Unitary Semi-Ring on Some Nexuses

1. Introduction

In 1980, Haristchain [6] introduced the concept of a plenix to handle large amounts of structured spatial data. In 1984, Nooshin [12] defined a nexus as a mathematical object that captures the constitution of a plenix, using the notion of an address set (also, see [7]). In 2009, Bolourian [2] introduced nexus algebras as an abstract algebraic structure and investigated their properties.

Since then, many authors have studied nexuses and subnexuses. For instance, Norouzi [13] worked on subnexuses based on N-structures. Kamrani et al. [4, 5] developed the structure of a moduloid on a nexus and verified the concepts of submoduloids, finitely generated submoduloids, and prime submoduloids (also, see [3]). Recently, Nazemi Niya et al. [9, 10, 11] defined (fuzzy) semi-rings on a (finite) nexus N, characterized all prime ideals, and studied homomorphisms, quotients, and localizations.

However, the semi-rings constructed in those works are not unitary unless the nexus is cyclic. In this case, if N = ⟨a⟩ be a cyclic nexus, 1 = a. In a non-unitary semi-ring, several standard properties fail. For example, the polynomial ring R[x] over a non-unitary semi-ring does not contain the monomial x (since x = 1 · x would require a multiplicative identity). Moreover, units cannot be defined (let R be a unitary semi-ring with 1R. An element 0 = ̸ a ∈ R is unite if there exists b ∈ R such that ab = ba = 1R).

In this paper, we construct a unitary semi-ring on certain nexuses where the multiplicative identity is exactly the address (1). For this purpose, we define the function [ ]n, where n ∈ N and the semi-ring Nn. But this semi-ring is not unitary. Then by the semi-ring of polynomials on Nn, i.e. Nn[x], we define two custom binary operations, ⊕ and ⊙n, parameterized by a natural number n. The central construction is the generation of a semi-ring ⟨N⟩n from a given finite-order nexus N, where n = n(N). This generated structure is shown to be itself a nexus. Notably, ⟨N⟩n is unitary, with multiplicative identity given by the unit address (1). The nexus N with n(N) = n is closed, if ⟨N⟩n = N. Every closed nexus with rise(N) ̸= 1 is an infinite nexus. We begin by defining the necessary operations and then verify the algebraic structures such as homorphism and isomorphism theorems, quotient semi-ring, localization semi-ring, Noetherian and Artinian properties, zero-divisor elements, idempotent elements and prime elements.

2. Preliminary Concepts

Now we recall the basic definitions ([8]).

A groupoid (G, ◦) is a nonempty set with a binary operation. A semi-group is an associative groupoid. A monoid (G, ◦, 0) is a semi-group with an identity element 0 such that 0 ◦ a = a ◦ 0 = a for all a ∈ G.

A semi-ring (R, +, ◦, 0) is a nonempty set with two operations such that (R, +, 0) is a commutative monoid, (R, ◦) is a semi-group, and ◦ distributes over +: a ◦ (b + c) = a ◦ b + a ◦ c and (b + c) ◦ a = b ◦ a + c ◦ a for all a, b, c ∈ R. If there exists an element 1 ∈ R such that 1 · a = a · 1 = a for all a ∈ R, then R is called unitary. An element a ∈ R is a unit if it has a multiplicative inverse.

A subset X ⊆ R is a sub-semi-ring if 0 ∈ X and X is closed under + and ◦. A subset I ⊆ R is an ideal if 0 ∈ I, a + b ∈ I for all a, b ∈ I, and ra, ar ∈ I for all r ∈ R and a ∈ I. An ideal I is prime if ab ∈ I implies a ∈ I or b ∈ I.

An address is a sequence whose elements belong to N = {0} ∪ N, with the property that if ak = 0 then ai = 0 for all i ≥ k. The empty sequence is denoted by (). A non-empty address is written as (a1, a2, . . . , an) where ai ∈ N. A nexus N is a non-empty set of addresses satisfying: (a1, . . . , an−1, an) ∈ N implies (a1, . . . , an−1, t) ∈ N for all 0 ≤ t ≤ an.

For an infinite nexus {ai}i=1 ∈ N, ai ∈ N implies ∀n ∈ N, ∀ 0 ⩽ t ⩽ an, (a1, · · · , an−1, t) ∈ N. Hereafter, N denotes a nexus.

Let a ∈ N. The level of a is said to be:

(I) n, if a = (a1, a2, · · · , an), for some ak ∈ N,

(II) ∞, if a is an infinite sequence of N,

(III) 0, if a = ().

The level of a is denoted by l(a). If a = (a1, a2, · · · , an), we define stem a = a1.

Let N be a nexus, we define rise(N) = max{l(a) | a ∈ N}[1]. In this paper, if a nexus N is a semi-ring with rise(N) = 1 ̸ , then N is an infinite nexus.

3. Semi-ring on nexuses

For a finite nexus N, define n(N) := max{a ∈ N | ∃(a0, . . . , ak) ∈ N, ∃i (1 ≤ i ≤ k), such that ai = a}. We now introduce two binary operations on addresses.

Definition 1. For a ∈ N = N ∪ {0} and n ∈ N, define [a]n by [a]n = a if 0 ≤ a < n, and [a]n = n if a ≥ n.

Example 2. Let n = 5, then [0]5 = 0, [1]5 = 1, [2]5 = 2, [3]5 = 3, [4]5 = 4, 5 = [5]5 = [6]5 = [7]5 = [8]5 = · · · .

Lemma 3. If a ≤ b then [a]n ≤ [b]n.

Proof : Straightforward. □

Definition 4. For a, b ∈ N , set a + b = a ∨ b = max{a, b} and a · b = [ab]n. Denote this algebraic structure by Nn.

Example 5. Let 2, 3 ∈ N4, hence 2 + 3 = 3 and 2 · 3 = [2 · 3]4 = [6]4 = 4.

Let 2, 3 ∈ N7, hence 2 + 3 = 3 and 2 · 3 = [2 · 3]7 = [6]7 = 6.

Theorem 6. (Nn, +, ·, 0) is a semi-ring.

Proof : Clearly binary operations + and · are well-defined and + is associative and for every a ∈ N, a∨0 = 0∨a = a. We show that · is associative. Let a, b, c ∈ N. We consider two cases:

[︁ ]︁ [︁ ]︁ Case 1: Let abc < n. Hence ab < n and bc < n. So [ab]nc = [abc]n = abc and a[bc]n = [abc]n = abc. Then (a · b) · c = a · (b · c) [︁ ]︁ Case 2: Let abc ⩾ n. If a ⩾ n, then ab ⩾ n and so (a · b) · c = [ab]nc = [nc]n = n. Also we have a[bc]n ⩾ n [︁ ]︁ n and so a · (b · c) = a[bc]n = n. If b ⩾ n or c ⩾ n, similarly we have (a · b) · c = a · (b · c) = n. Let n [︁ ]︁ a < n, b < n and c < n. If ab ⩾ n, we have (a · b) · c = [ab]nc = [nc]n = n. Since bc ⩾ b, by Lemma 3, [bc]n ⩾ [b]n = b, hence a[bc]n ⩾ ab ⩾ n, then [a[bc]n]n = n. Hence (a · b) · c = a · (b · c) = n. If bc ⩾ n, similarly, [︁ ]︁ (a · b) · c = a · (b · c) = n. Let a < n, b < n, c < n, ab < n and bc < n. Hence (a · b) · c = [ab]nc = [abc]n = n [︁ ]︁ n and a · (b · c) = a[bc]n = [abc]n = n. Therefore (a · b) · c = a · (b · c) = n and . is associative.

Now we show that a · (b + c) = a · b + a · c. Let b ⩾ c. So ab ⩾ ac and by Lemma 3, [ab]n ⩾ [ac]n. Hence a · b + a · c = [ab]n ∨ [ac]n = [ab]n = a · b = a · (b ∨ c) = a · (b + c). Similarly, (b + c) · a = b · a + c · a. Therefore · on + is distributive and so (Nn, +, ·, 0) is a semi-ring. □

Hereafter, Nn is a semi-ring, it means, (Nn, +, ·, 0) is a semi-ring.

Corollary 7. For n > 1, Nn is not unitary because for every n < a ∈ N, we have 1 · a = [a]n = n = ̸ a.

Let R[x] be the polynomial semi-ring over R. Define R⌊x⌋ = { aix ∈ R[x] | 0 = ̸ ai ∈ R} ∪ {0}.

Proposition 8. If R is a semi-ring where a + b ̸= 0 and ab ̸= 0 for all nonzero a, b ∈ R, then R⌊x⌋ is a sub-semi-ring of R[x].

Proposition 9. Nn⌊x⌋ is a sub-semi-ring of Nn[x].

Now we connect nexuses to polynomials.

Definition 10. Let M be the set of all finite addresses (including the empty address ()). For n ∈ N, define the map Φn : M −→ Nn⌊x⌋ by

Φn((a0, a1, . . . , ak)) = aix ,

with Φn(()) = 0 (the zero polynomial). For addresses a = (a0, . . . , ak), b = (b0, . . . , bt) define:

a ⊕ b = (a0 ∨ b0, a1 ∨ b1, . . . , al ∨ bl), l = k ∨ t,

a ⊙n b = (c0, c1, . . . , ck+t), ci = [ai−jbj]n.

Example 11. Let a = (1, 6, 3, 1), b = (8, 4, 6) and N = ⟨a, b⟩, then n(N) = 8. Hence (︁ )︁ a ⊕ b = (1 ∨ 8), (6 ∨ 4), (3 ∨ 6), (1 ∨ 0) = (8, 6, 6, 1).

(︁ )︁ a ⊙8 b = (1.8), (1.4 + 6.8), (1.6 + 6.4 + 8.3), (1.0 + 6.6 + 3.4 + 1.8) (︁ )︁ = [8]8, ([4]8 ∨ [48]8), ([6]8 ∨ [24]8 ∨ [24]8), ([0]8 ∨ [36]8 ∨ [12]8 ∨ [8]8) (︁ )︁ = (8, (4 ∨ 8), (6 ∨ 8 ∨ 8), (0 ∨ 8 ∨ 8 ∨ 8) = (8, 8, 8, 8).

Proposition 12. The map Φn is a bijection. Moreover, for all addresses a, b ∈ M, Φn(a ⊕ b) = Φn(a) + Φn(b), Φn(a ⊙n b) = Φn(a) · Φn(b).

Hence Φn is an isomorphism of semi-rings between (M, ⊕, ⊙n, ()) and Nn⌊x⌋.

Proof : Injectivity is clear because different sequences yield different polynomials (with non-zero leading coef-ficients). Surjectivity follows because every polynomial aix comes from the address (a0, . . . , ak). The two addresses () and (1) are exactly the definitions of ⊕ and ⊙n (cf. Definition 10). Thus Φn preserves both operations and we have () ↔ 0 and (1) ↔ 1 (notice that M and Nn⌊x⌋ aren’t unitary). □ (︁ )︁ Theorem 13. Let N be a nexus with n(N) = n that is closed under ⊕ and ⊙n. Then N, ⊕, ⊙n, (), (1) is a unitary semi-ring.

Proof : Because N ⊆ M and is closed under the operations and by Proposition 12, (M, ⊕, ⊙n, ()) is semi-ring, it (︁ )︁ inherits the semi-ring structure and so N, ⊕, ⊙n, (), (1) is a semi-ring. Finally, for any a = (a0, . . . , am) ∈ N, we have (1) ⊙n a = ([1 · a0]n, . . . , [1 · am]n) = (a0, . . . , am) = a because ai ≤ n. Hence (1) is the multiplicative identity. □

Definition 14. Let N be a nexus with n(N) = n. We define ⟨N⟩n the sub-nexus generated by N is closed under ⊕ and ⊙n.

(︁ )︁ Corollary 15. Let N be a nexus with n(N) = n. Then ⟨N⟩n, ⊕, ⊙n, (), (1) is a unitary semi-ring.

Proof : It is clear by Definition 14 and Theorem 13. □

Example 16. Let N = {(), (1)}. Then n(N) = 1 and ⟨N⟩1 = {(), (1)}.

Example 17. Let N = {(), (1), (1, 1)}. Then n(N) = 1 and ⟨N⟩1 contains all addresses of the form (1, . . . , 1) ⏞ ⏟⏟ ⏞

for k ≥ 0.

Example 18. For N = {(), (1), (2), . . . , (m)}, n(N) = m and ⟨N⟩m = N.

Example 19. Let N = {(), (1), (1, 1), (1, 2)}. Then n(N) = 2 and ⟨N⟩2 consists of all addresses starting with 1 and whose entries are at most 2, with the nexus closure property.

Lemma 20. Let N be a nexus with n(N) = n. For non-empty addresses a, b ∈ ⟨N⟩n with l(a) = k ≥ 1 and l(b) = m ≥ 1, we have l(a ⊕ b) = k ∨ m and l(a ⊙n b) = k + m − 1. If either address is empty, the formulas are trivial.

∑︁k−1 Proof : Let a = (a0, a1, · · · , ak−1) and b = (b0, b1, · · · , bm−1). By the proof of Theorem 13, Φn(a) = aix ∑︁m−1 is a polynomial of degree k − 1 and Φn(b) = bix is a polynomial of degree m − 1. Hence Φn(a) + Φn(b) is a polynomial of degree (k ∨ m) − 1 and Φn(a)Φn(b) is a polynomial of degree k + m − 2. Since a ⊕ b =

−1 −1

Φn (Φn(a) + Φn(b)) and a ⊙n b = φn (Φn(a)Φn(b)), l(a ⊕ b) = k ∨ m and l(a ⊙n b) = k + m − 1. □

Example 21. Let a = (2, 3, 1) and b = (3, 5, 2, 1). By Lemma 20, l(a ⊕ b) = 4 and l(a ⊙5 b) = 3 + 4 − 1 = 6. In

2 2 3 2

fact, a ⊕ b = (3, 5, 2, 1) and a ⊙5 b = (2 + 3x + x )(3 + 5x + 2x + x ) = [6]5 + [10 ∨ 9]5x + [4 ∨ 15 ∨ 3]5x + [2 ∨

3 4 5

6 ∨ 5]5x + [3 ∨ 2 ∨ 5]5x + [1]5x = (5, 5, 5, 5, 5, 1).

Corollary 22. Let N be a nexus with n(N) = n and rise(N) = 1 ̸ . Then ⟨N⟩n is infinite.

Proof : It is clear by Lemma 20. □

Proposition 23. Let n(N) = n and s ≤ n. Then (1, s) ∈ ⟨N⟩n iff (1, s) ∈ N.

Proof : If (1, s) ∈ / N but is in ⟨N⟩n, by Lemma 20 it would be a product of an address of length 1 and an address of length 2, forcing (1, s) ∈ N – contradiction. □

Proposition 24. Let s1, s2 ≤ n and (1, s1) ∈ N. Then (1, s1, s2) ∈ ⟨N⟩n iff either (1, s1, s2) ∈ N or s2 ≤ s1.

Proof : If (1, s1, s2) ∈ N, clearly (1, s1, s2) ∈ ⟨N⟩n. Now let s2 ≤ s1. We have (1, s1) ⊙n (1, s1) = (1, s1, [s1]n) ∈ ⟨N⟩n. So for every s2 ≤ s1, (1, s1, s2) ∈ ⟨N⟩n. Conversely, let (1, s1, s2) ∈ ⟨N⟩. Since l((1, s1, s2)) = 3, by Lemma 20, there exist a, b ∈ N with l(a) = 2 and l(b) = 2 such that a ⊙n b = (1, s1, s2). Let a = (a0, a1) and b = (b0, b1). We have a ⊙n b = ([a0b0]n, [a0b1 ∨ a1b0]n, [a1b1]n) = (1, s1, s2). Hence a0 = b0 = 1 and a1 ∨ b1 = s1 and [a1b1]n = s2. So a1, b1 ≤ s1 and s2 = [a1b1]n ≤ s1 and the proof is complete. □ ⟨︁ ⟩︁ Example 25. Let N be cyclic nexus N = (1, 10) . By Proposition 24, we have:

(I) (1, 1, 1) ∈ ⟨N⟩10, but for every 2 ≤ j ≤ 10, (1, 1, j) ∈ ⟨ / N⟩10.

(II) For every 1 ≤ i ≤ 4, (1, 2, i) ∈ ⟨N⟩10, but for every 5 ≤ i ≤ 10, (1, 2, i) ∈ ⟨ / N⟩10.

(III) For every 1 ≤ i ≤ 9, (1, 3, i) ∈ ⟨N⟩10, but (1, 3, 10) ∈ ⟨ / N⟩10.

(IV) For every 4 ≤ k ≤ 10 and for every 1 ≤ i ≤ 10, (1, k, i) ∈ ⟨N⟩10.

′ ′′ ′ ′′

Proposition 26. (1, s1, s2, s3) ∈ ⟨N⟩n is a product of two addresses iff there exist s , s ∈ N with s , s ≤ n such that one of the following conditions holds:

′ ′′ ′2 ′ ′′

(I) s ≤ s1 and s ≤ s such that s2 = [s s1]n and s3 = [s s1]n.

′ ′′ ′ 2 ′ ′ ′′

(II) s ≤ s1 and s ≤ s s1 ≤ s1 such that s2 = [s s1]n and s3 = [s s ]n.

′ ′ ′′ ′2 ′′ ′ ′′

(III) s ≤ s1 and s s1 ≤ s ≤ s such that s2 = s and s3 = [s s ]n.

′′ ′2 ′ ′′ ′′ ′2

Proof : If part (I) holds. Since s ≤ s , by Proposition 24, (1, s , s ) ∈ ⟨N⟩n. Now since s ≤ s , we have

′′ ′ ′ ′′ ′ ′ ′′ ′′ ′ ′′

s ≤ s s1 and (1, s1)(1, s , s ) = (1, s1 ∨ s , [s s1 ∨ s ]n, [s1s ]n) = (1, s1, [s s1]n, [s s1]n) ∈ ⟨N⟩n.

′′ ′ 2 ′′ ′ ′′

If part (II) holds. Since s ≤ s s1 ≤ s1, by Proposition 24, (1, s1, s ) ∈ ⟨N⟩n. Now we have (1, s )(1, s1, s ) =

′ ′ ′′ ′ ′′ ′ ′ ′′

(1, s1 ∨ s , [s s1 ∨ s ]n, [s s ]n) = (1, s1, [s s1]n, [s s ]n) ∈ ⟨N⟩n.

′ ′′ ′2 ′′ ′ ′′

If part (III) holds. Since s s1 ≤ s ≤ s , by Proposition 24, (1, s1, s ) ∈ ⟨N⟩n. Now we have (1, s )(1, s1, s ) =

′ ′ ′′ ′ ′′ ′′ ′ ′′

(1, s1 ∨ s , [s s1 ∨ s ]n, [s s ]n) = (1, s1, s , s s ) ∈ ⟨N⟩n.

Conversely, let (1, s1, s2, s3) ∈ ⟨N⟩n. Since l((1, s1, s2, s3)) = 4, by Lemma 20, there exist a, b ∈ N with l(a) = 2 and l(b) = 3 such that a ⊙ b = (1, s1, s2, s3). Let a = (a0, a1) and b = (b0, b1, b2). We have a ⊙ b = (a0b0, [a0b1 ∨ a1b0]n, [a0b2 ∨ a1b1]n, [a1b2]n) = (1, s1, s2, s3). Hence a0 = b0 = 1 and a1 ∨ b1 = s1 and [a0b2 ∨ a1b1]n = s2 and [a1b2]n = s3.

We consider three cases:

′ ′′ ′2 ′′ ′2 ′′ ′ ′′ ′ ′

Case 1: a1 = s1 and b1 = s ≤ s1 and b2 = s ≤ s . Since s ≤ s , s ≤ s s1. Hence s2 = [s ∨ s s1]n = [s s1]n and s3 = [s s1]n.

′ ′′ ′ 2 ′′ ′ ′ ′ ′′

Case 2: a1 = s ≤ s1 and b1 = s1 and b2 = s ≤ s s1 ≤ s1. Hence s2 = [s ∨ s s1]n = [s s1]n and s3 = [s s ]n.

′ ′ ′′ 2 ′′ ′ ′′ ′′

Case 3: a1 = s ≤ s1 and b1 = s1 and b2 = s s1 ≤ s ≤ s1. Hence s2 = [s ∨ s s1]n = [s ]n = s and

′ ′′

s3 = [s s ]n. □

Proposition 27. If (1, s1, . . . , sk) ∈ ⟨N⟩n then for each i, si ≤ s1.

Proof : By induction using the polynomial representation: the degree-i coefficient in a product of polynomials of

r t r+t−2

degrees r and t is bounded by s1 · s1 = s1 after appropriate normalization. □

Example 28. Let N be a nexus, (1, 3) ∈ N and n = 27. All addresses of the form (1, 3, s2, s3) ∈ ⟨N⟩27 are:

(1, 3, 3, i), 1 ≤ i ≤ 3,

(1, 3, 6, i), 1 ≤ i ≤ 12,

(1, 3, 9, i), 1 ≤ i ≤ 27,

(1, 3, 4, i), 1 ≤ i ≤ 4,

(1, 3, 5, i), 1 ≤ i ≤ 5,

(1, 3, 7, i), 1 ≤ i ≤ 14,

(1, 3, 8, i), 1 ≤ i ≤ 16.

The following example shows that the converse of Proposition 27 is not true.

⟨︁ ⟩︁ Example 29. Let N = (1, 2) , then (1, 1, 2) ∈ / N.

4. Ideals of semi-ring ⟨N⟩n

Now we verify the ideals of semi-ring ⟨N⟩n.

Definition 30. Let N be a nexus with n(N) = n. We say that N is a closed nexus if ⟨N⟩n = N.

Hereafter, N denotes a closed nexus with n(N) = n.

Definition 31. Let N be a nexus. An ideal I of N satisfies 0 ∈ I, a ⊕ b ∈ I for a, b ∈ I, and ra ∈ I for r ∈ N, a ∈ I.

Given a ∈ N, the principal ideal generated by a is ⟨a⟩n = {b ⊙n a | b ∈ N}.

⟨︁ ⟩︁ Example 32. Let s ∈ N, N = (1, s) , then n(N) = s. If I is the ideal generated by (1, s) of semi-ring ⟨N⟩s, {︁ }︁ then I = (1, s), (1, s, s), (1, s, s, s), · · · ∪ {()}.

Definition 33. Let a = (a1, · · · , ak), b = (b1, · · · , bt) be two addresses with k ⩽ t. We say a ⩽ b, if for every i(1 ⩽ i ⩽ t), ai ⩽ bi. Clearly a ⩽ b if and only if a ⊕ b = b.

Example 34. Let a = (3, 3) and b = (3, 3, 3), then a ⩽ b.

Let a = (3, 2) and b = (3, 3), then a ⩽ b.

Let a = (3, 1) and b = (3, 3, 3), then a ⩽ b.

Let a = (3, 3) and b = (2, 3, 3), then a ⩽̸ b and b ⩽̸ a Definition 35. Let N be a nexus and a ∈ N. We put Ba = {b ∈ N | b ⩾ a} ∪ {0}. It is easy to see that Ba is an ideal of N.

⟨︁ ⟩︁ Proposition 36. Suppose that N = (1, 1) and I be the ideal generated by (1, · · · , 1) with n ≥ 2. Then 1 ⏞ ⏟⏟ ⏞ {︁ }︁ {︁ }︁ n times I = () (1, · · · , 1) : k ⩾ n is prime if and only if n = 2.

⏞ ⏟⏟ ⏞

k times

Remark 37. Let M = ⟨a⟩ be a cyclic nexus with n(M) = n. Then the nexus N = ⟨M⟩n is a closed nexus. We denoted this nexus by Na. Let s ∈ N and a = (1, s). Then n(M) = s and we denoted the nexus Na by Ns and Ba by Bs.

Proposition 38. For every s ∈ N, Bs is a prime ideal.

Proof : It is easy to see that if a ∈ Bs − {(), (1, s)}, there exist s1, · · · , sk ∈ N, such that a = (1, s, s1, s2, · · · , sk). Now let b, c ∈ N and (a1, a2, · · · , am) = b ⊙s c ∈ Bs. Then there exist r, t ∈ N with r + t − 1 = m and b1, · · · , br, c1, · · · , ct ∈ N, such that b = (b1, · · · , br) and c = (c1, · · · , ct). Clearly b1 = c1 = 1. If b2, c2 < s, then a2 < s - contradiction. Hence b2 = s or c2 = s and so b ∈ Bs or c ∈ Bs. Therefore Bs is a prime ideal. □

The following example shows that for a nexus N and a ∈ N, Ba is not prime ideal.

Example 39. For the nexus N(3,3), we have (3, 2) ⊙3 (3, 2) = (3, 3, 3) ∈ B(3,3), but (3, 2) ∈ / B(3,3).

Proposition 40. For every 2 ⩽ t ⩽ s ∈ N, the ideal B(s,s) of nexus N(s,t) is not prime.

Proof : Case 1: Let t = ̸ s. There exist k ∈ N such that (s, t) = ( s, · · · , s ) ∈ B(s,s), but (s, t) ∈ / B(s,s).

⏞ ⏟⏟ ⏞

(k+1) times

Case 2: Let s = t, so (s) ⊙s (s, 1) = (s, s) ∈ B(s,s), but (s), (s, 1) ∈ / B(s,s). □

In the followig proposition, we generalize Proposition 40.

Proposition 41. Let a = (s1, · · · , st) with s1 ̸= 1 and s = max{s1, · · · , st}. Then the ideal B(s, · · · , s) of nexus ⏞ ⏟⏟ ⏞

t times

Na is not prime.

Proof : Since s1 ̸= 1, (s) and (s, · · · , s) ∈ Na. If there exist i(1 ⩽ i ⩽ t), such that si ̸= s, hence (s) ⊙s ⏞ ⏟⏟ ⏞

t times

(s1, · · · , st) = (s, · · · , s) ∈ B(s, · · · , s), but (s), (s1, · · · , st) ∈ / B(s, · · · , s). If for every i(1 ⩽ i ⩽ t), such that ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞

t times t times t times

si = s, hence (s) ⊙s (s, · · · , s, 1) = (s, · · · , s) ∈ B(s, · · · , s), but (s), (s, · · · , s, 1) ∈ / B(s, · · · , s). Therefore the ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞

t−1 times t times t times t−1 times t times

ideal B(s, · · · , s) is not prime. □ ⏞ ⏟⏟ ⏞

t times

In the followig proposition, we show that we can’t generalize Propposition 40.

Proposition 42. Let a = (1, s, · · · , s). Then the ideal B(1,s, · · · , s) of nexus Na is prime iff t = 1.

⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞

t times t times

Proof : Let t = 1, by Proposition 38, Bs is prime ideal. Conversely let t ̸= 1, then (1, s) ⊙s (1, s, · · · , s ) = ⏞ ⏟⏟ ⏞

(t−1) times

(1, s, · · · , s) ∈ B(1,s, · · · , s), but (s), (1, s, · · · , s ) ∈ / B(1,s, · · · , s). Therefore B(1,s, · · · , s) is not prime ideal. □ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞ ⏞ ⏟⏟ ⏞

t times t times (t−1) times t times t times

Proposition 43. Let N be a nexus and a, b ∈ N. Then (I) B(a⊕b) = Ba ∩ Bb.

(II) Ba⊙ b ⊆ Ba ∩ Bb.

Proof : (I) Let x ∈ B(a⊕b). Since x ⩾ a⊕b, x ⩾ a, b. So x ∈ Ba and x ∈ Bb. Hence B(a⊕b) ⊆ Ba ∩Bb. Conversely let x ∈ (Ba ∩ Bb). Since x ⩾ a, b, x ⩾ a ⊕ b. So x ∈ Ba ∩ Bb. Therefore B(a⊕b) = Ba ∩ Bb.

(II) Let x ∈ B(a⊙ b). Since x ⩾ a ⊙n b, x ⩾ a, b. So x ∈ Ba and x ∈ Bb. Hence B(a⊙ b) ⊆ Ba ∩ Bb. □

The following example shows that the converse of Proposition 43 is not true.

Example 44. Let N = N(6,3), a = (6, 2) and b = (6, 3). We have a ⊙6 b = (6, 6, 6). Hence (6, 6, 6) ∈ B(6,6,6) but (6, 6, 6) ∈ / B(6,2) and (6, 6, 6) ∈ / B(6,3).

Definition 45. For an ideal I and a ∈ N, define a + I = {a ⊕ b | b ∈ I}.

Lemma 46. a + I = I iff I ⊆ Ba.

Proof : Let I ⊆ Ba. Suppose that x ∈ a + I, then there exists c ∈ I suth that x = a ⊕ c. Since c ⩾ a, x = a ⊕ c = c ∈ I. Hence a + I ⊆ I. Now let x ∈ I. Since x ⩾ a, x = a ⊕ x ∈ a + I. Hence I ⊆ a + I and so a + I = I. Conversely let a + I = I and x ∈ I. Then there exist c ∈ I such that x = a ⊕ c ⩾ a and so x ∈ Ba. Therefore I ⊆ Ba. □

Corollary 47. Let N be a nexus and a ∈ N. Then

(I) a + Ba = Ba;

(II) a + I ⊆ Ba.

Proposition 48. Let N be a nexus, I be an ideal and a, b ∈ N. Hence a + I = b + I iff a = b.

Proof : By Corollary 47, a + I = b + I iff Ba = Bb iff a = b. □

Proposition 49. Let N be a nexus and a, b ∈ N. Then a + Bb = b + Ba = B(a⊕b).

Proof : Let x ∈ a + Bb, so there exist c ∈ Bb such that x = a ⊕ c. Since c ⩾ b, a ⊕ c ⩾ a ⊕ b and so x ∈ B(a⊕b). Hence a + Bb ⊆ B(a⊕b). Conversely let x ∈ B(a⊕b). Since x ⩾ a ⊕ b, x ⩾ a, b. So x ∈ Bb and x = a ⊕ x ∈ a + Bb. Hence B(a⊕b) = a + Bb. Similarly, b + Ba = B(a⊕b). Therefore a + Bb = b + Ba = B(a⊕b). □

Definition 50. Define = {a + I | a ∈ N} with ⊞ defined by (a + I) ⊞ (b + I) = (a ⊕ b) + I and ⊡ defined by (a + I) ⊡ (b + I) = (a ⊙n b) + I.

Proposition 51. ( , ⊞, ⊡, I, (1) + I) is a semi-ring.

Proof : It is clear by Proposition 48. □

Corollary 52. Let N be a nexus, I be an ideal of N. Then = N.

Proof : It is clear by Proposition 48. □

4. Conclusion

We have shown that the semi-ring Nn is not unitary, but by embedding a suitable nexus into the polynomial semi-ring Nn⌊x⌋, we obtain a unitary semi-ring ⟨N⟩n with identity (1). This construction opens the door to further study of algebraic properties (ideals, homomorphisms, localizations) in a framework that combines combinatorics of addresses with ring theory.

Acknowledgments

The authors would like to thank the anonymous referees for their thorough and insightful comments, which significantly improved the quality and clarity of this paper.

Author Contribution

Conceptualization, A.R. and V.N.N.; methodology, V.N.N.; formal analysis, A.R.; investigation, A.R. and V.N.N.; writing-original draft, A.R.; writing-review and editing, V.N.N.; visualization, A.R.; supervision, A.R. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare no conflicts of interest.