (4)(0) ≥ (31)(0) = Re(α4) Im(α3). By Theorem 4, we obtain that f(x) = 31x4 + 4x3 + 3x2+9x+5 has no proper factorization in OK [x] and so is irreducible over K. By Theorem 9.29(3) in [6], one can verify that the rational primes 3, 5, and 31 are prime elements of OK . This implies that f(x) is irreducible in OK [x]. By taking ω ∈ U(OK) in Theorem 4, we obtain that the polynomial f(x) has no proper factorization in OK [x]. Since π is also an irreducible element, then δ ∤ f(x) for all δ ∈ OK\U(OK), by Lemma 6 in [8]. This shows that f(x) is irreducible in OK [x]. Therefore, Theorem 4 is a generalization of Theorem D by considering ωπ instead of π, where ω ∈ OK\{0} and π is a prime element. 4. Conclusion For any imaginary quadratic field K, we provide the explicit shapes of all base-β(C) representations for nonzero elements of OK . Using such a representation, irreducibility criteria for polynomials in OK [x] are established, which extend and generalize the authors’ earlier work. Acknowledgements This work was supported by the Science Achievement Scholarship of Thailand (SAST). P. Phetnun, N.R. Kanasri / Eur. J. Pure Appl. Math, 18 (1) (2025), 5709 15 of 15 References [1] S Alaca and K S Williams. Introductory Algebraic Number Theory. Cambridge Uni- versity Press, Cambridge, 2004. [2] J Brillhart, M Filaseta, and A Odlyzko. On an irreducibility theorem of A Cohn. Canadian Journal of Mathematics, 33(5):1055–1059, 1981. [3] M Filaseta. A further generalization of an irreducibility theorem of A Cohn. 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