The variety of commutative additively and multiplicatively idempotent semirings
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Abstract
The variety \({\mathcal Z}\) of commutative additively and multiplicatively idempotent semirings is studied. We prove that \({\mathcal Z}\) is generated by a single subdirectly irreducible threeelement semiring and it has a canonical form for its terms. Hence, \({\mathcal Z}\) is locally finite despite the fact that it is residually large. The word problem in \({\mathcal Z}\) is solvable.
Keywords
Semiring Commutative Additively idempotent Multiplicatively idempotent Variety Locally finite Residually large Word problemFor 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.
Definition 1

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

\((S,\cdot ,1)\) is a monoid.

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

\(x0=0x=0\) for all \(x\in S\)
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, nontrivial Boolean rings with unit 1 do not belong to \({\mathcal Z}\) since in every nontrivial 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.
Example 2
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]:
Definition 3
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 nonnegative integer and put \(N:=\{1,\ldots ,n\}\).
Lemma 4
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). \(\square \)
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 nary 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 wellknown that every semilattice is locally finite.
Within \({\mathcal Z}\) 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 \(I_1,\ldots ,I_k\) of A with \(I_1\cup \cdots \cup I_{k1}=I_k\). 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:
Lemma 7
Proof
In the following we are going to show that every term in \({\mathcal Z}\) can be reduced.
Theorem 8
To every \(A\subseteq 2^N\) there exists some \(B\in \mathbb A\) with \({\mathcal Z}\models t_A\approx t_B\).
Proof
Corollary 9
For \(n=0,1,2\) resp. 3 there are exactly 2, 4, 14 resp. 122 reduced nary terms in \({\mathcal Z}\).
Proof
The proof is evident. \(\square \)
Now we state and prove our main theorem saying that \({\mathcal Z}\) is generated by \(\mathbf S_3\).
Theorem 10
If \(A,B\in \mathbb A\) then \(\mathbf S_3\models t_A\approx t_B\) if and only if \(A=B\).
Proof
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.
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
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.
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. \(\square \)
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 \({\mathcal Z}\).
Notes
Acknowledgements
Open access funding provided by Austrian Science Fund (FWF). The authors are very grateful to an anonymous referee for valuable remarks improving the quality of the paper.
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