The variety of commutative additively and multiplicatively idempotent semirings

The variety $${\mathcal Z}$$Z of commutative additively and multiplicatively idempotent semirings is studied. We prove that $${\mathcal Z}$$Z is generated by a single subdirectly irreducible three-element semiring and it has a canonical form for its terms. Hence, $${\mathcal Z}$$Z is locally finite despite the fact that it is residually large. The word problem in $${\mathcal Z}$$Z is solvable.

the identity 1 + x + x ≈ 1. The variety 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, 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 C the number of subdirectly irreducible members can be reduced. In our previous paper we studied members of C satisfying x + y + x yz ≈ x + y. In the present paper we investigate the subvariety Z of C determined by x + x ≈ x and make frequent use of the identity x + y + x y ≈ x + y. That this identity indeed holds in Z can be seen as follows: x + y + x y ≈ x 2 + x y + yx + y 2 ≈ (x + y) 2 ≈ x + y.
We show that Z is locally finite and we provide an explicit upper bound for the cardinalities of finite free algebras in Z. Our main result is to prove that Z is generated by a single three-element subdirectly irreducible semiring. Moreover, it turns out that there are single semirings of arbitrary cardinality generating 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.
• The operation · is distributive with respect to +. • x0 = 0x = 0 for all x ∈ S S is called commutative if · is commutative, additively idempotent if + is idempotent and multiplicatively idempotent if · is idempotent. Let Z denote the variety of commutative additively and multiplicatively idempotent semirings.
The corresponding semiring of type (2, 2) has been called a ·-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 + x y ≈ x + y plays a role.
Every bounded distributive lattice L = (L , ∨, ∧, 0, 1) is a semiring in Z. On the contrary, non-trivial Boolean rings with unit 1 do not belong to Z since in every nontrivial Boolean ring with 1 we have 1 + 1 + 1 · 1 = 1 = 0 = 1 + 1. In the following we present an example of a semiring in Z which is not a lattice since it does not satisfy the absorption law.
It is easy to see that S 3 is the unique semiring in Z with cardinality less than 4 which is not a lattice. Now we repeat the definitions of certain members of Z introduced in [2]: Definition 3 For every integer n > 1 let S n be the semiring with universe S n = {1, . . . , n} and operations defined as follows: for x and y are odd y for x is odd and y is even x for x is even and y is odd min(x, y) for x and y are even and x y := min(x, y) (x, y ∈ S n ). Here the operation symbols 0 and 1 are interpreted as 1 and n, respectively. Moreover, for every infinite cardinal k let C k = (C k , ≤, 0, 1) be a bounded chain of cardinality k and U k be the semiring with universe U k := C k × {1, 2} and operations defined as follows: and Here the operation symbols 0 and 1 are interpreted as (0, 1) and (1, 2), respectively.
It is easy to see that the semiring S 3 of Example 2 is isomorphic to the semiring S 3 of Definition 3.
It was proved in [2] that S 3 is subdirectly irreducible. Moreover, in this paper it was shown that for every integer n > 1 the semiring S n is a subdirectly irreducible member of Z of cardinality n and that for every infinite cardinal k the semiring U k is a subdirectly irreducible member of Z of cardinality k. Hence Z is residually large.
We are going to derive a canonical form of terms in Z.
In the following let n denote an arbitrary, but fixed non-negative integer and put N := {1, . . . , n}. term t (x 1 , . . . , x n ) in Z can be written in the form

Lemma 4 Every
where the empty sum is defined as 0 and the empty product as 1.
Proof Let W denote the set of all sums of products of elements of {x 1 , . . . , x n }.
Obviously, the elements of W are terms in Z. It is easy to see that 0, 1 ∈ 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).

Corollary 5 There exist only finitely many different terms in Z of fixed finite arity which means that Z is locally finite. The number of different n-ary terms in Z is at most 2 2 n .
In fact, one can show that Z is locally finite by another method. Indeed, if S is a semiring whose additive semigroup is a semilattice, then S is locally finite if and only if (S, ·) is locally finite. It is well-known that every semilattice is locally finite.
Within Z we can write terms in a more economic way.

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 , . . . , I k of A with I 1 ∪ · · · ∪ I k−1 = I k . Let A denote the set of all reduced subsets of 2 N . The term t A is called reduced if A ∈ A.
At first, we show that the identity x + y + x y ≈ x + y can be extended to arbitrary products of variables as follows: In the following we are going to show that every term in Z can be reduced.

Theorem 8 To every
Proof Let A ⊆ 2 N . If A ∈ A then we are done. Hence assume A / ∈ A. Then there exist an integer k > 2 and pairwise different elements I 1 , . . . , I k of A with I 1 ∪· · ·∪ I k−1 = I k . Then by Lemma 7 we have Either A\{I k } ∈ A or again one element of A\{I k } can be cancelled. After a finite number of steps one finally ends up with some B ∈ A satisfying Z | t A ≈ t B (every subset of 2 N having less that three elements automatically belongs to A).
Proof The proof is evident. Now we state and prove our main theorem saying that Z is generated by S 3 .

Theorem 10 If A, B ∈ A then S 3 | t A ≈ t B if and only if
Then t A (a 1 , . . . , a n ) = a. Hence we have t B (a 1 , . . . , a n ) = a. From this we conclude that there exists some J i ∈ B with i ∈ J i ⊆ I . Since I / ∈ B we have J i I . Now define Then t B (b 1 , . . . , b n ) = a. Hence t A (b 1 , . . . , b n ) = a which shows that there exists some K i ∈ A with i ∈ K i ⊆ J i . Therefore i ∈ K i I . Obviously,  f (B). Thus, the word problem in Z is solvable.

From Theorem 10 we finally obtain
Theorem 13 Z is generated by S 3 . Moreover, Z is generated by every S n for each integer n > 2 and by every U k for each infinite cardinal k.
Proof Let A, B ⊆ 2 N and assume S 3 | t A ≈ t B . According to Theorem 8 there exist C, D ∈ A with Z | t C ≈ t A and Z | t D ≈ t B . We conclude S 3 | t C ≈ t D . According to Theorem 10 we obtain C = D and hence Z | t A ≈ t C = t D ≈ t B . Therefore S 3 generates Z. If n > 2 is an integer then S n has a subalgebra being isomorphic to S 3 , namely the subalgebra with base set {1, 2, n}. Hence, every identity holding in S n also holds in Z. Analogously, if k is an infinite cardinal then U k has a subalgebra being isomorphic to S 3 , namely the subalgebra with base set {(0, 1), (0, 2), (1, 2)}. Hence, every identity holding in U k also holds in Z. This completes the proof of the theorem.
Remark 14 As mentioned in Remark 12 and Corollary 9, the free algebra F Z (x) with one free generator x in Z consists just of the four elements 0, 1, x, 1 + x. It is easy to see that the equivalence relation with classes {0}, {x, 1 + x}, {1} is a congruence on F Z (x) and (F Z (x))/ ∼ = S 3 . This shows that also F Z (x) generates Z.