Fundamental relation on m-idempotent hyperrings

Definition 2.1

([8]). A commutative hypergroup (H, ∘) is called canonical if

1. there exists e ∈ H, such that e ∘ x = {x}, for every x ∈ H;

2. for all x ∈ H there exists a unique x⁻¹ ∈ H, such that e ∈ x ∘ x⁻¹;

3. x ∈ y ∘ z implies y ∈ x ∘ z⁻¹.

Definition 2.2

([8]). An algebraic system (R, +, ⋅) is said to be a (general) hyperring, if (R, +) is a hypergroup, (R, ⋅) is a semihypergroup, and ”⋅” is distributive with respect to ”+”.

Theorem 2.3

([8]). Consider the equivalence relation ρ on the hypergroup (H, ∘) and the hyperoperation ρ(x) ⊗ ρ(y) = {ρ (z) | z ∈ ρ(x)∘ρ(y)} on the quotient H/ρ = {ρ(x) | x ∈ H}. Then ρ is regular (strongly regular) on H if and only if (H/ρ, ⊗) is a hypergroup (group).

Example 3.1

Consider the following hyperoperations on the set R = {a, b, c, d}:

Then (R, +, ⋅) is a semihyperring. We have {b, c} = a + a and so bγc. But, b ε̸_mc for all 2 ≤ m ∈ ℕ. Also, bγ^*c and b /εm∗ c.

Proposition 3.2

Let (R, +, ⋅) be a hyperring satisfying the relation (2) such that there exists 0 ∈ R such that x + 0 = {x} and x ⋅ 0 = {0} for all x ∈ R. If A₁, …, A_n are hyperideals of R, then X is an equivalence class of εm∗ . Moreover, ε_m is transitive.

Proof

It is sufficient that we show X × X ⊆ εm∗ . Let a, b ∈ X. Then there are k, l ∈ ℕ and the hyperideals A₁, …, A_k, B₁, …, B_l of R such that a∈∑i=1kAim and b∈∑i=1lBim. By hypothesis, we have 0 ∈ A_i (1 ≤ i ≤ k) and 0 ∈ B_i (1 ≤ i ≤ l). Hence, {a, 0} ⊆ ∑i=1kAim and {b,0}⊆∑i=1lBim, and so {a, b} ⊆ ∑i=1kAim+∑i=1lBim by x ∈ x + 0 for all x ∈ R. Using the relation (2), we obtain {a, b} ⊆ ∑i=1k+lzim for a suitable (z₁, …, z_k+l) ∈ R^k+l. Then, aε_mb and therefor the proof is completed.□

Theorem 3.3

The relation εm∗ is a strongly regular relation on the hyperring (R, +, ⋅) satisfying the relation (2).

Proof

Let a′ εm∗ a and b′ εm∗ b, for a, b, a′, b′ ∈ R. Thus, there exist x₁, …, x_s+1, y₁, …, y_t+1 ∈ R, with x₁ = a′, x_s+1 = a, y₁ = b′ and y_t+1 = b, such that a′ = x₁ε_mx₂ε_m…ε_mx_sε_mx_s+1 = a and b′ = y₁ε_my₂ε_m…ε_my_tε_my_t+1 = b. It implies that {x_i, x_i+1} ⊆ zi1m+…+zinm, for z_i1, …, z_in ∈ R and i ∈ {1, …, s}, and similarly {y_j, y_j+1} ⊆ dj1m+…+djnm, for d_j1, …, d_jn ∈ R and j ∈ {1, …, t}. Thus, we have

Choose some elements z₁, …, z_s+t such that z_i ∈ x_i + y₁, for i ∈ {1, …, s}, and z_s+j ∈ x_s+1 + y_j+1, for j ∈ {1, …, t}. Then, we have

It means that a′ + b′ ∋ z₁ε_mz₂ … ε_mz_s−1ε_mz_sε_mz_s+1…ε_mz_s+t−1ε_mz_s+t ∈ a + b. Hence, for all z ∈ a + b and z′ ∈ a′ + b′, we have z εm∗ z′. Therefore, εm∗ is a strongly regular relation on (R, +). Similarly, one proves that εm∗ is strongly regular also on (R,⋅).□

Theorem 3.4

Under the same hypothesis, the quotient R/ εm∗ = { εm∗ (x)| x ∈ R} is a ring.

Proof

As in the general case when we are talking about the quotient hyperstructure, consider on R/ εm∗ the following hyperoperations:

Let c, d∈εm∗(a)+εm∗(b). Then there exist a′,a″∈εm∗(a) and b′,b″∈εm∗(b) such that c ∈ a′ + b′ and d ∈ a″ + b″. According to the proof of Theorem 3.3, we have cεm∗p and dεm∗q, for some p, q ∈ a + b. By the transitivity of the relation εm∗ , we get that c εm∗ d and thus εm∗(c)=εm∗(d). Hence, εm∗(a)⊕εm∗(b)={εm∗(c)}, for all c∈εm∗(a)+εm∗(b). Similarly, we can show that εm∗(x)⊙εm∗(y)={εm∗(z)}, for all z∈εm∗(x)⋅εm∗(y). Therefore, (R/ εm∗ , ⊕, ⊙) is a trivial hyperring, meaning it is a ring.□

Lemma 3.5

Let ρ be a strongly regular relation on the hyperring R. For every z ∈ R, there exists a ∈ R such that z^m ⊆ ρ(a).

Proof

In the quotient ring (R/ρ, ⊕, ⊙), for all a, b ∈ R, A ⊆ ρ(a) and B ⊆ ρ(b), we have

By induction, we can extend the above equalities to finite sums and products. Now, let y ∈ z^m, then y ∈ (ρ(z))^m and so

It implies that y ∈ ρ(y) = ρ(t), for t ∈ z^m ⊆ R. This completes the proof.□

Theorem 3.6

The relation εm∗ is the smallest equivalence relation on a hyperring R satisfying the relation (2), such that the quotient R/ εm∗ is a ring.

Proof

Let ρ be an equivalence relation on R such that R/ρ is a ring with the following operations:

Now, let xε_my, for arbitrary x, y ∈ R. Then there exist z₁, …, z_n ∈ R such that {x, y} ⊆ z1m+…+znm. By Lemma 3.5, we have zim ⊆ ρ(a_i), for some a_i ∈ R and 1 ≤ i ≤ n. Hence,

This implies that xρy. Thus ε_m ⊆ ρ and so εm∗ ⊆ ρ. Therefore, εm∗ is the smallest equivalence relation on a hyperring R satisfying condition (2), such that the quotient R/ εm∗ is a ring.□

Remark 3.7

According to Theorem 3.3, Theorem 3.4 and Theorem 3.6, the relation εm∗ is the smallest strongly regular relation on a hyperring R satisfying condition (2), such that R/ εm∗ is a ring. Hence, εm∗ is called a fundamental relation on such hyperrings.

Corollary 3.8

(R/ εm∗ , ⊕, ⊙) is a multiplication hyperring, where

Theorem 3.9

Let (R, +, ⋅) be a hyperring satisfying relation (2). Then the following properties hold.

1. ωm∗⋅R⊆ωm∗ and R⋅ωm∗⊆ωm∗.

2. If (R, +) is a regular hypergroup, then ωm∗ is a hyperideal of R.

Proof

1. Let a ∈ r ⋅ x such that r ∈ R and x∈ωm∗. Then, εm∗(x)=φm∗(x)=0R/εm∗. We have

   φm∗(a)=εm∗(a)=εm∗(r)⊙εm∗(x)=εm∗(r)⊙0R/εm∗=0R/εm∗.

   Hence, a ∈ ωm∗ and so R⋅ωm∗⊆ωm∗. Similarly, one proves the other inclusion.

2. Let a, b ∈ ωm∗ . For all x ∈ a + b, we have

   φm∗(x)=εm∗(x)=εm∗(a)⊕εm∗(b)=0R/εm∗⊕0R/εm∗=0R/εm∗.

   It implies that x ∈ ωm∗ . Thus x+ωm∗⊆ωm∗, for all x ∈ R. On the other hand, since (R, +) is regular, there exists an identity element e of (R, +) such that εm∗(e)⊕εm∗(x)=εm∗(x)=εm∗(x)⊕εm∗(e). It follows that εm∗(e)=0R/εm∗ and so e ∈ ωm∗ . Now, consider x ∈ ωm∗ . By the regularity of (R, +), there exists x′ ∈ R such that e ∈ x + x′. Thus,

   0R/εm∗=εm∗(e)=εm∗(x)⊕εm∗(x′)=0R/εm∗⊕εm∗(x′)=εm∗(x′)=φm∗(x′),

   meaning that x′ ∈ ωm∗ . Hence, we have y ∈ e + y ⊆ (x + x′) + y = x + (x′ + y) ⊆ x + ωm∗ , for all y ∈ ωm∗ . Therefore, ωm∗⊆x+ωm∗ and so x+ωm∗=ωm∗, for all x ∈ ωm∗ . By using (i), the proof is complete.

Definition 3.10

Let (R, +, ⋅) be a hyperring. We say that R is m-idempotent if there exists a constant m, 2 ≤ m ∈ ℕ, such that x ∈ x^m, for all x ∈ R.

Theorem 3.11

If (R, +, ⋅) is an m-idempotent hyperring such that R ⋅ x ⊇ R ⋅ R ⊆ x ⋅ R for all x ∈ R, then (R, ⋅) is a hypergroup.

Proof

By the definition of a hyperring, (R, ⋅) is a semihypergroup and it is clear that t ⋅ R ⊆ R ⊇ R ⋅ t, for all t ∈ R. Since R is m-idempotent, for every x ∈ R there exists 2 ≤ m ∈ ℕ such that x∈ x^m. Hence, we have x ∈ x^m ⊆ R ⋅ R ⊆ t ⋅ R for all t ∈ R. Thus R ⊆ t ⋅ R (and similarly R ⊆ R ⋅ t). The above reproducibility axiom completes the proof.□

Corollary 3.12

Let (R, +, ⋅) be a Krasner hyperdomain such that R ⋅ R ⊆ x ⋅ R for all x ∈ R. If R is m-idempotent, then R is a Krasner hyperfield.

Proof

By Theorem 3.11, we know that (R \ {0}, ⋅) is a semigroup such that x ⋅ R = R, for all x ∈ R \ {0}. Hence, for x ∈ R \ {0}, there exists y ∈ R \ {0} such that x = x ⋅ y. Now take z ∈ R \ {0} such that y = y ⋅ z. Then

Thus, 0 ∈ x ⋅ (y − z) and so we have x ⋅ t = 0, for some t ∈ y − z. Since R is a hyperdomain and x ≠ 0, it follows that t = 0. This implies that y = z, since (R, +) is a canonical hypergroup. Hence, there exists a unique element, such as 1 ∈ R \ {0}, such that x = x ⋅ 1, for all x ∈ R \ {0}. Similarly, we can show that, for all x ∈ R \ {0}, there exists a unique element y ∈ R \ {0} such that x ⋅ y = 1. Therefore, (R \ {0}, ⋅) is a group and so (R, +, ⋅) is a Krasner hyperfield.□

Example 3.13

The set R = {0, 1} with the following hyperoperations is a hyperring [8].

It is easy to see that R is m-idempotent, for all m, 2 ≤ m ∈ ℕ.

Example 3.14

Define on R = {0, a, b} two hyperoperations as follows [13]:

It is easy to verify that (R, +, ⋅) is an m-idempotent hyperring, for all m, 2 ≤ m ∈ ℕ.

Example 3.15

Define x ⊕ y = {x, y, x + y} and x ⊗ y = {x ⋅ y}, for any x, y ∈ ℤ, the set of all integers, where “+” and “⋅” are the ordinary addition and multiplication of integers. Then (ℤ, ⊕, ⊗) is a hyperring, but not a Krasner hyperring. For all 1 ≠ x ∈ ℤ and for all 2 ≤ m ∈ ℕ, we have x∉{xm}=x⊗⋯⊗x⏟m. Then, the hyperring (ℤ, ⊕, ⊗) is not m-idempotent, for all m, 2 ≤ m ∈ ℕ.

Example 3.16

Consider the Krasner hyperring R = {0, a, b} with the hyperaddition and the multiplication defined as follows [14]:

1. For every odd number m ∈ ℕ, we have 0^m = 0, a^m = a and b^m = b. Hence, R is m-idempotent, for all odd natural numbers m.

2. Besides, since a² = a ⋅ a = b, it follows that R is not an 2-idempotent hyperring. Similarly, one proves that, for all even numbers m ∈ ℕ, the hyperring R is not m-idempotent.

Example 3.17

The set R = {0, a, b, c} with the following hyperaddition and multiplication is an m-idempotent hyperring, for all 2 ≤ m ∈ ℕ [14].

Example 3.18

Let (R, +, ⋅) be a Krasner hyperring and M(R)={ab00|a,b∈R} be a collection of 2 × 2 matrices over R. Define the following hyperoperation and operation on M(R), for all a, b, c, d ∈ R:

Then, M(R) is a Krasner hyperring [14]. It can be seen that M(R) is m-idempotent if and only if R is m-idempotent, for every m, 2 ≤ m ∈ ℕ.

Example 3.19

Let (R, +, ⋅) be a nonzero ring. For all a, b ∈ R, define the hyperoperation a * b = {a ⋅ b, 2a ⋅ b, 3a ⋅ b, …}. Then (R, +, *) is a multiplicative hyperring [8].

1. Consider the set ℤ₃ = {0, 1, 2}, where ℤ is the set of all integers, with

   ⊕0¯1¯2¯0¯0¯1¯2¯1¯1¯2¯0¯2¯2¯0¯1¯∗0¯1¯2¯0¯{0¯}{0¯}{0¯}1¯{0¯}Z3Z32¯{0¯}Z3Z3

   Then (ℤ₃, ⊕, *) is a commutative multiplicative hyperring that is m-idempotent, for all m, 2 ≤ m ∈ ℕ.

2. But the multiplicative hyperring (ℤ₄ = {0, 1, 2, 3}, ⊕, *) defined as follows:

   ⊕0¯1¯2¯3¯0¯0¯1¯2¯3¯1¯1¯2¯3¯0¯2¯2¯3¯0¯1¯3¯3¯0¯1¯2¯∗0¯1¯2¯3¯0¯{0¯}{0¯}{0¯}{0¯}1¯{0¯}Z4{0¯,2¯}Z42¯{0¯}{0¯,2¯}{0¯}{0¯,2¯}3¯{0¯}Z4{0¯,2¯}Z4

   is not m-idempotent, since, for all m, 2 ≤ m ∈ ℕ, we have 2 ∉ {0} =2^m.

Theorem 3.20

Let (R, +, ⋅) be an m-idempotent hyperring satisfying relation (2). Then β+∗⊆εm∗ and β⋅∗⊆εm∗.

Proof

Let xβ₊y; then there exist z₁, …, z_n ∈ R such that {x, y} ⊆ z₁ + … + z_n. Since R is m-idempotent, for every z_i, (1 ≤ i ≤ n), we have z_i ∈ zim . Thus, {x, y} ⊆ z1m+…+znm and so xε_my. Hence, β₊ ⊆ ε_m. This implies that β+∗⊆εm∗.

Set now xβ_⋅y; then there exist z₁, …, z_n ∈ R such that {x, y} ⊆ ∏i=1n z_i. Because R is m-idempotent, we have {x,y}⊆∏i=1nzim=(∏i=1nzi)m. Since B⊆∑i=1nAim implies that, there exist x_i ∈ A_i, for 1 ≤ i ≤ n, such that B ⊆ ∑i=1nxim, for all B, A₁, …, A_n ⊆ R, it follows there exists t ∈ ∏i=1n z_i such that {x, y} ⊆ t^m. Hence, xε_my and thus β⋅∗⊆εm∗. □

Example 3.21

Consider the Krasner hyperring R = {0, a, b} defined in Example 3.16. Then R is an 3-idempotent hyperring. Hence, for all x, y ∈ R, we have

Hence, ε₃ = β₊ and so ε3∗=β+∗. Moreover, it can be seen that β⋅∗(x)=R=ε3∗(x), for all x ∈ R. Thus, β⋅∗=ε3∗=β+∗.

Example 3.22

It is routine to verify that in the 2-idempotent Krasner hyperring R = {0, a, b, c} described in Example 3.17, we have ε2∗(0)=ε2∗(b)={0,b} and ε2∗(a)=ε2∗(c)={a,c}. Moreover, it can be seen that β⋅∗ (x) = {x}, for all x ∈ R. Hence, β⋅∗(x)⫋ε2∗(x), for all x ∈ R, and so β⋅∗⫋ε2∗.

Example 3.23

Consider the m-idempotent hyperring (ℤ₃, ⊕, *) defined in Example 3.19 (1). We obtain that εm∗ (x) = ℤ₃ and β+∗ (x) = {x}, for all x ∈ ℤ₃. Then, we have β+∗⫋εm∗.

Theorem 3.24

If R is an m-idempotent Krasner hyperring, then R/β+∗=R/εm∗.

Proof

Let (R, +, ⋅) be an m-idempotent Krasner hyperring. Then, z = z^m, for all z ∈ R. Hence,

and so β₊ = ε_m. Then β+∗=εm∗. This completes the proof.□

Corollary 3.25

If R is an m-idempotent Krasner hyperring, then β+∗=εm∗ = γ^*.

Proof

We know that β+∗ = γ^* on Krasner hyperrings [8]. Hence, the proof is clearly by Theorem 3.24.□

Theorem 3.26

On m-idempotent hyperrings satisfying relation (2), εm∗ = γ^*.

Proof

Let (R, +, ⋅) be an m-idempotent hyperring. Clearly, we have εm∗ ⊆ γ^*. Now, let xγy, for x, y ∈ R. Then {x,y}⊆∑j∈J(∏i∈Ijzi), for finite sets of indices J and I_j, and the elements z_i ∈ R. Since R is m-idempotent, z_i ∈ zim , for all z_i ∈ R. It implies that

and so, for every j ∈ J, there exist tj∈∏i∈Ijzi such that {x,y}⊆∑j∈Jtjm. Thus, we have xε_my and so x εm∗ y. Hence, γ^* ⊆ εm∗ and therefore the proof is complete.□