# On the Irreducibility of Polynomials Associated with the Complete Residue Systems in any Imaginary Quadratic Fields

Authors: Phitthayathon Phetnun; Narakorn Rompurk Kanasri; Patiwat Singthongla
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

For a Gaussian primeπ and a nonzero Gaussian integer β=a+bi∈ℤi with a≥1 and β≥2+2, it was proved that if π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβ where n≥1, αn∈ℤi\0, α0,…,αn−1 belong to a complete residue system modulo β, and the digits αn−1 and αn satisfy certain restrictions, then the polynomial fx is irreducible in ℤix. For any quadratic field K≔ℚm, it is well known that there are explicit representations for a complete residue system in K, but those of the case m≡1mod4 are inapplicable to this work. In this article, we establish a new complete residue system for such a case and then generalize the result mentioned above for the ring of integers of any imaginary quadratic field.


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## Body

## 1. Introduction

According to Pólya and Szegö [1], the classical result of A. Cohn states that if a prime p is expressed in the decimal representation as(1)p=an10n+an−110n−1+⋯+a110+a0,then the polynomial fx=anxn+an−1xn−1+⋯+a1x+a0 is irreducible in ℤx. For example, since 4831 is prime and 4831=4⋅103+8⋅102+3⋅10+1, it follows that fx=4x3+8x2+3x+1 is irreducible in ℤx. This result was subsequently generalized to any base b by Brillhart et al. [2]. In 2002, Murty gave a proof of this fact [3] that was conceptually simpler than the one in [2]. Moreover, Brillhart et al. [2] generalized Cohn’s result in another direction as follows: if(2)fx=anxn+an−1xn−1+⋯+a1x+a0∈ℤx,where 0≤ai≤167 for all i, and f10 is prime, then fx is irreducible. In 1988, Filaseta improved this fact, that is, if coefficients of the polynomial fx in (2) satisfy 0≤ai≤an1030 for 0≤i≤n−1 and f10 is prime, then fx is irreducible [4].In another direction, letK=ℚm, with a unique squarefree integer m unequal to 1, be a quadratic field and OK be its ring of integers. We have seen in [5] that OK is an integral domain. The quadratic field K is said to be real if K⊆ℝ and imaginary if K⊆ℝ. Clearly, K is real if m>0 and imaginary if m<0. It is well known that(3)OK=a+bσm|a,b∈ℤ,where(4)σm≔mifm≡1mod4,1+m2ifm≡1mod4.We note that the ring of Gaussian integersℤi and the Eisenstein domain ℤ1+−3/2 are rings of integers of ℚi and ℚ−3, respectively.We emphasize that a prime element inOK is an irreducible element, and the converse holds if OK is a unique factorization domain. Moreover, K is the quotient field of OK [5], and the units in OKx are the units in OK [6]. We say that a nonzero polynomial px∈OKx is irreducible in OKx if px is not a unit, and if px=fxgx with fxandgx in OKx, then fx or gx is a unit in OK. Polynomials that are not irreducible are called reducible. Furthermore, if OK is a unique factorization domain and fx is irreducible in OKx, then fx is irreducible over K by using Gauss’s lemma.In 2017, Singthongla et al. established the result of A. Cohn inOKx [7], where K is an imaginary quadratic field such that OK is a Euclidean domain, namely, m=−1,−2,−3,−7, and −11 [5]. The results are as follows:(i)
Forβ∈OK\0, each element in OK has a base β-expansion whose digits are bounded by certain constants.(ii)
For a prime elementπ in OK, if(5)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβ,is its baseβ-expansion in OK and the digits αn and αn−1 satisfy some natural restrictions, then the polynomial fx is irreducible in OKx.(iii)
For Gaussian integers, a similar baseβ-expansion but with digits belonging to a complete residue system modulo β is also valid.(iv)
The irreducibility result similar to that in (ii) continues to hold for Gaussian integers with base expansion as described in (iii).At first thought, we are interested in establishing a generalization of the result in (iv) for the ring of integersOK of any general quadratic field. However, to prove this, we must use the property that β≥1 for all β∈OK\0 which is valid only in the imaginary quadratic field [8] but not in the real quadratic field. Thus, the goal of this work is to establish a generalization of the result in (iv) for the ring of integers OK of any imaginary quadratic field.LetK=ℚm be any general quadratic field. For α,β∈OK with α≠0, we say that α divides β, denoted by α|β, if and only 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, and we write α≡βmodγ if α−β∈γ, a principal ideal generated by γ or, equivalently, if γ|α−β [9].Forβ=a+bσm∈OK, we denote the norm of β by(6)Nβ=a2−mb2ifm≡1mod4,a2+ab+b21−m4ifm≡1mod4.We note from [5] that if α∈OK is such that Nα=±p, where p is a rational prime, then α is an irreducible element. By a complete residue system modulo β in OK, abbreviated by CRSβ [9, 10], we mean a set of Nβ elements C=α1,α2,…,αNβ in OK such that(i)
for eachα∈OK, there is αi∈C such that α≡αimodβ and(ii)
αi≡αjmodβ for all i,j∈1,2,…,Nβ, with i≠j.In [10], Tadee et al. derived three explicit representations for a complete residue system in any general quadratic field. We are interested in the first one which is as follows: for β=a+bσm∈OK\0 and d=gcda,b, we have the following:(i)
Casem≡1mod4: the set(7)C≔x+ym|x=0,1,…,Nβd−1andy=0,1,…,d−1,is aCRSβ.(ii)
Casem≡1mod4:(1)
Ifd is even, let(8)C1≔x+ym|x=0,1,…,Nβd−1andy=0,1,…,d−22,C2≔x+12+y+12m|x=0,1,…,Nβd−1andy=0,1,…,d−22.Then,C=C1∪C2 is a CRSβ.(2)
Ifd is odd, let(9)C1≔x+ym|x=0,1,…,Nβd−1andy=0,1,…,d−12,C2≔x+12+y+12m|x=0,1,…,Nβd−1andy=0,1,…,d−32.Then,C=C1∪C2 is a CRSβ.We observe that the complete residue system for the casem≡1mod4 in [10] is complicated and inapplicable for our work. In this paper, we first establish a complete residue system for the case m≡1mod4 which is similar to that in (7). Then, we determine the so-called base βC′-representation in OK and generalize the result in (iv) for the ring of integers OK of any imaginary quadratic field K by using such representation.


## 2. Complete Residue System for the Casem≡1mod4

In this section, we first establish a complete residue system for any quadratic fieldK=ℚm, with m≡1mod4 as in the following theorem.Theorem 1.
LetK=ℚm be a quadratic field, with a squarefree integer m unequal to 1 such that m≡1mod4. If β=a+bσm∈OK∖0, with d=gcda,b, then the set(10)C=x+yσm|x=0,1,…,Nβd−1andy=0,1,…,d−1is a CRSβ.Proof.
Sinced=gcda,b, there exist r,s∈ℤ such that ra+sb=d. We first note that(11)s−r+rσmβ=s−r+rσma+bσm=as−r+ra+sb−rbσm+rbσm2=as−r+ra+sb−rbσm+rbσm+m−14=as−r+rbm−14+ra+sbσm=as−r+rbm−14+dσm.
Letα=u+vσm∈OK. Then, by the division algorithm, there exist v′,v″∈ℤ such that v=dv′+v″, where 0≤v″≤d−1. Then,(12)α=u+dv′σm+v″σm.
It follows from (11) and (12) that(13)α−v′s−r+rσmβ=u−v′as−r+rbm−14+v″σm.
Again, by the division algorithm, there existt′,t″∈ℤ such that(14)u−v′as−r+rbm−14=Nβdt′+t″,where 0≤t″≤Nβ/d−1. It follows from (13) and (14) that(15)α=v′s−r+rσmβ+Nβdt′+t″+v″σm=v′s−r+rσmβ+ββ¯dt′+t″+v″σm=βv′s−r+rσm+β¯dt′+t″+v″σm,where β¯=a+b1−m/2=a+b−bσm∈OK. We consequently get v′s−r+rσm+β¯/dt′∈OK, and so, α≡t″+v″σmmodβ, where t″+v″σm∈C.
Next, we show thatα1≡α2modβ for all α1,α2∈C, with α1≠α2. To see this, let α1=s1+s2σm,α2=t1+t2σm∈C be such that α1≡α2modβ. We now show that α1=α2. Since d|β and β|α1−α2 in OK, we have d|α1−α2, and so, d|s2−t2. It follows that s2=t2 because 0≤s2,t2≤d−1. Consequently, α1−α2=s1−t1, and so, β|s1−t1.
Ifa=0 and b≠0, we obtain that β=bσm, d=b, and Nβ/d=b1−m/4. Thus, there exist k,l∈ℤ such that(16)s1−t1=bσmk+lσm=bkσm+blσm2=bkσm+blσm+m−14=blm−14+bk+blσm.
Hence,s1−t1=blm−1/4, and so, bm−1/4|s1−t1. It follows that s1=t1 because 0≤s1,t1<b1−m/4=bm−1/4. Similarly, if a≠0 and b=0, then β=a, d=a, and Nβ/d=a. Since a|s1−t1, we have s1=t1 because 0≤s1,t1≤a−1.
For the casea≠0 and b≠0, since d=gcda,b, there exist a0,b0∈ℤ such that a=da0, b=db0, and gcda0,b0=1. From β|s1−t1 in OK, there exist c0,d0∈ℤ such that(17)s1−t1=a+bσmc0+d0σm=da0+b0σmc0+d0σm=da0c0+b0d0m−14+a0d0+b0c0+b0d0σm.
It follows thatda0d0+b0c0+b0d0=0, that is,(18)a0d0=−b0c0+d0.
Sincegcda0,b0=1, we have a0|c0+d0 and b0|d0. Thus, there exist a0′,b0′∈ℤ such that(19)c0+d0=a0a0′andd0=b0b0′.
It follows from (18) and (19) that a0b0b0′=−b0a0a0′, and so, b0′=−a0′. Now, we have that(20)s1−t1=a+bσmc0+d0σm=a+bσmc0+d0−d01−m2=a+bσma0a0′−b0b0′1−m2=a+bσma0a0′−b0−a0′1−m2=a+bσmda0+db01−m2a0′d=a+bσma+b1−m2a0′d=ββ¯a0′d=Nβd⋅a0′,so Nβ/d|s1−t1. Since 0≤s1,t1≤Nβ/d−1, it follows that s1=t1.
From all cases, we deduce thats1=t1, and so, α1=α2, as desired.Remark 1.
Keeping the notation of Theorem1 with m<0, we have that(21)C′≔x+yσm|x=0,1,…,maxa,b−1andy=0,1,…,d−1⊆C.Proof.
To show thatC′⊆C, we must show that(22)maxa,b≤a2+ab+b21−m/4d,=Nβd.
It is easy to see that inequality (22) holds if a=0 or b=0. We now assume that a≠0 and b≠0. Note that d≤a, d≤b, and a2−|ab|+b2≤a2+ab+b2. Since dmaxa,b≤|ab| and a−b2≥0, we have(23)dmaxa,b≤|ab|+a−b2=a2−|ab|+b2≤a2+ab+b2≤a2+ab+b21−m4,and (22) is proved.
For a quadratic fieldK=ℚm, with m≡1mod4 and β=a+bσm∈OK\0, we recall that(24)C=x+yσm|x=0,1,…,Nβd−1andy=0,1,…,d−1is a CRSβ, where d=gcda,b. It is clear that maxa,b≤a2−mb2/d whenever m<0. It follows that(25)C′≔x+yσm|x=0,1,…,maxa,b−1andy=0,1,…,d−1⊆C.
From both cases, we observe that ifd=1, then C′=0,1,…,maxa,b−1, while if b=0, then C′=x+yσm|x,y=0,1,…,a−1=C.
We end this section by determining the so-called baseβC-representation.Definition 1.
LetK=ℚm be an imaginary quadratic field. Let β∈OK\0 and C be a CRSβ. We say that η∈OK\0 has a base βC-representation if(26)η=αnβn+αn−1βn−1+⋯+α1β+α0,for some n≥1, αn∈OK\0, and αi∈Ci=0,1,…,n−1. If αi∈C′i=0,1,…,n−1, then (26) is called a base βC′-representation of η.Example 1.
LetK=ℚ−2, β=3+2−2, and η=−19+15−2. Then, d=1 and Nβ=17, and so,(27)C=0,1,…,16is a CRSβ. Since η∈OK, there exists a unique α0∈C such that η≡α0modβ. We see that α0=1 and(28)η=5−2β+1.
Since5−2∈OK, there exists a unique α1∈C such that 5−2≡α1modβ. We see that α1=1 and(29)5−2=1+−2β+1.
It follows from (28) and (29) that(30)η=1+−2β2+β+1.
Since1+−2∈OK, there exists a unique α2∈C such that 1+−2≡α2modβ. We see that α2=8 and(31)1+−2=−1+−2β+8.
It follows from (30) and (31) that(32)η=−1+−2β3+8β2+β+1.
Since−1+−2∈OK, there exists a unique α3∈C such that −1+−2≡α3modβ. We see that α3=6 and(33)−1+−2=−1+−2β+6.
It follows from (32) and (33) that(34)η=−1+−2β4+6β3+8β2+β+1.
Continuing the process, we conclude thatη can be written as a base βC-representation in infinite ways as (28), (30), and (32), and(35)η=−1+−2βn+6βn−1+⋯+6β3+8β2+β+1,for all n≥4.
SinceC′=0,1,2, we note that (28) and (30) are two base βC′-representations of η.Example 2.
LetK=ℚ−5, β=7, and η=4427+12504−5. Then, d=7 and Nβ=49, and so,(36)C=x+y−5|x,y=0,1,2,3,4,5,6=C′is a CRSβ. By using the process as in the previous example, we obtain that(37)η=632+1786−5β+3+2−5,(38)η=90+255−5β2+2+−5β+3+2−5,(39)η=12+36−5β3+6+3−5β2+2+−5β+3+2−5,(40)η=1+5−5β4+5+−5β3+6+3−5β2+2+−5β+3+2−5.
Since1+5−5∈C, we get that the process stops, and (37)–(40) are four base βC-representations (base βC′-representations) of η.Example 3.
LetK=ℚ−3, β=10−5σ−3, and η=7184−4667σ−3. Then, d=5 and Nβ=75, and so,(41)C=x+yσ−3|x=0,1,…,14andy=0,1,2,3,4is a CRSβ. By using the process as in the previous examples, we conclude that η can be written as a base βC-representation in infinite ways as(42)η=790−144σ−3β+4+3σ−3,(43)η=62+33σ−3β2+5+σ−3β+4+3σ−3,(44)η=2+8σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3,(45)η=−β4+12+3σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3,η=−σ−3β5+4+5σ−3β4+12+3σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3,η=−σ−3βn+5+4σ−3βn−1+⋯+5+4σ−3β5+4+5σ−3β4+12+3σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3,n≥6.
SinceC′=x+yσ−3|x=0,1,…,9andy=0,1,2,3,4, we note that (42)–(44) are three base βC′-representations of η.


## 3. Irreducibility Criteria for Polynomials over Imaginary Quadratic Fields

We first recall the result in (iv) as in the following theorem [7].Theorem 2.
Letβ∈2±2i,1±3i,3±i or β=a+bi∈ℤi be such that β≥2+2 and a≥1. For a Gaussian prime π, if(46)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβis its base βC′-representation with n≥1 and Reαn≥1 satisfying the condition Reαn−1Imαn≥ReαnImαn−1, then fx is irreducible in ℤix.In this section, we establish a generalization of Theorem2 for the ring of integers OK of any imaginary quadratic field. To prove this, we next recall the two essential lemmas from [7, 8] as follows.Lemma 1.
LetK=ℚm be an imaginary quadratic field. If β∈OK\0, then β≥1. We note for an imaginary quadratic field K that α=1 for all α∈UOK, the group of units in OK.Lemma 2.
Letfx=αnxn+αn−1xn−1+⋯+α1x+α0∈ℂx be such that n≥2 and αi≤M0≤i≤n−2 for some positive real number M. If fx satisfies(i)
Reαn≥1,Reαn−1≥0,and Imαn−1≥0 and(ii)
Reαn−1Imαn≥ReαnImαn−1,then any complex zeroα of fx satisfies either Reα<0 or α<1+1+4M/2.


### 3.1. Irreducibility Criterion form≡1mod4

To obtain an irreducibility criterion for the casem≡1,mod4, we begin with the following lemmas.Lemma 3.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bm∈OK be such that a≥1+1−m, and let(47)M≔maxa,b−12−md−12,where d=gcda,b. Then, 1−mβ−1≥M.Proof.
Ifb=0, then β=a and d=a. It follows that(48)1−mβ−1=1−ma−1=a−12−ma−12=M.
Now, we assume thatb≠0, so there are two possible cases.Case 1.
a≥b. We have a≥1+1−m>2 and b≥1. Since d=gcda,b, we obtain 1−m−ma−12−1−m−md−12+21−m2−ma−1b+1−m−m2b2>0, and so,(49)1−m2a−12+21−m2−ma−1b+1−m−m2b2>1−ma−12+1−m−md−12.
It follows from (49) that(50)1−m−m+1−ma−1+1−m−mb−12=1−ma−1+1−m−mb2=1−m2a−12+21−m2−ma−1b+1−m−m2b2>1−ma−12+1−m−md−12.
Thus,(51)1−m−m+1−ma−1+1−m−mb−1>1−ma−12−md−12,which implies that(52)△≔21−m−m+21−ma−1+21−m−mb−1−21−ma−12−md−12>0.
Letδ≔1−m−mb−12−1−m−m−ma−12+md−12.
We will show thatδ≥0. If b=1, then d=1. Since a≥1+1−m, we obtain that a−12≥1−m, and so, δ=−ma−12−1−m−m=−ma−12−1−m≥0. If b>1, then 1−m−mb−12−1−m−m≥0, and so, δ≥0. Thus, δ+△>0, which implies that 1−ma2−mb2>a−12−md−12+21−ma−12−md−12+1−m=a−12−md−12+1−m2.
Hence,(53)1−mβ−1=1−ma2−mb2−1>a−12−md−12=M.Case 2.
a<b. Since d=gcda,b and a≥1+1−m, we obtain that |b|>3 and 1−m−mb−12−1−m−md−12+21−m2−mab−1+1−m−m2a2>0. Hence,(54)1−m2b−12+21−m2−ma|b|−1+1−m−m2a2>1−mb−12+1−m−md−12.
It follows from (54) that(55)1−m−m+1−mb−1+1−m−ma−12=1−mb−1+1−m−ma2=1−m2b−12+21−m2−mab−1+1−m−m2a2>1−mb−12+1−m−md−12.
Thus,(56)1−m−m+1−mb−1+1−m−ma−1>1−mb−12−md−12,which implies that(57)△′≔21−m−m+21−mb−1+21−m−ma−1−21−mb−12−md−12>0.
Letδ′≔1−m−ma−12−1−m−m−mb−12+md−12.
We will show thatδ′>0. Since d=gcda,b, it follows that 1−m−ma−12+md−12=−m1−ma−12−d−12≥0. Since b>a≥1+1−m, we obtain that b−12>1−m, and so, −1−m−m−mb−12=−mb−12−1−m>0. Thus, δ′>0, and so, △′+δ′>0, which implies that(58)1−ma2−mb2=1−mb2+−1−mb2+a2≥1−mb2+−1−ma2+a2=1−mb2−ma2>b−12−md−12+21−mb−12−md−12+1−m=b−12−md−12+1−m2.
Hence,(59)1−mβ−1=1−ma2−mb2−1>b−12−md−12=M,as desired.Lemma 4.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4, and let π be an irreducible element in OK. Let β=a+bm∈OK\0 be such that a≥1, β≥1+1−m, π≥β, and(60)π=αnβn+αn−1βn−1+⋯+α1β+α0is a base βC′-representation, with Reαn≥1 satisfying condition (ii) of Lemma 2. If there exists δ∈OK such that δ|αi for all i=0,1,…,n, then δ∈UOK.Proof.
Ifα0=0, then π=βαnβn−1+⋯+α1. Hence, αnβn−1+⋯+α1∈UOK because β∉UOKβ≠1. Since δ|αi for all i=0,1,…,n, we have δ|αnβn−1+⋯+α1, and so, δ∈UOK. Now, assume that α0≠0. We first show that δ<π. We treat two separate cases:(i)
Case1: b≠0. Since δ|α0 and α0≠0, it follows from Lemma 1 that δ≤α0. Note that d=gcda,b and a≥1. Since α0∈C′, we have that α0=x+ym, where 0≤x≤maxa,b−1 and 0≤y≤d−1. There are two further subcases:(1)
Subcase 1:a≥b. Then, 0≤x≤a−1, and(61)δ≤α0=x2−my2≤a−12−md−12≤a−12−mb−12=a2−mb2−2a−1+m2b−1<a2−mb2=β≤π.(2)
Subcase 2:a<b. Then, 0≤x≤b−1, and(62)δ≤α0=x2−my2≤b−12−md−12≤b−12−ma−12=b2−ma2−2b−1+m2a−1<b2−ma2≤a2−mb2=β≤π.(ii)
Case2: b=0. Then, β=a and d=a. Since a=β≥1+1−m, we have a≥3. We consider two further subcases:(1)
Subcase 1:n=1. Then, π=α1a+α0. Let α1=x1+y1m∈OK\0 and α0=x0+y0m∈C′, that is, 0≤x0,y0≤a−1. Now, we obtain that(63)π=x1+y1ma+x0+y0m=ax1+x0+ay1+y0m.By the assumption, we have thatx1≥1andx0y1≥x1y0. If x0=0, then y0=0, which is impossible because α0≠0. Thus, x0>0, and so, y1≥0. Hence,(64)π=ax1+x02−may1+y02>ax12−may12=aα1>α1≥δ.(2)
Subcase 2:n≥2. Then, π=αnan+⋯+α1a+α0, where αn∈OK\0 and αi∈C′ for all i=0,1,…,n−1. Let αi=xi+yim for all i=0,1,…,n. Thus,(65)π=xnan+⋯+x1a+x0+ynan+⋯+y1a+y0m.
Note thatxn≥1 and 0≤xi,yi≤a−1 for all i=0,1,…,n−1.
Ifyn≥0, then(66)π=xnan+⋯+x1a+x02−mynan+⋯+y1a+y02>xn2−myn2=αn≥δ.
Ifyn<0, then by the property of a base a-representation of any positive integer, we must have 0≤y0+y1a+⋯+yn−1an−1<an≤−ynan. Hence, y0+y1a+⋯+ynan<0 so that(67)ynan+⋯+y1a+y02≥1.
Sincea=β≥1+1−m, we have a−12≥1−m, and so,(68)a4−m>a4>a−14≥1−ma−12.
It follows from (67) and (68), xn≥1, and n≥2 that(69)π=xnan+⋯+x1a+x02−mynan+⋯+y1a+y02≥a4−m>1−ma−12=a−12−ma−12≥x02−my02=α0≥δ.
From all cases, we now have thatδ<π, as required.
Asδ|αi for all i=0,1,…,n, we have δ|π, and so, π=δγ for some γ∈OK. Since π is an irreducible element in OK, either δ or γ is a unit. If δ is not a unit, then γ=1, and hence, π=|δγ|=δ<π, which is a contradiction. This proves the lemma.
By applying Lemmas2–4, we obtain an irreducibility criterion for the case m≡1mod4 as in the following theorem.Theorem 3.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bm∈OK be such that β≥2+1−m and a≥1+1−m. For an irreducible element π in OK, with π≥β, if(70)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβis a base βC′-representation with Reαn≥1 satisfying condition (ii) of Lemma 2, then fx is irreducible in OKx.Proof.
Ifdegfx=1, then fx=α1x+α0 is irreducible in OKx by using Lemma 4. We now suppose that degfx≥2 and fx is reducible in OKx. Again, Lemma 4 implies that fx=gxhx for some positive-degree polynomials gx and hx in OKx so that π=fβ=gβhβ. Since π is an irreducible element, either gβ or hβ is a unit, and so, either gβ=1 or hβ=1. Without loss of generality, we may assume that gβ=1. Since αi∈C′ for all i∈0,1,…,n−1, we have that αi≤M for all i∈0,1,…,n−1, where M is defined as in Lemma 3. Since β≥2+1−m, we have that 4β−1β−2+1−m≥0, and so, 2β−32≥1+41−mβ−1. It follows from Lemma 3 that(71)β≥3+1+41−mβ−12≥3+1+4M2,so(72)β−1+1+4M2≥1.
Sincedeggx≥1, gx can be expressed in the form(73)gx=ε∏ix−γi,where ε∈OK is the leading coefficient of gx and the product is over the set of complex zeros of gx. It follows from Lemma 2 that any zero γ of gx satisfies either Reγ<0 or(74)γ<1+1+4M2.
IfReγ<0, then β−γ≥Reβ−γ=Reβ−Reγ=a−Reγ>1. In the latter case, we obtain by (72) and (74) that(75)β−γ≥β−γ>β−1+1+4M2≥1.
From both cases, we deduce by Lemma1 that(76)1=gβ=ε∏iβ−γi≥∏iβ−γi>1,which is a contradiction. This completes the proof.
We note that the conditiona≥1+1−m in Theorem 3 can be replaced by a≥1 for the case of Gaussian integers. Moreover, it is known that an irreducible element is a prime element (Gaussian prime) in ℤi. Therefore, Theorem 3 is a generalization of Theorem 2.
By applying Theorem3, we can find an irreducible polynomial in OKx as in the following example.Example 4.
LetK=ℚ−5 and β=7∈OK. Then, C′=x+y−5|x,y=0,1,2,3,4,5,6. Note that β=7≥2+6 and a=7>1+6. Let π=4427+12504−5∈OK. Clearly, π≥β. Since Nπ=44272+5125042=801348409 which is a rational prime, we deduce that π is an irreducible element. By Example 2, we have that π=1+5−5β4+5+−5β3+6+3−5β2+2+−5β+3+2−5 is its base βC′-representation and Reαn=1≥1 satisfying condition (ii) of Lemma 2, i.e., Reαn−1Imαn=555≥15=ReαnImαn−1. By using Theorem 3, we get that(77)fx=1+5−5x4+5+−5x3+6+3−5x2+2+−5x+3+2−5is irreducible in OKx.


### 3.2. Irreducibility Criterion form≡1mod4

To establish an irreducibility criterion for the casem≡1mod4, we start with the following two lemmas.Lemma 5.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bσm∈Ok be such that a≥1, and let(78)M≔maxa,|b|−12+maxa,|b|−1d−1+d−121−m4,where d=gcda,b. Then, 9−m/4β−1≥M.Proof.
We first prove the following:(79)Claim.9−m4a2−ab+b21−m4≥maxa,b−12+maxa,|b|−1d−1+d−121−m4+9−m42.Proof.
of the claim: Ifb=0, then we are done. Now, assume that b≥1, and we treat two separate cases:(i)
Case1: a≥b. Since m2−6m−11b−12/16−1−md−12/4≥0 and 1−ma−1b−1/4−a−1d−1≥0, we have(80)△≔m2−6m−1116b−12−d−121−m4+1−m4a−1b−1−a−1d−1+m2−6m−2716≥0.Now, we let(81)δ≔5−m4a−b2+9−m4a−1+m2−8m−98b−1−29−m4a−12+a−1d−1+d−121−m4,and we show thatδ≥0. Since m≡1mod4 and m<0, we consider two separate cases, that is, m=−3 and m≤−7.Ifm=−3, then(82)5−m4a−b2+9−m4a−1+m2−8m−98b−12=2a−b2+3a−1+3b−12=4a−b4+9a−12+9b−12+12a−b2a−1+18a−1b−1+12a−b2b−1=4a−b4−3a−12−2a−1b−1+b−12+12a−12+12b−12+12a−1b−1+12a−b2a−1+12a−b2b−1=4a−b4−3a−b2+12a−12+12b−12+12a−1b−1+12a−b2a−1+12a−b2b−1≥12a−12+12a−1b−1+12b−12≥12a−12+12a−1d−1+12d−12=9−ma−12+9−ma−1d−1+9−md−121−m4=9−ma−12+a−1d−1+d−121−m4.Form≤−7, we see that 9−m/42≥9−m, 29−m/4m2−8m−9/8>9−m, and m2−8m−9/82>9−m1−m/4. Thus,(83)5−m4a−b2+9−m4a−1+m2−8m−98b−12=5−m42a−b4+9−m42a−12+m2−8m−982b−12+25−m49−m4a−b2a−1+29−m4m2−8m−98a−1b−1+25−m4m2−8m−98a−b2b−1≥9−ma−12+9−ma−1b−1+9−mb−121−m4≥9−ma−12+9−ma−1d−1+9−md−121−m4=9−ma−12+a−1d−1+d−121−m4.It follows from both cases that(84)5−m4a−b2+9−m4a−1+m2−8m−98b−1≥29−m4a−12+a−1d−1+d−121−m4,which implies that(85)δ=5−m4a−b2+9−m4a−1+m2−8m−98b−1−29−m4a−12+a−1d−1+d−121−m4≥0.Hence,△+δ≥0, which implies that(86)a−12+m2−6m−1116b−12+1−m4a−1b−1+5−m4a−b2+9−m4a−1+m2−8m−98b−1+m2−6m−2716+9−m4≥a−12+a−1d−1+d−121−m4+29−m4a−12+a−1d−1+d−121−m4+9−m4.Thus,(87)9−m4a2−ab+b21−m4≥a−12+a−1d−1+d−121−m4+9−m42.(ii)
Case2: a<b. Since m2−6m−11a−12/16−1−md−12/4≥0 and 1−ma−1b−1/4−b−1d−1≥0, it follows that(88)△′≔m2−6m−1116a−12−d−121−m4+1−m4a−1b−1−b−1d−1+m2−6m−2716≥0.
We now let(89)δ′≔5−m4b−a2+9−m4b−1+m2−8m−98a−1−29−m4b−12+b−1d−1+d−121−m4,and we show that δ′>0. If m=−3, then(90)5−m4b−a2+9−m4b−1+m2−8m−98a−12=2b−a2+3b−1+3a−12=2b−a2+3a−1+3b−12=4b−a4+9a−12+9b−12+12b−a2a−1+18a−1b−1+12b−a2b−1=4b−a4−3a−12−2a−1b−1+b−12+12a−12+12b−12+12a−1b−1+12b−a2a−1+12b−a2b−1=4b−a4−3b−a2+12a−12+12b−12+12a−1b−1+12b−a2a−1+12b−a2b−1>12b−12+12b−1a−1+12a−12≥12b−12+12|b|−1d−1+12d−12=9−mb−12+9−mb−1d−1+9−md−121−m4=9−mb−12+b−1d−1+d−121−m4.
Form≤−7, we see that 9−m/42≥9−m, 29−m/4m2−8m−9/8>9−m, and m2−8m−9/82>9−m1−m/4. Thus,(91)5−m4b−a2+9−m4b−1+m2−8m−98a−12=5−m42b−a4+9−m42b−12+m2−8m−982a−12+25−m49−m4b−a2b−1+29−m4m2−8m−98a−1b−1+25−m4m2−8m−98b−a2a−1>9−mb−12+9−ma−1b−1+9−ma−121−m4≥9−mb−12+9−mb−1d−1+9−md−121−m4=9−mb−12+b−1d−1+d−121−m4.
It follows from both cases that(92)5−m4b−a2+9−m4b−1+m2−8m−98a−1>29−m4b−12+b−1d−1+d−121−m4,so(93)δ′=5−m4b−a2+9−m4b−1+m2−8m−98a−1−29−m4b−12+b−1d−1+d−121−m4>0.
Hence,△′+δ′>0, which implies that(94)b−12+m2−6m−1116a−12+1−m4a−1b−1+5−m4b−a2+9−m4b−1+m2−8m−98a−1+m2−6m−2716+9−m4>b−12+b−1d−1+d−121−m4+29−m4b−12+b−1d−1+d−121−m4+9−m4,and so,(95)9−m4b2−ab+a21−m4>b−12+b−1d−1+d−121−m4+9−m42.
Because ofb2−a2<b2−a21−m/4 and (95), we have(96)9−m4a2−ab+b21−m4>b−12+b−1d−1+d−121−m4+9−m42,which proves the Claim.
Sincea2+ab+b21−m/4≥a2−ab+b21−m/4, it follows from the Claim that(97)9−m4β−1=9−m4a2+ab+b21−m4−1≥9−m4a2−ab+b21−m4−1≥maxa,b−12+maxa,b−1d−1+d−121−m4=M,as desired.Lemma 6.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4, and let π be an irreducible element in OK. Let β=a+bσm∈OK\0 be such that a≥1, β≥2+9−m/4, |π|>9−m/4β−1, and(98)π=αnβn+αn−1βn−1+⋯+α1β+α0is a base βC′-representation, with Reαn≥1 satisfying condition (ii) of Lemma 2. If there exists δ∈OK such that δ|αi for all i=0,1,…,n, then δ∈UOK.Proof.
Ifα0=0, then π=βαnβn−1+⋯+α1. Since π is an irreducible element, we have αnβn−1+⋯+α1∈UOK because β∉UOK. Since δ|αi for all i=0,1,…,n, it follows that δ|αnβn−1+⋯+α1, and so, δ∈UOK. Now, assume that α0≠0. Let M be as in Lemma 5. Since αi∈C′ for all i∈0,1,…,n−1, we obtain that αi≤M for all i∈0,1,…,n−1. Since δ|α0 and α0≠0, we get(99)δ≤α0≤M≤9−m4β−1<π,by using Lemmas 1 and 5. Since δ|π, it follows that π=δγ for some γ∈OK. As π=δγ is an irreducible element in OK, either δ∈UOK or γ∈UOK. If δ∉UOK, then γ∈UOK, and so, γ=1. Hence,(100)π=|δγ|=δ<π,which is a contradiction. Thus, δ∈UOK, as desired.
By applying Lemmas5 and 6, we obtain an irreducibility criterion for the case m≡1mod4 as in the following theorem.Theorem 4.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. 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 in OK such that π>9−m/4β−1, if(101)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβis a base βC′-representation with Reαn≥1 satisfying condition (ii) of Lemma 2, then fx is irreducible in OKx.Proof.
Ifdegfx=1, then fx is irreducible in OKx by using Lemma 6. Now, we suppose that degfx≥2 and fx is reducible in OKx. Again, Lemma 6 implies that fx=gxhx for some positive-degree polynomials gx and hx in OKx, and so, π=fβ=gβhβ. Since π is an irreducible element, either gβ or hβ is a unit so that either gβ=1 or hβ=1. Without loss of generality, we may assume that gβ=1. Since αi∈C′ for all i∈0,1,…,n−1, we have that αi≤M for all i∈0,1,…,n−1, where M is defined as in Lemma 5. Since β≥2+9−m/4, we obtain that 4β−1β−2+9−m/4≥0, and so, 2β−32≥1+49−m/4β−1. It follows from Lemma 5 that(102)β≥3+1+49−m/4β−12≥3+1+4M2.
This shows that(103)β−1+1+4M2≥1.
Sincedeggx≥1, gx can be expressed in the form(104)gx=ε∏ix−γi,where ε∈OK is the leading coefficient of gx and the product is over the set of complex zeros of gx. It follows from Lemma 2 that any zero γ of gx satisfies either Reγ<0 or(105)γ<1+1+4M2.
IfReγ<0, then β−γ≥Reβ−γ=Reβ−Reγ=a+b/2−Reγ>1. In another case, we get by (103) and (105) that(106)β−γ≥β−γ>β−1+1+4M2≥1.
From both cases, we deduce by Lemma1 that(107)1=gβ=ε∏iβ−γi≥∏iβ−γi>1,which is a contradiction. This completes the proof.
By applying Theorem4, we can find an irreducible polynomial in OKx as in the following example.Example 5.
LetK=ℚ−3 and β=10−5σ−3∈OK. Then, C′=x+yσ−3|x=0,1,…,9andy=0,1,2,3,4. Note that β=53≥2+3, a+b/2=10−5/2≥1, and a=10≥1. Let π=7184−4667σ−3∈OK. We see that π=39863017≈6313.72>353−1=3β−1. Since Nπ=71842−7184⋅4667+46672=39863017 which is a rational prime, we deduce that π is an irreducible element. By Example 3, we have that π=2+8σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3 is its base βC′-representation and Reαn=6≥1 satisfying condition (ii) of Lemma 2, i.e., Reαn−1Imαn=3.543≥61.53=ReαnImαn−1, and it follows from Theorem 4 that(108)fx=2+8σ−3x3+2+3σ−3x2+5+σ−3x+4+3σ−3is irreducible in OKx and so is irreducible over K because OK is a unique factorization domain.


## 3.1. Irreducibility Criterion form≡1mod4

To obtain an irreducibility criterion for the casem≡1,mod4, we begin with the following lemmas.Lemma 3.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bm∈OK be such that a≥1+1−m, and let(47)M≔maxa,b−12−md−12,where d=gcda,b. Then, 1−mβ−1≥M.Proof.
Ifb=0, then β=a and d=a. It follows that(48)1−mβ−1=1−ma−1=a−12−ma−12=M.
Now, we assume thatb≠0, so there are two possible cases.Case 1.
a≥b. We have a≥1+1−m>2 and b≥1. Since d=gcda,b, we obtain 1−m−ma−12−1−m−md−12+21−m2−ma−1b+1−m−m2b2>0, and so,(49)1−m2a−12+21−m2−ma−1b+1−m−m2b2>1−ma−12+1−m−md−12.
It follows from (49) that(50)1−m−m+1−ma−1+1−m−mb−12=1−ma−1+1−m−mb2=1−m2a−12+21−m2−ma−1b+1−m−m2b2>1−ma−12+1−m−md−12.
Thus,(51)1−m−m+1−ma−1+1−m−mb−1>1−ma−12−md−12,which implies that(52)△≔21−m−m+21−ma−1+21−m−mb−1−21−ma−12−md−12>0.
Letδ≔1−m−mb−12−1−m−m−ma−12+md−12.
We will show thatδ≥0. If b=1, then d=1. Since a≥1+1−m, we obtain that a−12≥1−m, and so, δ=−ma−12−1−m−m=−ma−12−1−m≥0. If b>1, then 1−m−mb−12−1−m−m≥0, and so, δ≥0. Thus, δ+△>0, which implies that 1−ma2−mb2>a−12−md−12+21−ma−12−md−12+1−m=a−12−md−12+1−m2.
Hence,(53)1−mβ−1=1−ma2−mb2−1>a−12−md−12=M.Case 2.
a<b. Since d=gcda,b and a≥1+1−m, we obtain that |b|>3 and 1−m−mb−12−1−m−md−12+21−m2−mab−1+1−m−m2a2>0. Hence,(54)1−m2b−12+21−m2−ma|b|−1+1−m−m2a2>1−mb−12+1−m−md−12.
It follows from (54) that(55)1−m−m+1−mb−1+1−m−ma−12=1−mb−1+1−m−ma2=1−m2b−12+21−m2−mab−1+1−m−m2a2>1−mb−12+1−m−md−12.
Thus,(56)1−m−m+1−mb−1+1−m−ma−1>1−mb−12−md−12,which implies that(57)△′≔21−m−m+21−mb−1+21−m−ma−1−21−mb−12−md−12>0.
Letδ′≔1−m−ma−12−1−m−m−mb−12+md−12.
We will show thatδ′>0. Since d=gcda,b, it follows that 1−m−ma−12+md−12=−m1−ma−12−d−12≥0. Since b>a≥1+1−m, we obtain that b−12>1−m, and so, −1−m−m−mb−12=−mb−12−1−m>0. Thus, δ′>0, and so, △′+δ′>0, which implies that(58)1−ma2−mb2=1−mb2+−1−mb2+a2≥1−mb2+−1−ma2+a2=1−mb2−ma2>b−12−md−12+21−mb−12−md−12+1−m=b−12−md−12+1−m2.
Hence,(59)1−mβ−1=1−ma2−mb2−1>b−12−md−12=M,as desired.Lemma 4.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4, and let π be an irreducible element in OK. Let β=a+bm∈OK\0 be such that a≥1, β≥1+1−m, π≥β, and(60)π=αnβn+αn−1βn−1+⋯+α1β+α0is a base βC′-representation, with Reαn≥1 satisfying condition (ii) of Lemma 2. If there exists δ∈OK such that δ|αi for all i=0,1,…,n, then δ∈UOK.Proof.
Ifα0=0, then π=βαnβn−1+⋯+α1. Hence, αnβn−1+⋯+α1∈UOK because β∉UOKβ≠1. Since δ|αi for all i=0,1,…,n, we have δ|αnβn−1+⋯+α1, and so, δ∈UOK. Now, assume that α0≠0. We first show that δ<π. We treat two separate cases:(i)
Case1: b≠0. Since δ|α0 and α0≠0, it follows from Lemma 1 that δ≤α0. Note that d=gcda,b and a≥1. Since α0∈C′, we have that α0=x+ym, where 0≤x≤maxa,b−1 and 0≤y≤d−1. There are two further subcases:(1)
Subcase 1:a≥b. Then, 0≤x≤a−1, and(61)δ≤α0=x2−my2≤a−12−md−12≤a−12−mb−12=a2−mb2−2a−1+m2b−1<a2−mb2=β≤π.(2)
Subcase 2:a<b. Then, 0≤x≤b−1, and(62)δ≤α0=x2−my2≤b−12−md−12≤b−12−ma−12=b2−ma2−2b−1+m2a−1<b2−ma2≤a2−mb2=β≤π.(ii)
Case2: b=0. Then, β=a and d=a. Since a=β≥1+1−m, we have a≥3. We consider two further subcases:(1)
Subcase 1:n=1. Then, π=α1a+α0. Let α1=x1+y1m∈OK\0 and α0=x0+y0m∈C′, that is, 0≤x0,y0≤a−1. Now, we obtain that(63)π=x1+y1ma+x0+y0m=ax1+x0+ay1+y0m.By the assumption, we have thatx1≥1andx0y1≥x1y0. If x0=0, then y0=0, which is impossible because α0≠0. Thus, x0>0, and so, y1≥0. Hence,(64)π=ax1+x02−may1+y02>ax12−may12=aα1>α1≥δ.(2)
Subcase 2:n≥2. Then, π=αnan+⋯+α1a+α0, where αn∈OK\0 and αi∈C′ for all i=0,1,…,n−1. Let αi=xi+yim for all i=0,1,…,n. Thus,(65)π=xnan+⋯+x1a+x0+ynan+⋯+y1a+y0m.
Note thatxn≥1 and 0≤xi,yi≤a−1 for all i=0,1,…,n−1.
Ifyn≥0, then(66)π=xnan+⋯+x1a+x02−mynan+⋯+y1a+y02>xn2−myn2=αn≥δ.
Ifyn<0, then by the property of a base a-representation of any positive integer, we must have 0≤y0+y1a+⋯+yn−1an−1<an≤−ynan. Hence, y0+y1a+⋯+ynan<0 so that(67)ynan+⋯+y1a+y02≥1.
Sincea=β≥1+1−m, we have a−12≥1−m, and so,(68)a4−m>a4>a−14≥1−ma−12.
It follows from (67) and (68), xn≥1, and n≥2 that(69)π=xnan+⋯+x1a+x02−mynan+⋯+y1a+y02≥a4−m>1−ma−12=a−12−ma−12≥x02−my02=α0≥δ.
From all cases, we now have thatδ<π, as required.
Asδ|αi for all i=0,1,…,n, we have δ|π, and so, π=δγ for some γ∈OK. Since π is an irreducible element in OK, either δ or γ is a unit. If δ is not a unit, then γ=1, and hence, π=|δγ|=δ<π, which is a contradiction. This proves the lemma.
By applying Lemmas2–4, we obtain an irreducibility criterion for the case m≡1mod4 as in the following theorem.Theorem 3.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bm∈OK be such that β≥2+1−m and a≥1+1−m. For an irreducible element π in OK, with π≥β, if(70)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβis a base βC′-representation with Reαn≥1 satisfying condition (ii) of Lemma 2, then fx is irreducible in OKx.Proof.
Ifdegfx=1, then fx=α1x+α0 is irreducible in OKx by using Lemma 4. We now suppose that degfx≥2 and fx is reducible in OKx. Again, Lemma 4 implies that fx=gxhx for some positive-degree polynomials gx and hx in OKx so that π=fβ=gβhβ. Since π is an irreducible element, either gβ or hβ is a unit, and so, either gβ=1 or hβ=1. Without loss of generality, we may assume that gβ=1. Since αi∈C′ for all i∈0,1,…,n−1, we have that αi≤M for all i∈0,1,…,n−1, where M is defined as in Lemma 3. Since β≥2+1−m, we have that 4β−1β−2+1−m≥0, and so, 2β−32≥1+41−mβ−1. It follows from Lemma 3 that(71)β≥3+1+41−mβ−12≥3+1+4M2,so(72)β−1+1+4M2≥1.
Sincedeggx≥1, gx can be expressed in the form(73)gx=ε∏ix−γi,where ε∈OK is the leading coefficient of gx and the product is over the set of complex zeros of gx. It follows from Lemma 2 that any zero γ of gx satisfies either Reγ<0 or(74)γ<1+1+4M2.
IfReγ<0, then β−γ≥Reβ−γ=Reβ−Reγ=a−Reγ>1. In the latter case, we obtain by (72) and (74) that(75)β−γ≥β−γ>β−1+1+4M2≥1.
From both cases, we deduce by Lemma1 that(76)1=gβ=ε∏iβ−γi≥∏iβ−γi>1,which is a contradiction. This completes the proof.
We note that the conditiona≥1+1−m in Theorem 3 can be replaced by a≥1 for the case of Gaussian integers. Moreover, it is known that an irreducible element is a prime element (Gaussian prime) in ℤi. Therefore, Theorem 3 is a generalization of Theorem 2.
By applying Theorem3, we can find an irreducible polynomial in OKx as in the following example.Example 4.
LetK=ℚ−5 and β=7∈OK. Then, C′=x+y−5|x,y=0,1,2,3,4,5,6. Note that β=7≥2+6 and a=7>1+6. Let π=4427+12504−5∈OK. Clearly, π≥β. Since Nπ=44272+5125042=801348409 which is a rational prime, we deduce that π is an irreducible element. By Example 2, we have that π=1+5−5β4+5+−5β3+6+3−5β2+2+−5β+3+2−5 is its base βC′-representation and Reαn=1≥1 satisfying condition (ii) of Lemma 2, i.e., Reαn−1Imαn=555≥15=ReαnImαn−1. By using Theorem 3, we get that(77)fx=1+5−5x4+5+−5x3+6+3−5x2+2+−5x+3+2−5is irreducible in OKx.


## 3.2. Irreducibility Criterion form≡1mod4

To establish an irreducibility criterion for the casem≡1mod4, we start with the following two lemmas.Lemma 5.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. Let β=a+bσm∈Ok be such that a≥1, and let(78)M≔maxa,|b|−12+maxa,|b|−1d−1+d−121−m4,where d=gcda,b. Then, 9−m/4β−1≥M.Proof.
We first prove the following:(79)Claim.9−m4a2−ab+b21−m4≥maxa,b−12+maxa,|b|−1d−1+d−121−m4+9−m42.Proof.
of the claim: Ifb=0, then we are done. Now, assume that b≥1, and we treat two separate cases:(i)
Case1: a≥b. Since m2−6m−11b−12/16−1−md−12/4≥0 and 1−ma−1b−1/4−a−1d−1≥0, we have(80)△≔m2−6m−1116b−12−d−121−m4+1−m4a−1b−1−a−1d−1+m2−6m−2716≥0.Now, we let(81)δ≔5−m4a−b2+9−m4a−1+m2−8m−98b−1−29−m4a−12+a−1d−1+d−121−m4,and we show thatδ≥0. Since m≡1mod4 and m<0, we consider two separate cases, that is, m=−3 and m≤−7.Ifm=−3, then(82)5−m4a−b2+9−m4a−1+m2−8m−98b−12=2a−b2+3a−1+3b−12=4a−b4+9a−12+9b−12+12a−b2a−1+18a−1b−1+12a−b2b−1=4a−b4−3a−12−2a−1b−1+b−12+12a−12+12b−12+12a−1b−1+12a−b2a−1+12a−b2b−1=4a−b4−3a−b2+12a−12+12b−12+12a−1b−1+12a−b2a−1+12a−b2b−1≥12a−12+12a−1b−1+12b−12≥12a−12+12a−1d−1+12d−12=9−ma−12+9−ma−1d−1+9−md−121−m4=9−ma−12+a−1d−1+d−121−m4.Form≤−7, we see that 9−m/42≥9−m, 29−m/4m2−8m−9/8>9−m, and m2−8m−9/82>9−m1−m/4. Thus,(83)5−m4a−b2+9−m4a−1+m2−8m−98b−12=5−m42a−b4+9−m42a−12+m2−8m−982b−12+25−m49−m4a−b2a−1+29−m4m2−8m−98a−1b−1+25−m4m2−8m−98a−b2b−1≥9−ma−12+9−ma−1b−1+9−mb−121−m4≥9−ma−12+9−ma−1d−1+9−md−121−m4=9−ma−12+a−1d−1+d−121−m4.It follows from both cases that(84)5−m4a−b2+9−m4a−1+m2−8m−98b−1≥29−m4a−12+a−1d−1+d−121−m4,which implies that(85)δ=5−m4a−b2+9−m4a−1+m2−8m−98b−1−29−m4a−12+a−1d−1+d−121−m4≥0.Hence,△+δ≥0, which implies that(86)a−12+m2−6m−1116b−12+1−m4a−1b−1+5−m4a−b2+9−m4a−1+m2−8m−98b−1+m2−6m−2716+9−m4≥a−12+a−1d−1+d−121−m4+29−m4a−12+a−1d−1+d−121−m4+9−m4.Thus,(87)9−m4a2−ab+b21−m4≥a−12+a−1d−1+d−121−m4+9−m42.(ii)
Case2: a<b. Since m2−6m−11a−12/16−1−md−12/4≥0 and 1−ma−1b−1/4−b−1d−1≥0, it follows that(88)△′≔m2−6m−1116a−12−d−121−m4+1−m4a−1b−1−b−1d−1+m2−6m−2716≥0.
We now let(89)δ′≔5−m4b−a2+9−m4b−1+m2−8m−98a−1−29−m4b−12+b−1d−1+d−121−m4,and we show that δ′>0. If m=−3, then(90)5−m4b−a2+9−m4b−1+m2−8m−98a−12=2b−a2+3b−1+3a−12=2b−a2+3a−1+3b−12=4b−a4+9a−12+9b−12+12b−a2a−1+18a−1b−1+12b−a2b−1=4b−a4−3a−12−2a−1b−1+b−12+12a−12+12b−12+12a−1b−1+12b−a2a−1+12b−a2b−1=4b−a4−3b−a2+12a−12+12b−12+12a−1b−1+12b−a2a−1+12b−a2b−1>12b−12+12b−1a−1+12a−12≥12b−12+12|b|−1d−1+12d−12=9−mb−12+9−mb−1d−1+9−md−121−m4=9−mb−12+b−1d−1+d−121−m4.
Form≤−7, we see that 9−m/42≥9−m, 29−m/4m2−8m−9/8>9−m, and m2−8m−9/82>9−m1−m/4. Thus,(91)5−m4b−a2+9−m4b−1+m2−8m−98a−12=5−m42b−a4+9−m42b−12+m2−8m−982a−12+25−m49−m4b−a2b−1+29−m4m2−8m−98a−1b−1+25−m4m2−8m−98b−a2a−1>9−mb−12+9−ma−1b−1+9−ma−121−m4≥9−mb−12+9−mb−1d−1+9−md−121−m4=9−mb−12+b−1d−1+d−121−m4.
It follows from both cases that(92)5−m4b−a2+9−m4b−1+m2−8m−98a−1>29−m4b−12+b−1d−1+d−121−m4,so(93)δ′=5−m4b−a2+9−m4b−1+m2−8m−98a−1−29−m4b−12+b−1d−1+d−121−m4>0.
Hence,△′+δ′>0, which implies that(94)b−12+m2−6m−1116a−12+1−m4a−1b−1+5−m4b−a2+9−m4b−1+m2−8m−98a−1+m2−6m−2716+9−m4>b−12+b−1d−1+d−121−m4+29−m4b−12+b−1d−1+d−121−m4+9−m4,and so,(95)9−m4b2−ab+a21−m4>b−12+b−1d−1+d−121−m4+9−m42.
Because ofb2−a2<b2−a21−m/4 and (95), we have(96)9−m4a2−ab+b21−m4>b−12+b−1d−1+d−121−m4+9−m42,which proves the Claim.
Sincea2+ab+b21−m/4≥a2−ab+b21−m/4, it follows from the Claim that(97)9−m4β−1=9−m4a2+ab+b21−m4−1≥9−m4a2−ab+b21−m4−1≥maxa,b−12+maxa,b−1d−1+d−121−m4=M,as desired.Lemma 6.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4, and let π be an irreducible element in OK. Let β=a+bσm∈OK\0 be such that a≥1, β≥2+9−m/4, |π|>9−m/4β−1, and(98)π=αnβn+αn−1βn−1+⋯+α1β+α0is a base βC′-representation, with Reαn≥1 satisfying condition (ii) of Lemma 2. If there exists δ∈OK such that δ|αi for all i=0,1,…,n, then δ∈UOK.Proof.
Ifα0=0, then π=βαnβn−1+⋯+α1. Since π is an irreducible element, we have αnβn−1+⋯+α1∈UOK because β∉UOK. Since δ|αi for all i=0,1,…,n, it follows that δ|αnβn−1+⋯+α1, and so, δ∈UOK. Now, assume that α0≠0. Let M be as in Lemma 5. Since αi∈C′ for all i∈0,1,…,n−1, we obtain that αi≤M for all i∈0,1,…,n−1. Since δ|α0 and α0≠0, we get(99)δ≤α0≤M≤9−m4β−1<π,by using Lemmas 1 and 5. Since δ|π, it follows that π=δγ for some γ∈OK. As π=δγ is an irreducible element in OK, either δ∈UOK or γ∈UOK. If δ∉UOK, then γ∈UOK, and so, γ=1. Hence,(100)π=|δγ|=δ<π,which is a contradiction. Thus, δ∈UOK, as desired.
By applying Lemmas5 and 6, we obtain an irreducibility criterion for the case m≡1mod4 as in the following theorem.Theorem 4.
LetK=ℚm be an imaginary quadratic field, with m≡1mod4. 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 in OK such that π>9−m/4β−1, if(101)π=αnβn+αn−1βn−1+⋯+α1β+α0≕fβis a base βC′-representation with Reαn≥1 satisfying condition (ii) of Lemma 2, then fx is irreducible in OKx.Proof.
Ifdegfx=1, then fx is irreducible in OKx by using Lemma 6. Now, we suppose that degfx≥2 and fx is reducible in OKx. Again, Lemma 6 implies that fx=gxhx for some positive-degree polynomials gx and hx in OKx, and so, π=fβ=gβhβ. Since π is an irreducible element, either gβ or hβ is a unit so that either gβ=1 or hβ=1. Without loss of generality, we may assume that gβ=1. Since αi∈C′ for all i∈0,1,…,n−1, we have that αi≤M for all i∈0,1,…,n−1, where M is defined as in Lemma 5. Since β≥2+9−m/4, we obtain that 4β−1β−2+9−m/4≥0, and so, 2β−32≥1+49−m/4β−1. It follows from Lemma 5 that(102)β≥3+1+49−m/4β−12≥3+1+4M2.
This shows that(103)β−1+1+4M2≥1.
Sincedeggx≥1, gx can be expressed in the form(104)gx=ε∏ix−γi,where ε∈OK is the leading coefficient of gx and the product is over the set of complex zeros of gx. It follows from Lemma 2 that any zero γ of gx satisfies either Reγ<0 or(105)γ<1+1+4M2.
IfReγ<0, then β−γ≥Reβ−γ=Reβ−Reγ=a+b/2−Reγ>1. In another case, we get by (103) and (105) that(106)β−γ≥β−γ>β−1+1+4M2≥1.
From both cases, we deduce by Lemma1 that(107)1=gβ=ε∏iβ−γi≥∏iβ−γi>1,which is a contradiction. This completes the proof.
By applying Theorem4, we can find an irreducible polynomial in OKx as in the following example.Example 5.
LetK=ℚ−3 and β=10−5σ−3∈OK. Then, C′=x+yσ−3|x=0,1,…,9andy=0,1,2,3,4. Note that β=53≥2+3, a+b/2=10−5/2≥1, and a=10≥1. Let π=7184−4667σ−3∈OK. We see that π=39863017≈6313.72>353−1=3β−1. Since Nπ=71842−7184⋅4667+46672=39863017 which is a rational prime, we deduce that π is an irreducible element. By Example 3, we have that π=2+8σ−3β3+2+3σ−3β2+5+σ−3β+4+3σ−3 is its base βC′-representation and Reαn=6≥1 satisfying condition (ii) of Lemma 2, i.e., Reαn−1Imαn=3.543≥61.53=ReαnImαn−1, and it follows from Theorem 4 that(108)fx=2+8σ−3x3+2+3σ−3x2+5+σ−3x+4+3σ−3is irreducible in OKx and so is irreducible over K because OK is a unique factorization domain.


## 4. Conclusion

For any quadratic fieldK≔ℚm, we establish a new complete residue system for the case m≡1mod4. Then, we determine the so-called base βC′-representation and generalize the result in Theorem 2 for the ring of integers of any imaginary quadratic field.



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