Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) Extension of primary and semiprimary bi-ideals of semirings G. Shanmugam1, P. Lakshmi Pallavi2, K.Arulmozhi3, Aiyared Iampan4,∗ 1Department of Mathematics, Saveetha School of Engineering, Saveetha Institute of Medical and Technical Sciences, Chennai-602105, India. 2B V Raju Institute of Technology, Narsapur Medak Dist, Telangana state-502313, India. 3Department of Mathematics, Bharath Institute of Higher Education and Research, Tamil Nadu, Chennai 600073, India. 4Department of Mathematics, School of Science, University of Phayao, 19 Moo 2, Tambon Mae Ka, Amphur Mueang, Phayao 56000, Thailand. E-mails:1gsm.maths@gmail.com, 2lakshmipallavi.p@bvrit.ac.in, 3arulmozhiems@gmail.com, 4aiyared.ia@up.ac.th, ∗Corresponding author: Aiyared Iampan. Received: 04-11-2024 Revised: 12-12-2024 Accepted: 01-01-2025. Abstract The 1PBID, 2PBID, and 3PBID of semirings are introduced. Additionally, we communicate with a number of the various SPBIs’ attributes. 2PBID and 3PBID are generalizations of 1PBID and 2PBID, respectively, which we explain. We go over the m1, m2 and m3-systems and BI generator. It is possible to generalize the m1-system to the m2-system and the m2-system to the m3-system. Keywords: 1PBID; 2PBID; 3PBID; m1-system, m2-system and m3-system. 1 Introduction Numerous studies have looked at different kinds of ideals in mathematical structures, such as rings and semir- ings,3, 4 respectively. Associative rings made up the notion of ideals that Dedekind introduced into the theory of algebraic numbers. In this approach the notion was expanded to include algebraic numbers. Moreover, it is a particular case of Lajos (m,n)-ideal. In order to analyze regular and intra-regular semigroups, Lajos used generalized BIDs and quasi-ideals. To reference5 while discussing different types of semigroups is a BID. The associative rings are somewhat arbitrary, however they are described in terms of BIDs. It would be almost perfect to expand LIs and RIs, which are specific types of BIDs. Semigroups and rings now referred to as quasi ideals were introduced by Otto Steinfeld. Quote3 states that semirings provide a range of methods for elucidating prime ideals. Commutative ring theory has been extensively influenced by prime ideal theory. It has been less frequently used for non-commutative rings than for commutative rings. Palanikumar et al.6 investigated different prime partial BIs in non-commutative partial rings. Walt9 studied the prime and semiprimate BIs of associative rings with unity. Associative rings without unity were extended to prime and semiprime BIs by Roux.10 Some descriptions of BIs in basic semir- ings were given by Flaska et al.11 Atani12 also provided some results for the ideal theory of commutative semirings with non-zero identities. McCoy gives some details on prime ideals in general rings.4 Informa- tion on the PID for rings and semirings was given in.3, 13, 14 Van der Walt coined the words prime BID and semiprime BID.9 The subsets X1 and X2 of § and the product X1 · X2 may be understood as follows: the subring of § is produced by the set of all products x1 · x2, where x1 ∈ X1, x2 ∈ X2. A BID a1 of a ring § is defined as a subring a1 of § that satisfies a1§a1 ⊆ a1. For ideals a and a1 of §, then a ⊆ ℵ or a1 ⊆ ℵ means that an ID ℵ of a ring § is PID if and only if whenever aa1 ⊆ ℵ.4 Each of the five components that make up this document is arranged differently. In section 2, we address the various kinds of main BIDs and their expansions. Section 3 offers a discussion of the semiprimary BIDs. https://internationalpubls.com 755 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) List of Abbreviations RID right ideal LID left ideal ID ideal primary BID primary bi-ideal primary ID primary ideal TID two sided ideal semi primary BID semi primary bi-ideal semi primary ID semi primary ideal 2 Characterization of PBIDs Definition 2.1. A BID ℵ of § is said to be (1) 1PBID if a1a2 ⊆ ℵ implies a1 ⊆ ℵ or a2 ⊆ √ ℵ for any BIDs a1 and a2 of §. (ii) 2PBID if ∝ §℘ ⊆ ℵ implies ∝∈ ℵ or ℘ ∈ √ ℵ . (iii) 3PBID if £1£2 ⊆ ℵ implies £1 ⊆ ℵ or £2 ⊆ √ ℵ for any IDs £1 and £2 of §. Theorem 2.2. Every 1PBID is a 2PBID. Proof. Let ℵ be an 1PBID of §. Let ∝, ℘ ∈ § and ∝ §℘ ⊆ ℵ. Now, (∝ §) · (§℘) ⊆∝ §℘ ⊆ ℵ, since ∝ § and §℘ are BIDs. Hence ∝ § ⊆ ℵ or §℘ ⊆ √ ℵ. Suppose that ∝ § ⊆ ℵ. Consider <∝>b · <∝>b⊆∝ § ⊆ ℵ. Then ∝∈ ℵ. Similarly if §℘ ⊆ √ ℵ then ℘ ∈ √ ℵ. Thus ℵ is a 2PBID of §. Converse is does not hold. Example 2.3. Consider the semiring § = D2(Z2) and ℵ = {( 0 0 0 0 )} is a 2PBID, but not 1PBID. Theorem 2.4. Every 2PBID is a 3PBID. Proof. Let ℵ be an 2PBID of §. For the IDs £1 and £2 of § such that £1 ·£2 ⊆ ℵ. If £1 6⊆ ℵ, let ∝∈ £1 \ ℵ. For any ℘ ∈ £2, ∝ §℘ ⊆<∝> · < ℘ >⊆ £1 ·£2 ⊆ ℵ. Hence ℘ ∈ √ ℵ. Then £2 ⊆ √ ℵ. Thus ℵ is a 3PBID of §. Definition 2.5. A subset D of § is said to be (i) mp1 -sys if for any ∝, ℘ ∈ D, ∃ ∝1∈<∝>b and ℘1 < ℘ >b such that ∝1 ℘1 ∈ D. (ii) mp2 -sys if for any ∝, ℘ ∈ D, ∃ ∝1∈<∝>r and ℘1 ∈< ℘ >l such that ∝1 ℘1 ∈ D. (iii) mp3 -sys if for any ∝, ℘ ∈ D, ∃ ∝1∈<∝> and ℘1 ∈< ℘ > such that ∝1 ℘1 ∈ D. Theorem 2.6. If ℵ is a BID of §, then ℵ is a 1PBID (2PBID,3PBID) if and only if §\ℵ is anmp1 -sys (mp2 -sys, mp3 -sys) of §. Proof. Let ς, ε ∈ § \ ℵ. Hence ς, ε ∈ § but ς, ε /∈ ℵ. So < ς >b · < ε >b 6⊆ ℵ. There exists ς ′ ∈< ς >b and ε ′ ∈< ε >b such that ς ′ · ε′ /∈ ℵ. Hence ς ′ · ε′ ∈ § \ ℵ. So we have proved that for ς, ε ∈ § \ ℵ ∃ ς ′ ∈< ς >b and ε ′ ∈< ε >b such that ς ′ · ε′ ∈ § \ ℵ. So § \ ℵ is an mp1 -sys. Conversely, let §\ℵ is anmp1 -sys. Let us shows that a1 ⊆ ℵ or a2 ⊆ √ ℵ. Let us arrive at a contradiction. If a1 6⊆ ℵ and a2 6⊆ √ ℵ, let ℘1 ∈ a1 \ ℵ and let ℘2 ∈ a2 \ √ ℵ. Since ℘2 /∈ √ ℵ, so ∃ an mp1 -sys § \ ℵ in § such that ℘2 ∈ § \ ℵ and (§ \ ℵ)∩ ℵ = φ. Thus ℘1, ℘2 ∈ § \ ℵ implies < ℘1 >b · < ℘2 >b 6⊆ ℵ. Thus a1 ⊆ ℵ or a2 ⊆ √ ℵ. Hence ℵ is a 1PBID of §. Corollary 2.7. Every mp1 -sys is an mp2 -sys. https://internationalpubls.com 756 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) Proof. Given that D be an mp1 -sys of §. For any ∝, ℘ ∈ D, ∃ ∝1∈<∝>b and ℘1 ∈< ℘ >b such that ∝1 ·℘1 ∈ D. Let us shows that D is an mp2 -sys. For ∝, ℘ ∈ D, ∃ ∝1∈<∝>r and ℘1 ∈< ℘ >l. Since right and LIDs are BIDs also, we have ∝1 ·℘1 ∈ D. Hence D is an mp2 -sys of §. Corollary 2.8. Every mp2 -sys is an mp3 -sys. Proof. Given that D be an mp2 -sys of §. For any ∝, ℘ ∈ D, ∃ ∝1∈<∝>r and ℘1 ∈< ℘ >l such that ∝1 ℘1 ∈ D. Let us shows that D is an mp3 -sys. For ∝, ℘ ∈ D, ∃ ∝1∈<∝> and ℘1 ∈< ℘ >. Since IDs are RID and LIDs also, we have ∝1 ℘1 ∈ D. Hence D is an mp3 -sys of §. Remark 2.9. Let √ a be any BID of a ring §. Then √ ßa = {ς ∈ √ a | §ς ⊆ √ a} and √ `a = {ε ∈√ ßa | ε§ ⊆ √ ßa}. Corollary 2.10. Let √ a be a BID of §. Then √ ßa is a LID of § such that √ ßa ⊆ √ a. Proof. Let ς, ε ∈ √ ßa. Then ς, ε ∈ √ a and §ς ⊆ √ a and §ε ⊆ √ a. Since √ a is a BID of §, ς + ε ∈ √ a and ςε ∈ √ a. Now, §(ς + ε) ⊆ §ς + §ε ⊆ √ a. Thus, ς + ε ∈ √ ßa. Now, §(ςε) ⊆ (§ς)(§ε) ⊆ √ a. Thus, ςε ∈ √ ßa. Hence √ ßa is a SSR of §. Let ς ∈ √ ßa and ~ ∈ §. Since ~ς ∈ §ς ⊆ √ a, we have ~ς ∈ √ a and §~ς ⊆ §§ς ⊆ §ς ⊆ √ a. Thus, ~ς ∈ √ ßa. Hence √ ßa is a LID of § and √ ßa ⊆ √ a. Corollary 2.11. Let √ a be a BID of §. Then √ `a is a SSR of §. Proof. Let ς, ε ∈ √ `a. Then ς, ε ∈ √ ßa and ς§ ⊆ √ ßa and ε§ ⊆ √ ßa. Since ς ∈ √ ßa, ς ∈ √ a and §ς ⊆ √ a. Since ε ∈ √ ßa, ε ∈ √ a and §ε ⊆ √ a. Since ς, ε ∈ √ a and √ a is a SSR of §. We have ς+ ε ∈ √ a and ςε ∈ √ a. Now, §(ς + ε) ⊆ §ς + §ε ⊆ √ a implies ς + ε ∈ √ ßa. Now, (ς + ε)§ ⊆ ς§ + ε§ ⊆ √ ßa. Hence ς + ε ∈ √ `a. Now, §(ςε) ⊆ (§ς)(§ε) ⊆ √ a implies ςε ∈ √ ßa and (ςε)§ ⊆ (ς§)(ε§) ⊆ √ ßa. That is ςε ∈ √ `a. Hence √ `a is a SSR of §. Corollary 2.12. Let √ a be a LID of §. Then √ ßa = √ a. Proof. Clearly, √ ßa ⊆ √ a. Let ς ∈ √ a, since √ a is a LID of §. We have §ς ⊆ √ a implies ς ∈ √ ßa. Thus,√ a ⊆ √ ßa. Hence √ ßa = √ a. Theorem 2.13. Let √ a is a BID of §. Then √ `a is the unique largest TID of § contained in √ a. Proof. Let √ a is any BID of §. To prove that √ `a is the TID of §. Since √ ßa ⊆ √ a and √ `a ⊆ √ ßa. Therefore √ `a ⊆ √ ßa ⊆ √ a. Let ς ∈ √ `a and 1∈ §. Then ς ∈ √ `a ⊆ √ a =⇒ ς ∈ √ a. Since ς is an element of √ ßa. We have §ς ⊆ √ a and ς§ ⊆ √ ßa. Then 1 ς ∈ §ς ⊆ √ a implies 1 ς ∈ √ a and § 1 ς ⊆ §§ς ⊆ §ς ⊆ √ a =⇒ 1 ς ∈ √ ßa. Now, ς 1∈ ς§ ⊆ √ ßa. Hence ς 1∈ √ ßa and 1 ς ∈ √ ßa. First to prove that ς 1∈ √ `a and 1 ς ∈ √ `a. Now, ς 1 § ⊆ ς§§ ⊆ ς§ ⊆ √ ßa. Hence ς 1 § ⊆ √ ßa implies ς 1∈ √ `a. Now, 1 ς§ ⊆ §ς§ ⊆ § √ ßa ⊆ √ ßa. Since √ ßa is a LID of §, 1 ς ∈ √ `a. Hence √ `a is a TID of §. It enough to prove √ `a is a largest two sided ID of §. Let √ § be any ID of § and √ § ⊆ √ a. Let % ∈ √ §, then % ∈ √ a and §% ⊆ √ § ⊆ √ a. Hence §% ⊆ √ a =⇒ % ∈ √ ßa. Hence √ § ⊆ √ ßa. Next, % ∈ √ ßa and %§ ⊆ √ § ⊆ √ ßa. Therefore %§ ⊆ √ ßa. Thus, % ∈ √ `a. Hence √ § ⊆ √ `a. Theorem 2.14. A BID a of a semiring § is 2PBID if and only if £1£2 ⊆ a, with £1 is a RID of § and £2 is a LID of § implies £1 ⊆ a or £2 ⊆ √ a. Proof. Let a be a 2PBID and £1£2 ⊆ a. Suppose £1 6⊆ a. For all ℘ ∈ £2 and ∝∈ £1 \ a, we have ∝ §℘ ⊆ £1£2 ⊆ a. Since a is primary and ∝6∈ a and ℘ ∈ √ a for all ℘ ∈ £2. So £2 ⊆ √ a. Conversely, suppose that ∝ §℘ ⊆ a. Now, (∝ §)(§℘) ⊆∝ §℘ implies ∝ § ⊆ a or §℘ ⊆ √ a. If ∝ § ⊆ a, then <∝>r< ℘ >l= {n ∝ + ∝ §|n ∈ Z+} · {m℘+ §℘|m ∈ Z+} = n ∝ m℘+n ∝ §℘+ ∝ ℘+ ∝ §§℘ ⊆∝ § ⊆ a. Thus, ∝∈ a or ℘ ∈ √ a. Similarly, suppose that §℘ ⊆ √ a =⇒ <∝>r< ℘ >l⊆ §℘ ⊆ √ a. Thus, ∝∈ a or ℘ ∈ √ a. Theorem 2.15. A BID a is a 3PBID of § if and only if `a is a PID of §. https://internationalpubls.com 757 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) Proof. Let a be an 3PBID of §. To show that `a is a PID of §. Let £1 and £2 be the IDs of § such that £1 ·£2 ⊆ `a. By thm 2.13, 2.14 and Proposition 6,10 `a and √ `a are unique largest TID contained in a and√ a respectively. Thus £1 ⊆ `a or £2 ⊆ √ `a. Conversely, suppose that £1 and £2 are IDs of § such that £1 · £2 ⊆ a. Then £1 · £2 ⊆ `a, implies £1 ⊆ `a ⊆ a or £2 ⊆ √ `a ⊆ √ a. Hence a is a 3PBIDs of §. Corollary 2.16. If a is a 1PBID of §, then `a is a PID of §. Proof. Let a be an 1PBID of §. Let us show that `a is a PID of §. Let £1 and £2 be an IDs of § such that £1£2 ⊆ `a. To show that £1 ⊆ `a or £2 ⊆ √ `a. Since `a ⊆ a and √ `a ⊆ √ a. Hence £1£2 ⊆ a. Since £1 and £2 are IDs of § is a BIDs also and a is an 1PBID of §. Hence £1 ⊆ a or £2 ⊆ √ a. By Proposition 6,10 `a is the largest ID of § such that `a ⊆ a and by thm 2.13, √ `a is the largest ID of § such that √ `a ⊆ √ a. Thus £1 ⊆ `a or £2 ⊆ √ `a. Hence `a is a PID of §. Corollary 2.17. If a is a 2PBID of §, then `a is a PID of §. Proof. Let a be an 2PBID of §. Let us show that `a is a PID of §. Let £1 and £2 be an IDs of § such that £1£2 ⊆ `a. To show that £1 ⊆ `a or £2 ⊆ √ `a. Since `a ⊆ a and √ `a ⊆ √ a. Hence £1£2 ⊆ a. Since £1 is an ID of § is an RID also and £2 is an ID of the ring § is an LID also. Since a is an 2PBID of §. Hence £1 ⊆ a or £2 ⊆ √ a. By Proposition 6,10 `a is the largest ID of § such that `a ⊆ a and by thm 2.13, √ `a is the largest ID of § such that √ `a ⊆ √ a. Thus £1 ⊆ `a or £2 ⊆ √ `a. Hence `a is a PID of §. Theorem 2.18. Let D be a mp3 - sys and a be a BID of § with a ∩D = φ. Then ∃ a 3PBID ℵ of § containing a with ℵ ∩D = φ. Proof. Let X = { £2|£2 is a BID with a ⊆ £2 and £2 ∩D = φ } . Clearly X is non-empty. By Zorn’s lem, ∃ an maximal element ℵ in X . We claim that ℵ is a 3PBID of §. It is enough if we show that `ℵ is a PID in §. Since `ℵ ⊆ ℵ and ℵ ∩D = φ, this =⇒ `ℵ ∩D = φ. Then `ℵ is a largest ID in § such that `ℵ ∩D = φ. We claim that <∝>< ℘ >⊆ `ℵ. Then <∝>⊆ `(Q) or < ℘ >⊆ `(Q). If <∝>6⊆ `ℵ and < ℘ > 6⊆ √ `ℵ, then ς ∈<∝> \`ℵ and ε ∈< ℘ > \ √ `ℵ. Then < ς >⊆<∝> and < ε >⊆< ℘ >. If <∝>< ℘ >⊆ `ℵ then < ς >< ε >⊆<∝>< ℘ >⊆ `ℵ. Since < ℘ >6⊆ √ `ℵ and hence (< ℘ >)n 6⊆ `ℵ =⇒ < ℘ >6⊆ `ℵ. Then (`ℵ+ < ς >) ∩D 6= φ and (`ℵ+ < ε >) ∩D 6= φ. Thus (`ℵ+ < ς >)(`ℵ+ < ε >) ⊆ `ℵ. Then the BID (`(Q) + ς) contains an element mp1 of D. Then ∃ $1 ∈ (`ℵ+ < ς >) ∩D. Similarly the BID (`(Q) + ε) contains an element mp2 of D. Then ∃$2 ∈ (`ℵ+ < ε >)∩D. Since D is mp3 -sys of A, $1 ′ ∈< $1 > and $2 ′ ∈< $2 > $1 ′ $2 ′ ∈ D for some $1 ′ ∈< $1 >⊆ (`ℵ+ < ς >) and $2 ′ ∈< $2 >⊆ (`ℵ+ < ε >). Hence $1 ′ $2 ′ ∈ (`ℵ+ < ς >)(`ℵ+ < ε >) ⊆ `ℵ. Which is a contradiction. Thus <∝>< ℘ > 6⊆ `ℵ. Hence `ℵ is a primary ID of §. By thm 2.15, then there is an maximal ID ℵ′ in § such that `ℵ ⊆ ℵ′ and ℵ′ ∩D = φ. Hence ℵ′ is the BID of §. 3 Characterization of SPBIDs Definition 3.1. A BID ℵ of § is said to be (i) 1SPBID if a2 ⊆ ℵ implies a ⊆ ℵ or a ⊆ √ ℵ for any BID a of §. (ii) 2SPBID if ∝ § ∝⊆ ℵ implies ∝∈ ℵ or ∝∈ √ ℵ. (iii) 3SPBID if £2 1 ⊆ ℵ implies £1 ⊆ ℵ or £1 ⊆ √ ℵ for any ID £1 of §. Theorem 3.2. Every 1SPBID is a 2SPBID of §. Proof. Let ℵ is a 1SPBID of §. Let ∝∈ § and ∝ § ∝⊆ ℵ. Now, (∝ §) · (§ ∝) ⊆∝ § ∝⊆ ℵ, since ∝ § and § ∝ are BIDs. Hence ∝ § ⊆ ℵ or § ∝⊆ √ ℵ. Suppose that ∝ § ⊆ ℵ. Consider <∝>b · <∝>b⊆∝ § ⊆ ℵ. Then ∝∈ ℵ. Similarly if § ∝⊆ √ ℵ then ∝∈ √ ℵ. Thus ℵ is a 2SPBID of §. Theorem 3.3. Every 2SPBID(2PBID) is a 3SPBID of §. https://internationalpubls.com 758 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) Proof. Suppose that ℵ is a 2SPBID and £2 ⊆ ℵ for an ID £ of §. To show that £ ⊆ ℵ or £ ⊆ √ ℵ. If £ 6⊆ ℵ and £ 6⊆ √ ℵ. For ∝∈ £, but ∝/∈ ℵ and ∝/∈ √ ℵ. Now ∝ § ∝⊆<∝> · <∝>⊆ £2 ⊆ ℵ. Since ℵ is a 2SPBID of §, then ∝∈ ℵ or ∝∈ √ ℵ. Which is contradiction, hence £ ⊆ ℵ or £ ⊆ √ ℵ. Thus ℵ is a 3SPBID of §. Definition 3.4. A subset N of § is said to be (i) np1 -sys if for any ∝∈ N , ∃ ∝1,∝2∈<∝>b such that ∝1∝2∈ N . (ii) np2 -sys if for any ∝∈ N , ∃ ∝1,∝2∈<∝>r (∝1,∝2∈<∝>l) such that ∝1∝2∈ N . (iii) np3 -sys if for any ∝∈ N , ∃ ∝1,∝2∈<∝> such that ∝1∝2∈ N . Theorem 3.5. If ℵ is a BID of §, then ℵ is a 1SPBID (2SPBID, 3SPBID) if and only if § \ ℵ is an np1 -sys (np2 -sys, np3 -sys). Proof. Let ℵ be a 1SPBID of §. Let ∝∈ § \ ℵ. Hence ∝∈ § but ∝/∈ ℵ. So <∝>b · <∝>b 6⊆ ℵ. There exists ∝′ ,∝′′∈<∝>b such that ∝′ · ∝′′ /∈ ℵ. Hence ∝′ · ∝′′∈ § \ ℵ. So we have proved that for ∝∈ § \ ℵ ∃ ∝′ ,∝′′∈<∝>b such that ∝′ · ∝′′∈ § \ ℵ. So § \ ℵ is an np1 -sys. Conversely, let § \ ℵ is an np1 -sys. Let a2 ⊆ ℵ for the BID a of §. Let us shows that a ⊆ ℵ or a ⊆ √ ℵ. If a 6⊆ ℵ and a 6⊆ √ ℵ, let ℘1 ∈ a \ ℵ and ℘1 ∈ a \ √ ℵ. Since ℘1 /∈ √ ℵ, so ∃ an np1 -sys § \ ℵ in § such that ℘1 ∈ § \ ℵ and (§ \ ℵ) ∩ ℵ = φ. Thus ℘1 ∈ § \ ℵ implies < ℘1 >b · < ℘1 >b 6⊆ ℵ, which is a contradiction. Thus a ⊆ ℵ or a ⊆ √ ℵ. Hence ℵ is a 1SPBID of §. Corollary 3.6. Every np1 -sys is an np2 -sys. Theorem 3.7. Let a be a 2SPBID of a ring §. Then £2 ⊆ a implies £ ⊆ a or £ ⊆ √ a for any LID (RID)£ of §. Theorem 3.8. A BID a is a 3SPBID of § if and only if `a is a SPID of §. Corollary 3.9. If a is a 1SPBID (2SPBID) of §, then `a is a SPID of §. Acknowledgment. This research was supported by University of Phayao and Thailand Science Research and Innovation Fund (Fundamental Fund 2025, Grant No. 5027/2567). 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