ON HOMOMORPHISMS of KRASNER HYPERRINGS 1. Introduction a Hyperoperation on a Nonempty Set H Is a Function

ANALELE S¸TIINT¸IFICE ALE UNIVERSITAT¸II˘ “AL.I. CUZA” DIN IAS¸I (S.N.) MATEMATICA,˘ Tomul LVII, 2011, f.2 DOI: 10.2478/v10157-011-0023-2

ON HOMOMORPHISMS OF KRASNER HYPERRINGS

W. PHANTHAWIMOL, Y. PUNKLA, K. KWAKPATOON and Y. KEMPRASIT

Abstract. By a homomorphism from a Krasner hyperring (A, +, ·) into a Krasner ′ ′ ′ ′ hyperring (A , + , · ) we mean a function f : A → A satisfying f(x+y) ⊆ f(x)+f(y) and ′ f(x · y) = f(x) · f(y) for all x, y ∈ A. The kernel of f, ker f, is deﬁned by ker f = {x ∈ ′ ′ ′ ′ ′ A | f(x) = 0 } where 0 is the zero of (A , + , · ). In fact, ker f may be empty. In this paper, some general properties of a Krasner hyperring homomorphism with nonempty kernel are given. Various examples are also provided. Mathematics Subject Classiﬁcation 2000: 20N20. Key words: Krasner hyperring, homomorphism.

1. Introduction A hyperoperation on a nonempty set H is a function ◦ : H × H → P(H)\{∅} where P(H) is the power set of H and (H, ◦) is called a hypergroupoid. For nonempty subsets A, B of H and x ∈ H, let ∪ A ◦ B = a ◦ b, A ◦ x = A ◦ {x} and x ◦ A = {x} ◦ A. a∈A b∈B An element e ∈ H is called an identity of (H, ◦) if x ∈ x ◦ e ∩ e ◦ x, for all x ∈ H and it is called a scalar identity of (H, ◦) if x ◦ e = e ◦ x = {x}, for all x ∈ H. If e is a scalar identity of (H, ◦), then e is the unique identity of (H, ◦). The hypergroupoid is said to be commutative if x ◦ y = y ◦ x, for all x, y ∈ H. A hypergroupoid (H, ◦) is called a semihypergroup if (x◦y)◦z = x◦(y◦z), for all x, y, z ∈ H. 240 W. PHANTHAWIMOL, Y. PUNKLA, K. KWAKPATOON, Y. KEMPRASIT 2 A semihypergroup is called a hypergroup if H ◦ x = x ◦ H = H, for all x ∈ H. An element y of a hypergroupoid (H, ◦) is called an inverse of x ∈ H if (x ◦ y) ∩ (y ◦ x) contains at least one identity. A hypergroup (H, ◦) is said to be regular if every element of (H, ◦) has at least one inverse. A regular hypergroup (H, ◦) is said to be reversible if for all x, y, z ∈ H, ′ ′ ′ x ∈ y ◦ z ⇒ z ∈ y ◦ x and y ∈ x ◦ z , for some inverse y of y and some ′ inverse z of z. A hypergroup (H, ◦) is called a canonical hypergroup if:

1. (H, ◦) is commutative;

2. (H, ◦) has a scalar identity;

3. every element of (H, ◦) has a unique inverse;

4. (H, ◦) is reversible.

A canonical subhypergroup of a canonical hypergroup (H, ◦) is a nonempty subset K of H containing the scalar identity of (H, ◦) which forms a canonical hypergroup under the hyperoperation ◦ on H. A Krasner hyperring is a triple (A, +, ·) such that 1. (A, +) is a canonical hypergroup, 2. (A, ·) is a semigroup with zero 0 where 0 is the scalar identity of (A, +), 3. for all x, y, z ∈ A, x · (y + z) = x · y + x · z and (y + z) · x = y · x + z · x. We call 0 the zero of the Krasner hyperring (A, +, ·). For x ∈ A, let −x denote the unique inverse of x in (A, +). Then −(−x) = x, for all x ∈ A. In addition, we have (x + y) · (z + w) ⊆ x · z + x · w + y · z + y · w, (−x) · y = x · (−y) = −(x · y), for all x, y, z, w ∈ A, ([1], p.167). A subhyperring of a Krasner hyperring (A, +, ·) is a nonempty subset S of A which forms a Krasner hyperring containing 0 under the hyperoperation + and the operation · on A, that is, S is a canonical subhypergroup of (A, +) and S · S ⊆ S. Then a nonempty subset S of A is a subhyperring of (A, +, ·) if and only if, for all x, y ∈ S, x + y ⊆ S, −x ∈ S and x · y ∈ S. A nonempty subset I of A is called a hyperideal of (A, +, ·) if I is a canonical subhypergroup of (A, +) and A · I ∪ I · A ⊆ I (see [1], p.168 or [2], p.20), that is, x + y ⊆ I and −x ∈ I, for all x, y ∈ I and xy, yx ∈ I, for all x ∈ I and y ∈ A. 3 ON HOMOMORPHISMS OF KRASNER HYPERRINGS 241

Let I be a hyperideal of A and A/I = {x + I | x ∈ A}. We can see that for x ∈ I, x + I = I and for x, y ∈ A, (x + I) ∩ (y + I) ≠ ∅ implies that x + I = y + I. Define (x + I) + (y + I) = {z + I | z ∈ x + y}, (x + I) · (y + I) = x · y + I, for all x, y ∈ I. Then (A/I, +, ·) is a Krasner hyperring ([1], p.169). It may be called the quotient hyperring of A by I. Notice that 0 + I is the zero of the Krasner hyperring (A/I, +, ·) and for every x ∈ A, −x + I is the unique inverse of x + I in the canonical hypergroup (A/I, +). A homomorphism from a Krasner hyperring (A, +, ·) into a Krasner ′ ′ ′ ′ ′ hyperring (A , + , · ) is a function f : A → A such that f(x + y) ⊆ f(x) + ′ f(y) and f(x · y) = f(x) · f(y), for all x, y ∈ A. A homomorphism f from ′ ′ ′ (A, +, ·) into (A , + , · ) is said to be a good homomorphism if f(x + y) = ′ f(x) + f(y), for all x, y ∈ A. ′ ′ ′ An isomorphism from (A, +, ·) into (A , + , · ) is a bijective good homo- ′ ′ ′ morphism from (A, +, ·) onto (A , + , · ). The Krasner hyperrings (A, +, ·) ′ ′ ′ ∼ ′ ′ ′ and (A , + , · ) are said to be isomorphic, and we write (A, +, ·) = (A , + , · ) ′ ′ ′ if there is an isomorphism from (A, +, ·) onto (A , + , · ). Notice that if f is ′ ′ ′ an isomorphism from (A, +, ·) onto (A , + , · ), then f −1 is an isomorphism ′ ′ ′ from (A , + , · ) onto (A, +, ·). Example 1.1. Let (A, ·) be a semigroup with zero 0 such that (A\{0}, ·) is a group. Define the hyperoperation + on A by   y + x = {x}, if y = 0, x + y = A\{x}, if x = y ≠ 0,  {x, y}, if x, y ∈ A\{0} and x ≠ y. Then (A, +, ·) is a Krasner hyperring ([1], p.170). Notice that 0 is the zero of (A, +, ·) and for every x ∈ A, x is the inverse of x in (A, +). Since Ax = A, for all x ∈ A\{0}, it follows that {0} and A are the only hyperideals of (A, +, ·). Example 1.2. Define the hyperoperation ⊕ on the unit interval [0, 1] by { {max {x, y}}, if x ≠ y, x ⊕ y = [0, x], if x = y. Then ([0, 1], ⊕, ·) is a Krasner hyperring where · is the usual multiplication ([3]). We can see that 0 is the zero of ([0, 1], ⊕, ·) and for every x ∈ [0, 1], x is the inverse of x in ([0, 1], ⊕). 242 W. PHANTHAWIMOL, Y. PUNKLA, K. KWAKPATOON, Y. KEMPRASIT 4

Deﬁne f : [0, 1] → [0, 1] by f(x) = 1, for all x ∈ [0, 1]. Since 1 ⊕ 1 = [0, 1] and 1 · 1 = 1, it follows that f is a homomorphism from the Krasner hyperring ([0, 1], ⊕, ·) into itself. Observe that ker f = ∅. It is easily seen that for a ∈ (0, 1], [0, a] and [0, a) are nonzero hyperideals of ([0, 1], ⊕, ·). Let I be an hyperideal of ([0, 1], ⊕, ·). Then 0 ∈ I. Let a be the supremum of I. If a ∈ I, then [0, a] ⊇ I ⊇ a ⊕ a = [0, a], so I = [0, a]. Assume that a∈ / I. Then a > 0 and I ⊆ [0, a). Let N be the set ∈ N − 1 of positive integers. Let N be such that a N > 0. Since a = sup I, it follows that for every [n ≥ N, there] exists an element xn ∈ I such that − 1 ≤ − 1 ⊆ ⊕ ⊆ ≥ a n xn < a. Hence 0, a n [0, xn] = xn xn I, for all n N. Thus ∪ [ 1 ] [0, a) = 0, a − ⊆ I. n n∈N n≥N It follows that I = [0, a). This shows that {[0, a] | a ∈ [0, 1]} ∪ {[0, a) | a ∈ (0, 1]} is the set of all hyperideals of the Krasner hyperring ([0, 1], ⊕, ·). Notice from the proof that it is also the set of all canonical subhypergroups of the canonical hypergroup ([0, 1], ⊕). For a ∈ (0, 1) and x, y ∈ [0, 1], we have { { [0, a], if x ≤ a, [0, a), if x < a, x ⊕ [0, a] = , x ⊕ [0, a) = {x}, if x > a, {x}, if x ≥ a,

{ } [0, 1]/[0, a] = {z ⊕ [0, a] | z ∈ [0, 1]} = {[0, a]} ∪ {z} | z ∈ (a, 1] , { } [0, 1]/[0, a) = {z ⊕ [0, a) | z ∈ [0, 1]} = {[0, a)} ∪ {z} | z ∈ [a, 1] ,

(x ⊕ [0, a]) ⊕ (y ⊕ [0, a]) = {z ⊕ [0, a] | z ∈ x ⊕ y} { }  { }  x , { } if x > y > a, = {[0, a]} ∪ {z} | z ∈ (a, x] , if x = y > a,  {[0, a]} if x, y ≤ a,

(x ⊕ [0, a)) ⊕ (y ⊕ [0, a)) = {z ⊕ [0, a) | z ∈ x ⊕ y} { }  { } ≥  x , { } if x > y a, = {[0, a)} ∪ {z} | z ∈ [a, x] , if x = y ≥ a,  {[0, a)}, if x, y ≤ a, 5 ON HOMOMORPHISMS OF KRASNER HYPERRINGS 243

{ [0, a], if xy ≤ a, (x ⊕ [0, a]) · (y ⊕ [0, a]) = xy ⊕ [0, a] = {xy}, if xy > a, { [0, a), if xy < a, (x ⊕ [0, a)) · (y ⊕ [0, a)) = xy ⊕ [0, a) = {xy}, if xy ≥ a.

2. Homomorphisms of Krasner hyperrings As shown in Example 1.2, the kernel of a Krasner hyperring homomorphism may be empty. In this section, we provide general properties of a Krasner hyperring homomorphism with nonempty kernel.

Theorem 2.1. Let f be a homomorphism from a Krasner hyperring A into a Krasner hyperring B such that ker f ≠ ∅. Then the following statements hold.

(i) f(0) = 0;

(ii) f(−x) = −f(x), for all x ∈ A;

(iii) ker f is a hyperideal of A;

(iv) If f is 1-1, then ker f = {0};

(v) If f is a good homomorphism and ker f = {0}, then f is 1-1;

(vi) If f is a good homomorphism, then f(A) is a subhyperring of B.

Proof. (i) Let a ∈ ker f. Then f(a) = 0, so {f(0)} = f(0) + 0 = f(0) + f(a) ⊇ f(0 + a) = f({a}) = {f(a)}. This implies that f(0) = f(a) = 0. (ii) Let x ∈ A. Since 0 ∈ x−x, we have f(0) ∈ f(x−x) ⊆ f(x)+f(−x). But f(0) = 0 by (i), so 0 ∈ f(x) + f(−x). Thus f(−x) is the inverse of f(x) in the canonical hypergroup (B, +). Hence f(−x) = −f(x). (iii) By (i), 0 ∈ ker f. Let x, y ∈ ker f and z ∈ A. Then f(x) = 0 = f(y). Since f(x + y) ⊆ f(x) + f(y) = 0 + 0 = {0}, it follows that x + y ⊆ ker f. From (ii), f(−x) = −f(x), so f(−x) = −0 = 0. Also, f(zx) = f(z)f(x) = f(z)0 = 0, f(xz) = f(x)f(z) = 0f(z) = 0. This proves that ker f is a hyperideal of A, as desired. (iv) It is evident by (i). 244 W. PHANTHAWIMOL, Y. PUNKLA, K. KWAKPATOON, Y. KEMPRASIT 6

(v) Assume that f is a good homomorphism and ker f = {0}. To show that f is 1-1, let x, y ∈ A be such that f(x) = f(y). Then 0 ∈ f(x)−f(x) = f(x) − f(y) = f(x) + f(−y) = f(x − y), so 0 = f(z), for some z ∈ x − y. Thus z ∈ ker f = {0}. It follows that 0 ∈ x − y which implies that −y is the inverse of x in the canonical hypergroup (A, +). Hence −y = −x and thus x = y. (vi) Assume that f is a good homomorphism. To show that f(A) is a subhyperring of B, let x, y ∈ A. Then f(x) + f(y) = f(x + y) ⊆ f(A), −f(x) = f(−x) ⊆ f(A) from (ii), f(x)f(y) = f(xy) ∈ f(A). Therefore f(A) is a subhyperring of B. The statements (v) and (vi) are not generally true if f is only a homomorphism as the following example shows. Example 2.2. Let ([0, 1], ⊕, ·) be the Krasner hyperring defined as in Example 1.2. Define f : [0, 1] → [0, 1] by f(0) = 0 and f(x) = 1, for all x ∈ (0, 1]. Since   f({0}) = {0} = 0 ⊕ 0 = f(x) ⊕ f(y), if x = y = 0,  f({x}) = {1} = 1 ⊕ 0 = f(x) ⊕ f(y), if x > y = 0, ⊕ f(x y) =  f({x}) = {1} ( [0, 1] = 1 ⊕ 1 = f(x) ⊕ f(y), if x > y > 0,  f([0, x]) = {0, 1} ( [0, 1] = 1 ⊕ 1 = f(x) ⊕ f(y), if x = y > 0, { 0, if x = 0 or y = 0, f(xy) = f(x)f(y) = 1, if x ≠ 0 and y ≠ 0, it follows that f is a homomorphism from ([0, 1], ⊕, ·) into itself which is not a good homomorphism. Also, ker f = {0} but f is not 1-1. Moreover, f([0, 1]) = {0, 1} which is not a subhyperring of the Krasner hyperring ([0, 1], ⊕, ·). We note that if f is a homomorphism from a Krasner hyperring A onto a Krasner hyperring B, then f(x) = 0 for some x ∈ A, that is, ker f ≠ ∅, so by Theorem 2.1(iii), ker f is a hyperideal of A. Theorem 2.3. Let f be a homomorphism from a Krasner hyperring A onto a Krasner hyperring B. Define θ : A/ ker f → B by θ(x+ker f) = f(x), for all x ∈ A. Then the following statements hold. 7 ON HOMOMORPHISMS OF KRASNER HYPERRINGS 245

(i) θ is a homomorphism from A/ ker f onto B.

(ii) If f is a good homomorphism, then θ is an isomorphism and hence ∼ A/ ker f = B. Proof. (i) First, to show that θ is well-deﬁned, let x, y ∈ A be such that x + ker f = y + ker f. Then y ∈ x + ker f, so y ∈ x + z for some z ∈ ker f. Thus f(y) ∈ f(x + z) ⊆ f(x) + f(z) = f(x) + 0 = {f(x)}, which implies that f(x) = f(y). Since f(A) = B, it follows that θ(A/ ker f) = B. To show that θ is a homomorphism, let x, y ∈ A. Then ( ) ( ) θ (x + ker f) + (y + ker f) = θ {z + ker f | z ∈ x + y} = {θ(z + ker f) | z ∈ x + y} = {f(z) | z ∈ x + y} = f(x + y) ⊆ f(x) + f(y) = θ(x + ker f) + θ(y + ker f),

( ) θ (x + ker f)(y + ker f) = θ(xy + ker f) = f(xy) = f(x)f(y) = θ(x + ker f)θ(y + ker f).

(ii) Assume that f is a good homomorphism. It can be seen from the proof of (i) that θ is a good homomorphism. Then it remains to show that θ is 1-1. By Theorem 2.1(v), it suffices to show that ker θ = {0+ker f}. From Theorem 2.1(i), we know that 0+ker f ∈ ker θ. Next, let x ∈ A be such that θ(x + ker f) = 0. Then f(x) = 0, so x ∈ ker f. Hence x + ker f = 0 + ker f. Finally, we show by the following example that for a homomorphism from a Krasner hyperring A onto a Krasner hyperring B, A/ ker f and B are not necessarily isomorphic. Example 2.4. Let ([0, 1], ⊕, ·) be the Krasner hyperring defined as in Example 1.2. It can be easily seen that ({0, 1}, , ·) is also a Krasner hyperring where 0 0 = {0}, 1 0 = 0 1 = {1}, 1 1 = {0, 1} and · is the usual multiplication. Define f : [0, 1] → {0, 1} by f(0) = 0 and f(x) = 1, for all x ∈ (0, 1]. Then    f({0}) = {0} = 0 0 = f(x) f(y), if x = y = 0,  f({x}) = {1} = 1 0 = f(x) f(y), if x > y = 0, ⊕ f(x y) =   f({x}) = {1} ( {0, 1} = 1 1 = f(x) f(y), if x > y > 0,  f([0, x]) = {0, 1} = 1 1 = f(x) f(y), if x = y > 0, 246 W. PHANTHAWIMOL, Y. PUNKLA, K. KWAKPATOON, Y. KEMPRASIT 8 { 0, if x = 0 or y = 0, f(xy) = f(x)f(y) = 1, if x ≠ 0 and y ≠ 0, It follows that f is a homomorphism from ([0, 1], ⊕, ·) onto ({0, 1}, , ·) {which is not a good} homomorphism. Since ker f = {0}, we have [0, 1]/ ker f = {x} | x ∈ [0, 1] . Then [0, 1]/ ker f is infinite. It follows that the quotient hyperring [0, 1]/ ker f and the Krasner hyperring ({0, 1}, , ·) are not isomorphic.

REFERENCES

1. Corsini, P. – Prolegomena of Hypergroup Theory, Supplement to Riv. Mat. Pura Appl. Aviani Editore, Tricesimo, 1993. 2. Corsini, P.; Leoreanu, V. – Applications of Hyperstructure Theory, Advances in Mathematics (Dordrecht), 5, Kluwer Academic Publishers, Dordrecht, 2003. 3. Kemprasit, Y. – Multiplicative interval semigroups on R admitting hyperring struc- ture, Ital. J. Pure Appl. Math., 11 (2002), 205–212.

Received: 30.VI.2009 Department of Mathematics, Revised: 22.VII.2009 Faculty of Science, Chulalongkorn University, Bangkok 10330, THAILAND [email protected] [email protected] [email protected] [email protected]