On $n$-ary ring congruences of $n$-ary semirings

1.   Introduction

A semiring which is a common generalization of rings and distributive lattices was introduced first by Vandiver ^[26] in 1934. This algebraic structure appears in a natural manner in some applications to the theory of automata, formal languages, optimization theory and other branches of applied mathematics (for example, see ^{[4,5,11,12,13,14,18]}). In algebraic structure point of view, we are able to study the concept of an $n$-ary semiring as a generalization of a semiring because a semiring is a special kind of an $n$-ary semiring where $n = 2$ and so every results on an $n$-ary semiring is also true on a semiring but not conversely.

In modern algebra, it is well-known that the kernel of a ring homomorphism is an ideal ^[2] and also true for a semiring homomorphism. Conversely, each ideal of a ring can be considered as the kernel of a ring homomorphism. Notwithstanding, this condition is not generally true in case of an ideal of a semiring ^[1]. However, this condition can be true if we replace the ideal by a special ideal which is called a $k$-ideal defined by Henriksen ^[15]. A more restrict class of ideals of a semiring which is called an $h$-ideal was introduced by Iizuka ^[16].

To generalize the algebraic system of an algebra from a binary operation to an $n$-ary operation, Dörnte ^[8] first defined the notion of an $n$-ary group in 1928. Later, Timm ^[25] studied an $n$-ary group with commutative property. As a generalization of a semigroup and an $n$-ary group, the notion of an $n$-ary semigroup was introduced by Siosson ^[22,23]. Some properties of idempotent elements of an $n$-ary semigroup were studied by Dudek ^[10]. The concepts of homomorphism, quotient structures, and some ideal theoretic were studied by Crombez and Timm ^[6,7]. Later, in 1981, Dudek ^[9] studied the divisibility property of an $(m, n)$-ring. As a generalization of a semiring and an $(m, n)$-ring, Alam, Rao and Davvaz ^[3,19] introduced the notion of an $(m, n)$-semiring.

In 1992, Sen and Adhikari ^[20] studied the notion of a full $k$-ideal which is a $k$-ideal containing the set of all additively idempotent elements of a semiring and use it to construct a congruence relation in order to make the quotient semiring to be a ring. More results of full $k$-ideals of a semiring were also investigated by Sen and Maity ^[21] in 2021. As a similar way of Sen and Adhikari ^[20], Sunitha, Nagi Reddy, and Shobhalatha ^[24] studied full $k$-ideals of ternary semirings.

It is well-known that if we have a congruence relation $\rho$ on an $n$-ary semiring $S$, then we can immediately obtain that $S/\rho$ is also an $n$-ary semiring. It is interesting that what is the kind of a congruence relation $\rho$ affecting $S/\rho$ to be an $n$-ary ring. In this work, we study the notions of $k$-ideals and $h$-ideals of $n$-ary semirings and also investigate their connections. Finally, we use a full $k$-ideal to construct a congruence relation in order to make the quotient $n$-ary semiring to be an $n$-ary ring.

2.   Preliminaries

Let $\mathbb{N}$ be the set of all natural numbers and $i, j, n\in\mathbb{N}$. An algebra $\langle S; f\rangle$ consisting of a nonempty set $S$ together with an $n$-ary operation $f:S^{n}\rightarrow S$ is called an $n$-ary groupoid ^[8]. For $1\leq i < j\leq n$, the sequence $y_i, y_{i+1}, y_{i+2}, \ldots, y_{j}$ of elements of $S$ is denoted by $y_{i}^{j}$. If $j < i$, then we denote it to be the empty symbol. If $x_{1} = x_{2} = \cdots = x_{i-1} = x$ where $x_{1}, x_{2}, \ldots, x_{i-1}, x\in S$, we write $\overset{(i-1)}{x}$ instead of $x_{1}^{i-1}$. So, the term

where $z_{j+1}, z_{j+2}, \ldots, z_{n}\in S$ can be simply represented by

Similarly, for $1\leq i < j\leq n$, we also denote the sequence $A_{i}$, $A_{i+1}$, $A_{i+2}, \dots, A_{j}$ of nonempty subsets of $S$ by $A_{i}^{j}$. If $A_{1} = A_{2} = \cdots = A_{k} = A$, where $1 < k\leq n$ and $A_{1}, A_{2}, \ldots, A_{k}$, $A$ are nonempty subsets of $S$, then we write $\overset{(k)}{A}$ instead of $A_{1}^{k}$.

Let $x_{1}^{2n-1}\in S$. The associative law ^[10] for the $n$-ary operation $f$ on $S$ is defined by for all $1\leq i < j\leq n$,

If this law holds for all elements $x_{1}^{2n-1}\in S$, an $n$-ary groupoid $\langle S; f\rangle$ is called an $n$-ary semigroup.

An $n$-ary semiring is an algebra $\langle S; +, f\rangle$ type $(2, n)$ for which $\langle S; +\rangle$ is a semigroup, $\langle S; f\rangle$ is an $n$-ary semigroup and for all $x_1^{n}, a, b\in S$, $1\leq i\leq n$,

Indeed, an $n$-ary semiring is a $(2, n)$-semiring ^[3]. An $n$-ary semiring $\langle S; +, f\rangle$ is said to be additively commutative if $a+b = b+a$ for all $a, b\in S$.

In this work, we simply write $S$ instead of an additively commutative $n$-ary semiring $\langle S; +, f\rangle$.

For any nonempty subsets $A$, $B$, $A_{1}^{n}$ of an $n$-ary semiring $S$, we denote

and

A nonempty subset $T$ of an $n$-ary semiring $S$ is called a subalgebra of $S$ if $T+T\subseteq T$ and $f(\overset{(n)}{T})\subseteq T$.

Definition 1. ^[3] Let $1\leq i\leq n$. A nonempty subset $A$ of an $n$-ary semiring $S$ is called an $i$-ideal of $S$ if $A+A\subseteq A$ and $f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S}) \subseteq A$. If $A$ is an $i$-ideal of $S$ for all $1\leq i\leq n,$ then $A$ is called an ideal of $S$.

An element $a$ of an $n$-ary semiring $S$ is called additively regular if $a = a+b+a$ for some $b\in S$. If in addition, the element $b$ is unique and satisfies $b = b+a+b$, then $b$ is called the additively inverse of $a$ in $S$ and will be denoted by the notation $a^{\prime}$. Particularly, if every element of $S$ is additively regular, then $S$ is called an additively regular $n$-ary semiring. Furthermore, if every additively regular element of $S$ has the unique additively inverse, then $S$ is called an additively inverse $n$-ary semiring.

Let $S$ be an additively inverse $n$-ary semiring. It is obvious that $x = (x^{\prime})^{\prime}$ and $(x+y)^{\prime} = x^{\prime}+y^{\prime}$ for all $x, y\in S$.

Lemma 1. Let $S$ be an additively inverse $n$-ary semiring. Then for any $x_{1}^{n}\in S, \; (f(x_{1}^{n}))^{\prime} = f(x_{1}^{i-1}, x_{i}^{\prime}, x_{i+1}^{n}),$ for all $1\leq i\leq n$.

Proof. Let $x_{1}^{n}\in S$ and $1\leq i\leq n$. Since

and

we obtain that

An element $x$ of an $n$-ary semiring $S$ is called additively idempotent if $x+x = x$. We define the set of all additively idempotent elements of $S$ by $E^{+} = \{x\in S\mid x+x = x\}.$ It is not difficult to verify that $E^{+}$ is an ideal of $S$.

A partially ordered set $(L, \prec)$ is said to be a lattice if every pair of elements $a, b$ of $L$ has both greatest lower bound and least upper bound. If every subset $A$ of a lattice $L$ has both greatest lower bound and least upper bound, then $L$ is called a complete lattice. It is not difficult to show that a partially ordered set $(L, \prec)$ has the greatest element and every subset of $L$ has the greatest lower bound if and only if $L$ is a complete lattice.

A lattice $L$ is called modular ^[17] if $L$ satisfies the following law; for all $a, b\in L, \; a\leq b$ implies $a\vee (x\wedge b) = (a\vee x)\wedge b,$ for every $x\in L,$ where $x\vee y$ and $x\wedge y$ is the least upper bound and the greatest lower bound of $x, y\in L$, respectively.

Lemma 2. ^[17] A lattice $L$ is modular if and only if for any $a, b, c\in L,$ $a\wedge b = a\wedge c,$ $a\vee b = a\vee c,$ and $b\leq c$ implies $b = c.$

3.   $k$-ideals and $h$-ideals of $n$-ary semirings

In this section, we introduce the notions of $k$-ideals and $h$-ideals of $n$-ary semirings and study some of their properties.

Definition 2. A nonempty subset $A$ of an $n$-ary semiring $S$ is called a $k$-ideal of $S$ if $A + A\subseteq A, \; f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})\subseteq A$ for all $1\leq i\leq n$ and the following condition is satisfied: for any $x\in S, \; x+a = b$ for some $a, b\in A$ implies $x\in A$. If $A$ is a $k$-ideal of $S$ and $E^{+}\subseteq A,$ then $A$ is said to be a full $k$-ideal.

According to Definition 2, it is clear that every $k$-ideal of an $n$-ary semiring is an ideal. However, the converse is not generally true as the following example shows.

Example 1. Define an $n$-ary operation $f$ on $\mathbb{N}$ by $f(a_{1}^{n}) = a_{1}\cdot a_{2}\cdot a_{3}\cdots a_{n}$ for any $a_{1}^{n}\in\mathbb{N}$. Then $\langle\mathbb{N}; \max, f\rangle$ is an $n$-ary semiring. We have that $2\mathbb{N}$ is an ideal of $\langle\mathbb{N}; \max, f\rangle$ but not a $k$-ideal because $\max\{1, 2\} = 2$ but $1\notin 2\mathbb{N}$.

The following example is an example of a $k$-ideal of an $n$-ary semiring which is not a full $k$-ideal.

Example 2. Define an $n$-ary operation $f$ on $\mathbb{N}$ by $f(a_{1}^{n}) = \min\{a_{1}, a_{2}, a_{3}\ldots, a_{n}\}$ for any $a_{1}^{n}\in\mathbb{N}$. Then $\langle\mathbb{N}; \max, f\rangle$ is an $n$-ary semiring and $E^{+} = \mathbb{N}$. It is easy to obtain that the set $\mathbb{I}_{m} = \{1, 2, 3, \ldots, m\}$ is a $k$-ideal of $\langle\mathbb{N}; \max, f\rangle$ but not a full $k$-ideal because $E^{+}\not\subseteq\mathbb{I}_{m}$.

The following example is an example of a $k$-ideal of a finite $n$-ary semiring which is not a full $k$-ideal.

Example 3. Let $S = \{a, b\}$. Then $\langle P(S); \cup, f\rangle$ is an $n$-ary semiring where $P(S)$ is the power set of $S$ and $f$ is the $n$-ary operation on $P(S)$ defined by $f(A_{1}^{n}) = \bigcap_{i = 1}^{n} A_{i}$ for any $A_{i}\in P(S)$. It is easy to show that $\{\emptyset, \{a\}\}$ is a $k$-ideal of $\langle P(S); \cup, f\rangle$ but not full because $E^{+} = P(S) \not\subseteq \{\emptyset, \{a\}\}$.

We give an example of a proper full $k$-ideal of an $n$-ary semiring as follows.

Example 4. Consider the $n$-ary semiring $\langle \mathbb{N}\cup\{0\}; +, f\rangle$ where $+$ is the usual addition and $f$ is the $n$-ary operation defined in Example 1. We have that the set of all additively idempotent elements of $\langle \mathbb{N}\cup\{0\}; +, f\rangle$ is $\{0\}$ and $2\mathbb{N}\cup\{0\}$ is a full $k$-ideal.

Remark 1. Let $\{A\}_{i\in I}$ be a family of full $k$-ideals of an $n$-ary semiring $S$. Then $\bigcap_{i\in I} A_{i}$ is a full $k$-ideal as well if it is not empty.

Remark 2. Every $k$-ideal of an additively inverse $n$-ary semiring $S$ is an additively inverse subalgebra of $S$.

Proof. Let $K$ be a $k$-ideal of $S$. Clearly, $K$ is a subalgebra of $S$. Let $a\in K$. Then $(a+a^{\prime})+a = a\in K$ and so $a+a^{\prime}\in K$. This implies that $a^{\prime}\in K$. Hence, $K$ is additively inverse.

The $k$-closure of a nonempty subset $A$ of an $n$-ary semiring $S$ is defined by

It is easy to prove that for any $\emptyset\neq A\subseteq S$, $A\subseteq [A]_{k}$ if $A+A\subseteq A$. Furthermore, if $A$ is closed under the addition, then $[A]_{k}$ is also closed. Now, we give some necessary properties of $k$-closure of nonempty subsets of an $n$-ary semiring as follows.

Lemma 3. Let $A, B,$ and $A_{1}^{n}$ be nonempty subsets of an $n$-ary semiring $S$. Then the following statements hold:

(1) if $A+A\subseteq A,$ then $[A]_{k} = [[A]_{k}]_{k};$

(2) if $A\subseteq B,$ then $[A]_{k}\subseteq [B]_{k};$

(3) $[A]_{k}+[B]_{k}\subseteq [A+B]_{k};$

(4) if $A_{1}^{n}$ are closed under the addition, then $f(A_{i}^{i-1}, [A_{i}]_{k}, A_{i+1}^{n})\subseteq [f(A_{1}^{n})]_{k}$ for all $1\leq i\leq n$.

Proof. (1) Let $\emptyset\neq A\subseteq S$ be such that $A+A\subseteq A$. Obviously, $[A]_{k}\subseteq [[A]_{k}]_{k}$. If $x\in [[A]_{k}]_{k}$, then $x+y = z$ for some $y, z\in [A]_{k}$ such that $y+a_{1} = b_{1}$ and $z+a_{2} = b_{2}$ for some $a_{1}, a_{2}, b_{1}, b_{2}\in A$. Then

We have $y+a_{1}+a_{2} = b_{1}+a_{2}\in A+A\subseteq A$ and $b_{2}+a_{1}\in A+A\subseteq A$. Using (3.1), we get $x\in [A]_{k}$ and so $[[A]_{k}]_{k}\subseteq [A]_{k}$.

(2)–(4) are straightforward.

Lemma 4. If $A$ is an ideal of an $n$-ary semiring $S,$ then $[A]_{k}$ is a $k$-ideal of $S$.

Proof. Let $A$ be an ideal of $S$. It is clear that $[A]_{k}$ is closed under the addition. Using $A$ being an ideal of $S$ and Lemma 3(2) and (4), we obtain that $f(\overset{(i-1)}{S}, [A]_{k}, \overset{(n-i)}{S}) \subseteq [f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})]_{k} \subseteq [A]_{k}$. If $x\in S$ is such that $x+a = b$ for some $a, b\in [A]_{k}$, then by Lemma 3(1), we get $x\in [[A]_{k}]_{k} = [A]_{k}$. Therefore, $[A]_{k}$ is a $k$-ideal of $S$.

The following corollary is directly obtained by Lemma 4.

Corollary 1. Let $S$ be an $n$-ary semiring. The following statements hold:

(1) an ideal $A$ of $S$ is a $k$-ideal if and only if $A = [A]_{k};$

(2) $[E^{+}]_{k}$ is a full $k$-ideal of $S$.

Lemma 5. Let $A$ and $B$ be two full $k$-ideals of an additively inverse $n$-ary semiring $S$. Then $[A+B]_{k}$ is a full $k$-ideal of $S$ such that $A\subseteq [A+B]_{k}$ and $B\subseteq [A+B]_{k}$.

Proof. Clearly, $A+B$ is closed under the addition. It holds that

for all $1\leq i\leq n$. Now, $A+B$ is an ideal of $S$. Using Lemma 4, we immediately get that $[A+B]_{k}$ is a $k$-ideal. Since $E^{+}\subseteq A$ and $E^{+}\subseteq B$, $E^{+} = E^{+}+E^{+}\subseteq A+B\subseteq [A+B]_{k}$. Hence, $[A+B]_{k}$ is a full $k$-ideal of $S$.

Let $a\in A$. Then

Hence, $A\subseteq [A+B]_{k}$. Similarly, we are able to get that $B\subseteq [A+B]_{k}$.

Theorem 1. Let $K(S)$ be the set of all full $k$-ideals of an additively inverse $n$-ary semiring $S$. Then $K(S)$ is a complete lattice which is also modular.

Proof. We have that $K(S)$ is a partially ordered set with respect to usual set inclusion. Let $A, B\in K(S)$. By Remark 1 and Lemma 5, we obtain that $A\cap B\in K(S)$ and $[A+B]_{k}\in K(S)$, respectively. Define $A\wedge B = A\cap B$ and $A\vee B = [A+B]_{k}$. Obviously, $A\cap B$ is the greatest lower bound of $A$ and $B$. Let $C\in K(S)$ such that $A\subseteq C$ and $B\subseteq C$. Then $A + B\subseteq C + C\subseteq C$. By Remark 3(2) and Corollary 1(1), we get $[A+B]_{k}\subseteq [C]_{k} = C$. Hence, $[A+B]_{k}$ is the least upper bound of $A$ and $B$. Now, $K(S)$ is a lattice. Clearly, $S$ is the greatest element of $K(S)$. Let $\{C_i\}_{i\in I}$ be a family of elements in $K(S)$. By Remark 1, we get that $\bigcap\{C_i\}_{i\in I}\in K(S)$. These imply that $K(S)$ is a complete lattice.

Finally, let $A, B, C\in K(S)$ such that

Let $x\in C$. Then $x\in C\subseteq A\vee C = A\vee B = [A+B]_{k}$. It follows that there exist $a_{1}, a_{2}\in A$ and $b_{1}, b_{2}\in B$ such that $x + a_{1} + b_{1} = a_{2}+ b_{2}$. Then

Now, $x\in C$, $a_{1} + a_{1}^{\prime}\in E^{+}\subseteq C$ and $b_{1}, b_{2}\in B\subseteq C$. Using (3.2), $a_{2} + a_{1}^{\prime}\in [C]_{k} = C$. At this point, $a_{1} + a_{1}^{\prime}, a_{2} + a_{1}^{\prime} \in A\cap C = A\wedge C = A\wedge B = A\cap B\subseteq B$. It follows that $a_{1} + a_{1}^{\prime} + b_{1}\in B$ and $a_{2} + a_{1}^{\prime} + b_{2}\in B$. Using (3.2) again, we obtain that $x\in [B]_{k} = B$ and so $C\subseteq B$. Hence, $B = C$. By Lemma 2, $K(S)$ is a modular lattice.

Now, we introduce a more restricted class of $k$-ideals of an $n$-ary semiring as follows.

Definition 3. A nonempty subset $A$ of an $n$-ary semiring $S$ is called an $h$-ideal of $S$ if $A+A\subseteq A, \; f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})\subseteq A$ for all $1\leq i\leq n$ and the following condition is satisfied: for any $x\in S, \; x + a + s = b + s$ for some $a, b\in A$ and $s\in S$ implies $x\in A$.

It is unnecessary to define a full $h$-ideal of an $n$-ary semiring because every $h$-ideal is immediately full, i.e., if $A$ is an $h$-ideal of $S$ and $x\in E^{+}$, then for any $a\in A$, $x + a + x = a + x$ implies $x\in A$.

It is obvious that every $h$-ideal of an $n$-ary semiring is a $k$-ideal. In general, the converse is not true as it is shown by the following example.

Example 5. Let $S = \{a, b, c\}$.

Define an $n$-ary operation $f$ on the power set $P(S)$ of $S$ by $f(A_{1}^{n}) = \bigcap_{i = 1}^{n}A_{i}$ for any $A_{1}^{n}\in P(S)$. Then $\langle P(S); \cup, f\rangle$ is an $n$-ary semiring. We have that $T = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$ is a $k$-ideal of $\langle P(S); \cup, f\rangle$. However, $T$ is not an $h$-ideal because $\{c\}\cup \{a, b\}\cup \{a, c\} = S = \{b\}\cup \{a, c\}$ where $\{a, b\}, \{b\}\in T$ but $\{c\}\notin T$.

Remark 3. Let $\{A\}_{i\in I}$ be a family of $h$-ideals of an $n$-ary semiring $S$. Then $\bigcap_{i\in I} A_{i}$ is an $h$-ideal as well if it is not empty.

Remark 4. Every $h$-ideal of an additively inverse $n$-ary semiring $S$ is an additively inverse subalgebra of $S$.

Proof. Let $H$ be an $h$-ideal of $S$. Clearly, $H$ is a subalgebra of $S$. Let $a\in H$. Then $(a + a^{\prime}) + a + s = a + s$ for all $s\in S$. So, $a + a^{\prime} \in H$. This means that $a^{\prime} + a = b$ for some $b\in H$ and thus $a^{\prime} + a + t = b + t$ for any $t\in S$. This implies that $a^{\prime}\in H$. Hence, $H$ is additively inverse.

The $h$-closure of a nonempty subset $A$ of an $n$-ary semiring $S$ is defined by

It is obvious that $[A]_{k}\subseteq [A]_{h}$ for any $\emptyset\neq A\subseteq S$. Moreover, it is not difficult to verify that for any $\emptyset\neq A\subseteq S$, $A\subseteq [A]_{h}$ if $A+ A \subseteq A$. Furthermore, if $A$ is closed under the addition, then $[A]_{h}$ is also closed. Now, we give some necessary properties of $h$-closure of nonempty subsets on an $n$-ary semiring as follows.

Lemma 6. Let $A, B$ and $A_{1}^{n}$ be nonempty subsets of an $n$-ary semiring $S$. Then the following statements hold:

(1) if $A + A\subseteq A,$ then $[A]_{h} = [[A]_{h}]_{h};$

(2) if $A\subseteq B,$ then $[A]_{h} \subseteq [B]_{h};$

(3) $[A]_{h} + [B]_{h}\subseteq [A+B]_{h};$

(4) if $A_{1}^{n}$ are closed under the addition, then $f(A_{i}^{i-1}, [A_{i}]_{h}, A_{i+1}^{n}) \subseteq [f(A_{1}^{n})]_{h}$ for all $1\leq i\leq n$.

Proof. (1) Let $\emptyset\neq A\subseteq S$ be such that $A + A\subseteq A$. Obviously, $[A]_{h}\subseteq [[A]_{h}]_{h}$. If $x\in [[A]_{h}]_{h}$, then $x + y + s = z + s$ for some $y, z\in [A]_{h}$ and $s\in S$ where $y + a_{1} + u = b_{1} + u$ and $z + a_{2} + v = b_{2} + v$ for some $a_{1}, a_{2}, b_{1}, b_{2}\in A$ and $u, v\in S$. Then

Using (3.3) and (3.4), we get that $x + (b_{1} + a_{2}) + u + s + v = (a_{1} + b_{2}) + u + s + v$ where $b_{1} + a_{2}, a_{1} + b_{2}\in A + A \subseteq A$ and $u + s + v\in S$ implies $x\in [A]_{h}$ and so $[[A]_{h}]_{h}\subseteq [A]_{h}$.

(2)–(4) are straightforward.

Lemma 7. If $A$ is an ideal of an $n$-ary semiring $S,$ then $[A]_{h}$ is an $h$-ideal of $S$.

Proof. Let $A$ be an ideal of $S$. Clearly, $[A]_{h}$ is closed under the addition. Using $A$ being an ideal of $S$ and Lemma 6(2) and (4), we obtain that $f(\overset{(i-1)}{S}, [A]_{h}, \overset{(n-i)}{S}) \subseteq [f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})]_{h} \subseteq [A]_{h}$. If $x\in S$ is such that $x + a + s = b + s$ for some $a, b\in [A]_{h}$ and $s\in S$, then by Lemma 6(1), we get $x\in [[A]_{h}]_{h} = [A]_{h}$. Therefore, $[A]_{h}$ is an $h$-ideal of $S$.

The following corollary is directly obtained by Lemma 7.

Corollary 2. Let $S$ be an $n$-ary semiring. The following statements hold:

(1) an ideal $A$ of $S$ is an $h$-ideal if and only if $A = [A]_{h};$

(2) $[E^{+}]_{h}$ is an $h$-ideal of $S$.

Lemma 8. Let $A$ and $B$ be two $h$-ideals of an additively inverse $n$-ary semiring $S$. Then $[A+B]_{h}$ is an $h$-ideal of $S$ such that $A\subseteq [A+B]_{h}$ and $B\subseteq [A+B]_{h}$.

Proof. Since $f(\overset{(i-1)}{S}, A + B, \overset{(n-i)}{S}) \subseteq f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S}) + f(\overset{(i-1)}{S}, B, \overset{(n-i)}{S}) \subseteq A + B$ for all $1\leq i\leq n$ and $A + B$ is closed under the addition, we get $A+B$ is an ideal of $S$. Using Lemma 7, we obtain that $[A+B]_{h}$ is an $h$-ideal. Let $a\in A$. Then $a = a + a^{\prime} + a = a + (a^{\prime} + a)\in A + E^{+} \subseteq A + B \subseteq [A+B]_{h}$. Hence, $A\subseteq [A+B]_{h}$. Similarly, we are able to get that $B\subseteq [A+B]_{h}$.

Theorem 2. Let $H(S)$ be the set of all $h$-ideals of an additively inverse $n$-ary semiring $S$. Then $H(S)$ is a complete lattice which is also modular.

Proof. We have that $H(S)$ is a partially ordered set with respect to the usual set inclusion. Let $A, B\in H(S)$. By Remark 3 and Lemma 8, we obtain that $A\cap B\in H(S)$ and $[A+B]_{h}\in H(S)$, respectively. Define $A\wedge B = A\cap B$ and $A\vee B = [A+B]_{h}$. Obviously, $A\cap B$ is the greatest lower bound of $A$ and $B$. Let $C\in H(S)$ such that $A\subseteq C$ and $B\subseteq C$. Then $A + B\subseteq C + C\subseteq C$. By Remark 6(2) and Corollary 2(1), we get $[A+B]_{h}\subseteq [C]_{h} = C$. Hence, $[A+B]_{h}$ is the least upper bound of $A$ and $B$. Now, $H(S)$ is a lattice. Clearly, $S$ is the greatest element of $H(S)$. Let $\{C_i\}_{i\in I}$ be a family of elements of $H(S)$. By Remark 3, we obtain that $\bigcap\{C_i\}\in H(S)$. These imply that $H(S)$ is a complete lattice.

Finally, let $A, B, C\in H(S)$ such that

Let $x\in C$. Then $x\in C\subseteq A\vee C = A\vee B = [A+B]_{h}$. It follows that there exist $a_{1}, a_{2}\in A$, $b_{1}, b_{2}\in B$ and $s\in S$ such that $x + a_{1} + b_{1} + s = a_{2} + b_{2} + s$. Then

Since, $x\in C$, $a_{1} + a_{1}^{\prime}\in E^{+}\subseteq C$ and $b_{1}\in B\subseteq C$, we have $x + a_{1} + a_{1}^{\prime} + b_{1}\in C$. Using (3.5) and $b_{2}\in B\subseteq C$, we get $a_{2} + a_{1}^{\prime}\in [C]_{h} = C$. At this point, $a_{1} + a_{1}^{\prime}, a_{2} + a_{1}^{\prime}\in A\cap C = A\wedge C = A\wedge B = A\cap B\subseteq B$. It follows that $a_{1} + a_{1}^{\prime}+b_{1}\in B$ and $a_{2} + a_{1}^{\prime} + b_{2}\in B$. Using (3.5) again, we obtain that $x\in [B]_{h} = B$ and so $C\subseteq B$. Hence, $B = C$. By Lemma 2, $H(S)$ is a modular lattice.

4.   $n$-ary ring congruences

In this section, we characterize an $n$-ary ring congruence with respect to a full $k$-ideal of an additively inverse $n$-ary semiring.

Definition 4. A binary relation $\rho$ on an $n$-ary semigroup $\langle S; f\rangle$ is said to be a congruence if $\rho$ is an equivalence relation and satisfies the following property; for any $a_{1}^{n}, \; b_{1}^{n}\in S,$

Lemma 9. An equivalence relation $\rho$ on an $n$-ary semigroup $\langle S; f\rangle$ is a congruence if and only if for any $a, b, x_{1}^{n}\in S,$

for each $1\leq i\leq n$.

An equivalence relation $\rho$ on an $n$-ary semiring $\langle S; +, f\rangle$ is a congruence if $\rho$ is a congruence on $\langle S; + \rangle$ and $\langle S; f\rangle$.

Definition 5. An $n$-ary semiring $\langle S; +, f\rangle$ is called an $n$-ary ring if $\langle S; +\rangle$ is a group. In other words, the following conditions are satisfied;

(1) there exists $0\in S$ such that $x + 0 = x = 0 + x$ for all $x\in S;$

(2) for each $x\in S,$ there is $y\in S$ such that $x + y = 0 = y + x$.

If $\langle S; +, f\rangle$ is an $n$-ary ring, then the element $y$ in the condition 2 is usually denoted by $-x$.

Definition 6. A congruence $\rho$ on an $n$-ary semiring $S$ is called an $n$-ary ring congruence if the quotient $n$-ary semiring $S/\rho: = \{[a]_\rho \mid a\in S\}$ is an $n$-ary ring.

Theorem 3. Let $A$ be a full $k$-ideal of an additively inverse $n$-ary semiring $S$. Then the relation $\rho_{A} = \{(a, b) \in S\times S\mid a+b^{\prime}\in A\}$ is an $n$-ary ring congruence such that $-[a]_{\rho_{A}} = [a^{\prime}]_{\rho_{A}}$.

Proof. Let $A$ be a full $k$-ideal of $S$.

Firstly, we show that $\rho$ is an equivalence relation on $S$. Let $a, b, c\in S$. Since $a + a^{\prime}\in E^{+} \subseteq A, \; \rho_{A}$ is reflexive. If $a + b^{\prime}\in A$, then by Remark 2, we get $b + a^{\prime} = (b^{\prime})^{\prime} + a^{\prime} = (a + b^{\prime})^{\prime}\in A$ and so $\rho_{A}$ is symmetric. Assume that $a + b^{\prime}\in A$ and $b + c^{\prime}\in A$. Then $a + c^{\prime} + b + b^{\prime}\in A$. Since $b + b^{\prime}\in E^{+}\subseteq A$, $a + c^{\prime}\in [A]_{k} = A$. So, $\rho_{A}$ is transitive. Now, $\rho_{A}$ is an equivalence relation.

Secondly, let $a, b, c, x_{1}^{n}\in S$. Assume that $(a, b)\in\rho_{A}$. Then $a + b^{\prime}\in A$ and so

Hence, $(a + c, b + c)\in\rho_{A}$. Using Lemma 1, we obtain that for each $1\leq i\leq n$,

Hence, $(f(x_{1}^{i-1}, a, x_{i+1}^{n}), f(x_{1}^{i-1}, b, x_{i+1}^{n}))\in\rho_{A}$. Therefore, we obtain that $\rho_{A}$ is a congruence on $S$.

Finally, we show that $S/\rho_{A}$ is an $n$-ary ring together with the operations $\oplus$ and $F$ on $S/\rho_{A}$ defined by

for any $a, b, a_{1}^{n}\in S$. It is immediately to obtain that $\langle S/\rho_{A}; \oplus, F\rangle$ is a quotient $n$-ary semiring of $\langle S; +, f\rangle$. Let $e\in E^{+}$ and $x\in S$. Then $(e + x) + x^{\prime} = e + (x + x^{\prime})\in E^{+} + E^{+} = E^{+}\subseteq A$ and so $(e + x, x) \in\rho_{A}$. It follows that

Since $e + (x + x^{\prime})^{\prime} = e + x^{\prime} + x \in A$, $(e, x + x^{\prime}) \in\rho_{A}$. It turns out that

Therefore, $S/\rho_{A}$ is an $n$-ary ring.

Theorem 4. Let $\rho$ be a congruence on an additively inverse $n$-ary semiring $S$ such that $S/\rho$ is an $n$-ary ring and $-[a]_{\rho} = [a^{\prime}]_{\rho}$. Then there exists a full $k$-ideal $A$ of $S$ such that $\rho_{A} = \rho$.

Proof. Let $A = \{a\in S\mid (a, e) \in \rho\text{ for some } e \in E^{+}\}$. Using the reflexivity of $\rho$, we get $E^{+} \subseteq A \neq \emptyset$. Let $a, b\in A$. Then there exist $e, f\in E^{+}$ such that $(a, e)\in\rho$ and $(b, f)\in\rho$. Then $(a + b, e + f) \in \rho$ and $e + f\in E^{+}$. Hence, $a + b\in A$ and thus $A + A\subseteq A$. Let $x\in f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})$ for each $1\leq i\leq n$. Then $x = f(x_{1}^{i-1}, c, x_{i+1}^{n})$ for some $x_{1}^{n}\in S$ and $c\in A$ such that $(c, g)\in\rho$ for some $g\in E^{+}$. It follows that $(x, f(x_{1}^{i-1}, g, x_{i+1}^{n})) = (f(x_{1}^{i-1}, c, x_{i+1}^{n}), f(x_{1}^{i-1}, g, x_{i+1}^{n})) \in\rho$. Since $E^{+}$ is an ideal of $S$, $f(x_{1}^{i-1}, g, x_{i+1}^{n}) + f(x_{1}^{i-1}, g, x_{i+1}^{n}) = f(x_{1}^{i-1}, g + g, x_{i+1}^{n}) \in f(\overset{(i-1)}{S}, E^{+}, \overset{(n-i)}{S}) \subseteq E^{+}$. So, $x\in A$ leads to $f(\overset{(i-1)}{S}, A, \overset{(n-i)}{S})\subseteq A$. Now, $A$ is an ideal of $S$.

Let $x\in [A]_{k}$. Then $x + a = b$ for some $a, b\in A$ where $(a, e)\in\rho$ and $(b, f)\in\rho$ for some $e, f\in E^{+}$. However, $[f]_{\rho}$ and $[e]_{\rho}$ are additively idempotent in the ring $S/\rho$. This obtains that $[e]_{\rho} = [f]_{\rho}$ is the zero element of $S/\rho$. It follows that $[f]_{\rho} = [b]_{\rho} = [x+a]_{\rho} = [x]_{\rho} \oplus [a]_{\rho} = [x]_{\rho} \oplus [e]_{\rho} = [x]_{\rho}$. Thus, $x\in A$ and so $[A]_{k} = A$. By Corollary 1(1), $A$ is a full $k$-ideal of $S$.

Finally, we show that $\rho = \rho_{A}$. Let $(a, b) \in \rho$. Then $(a + b^{\prime}, b + b^{\prime}) \in \rho$. Since $b + b^{\prime}\in E^{+}$, $a + b^{\prime}\in A$ and thus $(a, b)\in\rho_{A}$. Hence, $\rho\subseteq\rho_{A}$. If $(a, b)\in\rho_{A}$, then $a + b^{\prime}\in A$. Thus, $(a + b^{\prime}, e)\in\rho$ for some $e\in E^{+}$. We have that $[b]_{\rho} = [e]_{\rho} \oplus [b]_{\rho} = [a+b^{\prime}]_{\rho} \oplus [b]_{\rho} = [a]_{\rho} \oplus [b^{\prime}]_{\rho} \oplus [b]_{\rho} = [a]_{\rho} \oplus [b+b^{\prime}]_{\rho} = [a]_{\rho}$, since $b + b^{\prime}\in E^{+}$. This shows that $(a, b) \in \rho$ and so $\rho_{A} \subseteq \rho$. Therefore, $\rho = \rho_{A}$.

The concepts of full $k$-ideals and $h$-ideals of an additively inverse $n$-ary semiring are coincidence as the following remark.

Remark 5. If $S$ is an additively inverse $n$-ary semiring, then full $k$-ideals and $h$-ideals coincide in $S$.

Proof. Since every $h$-ideal is a full $k$-ideal, we show that every full $k$-ideal is an $h$-ideal. Let $A$ be a full $k$-ideal. Then $S/\rho_{A}$ is an $n$-ary ring and $A$ is its zero element by Theorem 3. Let $x\in S$ and $x + a + s = b + s$ for some $a, b\in A$ and $s\in S$. Then $[x]_{\rho} + [a]_{\rho} + [s]_{\rho} = [b]_{\rho} + [s]_{\rho}$. It follows that $[x]_{\rho} + 0 + [s]_{\rho} = 0 + [s]_{\rho}$ where $0$ is the zero of $S/\rho_{A}$. Hence, $[x]_{\rho} + [s]_{\rho} = [s]_{\rho}$ and so $[x]_{\rho} = 0$. It turns out that $x\in A$. Therefore, $A$ is an $h$-ideal.

5.   Discussion and conclusions

In algebraic structure point of view, we are able to say that an $n$-ary semiring is a generalization of a semiring and a ternary semiring, and any results on an $n$-ary semiring are also true on a semiring and a ternary semiring because a semiring is an $n$-ary semiring where $n = 2$ and a ternary semiring is an $n$-ary semiring where $n = 3$.

The notions of a $k$-ideal and a full $k$-ideal of an $n$-ary semiring were defined in Section 3. There is a $k$-ideal which is not full as it is shown by Example 2. However, every $h$-ideal of an $n$-ary semiring is immediately full. Moreover, $h$-ideals and full $k$-ideals are coincidence in an additively inverse $n$-ary semiring and the set of all of them forms a complete lattice and also a modular lattice.

A group (ring) congruence is such a congruence relation on a semigroup (semiring) that the quotient semigroup (semiring) is a group (ring). Similarly, an $n$-ary ring congruence is such a congruence relation on an $n$-ary semiring that the quotient $n$-ary semiring is an $n$-ary ring. Constructing a relation with respect to a full $k$-ideal of an additively inverse $n$-ary semiring is a way to obtain an $n$-ary ring congruence. Indeed, if $A$ is a full $k$-ideal of an additively inverse $n$-ary semiring $S$, then the relation $\rho_{A} = \{(a, b) \in S\times S\mid a+b^{\prime}\in A\}$ is an $n$-ary ring congruence where $b^{\prime}$ is the additively inverse of $b$. Conversely, if $\rho$ is an $n$-ary ring congruence on an additively inverse $n$-ary semiring $S$, then there exists a such full $k$-ideal $A$ of $S$ that $\rho = \rho_{A}$.

Acknowledgment

This research was supported by the Fundamental Fund of Khon Kaen University and has received funding support from the National Science, Research and Innovation Fund (NSRF).

Conflict of interest

All authors declare no conflicts of interest in this paper.