On some varieties of ai-semirings satisfying x^p+1 ≈ x

Lemma 2.1

IfSsatisfies the identityzxyz ≈ (zxzyz)^pzyxz(zxzyz)^p, thenSis a union of abelian groups which belong toA_p+1.

Lemma 2.2

SA_p+1 = Sℓ ∨ A_p+1.

Lemma 2.3

ROBAp+1∘satisfies the following identities

Proof

By (4), it follows immediately that (5), (6) and (7) hold in ROBAp+1∘. Also,

Thus, ROBAp+1∘. In the remainder we need only to prove that (8) holds in ROBAp+1∘. The following is derivable from the identities determining ROBAp+1∘ and identities (4), (5), (9):

Thus, ROBAp+1∘ satisfies the identity

In a left-right dual way we can show that ROBAp+1∘ satisfies the identity

Furthermore, we have

Thus, ROBAp+1∘ satisfies identity (8). □

Lemma 2.4

LetSbe an ai-semiring inSr(p + 1, 1). Then the following is true,

Proof

Suppose that S is an ai-semiring in Sr(p + 1, 1). Then for any a ∈ S, a^p+1 = a. Without loss of generality, suppose that i, j ∈ p and i < j. We have

This completes the proof. □

Lemma 3.1

LetSbe an ai-semiring inROBAp+1∘. ThenSis a member ofSr(2, 1) if and only if it does not contain a copy ofZp0.

Proof

The direct part is obvious. Conversely, suppose that S is an ai-semiring in ROBAp+1∘, but not in Sr(2, 1). Then there exists a ∈ S such that a is not equal to a^p. Thus by Lemma 2.4, we can show that {a, a², …, a^p, a + a²+ ⋯ +a^p} is a subsemiring of S and is a copy of Zp0. This completes the proof. □

Lemma 3.2

The interval [T, M_p] consists of the three varietiesT, MandM_p. □

Proof

Let V ∈ [T, M_p] and V ≠ T. Then M ⊆ V since M is minimal nontrivial subvariety of M_p. It follows that V ∈ [M, M_p] and so V = M or V = M_p. □

Let u = u₁+⋯+u_k, where u_i ∈ X⁺, i ∈ k. We put C(u) = ⋃_{i ∈ k}c(u_i). To solve the word problem of the ai-semiring Zp0 we need the following Lemmas, which are the special cases of [5: Corollary 4.15 and Lemma 5.1] when n = p+1.

Lemma 3.3

Let u ≈ u+q be anAI-identity, where u = u₁+⋯+u_k, u_i, q ∈ X⁺, i ∈ k. ThenZp0satisfies u ≈ u+q if and only if c(q) ⊆ C(u) and r_p+1(q) = r_p+1(u_i1⋯ u_i(p+1)ℓ) for some positive integer ℓ and some u_i1, …, u_i(p+1)ℓ ∈ {u_i | i ∈ k}.

Lemma 3.4

Let u ≈ u+q be anAI-identity, where u = u₁+⋯+u_k, u_i, q ∈ X⁺, i ∈ k. IfZp0satisfies u ≈ u+q, then there exists q₁in X⁺with r_p+1(q₁) = r_p+1(q) and c(q) ⊆ c(q₁) ⊆ C(u) such that u ≈ u+q₁is satisfied inROBAp+1∘.

For a semiring variety V, we denote by 𝓛(V) the lattice of all subvarieties of V. In the following we shall characterize the lattice 𝓛( ROBAp+1∘). Suppose that S be a member of ROBAp+1∘. It is shown in [20, Lemma 2.1] that E(S) forms a member of Sr(2, 1). Define a mapping φ as follows

Then it is easy to see that φ is subjective. For any V ∈ 𝓛( ROBAp+1∘), if S ∈ φ(V), then S ∈ V ∩ Sr(2, 1) and so S ∈ {E(S)|S ∈ V}. It follows that φ(V) ⊆ {E(S)|S ∈ V}. On the other hand, if E(S) ∈ {E(S)|S ∈ V}, then S ∈ V. Since E(S) is a subsemiring of S, we have that E(S) ∈ V and so E(S) ∈ V ∩ Sr(2, 1). It follows that {E(S)|S ∈ V} ⊆ φ(V). Therefore, φ(V) = {E(S)|S ∈ V}. We use t(x₁, …, x_n) to denote the AI-term t which contains no other variables than x₁, …, x_n (but not necessarily all of them). If W is the subvariety of Sr(2, 1) determined by the additional identities

then we denote by Wp^ the subvariety of ROBAp+1∘ determined by the additional identities

Then for any V ∈ [W, Ŵ_p],

it follows that V ∈ φ⁻¹(W) and so [W, Ŵ_p] ⊆ φ⁻¹(W). Conversely, if V ∈ φ⁻¹(W), then V ∈ [W, Ŵ_p]. Otherwise, φ(V) ≠ W, a contradiction. Thus, for any W ∈ 𝓛 (Sr(2, 1)), φ⁻¹(W) is the interval [W, Wp^] of 𝓛( ROBAp+1∘). Moreover, if W₁, W₂ ∈ 𝓛(Sr(2, 1)) such that W₁ ⊆ W₂, then it is easy to check that Ŵ₁ ⊆ Ŵ₂. Suppose that (V_i)_{i ∈ I} is a family of varieties in ROBAp+1∘. Then φ(V_i) ⊆ V_i ⊆ φ(Vi)^ for all i ∈ I. Furthermore,

This implies that φ(⋁_{i ∈ I}V_i) = ⋁_{i ∈ I}φ(V_i) and so φ is a complete ∨-epimorphism. Moreover, it is clear that φ is a complete ∧-epimorphism. It follows that φ is a complete epimorphism. Thus, 𝓛( ROBAp+1∘) = ⋃_{W ∈ 𝓛 (Sr(2, 1))}[W, Wp^].

Notice that M is the subvariety of Sr (2, 1) determined by the additional identities xy ≈ yx and x+y ≈ xy. Thus Mp^ is the subvariety of ROBAp+1∘ determined by the additional identities x^py^p ≈ y^px^p and x^p+y^p ≈ x^py^p. By [22, Lemma IV.2.3], ROBAp+1∘ ∩ [x^py^p ≈ y^px^p] = SAp+1∘ and so Mp^ = M_p.

Recall (see [1]) that the lattice 𝓛(Sr (2, 1)) can be divided into five intervals: [T, N ∩ P ∩ Sr (2, 1)], [D, N ∩ Sr (2, 1)], [M, P ∩ Sr (2, 1)], [D ∨ M, K ∩ Sr(2, 1)] and [Bi, Sr(2, 1)], where N ∩ Sr(2, 1) is the subvariety of Sr(2, 1) determined by the identity

P ∩ Sr(2, 1) is the subvariety of Sr(2, 1) determined by the identity

and K ∩ Sr(2, 1) is the subvariety of Sr(2, 1) determined by the identity

Thus we have

Theorem 3.5

LetV ∈ [T, N ∩ P ∩ Sr(2, 1)] ∪ [D, N ∩ Sr(2, 1)]. ThenVp^ = V.

Proof

From N ∩ Sr(2, 1) ⊨ (11), we have that N∩Sr(2,1)p^ satisfies the identity

Notice that Zp0 ⊭ (14). Then Zp0∉N∩Sr(2,1)p^ and so by Lemma 3.1, N∩Sr(2,1)p^ ⊆ Sr(2, 1). It follows from Vp^⊆N∩Sr(2,1)p^ that Vp^⊆Sr(2,1).□

Assume that V ∈ 𝓛(Sr(2, 1)) and V ⊨ u ≈ u+q, where u = u₁+⋯+u_k, u_i, q ∈ X⁺, i ∈ k. From ROBAp+1∘ ⊨ (5) it follows that Vp^ satisfies the identity

Lemma 3.6

LetV ∈ [M, P ∩ Sr(2, 1)]. ThenV ∨ HSP( Zp0) = Vp^.

Proof

Since P ∩ Sr(2, 1) ⊨ (12), it follows that P∩Sr(2,1)p^ satisfies the identity

We immediately have that this is the case for Vp^, too. Notice that HSP( Zp0) = Mp^. Thus V ∨ HSP( Zp0) ⊆ Vp^. It remains to show that every identity which is satisfied in V ∨ HSP( Zp0) can be deduced from the identities which hold in Vp^.

Let u ≈ u+q be an AI-identity which is satisfied in V ∨ HSP( Zp0), where u = u₁+⋯+u_k, u_i, q ∈ X⁺, i ∈ k. Since Zp0 satisfies this identity, by Lemma 3.4 there exists q₁ in X⁺ with r_p+1(q₁) = r_p+1(q) (and so r_i(q₁ = r_i(q) for all i ∈ p−1) and c(q) ⊆ c(q₁) such that u ≈ u+q₁ is satisfied in ROBAp+1∘. Thus the following are derivable from the identities which hold in Vp^:

This derives the identity

In a left-right dual way we have

Therefore the following are satisfied in Vp^:

Furthermore, we have the following are derivable from the identities which are satisfied in Vp^:

This shows that u ≈ u+q is satisfied in Vp^ and so V ∨ HSP( Zp0) = Vp^. □

By Lemma 3.1 and 3.6, we can establish the following result.

Theorem 3.7

LetV ∈ [M, P ∩ Sr(2, 1)]. Then the interval [V, Vp^] of 𝓛( ROBAp+1∘) consists of two varietiesVandV ∨ HSP( Zp0).

For an ai-semiring S we denote by S⁰ the ai-semiring obtained from S by adding an extra element 0, where a + 0 = a, a0 = 0a = 0 for every a ∈ S. Written B_p as Zp0, then we have

Lemma 3.8

Let S ∈ ROBAp+1∘. Then S satisfies the identity

if and only if it does not contain a copy ofBp0.

Proof

Necessary. It is obvious.

Sufficiency. Suppose that S ∈ ROBAp+1∘, but that S does not satisfy identity (17). Then there exist a, b ∈ S such that

It is easily verified that c, a+c, a²+c, …, a^p+c and a+a^p+c are not equal to each other, where c = (a+a^p)b^p(a+a^p). Put S₁ = {c, a+c, a²+c, …, a^p+c, a+a^p+c}. By identities (1) and (10), it is routine to verify that S₁ forms a subsemiring of S, which is a copy of Bp0. □

Let u = u₁+⋯+u_m, where u_i ∈ X⁺, i ∈ m, and Z ⊆ ⋃_{i ∈ m}c(u_i). If c(u_i) ∩ Z ≠ ∅ for every i, then we write D_Z(u) = ∅. Otherwise, D_Z(u) is the sum of terms u_i for which c(u_i) ∩ Z = ∅. The proof of the following result is easy and omitted.

Lemma 3.9

Let S ∈ ROBAp+1∘and u ≈ v be anAI-identity, where u = u₁+⋯+u_m, v = v₁+⋯+v_n, u_i, v_j ∈ X⁺, i ∈ m, j ∈ n. Then S⁰satisfies u ≈ v if and only if for every Z ⊆ C(u) ⋃ C(v), either D_Z(u) = D_Z(v) = ∅ orD_Z(u) ≠ ∅ ≠ D_Z(v) and D_Z(u) ≈ D_Z(v) is satisfied in S.

Lemma 3.10

HSP( Bp0) = SAp+1∘.

Proof

It is obvious that HSP( Bp0) ⊆ SAp+1∘. To prove that SAp+1∘ ⊆ HSP( Bp0), we need only to show that every identity satisfied in HSP( Bp0) can be derived from the identities determining SAp+1∘.

Let u ≈ u+q be any identity which is satisfied in HSP( Bp0), where u = u₁+⋯+u_n, u_i, q ∈ X⁺, i ∈ n. Choose Z = C(u) ∖ c(q). Since D_Z(u+q) ≠ ∅, by Lemma 3.9, D_Z(u) ≠ ∅. Assume that D_Z(u) = u₁+⋯+u_k, k ≤ n. By using Lemma 3.4 again, B_p ⊨ D_Z(u) ≈ D_Z(u)+q. It follows from Lemma 3.3 that C(D_Z(u)) = c(q). By Lemma 3.4, there exists q₁ in X⁺ with r_p+1(q₁) = r_p+1(q) (and so r_i(q₁) = r_i(q) for all i ∈ p − 1) and c(q₁) ⊆ C(D_Z(u)) such that D_Z(u) ≈ D_Z(u)+q₁ is satisfied in SAp+1∘. Thus, it is easy to check that

also is satisfied in SAp+1∘. Notice that u1p⋯ukpq1≈q is satisfied in SA_p+1. Thus D_Z(u) ≈ D_Z(u)+q is satisfied in SAp+1∘ and so u ≈ u+q is also satisfied in SAp+1∘. This shows that HSP( Bp0) = SAp+1∘. □

Suppose that S ∈ Bip^. Then S ∈ ROBAp+1∘ and E(S) ∈ Bi. This means that (S, ⋅) is a semilattice of abelian groups which belong to A_p+1 and so by [22, Exercise IV.2.16 (v), p177], S ∈ SAp+1∘. It follows that Bip^ ⊆ SAp+1∘. Thus, Bip^ = SAp+1∘ since SAp+1∘ ⊆ Bip^ is clear.

Lemma 3.11

LetV ∈ [Bi, Sr(2, 1)]. ThenV ∨ HSP( Bp0) = Vp^.

Proof

Let V ∈ [Bi, Sr(2, 1)]. Since it is easily seen that V ∨ HSP( Bp0) ⊆ Vp^, we only need to show that Vp^ ⊆ V ∨ HSP( Bp0). Let u ≈ u+q be an AI-identity which is satisfied in V ∨ HSP( Bp0), where u = u₁+⋯ +u_n, u_i, q ∈ X⁺, i ∈ n. In the following we show that this identity is derivable from the identities determining Vp^.

Let Z = C(u) ∖ c(q). Since D_Z(u+q) ≠ ∅, by Lemma 3.9, D_Z(u) ≠ ∅. Assume that D_Z(u) = u₁+⋯ +u_k, k ≤ n. By Lemma 3.9, B_p ⊨ D_Z(u) ≈ D_Z(u)+q and so ⋃_{i ∈ k}c(u_i) = c(q). By Lemma 3.4, there exists q₁ in X⁺ with r_p+1(q₁) = r_p+1(q) (and so r_i(q₁) = r_i(q) for all i ∈ p − 1) and c(q) ⊆ c(q₁) ⊆ ⋃_{i ∈ k}c(u_i) such that D_Z(u) ≈ D_Z(u)+q₁ is satisfied in ROBAp+1∘. Therefore, u ≈ u+q₁ is also satisfied in ROBAp+1∘. Proceeding as in the proof of Lemma 3.6, we have that u ≈ u+q^pq₁q^p is derivable from the identities determining Vp^. Notice that Vp^ ⊨ q^pq₁q^p ≈ q. Thus u ≈ u+q is derivable from the identities determining Vp^. Thus, Vp^ ⊆ V ∨ HSP( Bp0) and so Vp^ = V ∨ HSP( Bp0). □

Lemma 3.12

The subvariety ofROBAp+1∘determined by(17)satisfies the identity

Proof

We only need to show that identity (18) is derivable from (17) and the identities determining ROBAp+1∘. On one hand, we have

This derives the identity

On the other hand, we also have