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

Phetnun, Phitthayathon; Kanasri, Narakorn Rompurk; Singthongla, Patiwat

doi:https://doi.org/10.1155/2021/5564589

International Journal of Mathematics and Mathematical Sciences

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Abstract Introduction Conclusion Data Availability Conflicts of Interest Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2021 | Article ID 5564589 | https://doi.org/10.1155/2021/5564589

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

Phitthayathon Phetnun,¹Narakorn Rompurk Kanasri,¹and Patiwat Singthongla¹

Academic Editor: Sergejs Solovjovs

Received18 Feb 2021

Accepted25 Mar 2021

Published14 Apr 2021

Abstract

For a Gaussian prime and a nonzero Gaussian integer with and , it was proved that if where , , belong to a complete residue system modulo , and the digits and satisfy certain restrictions, then the polynomial is irreducible in . For any quadratic field , it is well known that there are explicit representations for a complete residue system in , but those of the case 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.

1. Introduction

According to Pólya and Szegö [1], the classical result of A. Cohn states that if a prime is expressed in the decimal representation asthen the polynomial is irreducible in . For example, since 4831 is prime and , it follows that is irreducible in . This result was subsequently generalized to any base 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: ifwhere for all , and is prime, then is irreducible. In 1988, Filaseta improved this fact, that is, if coefficients of the polynomial in (2) satisfy for and is prime, then is irreducible [4].

In another direction, let , with a unique squarefree integer unequal to 1, be a quadratic field and be its ring of integers. We have seen in [5] that is an integral domain. The quadratic field is said to be real if and imaginary if . Clearly, is real if and imaginary if . It is well known thatwhere

We note that the ring of Gaussian integers and the Eisenstein domain are rings of integers of and , respectively.

We emphasize that a prime element in is an irreducible element, and the converse holds if is a unique factorization domain. Moreover, is the quotient field of [5], and the units in are the units in [6]. We say that a nonzero polynomial is irreducible in if is not a unit, and if with in , then or is a unit in . Polynomials that are not irreducible are called reducible. Furthermore, if is a unique factorization domain and is irreducible in , then is irreducible over by using Gauss’s lemma.

In 2017, Singthongla et al. established the result of A. Cohn in [7], where is an imaginary quadratic field such that is a Euclidean domain, namely, , and [5]. The results are as follows:(i)For , each element in has a base -expansion whose digits are bounded by certain constants.(ii)For a prime element in , if is its base -expansion in and the digits and satisfy some natural restrictions, then the polynomial is irreducible in .(iii)For Gaussian integers, a similar base -expansion but with digits belonging to a complete residue system modulo 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 integers of any general quadratic field. However, to prove this, we must use the property that for all 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 of any imaginary quadratic field.

Let be any general quadratic field. For with , we say that divides , denoted by , if and only if there exists such that . For with , we say that is congruent to modulo , and we write if , a principal ideal generated by or, equivalently, if [9].

For , we denote the norm of by

We note from [5] that if is such that , where is a rational prime, then is an irreducible element. By a complete residue system modulo in , abbreviated by [9, 10], we mean a set of elements in such that(i)for each , there is such that and(ii) for all , with

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 and , we have the following:(i)Case : the set is a .(ii)Case :(1)If is even, let Then, is a .(2)If is odd, let

Then, is a .

We observe that the complete residue system for the case in [10] is complicated and inapplicable for our work. In this paper, we first establish a complete residue system for the case which is similar to that in (7). Then, we determine the so-called base -representation in and generalize the result in (iv) for the ring of integers of any imaginary quadratic field by using such representation.

2. Complete Residue System for the Case

In this section, we first establish a complete residue system for any quadratic field , with as in the following theorem.

Theorem 1. Let be a quadratic field, with a squarefree integer unequal to 1 such that . If , with , then the setis a .

Proof. Since , there exist such that . We first note thatLet . Then, by the division algorithm, there exist such that , where . Then,It follows from (11) and (12) thatAgain, by the division algorithm, there exist such thatwhere . It follows from (13) and (14) thatwhere . We consequently get , and so, , where .
Next, we show that for all , with . To see this, let be such that . We now show that . Since and in , we have , and so, . It follows that because . Consequently, , and so, .
If and , we obtain that , , and . Thus, there exist such thatHence, , and so, . It follows that because . Similarly, if and , then , , and . Since , we have because .
For the case and , since , there exist such that , , and . From in , there exist such thatIt follows that , that is,Since , we have and . Thus, there exist such thatIt follows from (18) and (19) that , and so, . Now, we have thatso . Since , it follows that .
From all cases, we deduce that , and so, , as desired.

Remark 1. Keeping the notation of Theorem 1 with , we have that

Proof. To show that , we must show thatIt is easy to see that inequality (22) holds if or . We now assume that and . Note that , , and . Since and , we haveand (22) is proved.
For a quadratic field , with and , we recall thatis a , where . It is clear that whenever . It follows thatFrom both cases, we observe that if , then , while if , then .
We end this section by determining the so-called base -representation.

Definition 1. Let be an imaginary quadratic field. Let and be a . We say that has a base -representation iffor some , , and . If , then (26) is called a base -representation of .

Example 1. Let , , and . Then, and , and so,is a . Since , there exists a unique such that . We see that andSince , there exists a unique such that . We see that andIt follows from (28) and (29) thatSince , there exists a unique such that . We see that andIt follows from (30) and (31) thatSince , there exists a unique such that . We see that andIt follows from (32) and (33) thatContinuing the process, we conclude that can be written as a base -representation in infinite ways as (28), (30), and (32), andfor all .
Since , we note that (28) and (30) are two base -representations of .

Example 2. Let , , and . Then, and , and so,is a . By using the process as in the previous example, we obtain thatSince , we get that the process stops, and (37)–(40) are four base -representations (base -representations) of .

Example 3. Let , , and . Then, and , and so,is a . By using the process as in the previous examples, we conclude that can be written as a base -representation in infinite ways asSince , we note that (42)–(44) are three base -representations of .

3. Irreducibility Criteria for Polynomials over Imaginary Quadratic Fields

We first recall the result in (iv) as in the following theorem [7].

Theorem 2. Let or be such that and . For a Gaussian prime , ifis its base -representation with and satisfying the condition , then is irreducible in .

In this section, we establish a generalization of Theorem 2 for the ring of integers of any imaginary quadratic field. To prove this, we next recall the two essential lemmas from [7, 8] as follows.

Lemma 1. Let be an imaginary quadratic field. If , then . We note for an imaginary quadratic field that for all , the group of units in .

Lemma 2. Let be such that and for some positive real number . If satisfies(i) and(ii), then any complex zero of satisfies either or .

3.1. Irreducibility Criterion for

To obtain an irreducibility criterion for the case , we begin with the following lemmas.

Lemma 3. Let be an imaginary quadratic field, with . Let be such that , and letwhere . Then, .

Proof. If , then and . It follows thatNow, we assume that , so there are two possible cases.

Case 1. . We have and . Since , we obtain , and so,It follows from (49) thatThus,which implies thatLet .
We will show that . If , then . Since , we obtain that , and so, . If , then , and so, . Thus, , which implies that .
Hence,

Case 2. . Since and , we obtain that and . Hence,It follows from (54) thatThus,which implies thatLet .
We will show that . Since , it follows that . Since , we obtain that , and so, . Thus, , and so, , which implies thatHence,as desired.

Lemma 4. Let be an imaginary quadratic field, with , and let be an irreducible element in . Let be such that , , , andis a base -representation, with satisfying condition (ii) of Lemma 2. If there exists such that for all , then .

Proof. If , then . Hence, because . Since for all , we have , and so, . Now, assume that . We first show that . We treat two separate cases:(i)Case 1: . Since and , it follows from Lemma 1 that . Note that and . Since , we have that , where and . There are two further subcases:(1)Subcase 1: . Then, , and(2)Subcase 2: . Then, , and(ii)Case 2: . Then, and . Since , we have . We consider two further subcases:(1)Subcase 1: . Then, . Let and , that is, . Now, we obtain that By the assumption, we have that . If , then , which is impossible because . Thus, , and so, . Hence,(2)Subcase 2: . Then, , where and for all . Let for all . Thus,Note that and for all .
If , thenIf , then by the property of a base -representation of any positive integer, we must have . Hence, so thatSince , we have , and so,It follows from (67) and (68), , and thatFrom all cases, we now have that , as required.
As for all , we have , and so, for some . Since is an irreducible element in , either or is a unit. If is not a unit, then , and hence, , which is a contradiction. This proves the lemma.
By applying Lemmas 2–4, we obtain an irreducibility criterion for the case as in the following theorem.

Theorem 3. Let be an imaginary quadratic field, with . Let be such that and . For an irreducible element in , with , ifis a base -representation with satisfying condition (ii) of Lemma 2, then is irreducible in .

Proof. If , then is irreducible in by using Lemma 4. We now suppose that and is reducible in . Again, Lemma 4 implies that for some positive-degree polynomials and in so that . Since is an irreducible element, either or is a unit, and so, either or . Without loss of generality, we may assume that . Since for all , we have that for all , where is defined as in Lemma 3. Since , we have that , and so, . It follows from Lemma 3 thatsoSince , can be expressed in the formwhere is the leading coefficient of and the product is over the set of complex zeros of . It follows from Lemma 2 that any zero of satisfies either orIf , then . In the latter case, we obtain by (72) and (74) thatFrom both cases, we deduce by Lemma 1 thatwhich is a contradiction. This completes the proof.
We note that the condition in Theorem 3 can be replaced by for the case of Gaussian integers. Moreover, it is known that an irreducible element is a prime element (Gaussian prime) in . Therefore, Theorem 3 is a generalization of Theorem 2.
By applying Theorem 3, we can find an irreducible polynomial in as in the following example.

Example 4. Let and . Then, . Note that and . Let . Clearly, . Since which is a rational prime, we deduce that is an irreducible element. By Example 2, we have that is its base -representation and satisfying condition (ii) of Lemma 2, i.e., . By using Theorem 3, we get thatis irreducible in .

3.2. Irreducibility Criterion for

To establish an irreducibility criterion for the case , we start with the following two lemmas.

Lemma 5. Let be an imaginary quadratic field, with . Let be such that , and letwhere . Then, .

Proof. We first prove the following:

Proof. of the claim: If , then we are done. Now, assume that , and we treat two separate cases:(i)Case 1: . Since and , we have Now, we let and we show that . Since and , we consider two separate cases, that is, and . If , then For , we see that , , and . Thus, It follows from both cases that which implies that Hence, , which implies that Thus,(ii)Case 2: . Since and , it follows thatWe now letand we show that . If , thenFor , we see that , , and . Thus,It follows from both cases thatsoHence, , which implies thatand so,Because of and (95), we havewhich proves the Claim.
Since , it follows from the Claim thatas desired.

Lemma 6. Let be an imaginary quadratic field, with , and let be an irreducible element in . Let be such that , , , andis a base -representation, with satisfying condition (ii) of Lemma 2. If there exists such that for all , then .

Proof. If , then . Since is an irreducible element, we have because . Since for all , it follows that , and so, . Now, assume that . Let be as in Lemma 5. Since for all , we obtain that for all . Since and , we getby using Lemmas 1 and 5. Since , it follows that for some . As is an irreducible element in , either or . If , then , and so, . Hence,which is a contradiction. Thus, , as desired.
By applying Lemmas 5 and 6, we obtain an irreducibility criterion for the case as in the following theorem.

Theorem 4. Let be an imaginary quadratic field, with . Let be such that , , and . For an irreducible element in such that , ifis a base -representation with satisfying condition (ii) of Lemma 2, then is irreducible in .

Proof. If , then is irreducible in by using Lemma 6. Now, we suppose that and is reducible in . Again, Lemma 6 implies that for some positive-degree polynomials and in , and so, . Since is an irreducible element, either or is a unit so that either or . Without loss of generality, we may assume that . Since for all , we have that for all , where is defined as in Lemma 5. Since , we obtain that , and so, . It follows from Lemma 5 thatThis shows thatSince , can be expressed in the formwhere is the leading coefficient of and the product is over the set of complex zeros of . It follows from Lemma 2 that any zero of satisfies either orIf , then . In another case, we get by (103) and (105) thatFrom both cases, we deduce by Lemma 1 thatwhich is a contradiction. This completes the proof.
By applying Theorem 4, we can find an irreducible polynomial in as in the following example.

Example 5. Let and . Then, . Note that , , and . Let . We see that . Since which is a rational prime, we deduce that is an irreducible element. By Example 3, we have that is its base -representation and satisfying condition (ii) of Lemma 2, i.e., , and it follows from Theorem 4 thatis irreducible in and so is irreducible over because is a unique factorization domain.

4. Conclusion

For any quadratic field , we establish a new complete residue system for the case . Then, we determine the so-called base -representation and generalize the result in Theorem 2 for the ring of integers of any imaginary quadratic field.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work was supported by the Science Achievement Scholarship of Thailand (SAST).

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Copyright

Copyright © 2021 Phitthayathon Phetnun et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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