The concept of semiring is surprisingly successful in applications both in mathematics and computer science as it was shown by Golan in [6] and Kuich and Salomaa in [8]. In this paper we concentrate on the variety \({\mathcal C}\) of commutative multiplicatively idempotent semirings which was investigated by the authors already in [2,3,4] and in the subvariety of this variety satisfying \(x+x\approx x\). The concept of a Boolean semiring was introduced by Guzmán in [7]. It is a commutative multiplicatively idempotent semiring satisfying the identity \(1+x+x\approx 1\). The variety \({\mathcal B}\) of Boolean semirings contains just two subdirectly irreducible members, namely the two-element (distributive) lattice and the two-element Boolean ring. On the contrary, \({\mathcal C}\) has a proper class of subdirectly irreducible members, i. e. it is residually large, as it was proved by the authors in [2]. The natural question arises if by adding another elementary identity to the identities defining \({\mathcal C}\) the number of subdirectly irreducible members can be reduced. In our previous paper we studied members of \({\mathcal C}\) satisfying \(x+y+xyz\approx x+y\). In the present paper we investigate the subvariety \({\mathcal Z}\) of \({\mathcal C}\) determined by \(x+x\approx x\) and make frequent use of the identity \(x+y+xy\approx x+y\). That this identity indeed holds in \(\mathcal Z\) can be seen as follows:

$$\begin{aligned} x+y+xy\approx x^2+xy+yx+y^2\approx (x+y)^2\approx x+y. \end{aligned}$$

We show that \({\mathcal Z}\) is locally finite and we provide an explicit upper bound for the cardinalities of finite free algebras in \({\mathcal Z}\). Our main result is to prove that \({\mathcal Z}\) is generated by a single three-element subdirectly irreducible semiring. Moreover, it turns out that there are single semirings of arbitrary cardinality generating \({\mathcal Z}\).

For the reader’s convenience, all the concepts concerning semirings are included. What concerns concepts from Universal Algebra necessary for understanding the paper, the reader is referred to [1].

We start with the definition of a semiring in the sense of the monograph [6] by Golan.

FormalPara Definition 1

A semiring is an algebra \(\mathbf{S}=(S,+,\cdot ,0,1)\) of type (2, 2, 0, 0) such that

  • \((S,+,0)\) is a commutative monoid.

  • \((S,\cdot ,1)\) is a monoid.

  • The operation \(\cdot \) is distributive with respect to \(+\).

  • \(x0=0x=0\) for all \(x\in S\)

\(\mathbf{S}\) is called commutative if \(\cdot \) is commutative, additively idempotent if \(+\) is idempotent and multiplicatively idempotent if \(\cdot \) is idempotent. Let \({\mathcal Z}\) denote the variety of commutative additively and multiplicatively idempotent semirings.

The corresponding semiring of type (2, 2) has been called a \(\cdot \)-distributive bisemilattice by Romanowska and a bisemilattice by Pastijn and Zhao. McKenzie and Romanowska [9] showed that the variety of all bisemilattices has exactly 5 subvarieties. Based on this result, Ghosh et al. [5] and Pastijn [10] have shown that the variety of all semirings with type (2, 2) whose additive semigroup is a semilattice and multiplicative semigroup is a band has exactly 78 subvarieties. In this process, the identity \(x+y+xy\approx x+y\) plays a role.

Every bounded distributive lattice \(\mathbf{L}=(L,\vee ,\wedge ,0,1)\) is a semiring in \({\mathcal Z}\). On the contrary, non-trivial Boolean rings with unit 1 do not belong to \({\mathcal Z}\) since in every non-trivial Boolean ring with 1 we have \(1+1+1\cdot 1=1\ne 0=1+1\). In the following we present an example of a semiring in \({\mathcal Z}\) which is not a lattice since it does not satisfy the absorption law.

FormalPara Example 2

The algebra \(\mathbf{S}_3:=(S_3,+,\cdot ,0,1)\) with \(S_3=\{0,a,1\}\),

$$\begin{aligned} \begin{array}{c|ccc} + &{}\quad 0 &{}\quad a &{}\quad 1 \\ \hline 0 &{} \quad 0 &{}\quad a &{}\quad 1 \\ a &{}\quad a &{}\quad a &{}\quad a \\ 1 &{}\quad 1 &{}\quad a &{}\quad 1 \end{array} \quad \text{ and }\quad \begin{array}{c|ccc} \cdot &{}\quad 0 &{}\quad a &{}\quad 1 \\ \hline 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ a &{}\quad 0 &{}\quad a &{}\quad a \\ 1 &{}\quad 0 &{}\quad a &{}\quad 1 \end{array} \end{aligned}$$

belongs to \({\mathcal {Z}}\). Because of \(1(1+a)=a\ne 1\), \(\mathbf{S}_3\) is not a lattice. Let us remark that \(\mathbf{S}_3\) satisfies the so-called weak-absorption law \(x(x+y)\approx x+xy\).

It is easy to see that \(\mathbf{S}_3\) is the unique semiring in \({\mathcal Z}\) with cardinality less than 4 which is not a lattice.

Now we repeat the definitions of certain members of \({\mathcal Z}\) introduced in [2]:

FormalPara Definition 3

For every integer \(n>1\) let \(\mathbf S_n\) be the semiring with universe \(S_n=\{1,\dots ,n\}\) and operations defined as follows:

$$\begin{aligned} x+y:= {\left\{ \begin{array}{ll} \max (x,y) &{}\quad \text {for }x\text { and }y\text { are odd} \\ y &{}\quad \text {for }x\text { is odd and }y\text { is even} \\ x &{}\quad \text {for }x\text { is even and }y\text { is odd} \\ \min (x,y) &{}\quad \text {for }x\text { and }y\text { are even} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} xy:=\min (x,y) \end{aligned}$$

(\(x,y\in S_n\)). Here the operation symbols 0 and 1 are interpreted as 1 and n, respectively. Moreover, for every infinite cardinal k let \(\mathbf C_k=(C_k,\le ,0,1)\) be a bounded chain of cardinality k and \(\mathbf U_k\) be the semiring with universe \(U_k:=C_k\times \{1,2\}\) and operations defined as follows:

$$\begin{aligned} (x,i)+(y,j):= {\left\{ \begin{array}{ll} (\max (x,y),1) &{}\quad \text {for }(i,j)=(1,1) \\ (y,2) &{}\quad \text {for }(i,j)=(1,2) \\ (x,2) &{}\quad \text {for }(i,j)=(2,1) \\ (\min (x,y),2) &{}\quad \text {for }(i,j)=(2,2) \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} (x,i)(y,j):= {\left\{ \begin{array}{ll} (x,i) &{}\quad \text {for }x<y \\ (x,\min (i,j)) &{}\quad \text {for }x=y \\ (y,j) &{}\quad \text {for }x>y \end{array}\right. } \end{aligned}$$

(\((x,i),(y,j)\in U_k\)). Here the operation symbols 0 and 1 are interpreted as (0, 1) and (1, 2), respectively.

It is easy to see that the semiring \(\mathbf S_3\) of Example 2 is isomorphic to the semiring \(\mathbf S_3\) of Definition 3.

It was proved in [2] that \(\mathbf S_3\) is subdirectly irreducible. Moreover, in this paper it was shown that for every integer \(n>1\) the semiring \(\mathbf S_n\) is a subdirectly irreducible member of \({\mathcal Z}\) of cardinality n and that for every infinite cardinal k the semiring \(\mathbf U_k\) is a subdirectly irreducible member of \({\mathcal Z}\) of cardinality k. Hence \({\mathcal Z}\) is residually large.

We are going to derive a canonical form of terms in \({\mathcal Z}\).

In the following let n denote an arbitrary, but fixed non-negative integer and put \(N:=\{1,\ldots ,n\}\).

FormalPara Lemma 4

Every term \(t(x_1,\ldots ,x_n)\) in \({\mathcal Z}\) can be written in the form

$$\begin{aligned} t_A(x_1,\ldots ,x_n)=\sum _{I\in A}\prod _{s\in I}x_s\text { with }A\subseteq 2^N \end{aligned}$$
(1)

where the empty sum is defined as 0 and the empty product as 1.

FormalPara Proof

Let W denote the set of all sums of products of elements of \(\{x_1,\ldots ,x_n\}\). Obviously, the elements of W are terms in \(\mathcal Z\). It is easy to see that \(0,1\in W\) and that the sum of two elements of W again belongs to W. Using the distributive law one can see that also the product of two elements of W again belongs to W. Idempotency of addition and multiplication shows that the elements of W can be written in the form (1). \(\square \)

FormalPara Corollary 5

There exist only finitely many different terms in \({\mathcal Z}\) of fixed finite arity which means that \({\mathcal Z}\) is locally finite. The number of different n-ary terms in \({\mathcal Z}\) is at most \(2^{2^n}\).

In fact, one can show that \(\mathcal Z\) is locally finite by another method. Indeed, if \(\mathbf S\) is a semiring whose additive semigroup is a semilattice, then \(\mathbf S\) is locally finite if and only if \((S,\cdot )\) is locally finite. It is well-known that every semilattice is locally finite.

Within \({\mathcal Z}\) we can write terms in a more economic way.

FormalPara Definition 6

A subset A of \(2^N\) is called reduced if there does not exist an integer \(k>2\) and pairwise different elements \(I_1,\ldots ,I_k\) of A with \(I_1\cup \cdots \cup I_{k-1}=I_k\). Let \(\mathbb A\) denote the set of all reduced subsets of \(2^N\). The term \(t_A\) is called reduced if \(A\in \mathbb A\).

At first, we show that the identity \(x+y+xy\approx x+y\) can be extended to arbitrary products of variables as follows:

FormalPara Lemma 7

We have

$$\begin{aligned} {\mathcal Z}\models \prod _{i\in I}x_i+\prod _{i\in J}x_i\approx \prod _{i\in I}x_i+\prod _{i\in J}x_i+\prod _{i\in I\cup J}x_i \end{aligned}$$

for all \(I,J\subseteq N\).

FormalPara Proof
$$\begin{aligned} {\mathcal Z}\models \prod _{i\in I}x_i+\prod _{i\in J}x_i\approx \prod _{i\in I}x_i+\prod _{i\in J}x_i+\prod _{i\in I}x_i\prod _{i\in J}x_i\approx \prod _{i\in I}x_i+\prod _{i\in J}x_i+\prod _{i\in I\cup J}x_i \end{aligned}$$

\(\square \)

In the following we are going to show that every term in \({\mathcal Z}\) can be reduced.

FormalPara Theorem 8

To every \(A\subseteq 2^N\) there exists some \(B\in \mathbb A\) with \({\mathcal Z}\models t_A\approx t_B\).

FormalPara Proof

Let \(A\subseteq 2^N\). If \(A\in \mathbb A\) then we are done. Hence assume \(A\notin \mathbb A\). Then there exist an integer \(k>2\) and pairwise different elements \(I_1,\ldots ,I_k\) of A with \(I_1\cup \cdots \cup I_{k-1}=I_k\). Then by Lemma 7 we have

$$\begin{aligned} {\mathcal Z}\models & {} \prod _{i\in I_1}x_i+\cdots +\prod _{i\in I_k}x_i\approx \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\cdots +\prod _{i\in I_k}x_i\approx \\\approx & {} \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\prod _{i\in I_1\cup I_2\cup I_3}x_i+\prod _{i\in I_4}x_i+\cdots +\prod _{i\in I_k}x_i\approx \cdots \approx \\\approx & {} \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\cdots +\prod _{i\in I_1\cup \cdots \cup I_{k-2}}x_i+\prod _{i\in I_{k-1}}x_i+\prod _{i\in I_k}x_i\approx \\\approx & {} \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\cdots +\prod _{i\in I_1\cup \cdots \cup I_{k-2}}x_i+\prod _{i\in I_{k-1}}x_i\approx \cdots \approx \\\approx & {} \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\prod _{i\in I_1\cup I_2\cup I_3}x_i+\prod _{i\in I_4}x_i+\cdots +\prod _{i\in I_{k-1}}x_i\approx \\\approx & {} \prod _{i\in I_1}x_i+\prod _{i\in I_2}x_i+\prod _{i\in I_1\cup I_2}x_i+\prod _{i\in I_3}x_i+\cdots +\prod _{i\in I_{k-1}}x_i\approx \prod _{i\in I_1}x_i+\cdots +\prod _{i\in I_{k-1}}x_i \end{aligned}$$

and hence

$$\begin{aligned} {\mathcal Z}\models t_A\approx t_{A{\setminus }\{I_k\}}. \end{aligned}$$

Either \(A{\setminus }\{I_k\}\in \mathbb A\) or again one element of \(A{\setminus }\{I_k\}\) can be cancelled. After a finite number of steps one finally ends up with some \(B\in \mathbb A\) satisfying \({\mathcal Z}\models t_A\approx t_B\) (every subset of \(2^N\) having less that three elements automatically belongs to \(\mathbb A\)). \(\square \)

FormalPara Corollary 9

For \(n=0,1,2\) resp. 3 there are exactly 2, 4, 14 resp. 122 reduced n-ary terms in \({\mathcal Z}\).

FormalPara Proof

The proof is evident. \(\square \)

Now we state and prove our main theorem saying that \({\mathcal Z}\) is generated by \(\mathbf S_3\).

FormalPara Theorem 10

If \(A,B\in \mathbb A\) then \(\mathbf S_3\models t_A\approx t_B\) if and only if \(A=B\).

FormalPara Proof

Assume \(t_A\approx t_B\) and \(A\ne B\). Without loss of generality \(A{\setminus } B\ne \emptyset \). Let \(I\in A{\setminus } B\). Since \(I=\emptyset \) would imply \(1=t_A(0,\ldots ,0)=t_B(0,\ldots ,0)=0\) we have \(I\ne \emptyset \). Let \(i\in I\). Put

$$\begin{aligned} a_j:=\left\{ \begin{array}{ll} 0 &{}\quad \text {if }j\in N{\setminus } I \\ a &{}\quad \text {if }j=i \\ 1 &{}\quad \text {if }j\in I{\setminus }\{i\} \end{array} \right. \end{aligned}$$

Then \(t_A(a_1,\ldots ,a_n)=a\). Hence we have \(t_B(a_1,\ldots ,a_n)=a\). From this we conclude that there exists some \(J_i\in B\) with \(i\in J_i\subseteq I\). Since \(I\notin B\) we have \(J_i\subsetneqq I\). Now define

$$\begin{aligned} b_j:=\left\{ \begin{array}{ll} 0 &{}\quad \text {if }j\in N{\setminus } J_i \\ a &{}\quad \text {if }j=i \\ 1 &{}\quad \text {if }j\in J_i{\setminus }\{i\} \end{array} \right. \end{aligned}$$

Then \(t_B(b_1,\ldots ,b_n)=a\). Hence \(t_A(b_1,\ldots ,b_n)=a\) which shows that there exists some \(K_i\in A\) with \(i\in K_i\subseteq J_i\). Therefore \(i\in K_i\subsetneqq I\). Obviously,

$$\begin{aligned} I=\bigcup _{i\in I}K_i. \end{aligned}$$

Let \(K_1,\ldots ,K_s\) denote the pairwise distinct elements of \(\{K_i\mid i\in I\}\). Then \(s>1\), \(K_1,\ldots ,K_s,I\) are pairwise distinct and \(K_1\cup \cdots \cup K_s=I\). This contradicts \(A\in \mathbb A\). Hence \(A=B\). \(\square \)

FormalPara Remark 11

It follows from Theorem 10 that B in Theorem 8 is uniquely determined by A. So we could write \(B=f(A)\) in Theorem 8. For \(A,B\subseteq 2^N\) we then have \({\mathcal Z}\models t_A\approx t_B\) if and only if \(f(A)=f(B)\). Thus, the word problem in \({\mathcal Z}\) is solvable.

FormalPara Remark 12

From Corollary 9 and Theorem 10 we conclude that the free algebras in \({\mathcal Z}\) with 0, 1, 2 resp. 3 free generators have 2, 4, 14 resp. 122 elements.

From Theorem 10 we finally obtain

FormalPara Theorem 13

\({\mathcal Z}\) is generated by \(\mathbf S_3\). Moreover, \({\mathcal Z}\) is generated by every \(\mathbf S_n\) for each integer \(n>2\) and by every \(\mathbf U_k\) for each infinite cardinal k.

FormalPara Proof

Let \(A,B\subseteq 2^N\) and assume \(\mathbf S_3\models t_A\approx t_B\). According to Theorem 8 there exist \(C,D\in \mathbb A\) with \(\mathcal Z\models t_C\approx t_A\) and \(\mathcal Z\models t_D\approx t_B\). We conclude \(\mathbf S_3\models t_C\approx t_D\). According to Theorem 10 we obtain \(C=D\) and hence \(\mathcal Z\models t_A\approx t_C=t_D\approx t_B\). Therefore \(\mathbf S_3\) generates \(\mathcal Z\). If \(n>2\) is an integer then \(\mathbf S_n\) has a subalgebra being isomorphic to \(\mathbf S_3\), namely the subalgebra with base set \(\{1,2,n\}\). Hence, every identity holding in \(\mathbf S_n\) also holds in \({\mathcal Z}\). Analogously, if k is an infinite cardinal then \(\mathbf U_k\) has a subalgebra being isomorphic to \(\mathbf S_3\), namely the subalgebra with base set \(\{(0,1),(0,2),(1,2)\}\). Hence, every identity holding in \(\mathbf U_k\) also holds in \({\mathcal Z}\). This completes the proof of the theorem. \(\square \)

FormalPara Remark 14

As mentioned in Remark 12 and Corollary 9, the free algebra \(\mathbf F_{\mathcal Z}(x)\) with one free generator x in \({\mathcal Z}\) consists just of the four elements \(0,1,x,1+x\). It is easy to see that the equivalence relation \(\Theta \) with classes \(\{0\},\{x,1+x\},\{1\}\) is a congruence on \(\mathbf F_{\mathcal Z}(x)\) and \((\mathbf F_{\mathcal Z}(x))/\Theta \cong \mathbf S_3\). This shows that also \(\mathbf F_{\mathcal Z}(x)\) generates \({\mathcal Z}\).