C-Simple and Bi-Ideal Simple Semirings

The Islamic University of Gaza Deanery of Higher Studies Faculty of Science Department of Mathematics

C-SIMPLE AND BI-IDEAL SIMPLE SEMIRINGS

Presented By Saeed Nouh Saeed Al-Zayyan

Supervised By Dr. Ahmed Al-Mabhouh

SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE AT THE ISLAMIC UNIVERSITY OF GAZA GAZA, PALESTINE JUNE 2006

°c Copyright by Saeed Nouh Saeed Al-Zayyan, 2006 To My parents, wife, sons, and my sincere friends, and to all knowledge seekers.

ii Table of Contents

Table of Contents iii

Abstract iv

Acknowledgements v

Introduction 1

1 Preliminary Notions 3 1.1 Groups, Rings, Semigroups, Semirings, and Relations ...... 3 1.2 Ideals and Bi-Ideals of a Semirings ...... 11 1.3 Congruence Relations ...... 13

2 Bi-Ideal-Simple Semirings 20 2.1 Bi-Ideal-Simple Semiring-Introduction ...... 20 2.2 Additively Zeropotent Semirings ...... 25 2.3 Bi-Ideal Simple Zeropotent Semirings ...... 30

3 Congruence-Simple Semirings 33 3.1 Congruence-Simple Commutative Semirings ...... 33 3.2 Congruence-Simple Additively Commutative Semirings of Finite Order 44

4 Smaradache Semirings 56 4.1 Deﬁnition of Smarandache Semirings and Examples ...... 56 4.2 Smarandache Special Elements in Semirings ...... 61 conclusions 66

References 67

iii Abstract

In this research we survey and introduce some concepts of semirings, as a bi-ideal simple semirings, congruence simple semirings, and bi-ideal simple zeropotent semirings. Finally we introduce smarandache semirings, smarandache zero divisors, and we classify all the zero divisors in spacial semirings.

iv Acknowledgements

My special thanks after thanking Allah almighty, to the Department of Mathematics in the Islamic University of Gaza, specially Dr: Ahmed Al-Mabhouh for his great concern and support. Also, I thank my family for providing a comfortable environ- ment to me, to my sincere friends for their endless support. Finally, I would like to thank everyone who contributed in achieving this thesis.

v Introduction

The notion of a semiring (i,e., a universal algebra with two associative binary operations, where one of them distributes over the other) was introduced by Vandiver

[6] in (1934). Needless to say, semiring found their full place in mathematics long before that year (e.g., the semirings of positive elements in ordered rings) and even more after (e.g., Various applications in theoretical computer science and algorithm theory).

Congruence simple algebras (i.e., those possessing just two congruence relations) serve a basic construction material for any algebraic structure. The study of congruence simple commutative semiring with unity was introduced by mitchell and Fenoglio

[11] in 1988. See also [12]

In this thesis, we study congruence simple semirings and bi-ideal simple semirings.

See [10], [13].

This thesis is divided into four chapters. In chapter1, we introduce some basic deﬁnition, and results about ideals, bi-ideal and congruence relations of semirings.

For more informations, see [1], [2], [8], [10], [11], [12], and [13].

Chapter2 is devoted for the study of bi-ideals simple semirings. See [13].

In Chapter3, Congruence-simple semirings are studied in more details. Also, we classify the congruence-simple commutative semirings. See [10], [1].

1 2

In chapter4, we introduce the concept of Smarandache semirings. We give some deﬁnitions and examples about this topic. See [14], [15], [16]. Chapter 1

Preliminary Notions

In this chapter we introduce basic deﬁnitions, theorems and results that would be useful for the rest of the thesis. More information can be found in [6], [8], [10], [11],

[12], [15].

1.1 Groups, Rings, Semigroups, Semirings, and Relations

Deﬁnition 1.1.1

Let S be a non-empty set, we call S a group under the

operation (∗), if the following properties hold

1. S is Closed under (∗), i.e for all a, b ∈ S we have

a ∗ b ∈ S.

2. The operation (∗) is associative, i.e a ∗ (b ∗ c) = (a ∗ b) ∗ c for all a, b, c ∈ S

3. There exists an auniqe element e in S such that e ∗ a = a ∗ e = a, for all a ∈ S

3 4

we call it an identity for (∗).

4. For each a ∈ S, there exists an element b ∈ S such that a ∗ b = e = b ∗ a,

b is called an inverse for a, and a is called an inverse for b.

In addition, if a ∗ b = b ∗ a for all a, b ∈ S, then S is called an abelian group.

In the above deﬁnition, if only conditions 1,2 are satisﬁed, then S is called a semigroup. (S, ∗) is a cancellative semigroup, if a ∗ b = a ∗ c, then b = c.

Deﬁnition 1.1.2

Let S be a group, let H ⊆ S, we call H is a subgroup if H is a group under the same operation of S.

Deﬁnition 1.1.3

Let (S, +,.) be a non-empty set with two binary operations, called addition and multiplication, and denoted by (+) and (.). Then S is called a ring with respect to these operations if the following properties hold.

1. (S, +) is an abelian group,

2. (S, .) is a semigroup,

3. For all a, b, c ∈ S, we have a.(b + c) = a.b + a.c, and (a + b).c = a.c + b.c.

A ring S has an identity element if there is an element in S, say 1 such that

1.x = x.1 = x for all x ∈ S 5

Let S be a commutative ring with identity 1. An element a ∈ S is said to be invertible if there exists an element b ∈ S such that ab = 1. The element a is called a unit of S, and its multiplicative inverse is usually denoted by a−1.

Deﬁnition 1.2.4

Let S be a commutative ring with identity, and let a ∈ S.

The ideal < a >= Sa = {x ∈ S : x = ra for some r ∈ S}

is called the principal ideal generated by a.

Deﬁnition 1.1.5

If S is commutative ring with identity 1, and each x ∈ S has a multiplicative inverse, we call it a ﬁeld.

Deﬁnition 1.1.6

Let S be a non-empty set, with two operations (+, .), we call S a semiring if the following properties hold.

1. (S, +), (S, .) are semigroups.

2. ∀a, b, c ∈ S, we have a.(b + c) = a.b + a.c and

(a + b).c = a.c + b.c.

A semiring S need not have additive or multiplicative identity. A semigroup (S, ∗) is commutative iﬀ ∀a, b ∈ S, a ∗ b = b ∗ a. We call a semiring S commutative if S(+),

S(.) are both commutative, |S| will denote the cardinality of S. 6

An element α of a semiring S is called additively absorbing if α + x = x + α = α

∀x ∈ S, and multiplicatively absorbing if α.x = x.α = α∀x ∈ S. An element ∞ of a semiring S is called an inﬁnity or (bi-absorbing element)if it is both additively and multiplicatively absorbing. An element x ∈ S is a multiplicative idempotent, if x.x = x2 = x, and an additive idempotent, if x + x = x.

Deﬁnition 1.1.7

A non-trivial commutative semiring S is said to be a semiﬁeld if there exists an element w ∈ S and (T,.) is a group such that Sw == {w} and T = S\{w}.

Theorem 1.1.8

Let (S, ∗) be a ﬁnite cancellative commutative semigroup, then (S, ∗) is a group.

Proof. Let |S| = n, S = {x1, x2, ..., xn}, let a ∈ S, and consider the set {a ∗ x1, a ∗ x2, ..., a ∗ xn}. Note that if a ∗ xi = a ∗ xj, then xi = xj, since (S, ∗) is cancellative, then S = {a ∗ x1, a ∗ x2, ..., a∗ xn}, since a ∈ S, then ∃xt ∈ S, t ≤ n, such that

a ∗ xt = a (1.1.1)

Now we will show the element xt is the identity, so let y ∈ S, then y = a ∗ xj, so y ∗ xt = (a ∗ xj) ∗ xt = (a ∗ xt) ∗ xj = a ∗ xj = y, then xt is the identity element in

(S, ∗), for any a ∈ S, we have xt = a ∗ xj, so (S, ∗) closed under inverse.

Note that an additive identity in a semiring need not be

multiplicatively absorbing, see the example below 7

Example 1.1.9

Let S = [1, ∞) ⊂ R+ (the set of positive real number), and (S, max, .), where (.) denotes the usual multiplication, a + b = max(a, b) , clearly that (S, max, .) is a semiring, note that 1 is the additive identity of S, since 1 + a = max(1, a) = a, for all 1 ≤ a, a ∈ R and not a multiplicatively absorbing, since 1.a = a, for all a ∈ S

If however a semiring has a multiplicatively absorbing additive

identity we call it a zero and denote it by o, If has additive identity (additive zero) we call it 0.

A semiring with additive zero is called zero-sum free if for all

a, b ∈ S, a + b = 0 implies a = b = 0.

Note that o=0 in the set of real numbers, rational numbers, and integers.

Deﬁnition 1.1.10

For a multiplication abelian group G, set V (G) = G∪{∞}. Extend the multiplication of G to V (G) by the rule x∞ = ∞x = ∞ for all x ∈ V (G). Deﬁne an addition on

V (G) by the rule x + x = x, x + y = ∞ for all x, y ∈ V (G) with x 6= y.

Example 1.1.11

+ Let Z◦ = Z ∪ {0}, note that (Z◦, +,.) is a zero sum-free semiring of inﬁnite cardinality.

Example 1.1.12

Let S be a semiring, Mn×n = {(aij), aij ∈ S} be the set of all n × n matrices. Then

Mn×n is a semiring under usual addition and multiplication of matrices. 8

Deﬁnition 1.1.13 [15]

Let S1 and S2 be two semirings. The direct product of S1×S2 = {(s1, s2): s1 ∈ S1 and s2 ∈ S2} is also a semiring with component-wise operations ( for any (x, y), (a, b) ∈

S1 × S2, we have (x, y).(a, b) = (xa, yb), and (x, y) + (a, b) = (x + a, y + b) ) Similarly if S1,S2, ...., Sn be a semirings, the direct product of these semirings denoted by

S1 × S2 × ... × Sn = {(s1, s2, ..., sn): si ∈ Si; i = 1, 2, ..., n} is a semiring also known as the direct product of semirings.

Deﬁnition 1.1.14

Let S be a semiring, P a subset of S, P is said to be a subsemiring of S if P itself is a semiring.

Example 1.1.15

Let Z◦ be a semiring, 2Z◦ = {0, 2, 4, ...} is a subsemiring of Z◦.

Deﬁnition 1.1.16

Let S be a semiring, x ∈ S\{o} is said to be a zero divisor of S if there exists y 6= o in S such that x.y = o.

Deﬁnition 1.1.17

Let A and B be non-empty sets. A relation R from A to B is a subset of A × B.

Relations from A to A are called relations on A, for short, if (a, b) ∈ R then we write aRb and say that a is in relation R to b. Also if a is not in relation R to b, we write a6Rb.

A relation R on a non-empty set A may have some of the following properties; 9

R is reﬂexive if for all a in A we have aRa.

R is symmetric if for a and b in A; aRb implies bRa.

R is anti symmetric if for all a and b in A; aRb and bRa imply a = b.

R is transitive if for a, b, c in A; aRb and bRc imply aRc.

A relation R on A is an equivalence relation if R is reﬂexive, symmetric and transitive.

In this case, [a] = {b ∈ A; aRb}, is called the equivalence class of a for any a ∈ A.

Deﬁnition 1.1.18 [15]

A relation R on a set A is called a partial order (relation) if R is reﬂexive, anti symmetric and transitive. In this case (A, R) is a partially ordered set or poset.

We denote the partial order relation by ≤ or ⊆ .

Deﬁnition 1.1.19 [15]

A partial order relation ≤ on A is called a total order if for each a, b ∈ A, either a ≤ b or b ≤ a.({A, ≤}) is called a chain or a totally ordered set.

Example 1.1.20

Let A = {1, 2, 3, 4, 7}, (A, ≤) is a total order, where (≤) is the usual (less than or equal to) relation, is a partial order relation.

Deﬁnition 1.1.21 [15]

Let (A, ≤) be a poset and B ⊆ A. i) a ∈ A is called an upper bound of B if and only if for all b ∈ B, b ≤ a. ii) a ∈ A is called a lower bound of B if and only if for all b ∈ B; a ≤ b. 10

iii) The greatest lower bound, whenever it exists is called the inﬁmum of B, and is denoted by infB, iv) The least upper bound of B whenever it exists is called the supremum of B and denoted by supB.

Deﬁnition 1.1.22 [15]

A poset (L, ≤) is called a semilattice order if for every pair of elements x, y in L the sup(x, y) exists (or equivalently we can say inf(x, y) exists).

Deﬁnition 1.1.23 [15]

A poset (L, ≤) is called a lattice ordered if for every pair of elements x, y in L the sup(x, y) and inf(x, y) exists.

Deﬁnition 1.1.24 [15]

An algebraic lattice (L, ∩, ∪) is a non-empty set L with two binary operation ∪(join) and ∩(meet) (also called union or sum and intersection or product respectively) which satisfy the following conditions for all x, y, z ∈ L.

L1.x ∩ y = y ∩ x, x ∪ y = y ∪ x,

L2.x ∩ (y ∩ z) = (x ∩ y) ∩ z, x ∪ (y ∪ z) = (x ∪ y) ∪ z,

L3.x ∩ (x ∪ y) = x, x ∪ (x ∩ y) = x,

L4.x ∩ x = x, x ∪ x = x. Remark

Let (L, ≤) be a lattice ordered set. If we deﬁne x∩y = inf(x, y) and x∪y = sup(x, y) then (L, ∪, ∩) is an 11

1.2 Ideals and Bi-Ideals of a Semirings

Deﬁnition 1.2.1

Let S be a semiring, I be a non-empty subset of S. I is a right(left) ideal of S if

1. a + b ∈ I for any a, b ∈ I,

2. For all i ∈ I and s ∈ S we have is ∈ I (si ∈ I).

Deﬁnition 1.2.2

Let S be a semiring. A non-empty subset I of S is said to be an ideal of S if I is simultaneously a right and left ideal of S.

Example 1.2.3

nZ◦ is an ideal of Z◦, for any positive integer n.

Deﬁnition 1.2.4

Let S be a semiring, I ⊆ S is a bi-ideal of S if

(SI) ∪ (IS) ∪ (I + S) ∪ (S + I) ⊆ I, such that SI = {xy : x ∈ S, y ∈ I},

S + I = {x + y, x ∈ S, y ∈ I}

That is I is an ideal of S under (+) and (.)

Note that if I an bi-ideal, then I is an ideal, but the converse not true.

Counter example, let I =< 2 >= {2n : n ∈ Z} be an a principal ideal generated by 2, let 1 ∈ Z, note that 1 + 2 = 3 ∈/ I 12

Deﬁnition 1.2.5

A non trivial semiring S is called an ideal-simple(bi-ideal-simple) if for any non-trivial ideal (bi-ideal) I of S we have I = S.

Note that any ideal simple semiring is a bi-ideal simple semiring, but the converse is not true.

Counter example, note that Z◦ bi-ideal simple, but not ideal-simple.

Lemma 1.2.6 [13]

Let S be an additively commutative semirings. a one-element subset {w} of S is a bi-ideal of S if and only if w is an inﬁnity element of S.

Proof. Suppose that {w} is a bi-ideal, then S + {w}∪ S{w} ∪ {w}S ⊂ {w}, then

S + {w} = S{w} = {w}S = {w}. Then w must be an inﬁnity element of S.

Conversely, suppose that w is an inﬁnity element, then S.w = S + w = w.S = {w}, so {w} is a bi-ideal of S.

Lemma 1.2.7 [13]

Let S be an additively commutative semirings, then:

1. The subsets S, S + S, SS + S, SSS + S,... are bi-ideal of S.

2. SaS + S is a bi-ideal of S for every a ∈ S.

In the remaining cases, assume that ∞ ∈ S.

3. The set {x ∈ S; xSx = {∞}} is a bi-ideal. 13

4. The set {x ∈ S; x + xSx = {∞}} is a bi-ideal.

5. The set {x ∈ S; 2x = ∞} is a bi-ideal.

6. The sets (∞ : S)l = {x ∈ S; xS = {∞}}, (∞ : S)r = {x ∈ S; Sx = {∞}} and

(∞ : S)m = {x ∈ S; SxS = {∞}} are bi-ideals.

Deﬁnition 1.2.8

A semiring is a bi-ideal-free semiring, iﬀ it has no a proper bi-ideal.

Proposition 1.2.9

A semiring S is bi-ideal-free if and only if S = SaS + S for every a ∈ S.

Proof. Suppose that S is a bi-ideal-free, so SaS + S is a bi-ideal from (1.2.7)part(2) and so S = SaS + S for every a ∈ S. Conversely Suppose that S = SaS + S for every a ∈ S, let I ⊆ S be a bi-ideal, then we have SI ⊆ S, S = SIS ⊂ IS ⊂ I,

S = SIS + S ⊆ I + S ⊆ I, and from assumption we have I ⊆ S = SaS + S, then

S = I, so S is a bi-ideal-free.

1.3 Congruence Relations

Deﬁnition 1.3.1

[1]

A congruence relation on a semiring S is an equivalence relation ∼ that also satisﬁes 14

  c + x ∼ c + x  1 2  x1 + c ∼ x2 + c x1 ∼ x2 =⇒  cx ∼ cx  1 2  x1c ∼ x2c

for all c ∈ S, for all x1 ∼ x2.

We call S a congruence simple semiring (C-simple semiring) when S has only two congruence relations ids and S × S (trivial congruence relation), where ids is the identity relation, i.e, ids = {(x, x): x ∈ S}

Note that the trivial semiring of order 1 and every semiring of order 2 are congruence simple semirings, because if S = {a, b} be a semiring, then ids = {(a, a), (b, b)} and S × S = {(a, a), (b, b), (a, b), (b, a)}, are the only equivalence relations on S.

Lemma 1.3.2 [1]

If B ⊆ S is a bi-ideal then ids ∪ (B × B) is a congruence relation.

Proof. Let r = ids ∪ (B × B), note that (x, x) ∈ r for all x ∈ S, r is reﬂexive. Let

(x, y) ∈ r. if (x, y) ∈ ids, then (y, x) ∈ r, if (x, y) ∈ B × B, then (y, x) ∈ r, so r is symmetric.

Now let (x, y) ∈ r,(y, z) ∈ r

1. If x = y = z ,then (x, z) ∈ r.

2. If x 6= y = z, then (x, y) = (x, z) ∈ r.

3. If x = y 6= z, then (x, z) = (y, z) ∈ r. 15

4. if x 6= y 6= z, then (x, y), (y, z) ∈ B × B, then (x, z) ∈ B × B, so (x, z) ∈ r.

then r is transitive, so r is an equivalence relation.

Now Let (x, y) ∈ r, if x = y, then (x+c, y +c), (c+x, c+y), (cx, cy), (xc, cy) ∈ ids, and if x 6= y, then (x + c, y + c), (c + x, c + y), (cx, cy), (xc, cy) ∈ B × B, since B is a bi-ideal then r = ids ∪ (B × B) is congruence relation.

Lemma 1.3.3 [1]

If B ⊆ S is a bi-ideal and S is c-simple, then |B| = 1 or B = S.

Proof. From (1.3.2) we have ids ∪B×B is congruence relation, and since S is c-simple semiring we have,

ids ∪ (B × B) = ids, so |B| = 1.

ids ∪ (B × B) = S × S, so B = S.

Deﬁnition 1.3.4

Let S and R be semirings, and f : R −→ S a function. Then f is called a semiring homomorphism if for all x, y ∈ R

f(x + y) = f(x) + f(y). and f(xy) = f(x)f(y).

A bijective semirings homomorphism is called a semiring isomorphism. 16

Deﬁnition 1.3.5 [2]

Let π : R −→ S be a semiring homomorphism. The set {a ∈ R : π(a) = o}

is called the kernel of π, denoted by ker(π).

Lemma 1.3.6

[2]

Let π : R −→ S be a semiring homomorphism. Then ker(π) is an ideal of R.

Proof. If a, b ∈ ker(π), then

π(a ± b) = π(a) ± π(b) = o ± o = o,

and so a ± b ∈ ker(π). If r ∈ R, then

π(ra) = π(r).π(a) = π(r).o = o,

so ra ∈ ker(π), so ker(π) is an ideal of R

Lemma 1.3.7 [2]

Let S be a semiring.

1. If ∼ is a congruence relation on S, then the set of equivalence classes S/∼, is a

semiring under the operations [a]+[b] = [a+b], [a][b] = [ab] , and π : S −→ S/ ∼

where π(s) = [s] is a semiring homomorphism, such that [a] = {x ∈ S : x ∼ a}

equivalence class of a 17

2. If f : S −→ T is a semiring homomorphism, then

x ∼ y iﬀ f(x) = f(y)

deﬁnes a congruence relation on S.

Proof. 1. We need to show that the operations

[a] + [b] = [a + b] and [a].[b] = [a.b]

are well deﬁned. For this, suppose a1 ∼ a2 and b1 ∼ b2. Then

a1 ∼ a2 =⇒ a1 + b2 ∼ a2 + b2,

b1 ∼ b2 =⇒ a1 + b1 ∼ a1 + b2.

It follows from transitivity of ∼ that a1 + b2 ∼ a2 + b2, so (+) is well deﬁned on equivalence classes. A similar argument shows that (.) is well deﬁned.

Associativity of both operations and distributivity then follow from associativity

and distributivity in S. And so π is a semiring homomorphism is obvious.

2. The second statement, we have (x, x) ∈∼, note that (y, x) ∈∼ for all (x, y) ∈∼,

and if (x, y) ∈∼, (y, z) ∈∼, then f(x) = f(y), f(y) = f(z), so f(x) = f(z), then

(x, z) ∈∼, then ∼ is reﬂexive, symmetric and transitive, so ∼ is equivalence

relation. Now let (x, y) ∈∼, for every c ∈ S, we have f(cx) = f(c).f(x) =

f(c).f(y) = f(cy), then (cx, cy) ∈∼, and by the same way we have (xc, yc) ∈∼,

see f(c + x) = f(c) + f(x) = f(c) + f(y) = f(c + y), so (c + x, c + y) ∈∼, and

by the same way (x + c, y + c) ∈∼, then ∼ is a congruence relation on S. 18

Lemma 1.3.8 [10]

Let S be a commutative ring, then S is a congruence simple ring iﬀ S is ideal simple.

Proof. Assume that S is a congruence simple, suppose that I is a proper ideal in a ring S, we deﬁne a relation ∼ by a ∼ b if a − b ∈ I, so ∼ is an equivalence relation.

Let a ∼ b, then a − b = a − b + (c − c) = (a + c) − (b + c) ∈ I, so a + c ∼ b + c.

And c(a − b) = ca − cb in I since I is an ideal, then ca ∼ cb, let x ∈ S, x∈ / I, then

2x − x = x∈ / I, then (2x, x) ∈/∼, and note that for any x ∈ S\{0}, we have 2x 6= x, so ∼ is a nontrivial congruence relation, is a contradiction.

For the converse, assume that S is ideal simple, let ∼ be a non trivial congruence relation of S, so the set of equivalence classes S/ ∼ form a ring. Furthermore, there is an induced homomorphism π : S −→ S/ ∼ , so the kernel of this homomorphism is a proper ideal in S, since there is (x, y) ∈/∼ then (0, y − x) ∈/∼, so π[y − x] 6= [0], so this a contradiction, then S is a congruence simple ring.

Theorem 1.3.9 [10]

Let S be a commutative ring, so S is an ideal simple iﬀ S is a ﬁeld or S is a zero multiplication ring ({S2 = {o}}) of prime order.

Proof. Suppose that S is an ideal simple, so < a > is an ideal for all a ∈ S, then

< a >= {o} or < a >= S. Now if < a >= {o} ∀a ∈ S, so S2 = {o}, let I be a subgroup of (S, +), we deﬁne a relation ∼ of S by a ∼ b if a − b ∈ I, so ∼ is a non-trivial congruence relation, see proof(1.3.8) a contradiction, so (S, +) has no non- trivial subgroup, then (S, +) has prime order. Assume < a >= S = {ra : r ∈ S} for some a ∈ S, from this we have a = ra for some r ∈ S, rx = rrta = rt(ra) = rta = x, 19

then r is the identity in (S, .), so | < x > | ≥ 2, ∀x ∈ S\{o}, so < x >= S, since

S is ideal simple, then for any x ∈ S\{o} there exist y ∈ S\{o}, such that r = xy, then (S, .) closed under inverse, then S is aﬁeld. Conversely if S is aﬁeld, assume I a non-trivial ideal of S, so I has the identity, that mean I = S, so S is an ideal simple.

If S is a zero multiplication ring ({S2 = {o}}) of prime order, then (S, +) has no any non-trivial subgroup, so S has no non-trivial ideal, then S is an ideal simple. Chapter 2

Bi-Ideal-Simple Semirings

2.1 Bi-Ideal-Simple Semiring-Introduction

In this chapter we introduce and study in details all concepts related to bi-ideal simple semirings with additively commutative semirings, see [10],[13]

Proposition 2.1.1 [13]

A semiring S is bi-deal-simple if and only if at least one (and then just one) of the following ﬁve cases takes place:

1. |S| = 2;

2. |S| ≥ 3 and S = SaS + S for every a ∈ S;

3. |S| ≥ 3, ∞ ∈ S and S = SaS + S for every a ∈ S, a 6= ∞;

4. |S| ≥ 3, ∞ ∈ S, S + S = {∞}, and SaS = S for every a ∈ S, a 6= ∞;

5. |S| ≥ 3, ∞ ∈ S, SS = {∞}, and S + a = S for every a ∈ S, a 6= ∞.

20 21

Proof. Assume S is a bi-ideal simple, suppose that |S| = 2 so (1) hold. Now assume that |S| ≥ 3 and if S is a bi-ideal-free, then (2) hold from (1.2.9).

If S is not bi-ideal-free, so there is a proper subset I such that I is a bi-ideal of S, so |I| = 1, and from (1.2.6), we have ∞ ∈ S. We distinguish the following four cases,

1. Let S +S = {∞}, SS = {∞}, then every subset of S containing ∞ is a bi-ideal,

let A = {x, ∞}, so A is an bi-ideal, then A = S, then |S| = 2, a contradiction.

2. Let S + S={∞},SS 6= {∞}, then there exists x, y ∈ S such that xy 6= ∞,

so xS 6= {∞}, Sy 6= {∞}, then (∞ : S)l, (∞ : S)r are proper subsets of S, so

(∞ : S)l = (∞ : S)r = ∞, and by combination, we get (∞ : S)m = {∞} . Now if a ∈ S, a 6= ∞, then SaS 6= {∞}, since S + S = {∞}, the set SaS is a

bi-ideal. Thus SaS = S and (4) is true.

3. Let S + S 6= {∞}, SS = {∞}. Then every ideal of S(+) is a bi-ideal, and so

the semigroup S(+) is ideal-simple, and since

S + a is an ideal of S(+), so bi-ideal, then S + a = S, so (5)is hold.

4. Let S + S 6= {∞} 6= SS, so we have (∞ : S)l = (∞ : S)r = (∞ : S)m = {∞} from (3), further more S + S = S by (1.2.7)part(1). Now, if a ∈ S, a 6= ∞,

then bac 6= ∞ for some b, c ∈ S, and b = d + e, d, e ∈ S. Then ∞ 6= bac =

dac + eac ∈ SaS + S, so |SaS + S| ≥ 2 and consequently, SaS + S = S by

(1.2.7)part(2). It means that (3) is true.

Conversely, assume that (1) hold, then S is a bi-ideal simple. If (2) hold, then

from (1.2.9), We done. If (3) hold, let I a bi-ideal of S, |I| ≥ 2, let a ∈ I\{∞},

then S = SaS + S ⊆ I, so I = S, then S is a bi-ideal simple. If (4) or (5) 22

holds, so by the same way for any a ∈ I\{∞}, we have S = SaS = S + a ⊆ I,

so I = S, then S is a bi-ideal simple.

In the rest of this section, we assume that ∞ ∈ S.

Deﬁnition 2.1.2 [13]

A semiring S is called a nil-semiring, if for every x ∈ S there exist n ≥ 1 with xn = ∞, we call a ∈ S, a 6= ∞, be a nilpotent if there exist n ≥ 1,such that xn = ∞

Proposition 2.1.3 [13]

Let S be a bi-ideal simple, if S is a nil-semiring, then SS = {∞}.

Proof. If |S| = 2 it is hold. And so let |S| ≥ 3. If a ∈ S, a 6= ∞, is such that either

S = SaS + S or S = SaS, then a = bac + w, b, c ∈ S, w ∈ S ∪ {0}, and we have, a = b2ac2 + bwc + w = b3ac3 + b2wc2 + bwc + w = ...,a contradiction, since bn = ∞.

Thus SaS + S 6= S 6= SaS, so from (2.1.1) we have case (5) is true, then SS = {∞}.

Lemma 2.1.4 [13]

Let S be a bi-ideal simple, if |S| ≥ 3 and SS 6= {∞}, then for every a ∈ S, a 6= ∞, there exist elements b, c, d ∈ S such that bac 6= ∞ 6= ada.

Proof. From (1.2.7)part(3) say M = {x ∈ S : xSx = {∞}} is a bi-ideal, from

(2.1.3) there exist x ∈ S such that xn 6= ∞ for all n ≥ 1. M = {∞} or M = S, since S is a bi-ideal simple, but M = S is false since xn 6= ∞ for all n ≥ 1, so 23

M = {∞}, then for all a ∈ S, a 6= ∞ there exist d ∈ S, ada 6= ∞. From (1.2.7)part(6), say B = {x ∈ S : SxS = {∞}} is a bi-ideal, so B 6= S, B = {∞}, then for all a ∈ S, a 6= ∞ there exist c, b ∈ S, bac 6= ∞.

Lemma 2.1.5 [13]

Let S be a bi-ideal simple, so Just one of the following two cases takes place:

1. x + xSx = {∞} for every x ∈ S

2. For every a ∈ S, a 6= ∞, there exists at least one b ∈ S with a + aba 6= ∞

Proof. From (1.2.7)part(3) say M = {x ∈ S : x+xSx = {∞}}is a bi-ideal, so M = S or M = {∞}. If M = S, then x + xSx = {∞} for all x ∈ S, if M = {∞}, then for all a ∈ S, a 6= ∞ there exist b ∈ S such that a + aba 6= ∞.

Lemma 2.1.6 [13]

Let S be a bi-ideal simple, so Just one of the following two cases takes place:

1. 2x = ∞ for every x ∈ S

2. 2a 6= ∞ for every a ∈ S, a 6= ∞.

Proof. From (1.2.7)part(5) say M = {x ∈ S : 2x = ∞} is a bi-ideal, so M = S or

M = {∞}, if M = S , then 2x = ∞ for all x ∈ S, if M = {∞}, then 2a 6= ∞ for every a ∈ S, a 6= ∞. 24

Lemma 2.1.7 [13]

If S is a bi-ideal-simple semiring of type (2.1.1(4)), then the multiplication semigroup of S is an ideal-simple.

Proof. Suppose that I ⊂ S(.) is an ideal, I 6= {∞}, then |I| ≥ 2, let a ∈ I, a 6= ∞, then S = SaS ⊂ I, then I = S, so S(.) is an ideal simple semiring.

Lemma 2.1.8 [13]

Let S be a multiplicative ideal-simple semigroup with |S| ≥ 3 and ∞ ∈ S. Setting

S + S = {∞} we get a bi-ideal-simple semiring of type (2.1.1(4)).

Proof. Suppose that I ⊂ S is a bi-ideal and |I| ≥ 2, then I(.) is an ideal, so I = S, since S(.) is an ideal simple, so S is a bi-ideal simple, and since S + S = {∞}, then from Prop.(2.1.1) S is a bi-ideal simple semiring of type (2.1.1(4))

Lemma 2.1.9 [13]

If S is a bi-ideal-simple semiring of type (2.1.1(5)), then T (+) is an (abelian) subgroup of S(+), where T = S\{∞}.

Proof. Note that S(+) is closed, since a + b = ∞ iﬀ a = ∞ or b = ∞, and T (+) is an associative. Now S+a = S for all a ∈ S, a 6= ∞, so there exist e ∈ S, e 6= ∞, e+a = a, so e is the identity for a, we show e is the identity for all x ∈ S, x = t + a for some t ∈ S, x 6= ∞, so x = t + (e + a) = (t + a) + e = x + e, so e is the identity of S(+), since S + a = S for all a 6= ∞, then there exist x ∈ S such that x + a = e, so S(+) 25

is closed under inverse, then S(+) is a group, and so T (+) is an abelian subgroup of

S(+).

Lemma 2.1.10 [13]

Let T (+) be an abelian group, |T | ≥ 2, ∞ ∈/ T and S = T ∪ {∞}. Setting SS =

S + ∞ = ∞ + S = {∞} we get a bi-ideal-simple semiring of type(2.1.1(5))

Proof. We show that S +a = S for all a 6= ∞, let x ∈ S, so x = x+e = x+(−a+a) =

(x − a) + a ∈ S + a, then S + a = S. Now let I is a bi-ideal of S, I 6= {∞}, let a ∈ I, a 6= ∞, then S = S + a ⊂ I, so I = S, so S is a bi-ideal simple semiring of type

2.1.1.(5)

2.2 Additively Zeropotent Semirings

Deﬁnition 2.2.1 [13]

We call S is additively zeropotent semiring(a zp-semiring for short). That is, ∞ ∈ S and 2S = {∞},(x + x = ∞ for all x ∈ S).

In the rest of this section, we will assume that S a zp − semiring.

Deﬁnition 2.2.2 [13]

The relation ≤ on S by a ≤ b iﬀ b ∈ (S + a) ∪ {a}. ≤ is a relation of order which is compatible with the two operation deﬁned on S. That is, ≤ is an ordering of the semiring S. Clearly, ∞ is a greatest element of S. 26

Lemma 2.2.3 [13]

If |S| ≥ 2, then an element a ∈ S, a 6= ∞, is maximal in S\{∞} if and only if

S + a = ∞.

Proof. Suppose that a is a maximal, that is x ≤ a for all x ∈ S\{∞}, then a ∈

(S + x) ∪ {x}. Now if |S| = 2, then S + a = {∞ + a, a + a} = {∞}, if a 6= x,

|S| ≥ 2 then a = t + x for some t ∈ S, assume that S + a 6= {∞}, so there exist r ∈ (S + a) ∪ {∞}, a 6= r, such that a ≤ r, a contradiction, so S + a = {∞}.

Conversely, assume that S + a = {∞}, a not a maximal, then there exist x 6= a 6=

∞, such that a ≤ x, then x ∈ (S + a) ∪ {a}, then x ∈ S + a, since x 6= a, so x = ∞ contradiction.

In the rest of this section, we will assume that S = S + S.

Lemma 2.2.4 [13]

If |S| ≥ 2, then S has no minimal elements.

Proof. suppose that S has minimal element a 6= ∞, so a ≤ x for all x ∈ S, but a = b + c, so b ≤ a. If b = a, then a = a + c = a + 2c = a + ∞ = ∞, a contradiction.

Corollary 2.2.5 [13]

Either |S| = 1 or S is inﬁnite. 27

Proof. suppose that S is ﬁnite, say |S| = n, for all x ∈ S, we have x = x1+x2, x1 6= x2, we continue in this we have x = x1 + x2 + x3 + ... + xn−1, but xs = xj + xi, i, j, s ∈ {1, 2, 3, ...n − 1}, thats mean x = ∞, and so |S| = 1.

Lemma 2.2.6 [13]

The only idempotent element of S is the bi-absorbing element ∞.

Proof. Let b2 = b for some b ∈ S. Then b = c + d and

b = b3 = b(c + d)b = bcb + bdb. (2.2.1)

Of course, c ≤ b, d ≤ b, and hence cd ≤ bd, cd ≤ cb, ∞ = 2cd ≤ bd + cb and bd + cb = ∞. Finally, from eq(2.2.1), we have ∞ = b∞b = bdb + bcb = b.

Corollary 2.2.7 [13]

If S contain a left(or right) unity, then |S| = 1.

Proof. Suppose that S has a left unity e, so e.x = x for all x ∈ S, note that e2 = e, so e is an idempotent, so e = ∞ from (2.2.6), and so |S| = 1. And by the same way when a right unity.

Lemma 2.2.8 [13]

If ak = al for some a ∈ S and 1 ≤ k ≤ l, then ak = ∞. 28

Proof. There are positive integers m, n such that m(l − k) = k + n. Now, If b = ak+n, then b = akan = alan = akal−kan = alal−kan = akal−kal−kan = ... = akam(l−k)an = a(2k+2n) = b2. By (2.2.6), b = ∞, and hence ak = al = akal−k = akal−kal−k = ... = akam(l−k) = aka(k+n) = akb = ∞

Lemma 2.2.9 [13]

Let a, b ∈ S and k, l ≥ 1 be such that ak = al + b. Then a2k = ∞. Moreover, if 2k ≤ l, then ak = ∞.

Proof. We have a2l + alb = al(al + b) = ak+l = (al + b)al = a2l + bal. Consequently, a2k = (al + b)2 = a2l + alb + bal = a2l + bal + b2 = ∞. If 2k ≤ l, then a2k = ∞ implies al = ∞ and hence ak = al + b = ∞.

Lemma 2.2.10 [13]

If a ∈ S is a non-nilpotent element. Then the powers a1, a2, a3, ... are pair-wise incomparable.

Proof. Suppose that an ≤ am for some n, m, then am ∈ (S + an) ∪ {an}. Now, if am = an, so am = ∞, if m ≤ n, or an = ∞ if n ≤ m, this a contradiction. If am = b + an, then a2m = ∞ from (2.2.8), a contradiction.

Lemma 2.2.11 [13]

If a, b ∈ S are such that aba ≤ a, then a = ∞ 29

Proof. If aba = a, then (ab)2 = ab, ab = ∞ by (2.2.5) and a = aba = ∞a = ∞. If aba + c = a, then ab = abab + cb = (ab)2 + cb, ab = ∞ by (2.2.9) and a = ∞, too.

Lemma 2.2.12 [13]

If a, b ∈ S are such that ab = a 6= ∞ (ab = b 6= ∞, resp.), then a b (b a, resp.).

Proof. Firstly, a 6= b by (2.2.6). Now if a ≤ b, then b = a + c, ac ≤ bc, c ≤ b, ac ≤ ab = a, ∞ = 2ac ≤ a+bc and a+bc = ∞. Thus ∞ = a(a+bc) = a2 +abc = a2 +ac = a(a + c) = ab = a, a contradiction. Similarly the second case.

Proposition 2.2.13 [13]

Let S be a zp-semiring with S + S = S and |S| ≥ 2. Then:

(i) S is inﬁnite.

(ii) The ordered set (S, ≤) has no minimal elements.

(iii) The inﬁnity element ∞ is the only idempotent element of S.

(iv) S contains neither a left nor a right unit.

(v) If a ∈ S is not nilpotent, then the elements ai, i ≥ 1, are pair-wise

incomparable in (S, ≤)

(vi) If a 6= ∞, then aba a for every b ∈ S.

(vii) If ∞ 6= a ≤ b, then ab 6= a.

(viii) If ∞ 6= b ≤ a, then ab 6= b.

Proof. Combine (2.2.4),(2.2.5),(2.2.6),(2.2.7),(2.2.10),(2.2.11),(2.2.12) 30

2.3 Bi-Ideal Simple Zeropotent Semirings

In this section let S be a zp-semiring with ∞ ∈ S

Proposition 2.3.1 [13]

Let S be a zp-semiring with |S| ≥ 3. Then S is a bi-ideal- simple if and only if at least one (and then just one) of the following two cases takes place:

1. S + S = {∞} and SaS = S for every a ∈ S, a 6= ∞;

2. S = S + SaS for every a ∈ S, a 6= ∞

Proof. From (2.1.1), we have 2.1.1(1),2.1.1.(2) not holds, because |S| ≥ 3, and ∞ ∈ S, and 2.1.1.(5) not hold, because if S + a = S, a 6= ∞, we have a = c + a for some c ∈ S, so a = c + a = c + (c + a) = 2c + a = ∞, a contradiction, so now remain cases

2.1.1(3) or 2.1.1(4). Conversely, from (2.1.1)

In the rest of this section, let S be a bi-ideal-simple semiring such that S + S 6=

{∞}. Then S + S = S and S is inﬁnite see (2.2.13) . Moreover, S is not nil, since if SS = {∞}, let x 6= ∞, x2 = ∞, a contradiction, because the only idempotent element is ∞, so SS 6= {∞}, then by (2.1.3), S is not nil.

Deﬁnition 2.3.2 [13]

Let V be a ﬁnite subset of S\{∞} a sequence v1, ..., vk, k ≥ 2, of elements from V will be called admissible in the sequel if these elements are pair-wise distinct and viaivi ≤ vi+1 for some ai ∈ S, 1 ≤ i ≤ k − 1. 31

Lemma 2.3.3 [13]

Let V be a ﬁnite subset of S\{∞}. Then there exists at least one element a ∈ S such that a 6= ∞ and a v for every v ∈ V.

Proof. Firstly, by (2.3.1), we have S = S + SbS for every b ∈ S, b 6= ∞. In particular,

SbS 6= {∞},SS 6= {∞} and, by (2.1.4), for every w ∈ V there is at least one aw ∈ S with waww 6= {∞}. Then aw 6= w and waww ∞ by (2.2.10). Now, if there is no admissible sequence, then, waww 6= ∞ and waww v for all w, v ∈ V. The result is proved in this case, and hence we can assume that v1, ..., vk, k ≥ 2, is an admissible sequence with maximal length k.

Let m be maximal with respect to 1 ≤ m ≤ k and vkbvk ≤ vm for at least one b ∈ S. Then m < k by (2.2.11) and vmamvm ≤ vm+1 implies vkbvkamvkbvk ≤ vm+1, a contradiction with the maximality of m. We have thus shown that vkcvk vi for all 1 ≤ i ≤ k and c ∈ S. In particular, ∞ 6= vkakvk vi for some ak ∈ S and all i, 1 ≤ i ≤ k. Finally, it follows from the maximality of k that vkakvk v for every v ∈ V .

Corollary 2.3.4 [13]

Denote by A the set of maximal elements of (S\{∞}, ≤), see (2.2.2) and assume that every element from S\{∞} is smaller or equal to an element from A. Then the set

A is inﬁnite.

Proof. Suppose that A is a ﬁnite, and from assumption we have x ≤ a for all x ∈

A, but from (2.3.3) there exist at least b ∈ S\{∞}, such that b a, b 6= ∞, a contradiction, so A is inﬁnite. 32

Proposition 2.3.5 [13]

Just one of the following two cases takes place:

1. x + xSx = {∞} and xm + xn = ∞ for every x ∈ S and all positive integers

m, n.

2. For every a ∈ S, a 6= ∞, there exists at least one b ∈ S such that a + aba 6= ∞.

Proof. Taking into account (2.1.5) we may assume that x + xSx = {∞} for every x ∈ S\{∞}. Then x + x3 = ∞ and we put I = {a ∈ S : a + a2 = ∞}. Clearly, I is an ideal of S(+). Moreover, if a ∈ I and b ∈ S, then ab + abab = (a + aba)b = ∞.

Thus ab ∈ I, similarly ba ∈ I and we see that I is a bi-ideal of S.

If I = {∞}, then a + a2 6= ∞ for every a ∈ S, a contradiction with (2.1.3), thus

I 6= {∞} and we get I = S and x + x2 = ∞ for every x ∈ S. Further, x + x = ∞ by the zp-property and x+xn = ∞ for every n ≥ 3, since x+xSx = {∞}. If 2 ≤ n ≤ m, then xn + xm = xn−1(x + xm−n+1) = xn−1∞ = ∞.

Proposition 2.3.6 [13]

Denote by A the set of maximal elements of the ordered set (S\{∞}, ≤). If A is non-empty, then x + xSx = {∞} for every x ∈ S (i.e., the case 2.3.5(1) takes place)

Proof. From (2.2.3), we have S + a = {∞}, and from (2.3.5) there exist b ∈ S, such that a + aba 6= ∞ a contradiction, since S + a = {∞}, and so x + xSx = {∞} for every x ∈ S. Chapter 3

Congruence-Simple Semirings

3.1 Congruence-Simple Commutative Semirings

The goal of this chapter is to derive some basic structure information for congruence simple commutative semirings and ﬁnite, additively commutatively, congruence- simple semirings. see [10], [1], [11], [12].

Theorem 3.1.1

Let S be a commutative semiring, all the following relation are congruence relations.

1. (x, y) ∈ r iﬀ 2x = 2y,

2. (x, y) ∈ s iﬀ there exist a non-negative i and elements u, v ∈ S ∪ {0} such that

2ix = y + u, 2iy = x + v,

3. (x, y) ∈ t iﬀ 3x = 3y,

4. (x, y) ∈ p iﬀ a + x = a + y,

5. (x, y) ∈ ha iﬀ ax = ay, for every a ∈ S,

33 34

6. qT = T × T ∪ ids,TT ⊂ T ,

7. (x, y) ∈ s if and only if {z1 ∈ S; ax + z1 = ∞} = {z2 ∈ S; ay + z2 = ∞} for every a ∈ S.

Proof. Part 1.

• (x, x) ∈ r ∀x ∈ S because 2x = 2x then r is reﬂexive

• If (x, y) ∈ r, then 2x = 2y, so 2y = 2x,we have (y, x) ∈ r then r is a symmetric

• If (x, y) ∈ r, (y, z) ∈ r, then 2x = 2y = 2z, so (x, z) ∈ r, then r is transitive.

from three cases above we have r is an equivalence relation. Now let (x, y) ∈ r,

then 2x = 2y, cx + cx = cy + cy, we have 2cx = 2cy, then (cx, cy) ∈ r, by the

same we have (xc, yc) ∈ r, 2x + 2c = 2y + 2c, then 2(x + c) = 2(y + c), then

(x + c, y + c) ∈ r, by the same we have (c + x, c + y) ∈ r, then r is a congruence

relation semiring.

And the remaining cases by the same way.

Theorem 3.1.2 [10]

Let S be a congruence-simple semiring. Then just one of the following three cases takes place:

1. S is a two-element semiring isomorphic either to T1orT2, table[I].

2. S is additively idempotent.

3. S is additively cancellative. 35

Proof. From (3.1.1) part (1) (x, y) ∈ r iﬀ 2x = 2y, r is a congruence relation on S, then r = ids or r = S × S since S is congruence simple semiring. Let r = ids then the mapping x −→ 2x is an injective transformation of S because if 2x = 2y, then

(x, y) ∈ r, then x = y, since r = ids. Now we deﬁne a congruence relations, see

(3.1.1) part (2), We have (x.2x) ∈ s ∀x ∈ S, because 2x = 2x + 0, 2(2x) = x + 3x, and if s = ids then x = 2x = x + x, then S(+) is an idempotent. If s = S × S, let a + b = a + c for some a, b, c ∈ S, then there exist i ≥ 0, w ∈ S ∪ {0} such that

2ib = a + w, then b + 2ib = b + a + w = c + a + w = c + 2ib, so

2b = c + b, i = 0. (3.1.1)

And by the same above there exist j ≥ 0 such that 2jc = a + v, v ∈ S ∪ {0}, so c + 2jc = c + a + v = b + a + v = b + 2jc, so,

2c = b + c, j = 0, (3.1.2)

so from eq(3.1.1), eq(3.1.2) we have 2b = 2c, then (b, c) ∈ r = ids then b = c. Now if i > 0, b + 2(2i−1)b = b + 2ib = c + 2ib = c + 2(2i−1b), and so 2b + 2(2i−1b) = b + c + 2(2i−1b), by add b to the two side, and so

2(b + 2i−1) = b + c + 2(2i−1b). (3.1.3)

And b + c + 2(2i−1b) = 2c + 2(2i−1b), by add c to the two side, and so,

b + c + 2(2i−1b) = 2(c + 2i−1b). (3.1.4)

We have from eq(3.1.3), eq(3.1.4) 2(b+2i−1b) = 2(c+2i−1b), so (b+2i−1b, c+2i−1b) ∈ r, so b + 2i−1b = c + 2i−1b. Now we show by induction 2b = c + b for all i ≥ 1. If i = 1, 36

we have 2b = c + b, suppose that the statement

[(b + 2i−1b) = (c + 2i−1b) =⇒( 2b=b+c)]

is true for i = k we have b + 2k−1b = c + 2k−1b =⇒ 2b = b + c, we note that,

b + 2kb = b + 2(2k−1)b = c + 2(2k−1)b = c + 2kb, then

b + 2k−1b + 2k−1b = c + 2k−1b + 2k−1b,

so 2b + 2k−1b + 2k−1b = b + c + 2k−1b + 2k−1b,

so 2(b + 2k−1b) = (b + 2k−1b) + (c + 2k−1b),

so 2(b + 2k−1b) = 2(c + 2k−1b), then b + 2k−1b = c + 2k−1b so,

2b = b + c. (3.1.5)

And by the same proof for the statement[c + 2jc = b + 2jc] we

have 2(c + 2j−1c) = 2(b + 2j−1c), so

(c + 2j−1c) = (b + 2j−1c), and so by induction we also have

2c = b + c. (3.1.6)

And so from (3.5),(3.6) we have 2b = 2c so, (b, c) ∈ r, so b = c, then we have proved that S(+) is cancellative.

Now let r = S × S, then we have 2x = 2y for all x, y ∈ S, then there exist g ∈ S such that 2x = g, ∀x ∈ S, and we let a congruence relation t from (3.1.1) part(3)

,(x, y) ∈ t iﬀ 3x = 3y and assume that a + b = a + c, if t = ids then x → 3x is an injective mapping, because if 3x = 3y then(x, y) ∈ t = ids, then x = y, and we have 3b = b + 2b = b + g = b + 2a = (b + a) + a = c + a + a = c + g = c + 2c = 3c, then b = c, so S(+) is cancellative. 37

If t = S × S, then we have x + g = x + 2x = 3x = 3g = g + 2g = g + g =

2g = g, ∀x ∈ S. We have to consider the following four cases,

1. Let S + S 6= {g},SS 6= {g}, then there are a, b, c, d ∈ S such that a + b = c 6=

g, cd 6= g, since x(g) = xg + xg = 2xg = g for every x ∈ S, we know that all the

elements a,b,c,d are diﬀerent from g. Put I = {x : x = cu+v, u ∈ S, v ∈ S∪{0}},

is a bi-ideal of S and g = cg + g, cd = cd + g and so g, cd ∈ I, g 6= cd then

|I| ≥ 2, then I = S, and in particular a = cu1 + v1, b = cu2 + v2 for suitable

u1, u2 ∈ S and v1, v2 ∈ S ∪ {0}. But then a = (a + b)u1 + v1 = au1 + bu1 + v1

= au1 + (cu2 + v2)u1 + v1 = au1 + cu2u1 + v2u1 + v1, b = (a + b)u2 + v2 =

bu2 + v2 + v1v2 + cu1u2, and o 6= c = a + b = au1 + v1 + v2u1 + bu2 + v2 + 2cu1u2,

let x = au1 + v1 + v2u1 + bu2 + v2, and y = cu1u2, since 2y = g∀y ∈ S, then

2cu1u2 = g, x + g = g ∀x ∈ S, then g 6= c = x + 2y = x + g = g, is a contradiction.

2. Let S + S 6= {g} = SS, we have a + b = c 6= g, for some a, b, c ∈ S, and from

(3.1.1) part (4) (x, y) ∈ p iﬀ a + x = a + y, so p is a congruence relation of S,

since a 6= g, a + a = a + g = g, we have (a, g) ∈ p, so p 6= ids and so p = S × S, (b, g) ∈ p, and g 6= c = a + b = a + g = g, this is a contradiction.

3. Let S + S = {g} 6= SS, put J = {x ∈ S : xS = g} is a bi-ideal of S, then

J = S or |J| = 1, if J = S then ∀x ∈ S we have xS = g, but this contradicts

with assumption so J 6= S then |J| = 1 so J = {g}. Now put T = S\{g},

and from (3.1.1) part (5) every a ∈ T , we have ha is a congruence relation

since (x, y) ∈ ha iﬀ ax = ay, then there exist at least (b, g) ∈/ ha, b ∈ T , and

consequently ha = ids. Thus TT ⊆ T, because if ab = g, then a = g or b = g, 38

and the relation qT = T × T ∪ids, is a congruence relation from (3.1.1) part (6),

and since ∃a ∈ T , g ∈ S such that a 6= g, we have (a, g) ∈/ T × T , so qT 6= S × S,

then qT = ids, then |T | = 1, so |S| = 2, say S = {g, x}. Now we show ∼ S = T2, suppose that the mapping p : S −→ T2, and we deﬁned the addition

and the multiplication on T2 from table(I), such that P (x) = 1,P (g) = g, so we haveP (xy) = P (x)P (y),P (x + y) = P (x) + P (y) ∀x, y ∈ S, and we saw P ∼ is a one-to-one, onto mapping then S = T2.

4. Let S + S = {g} = SS, then every equivalence deﬁned on S is a congruence of ∼ S, and therefor |S| = 2, and S = T1 from table (I).

Theorem 3.1.3 [10]

Let S be a congruence-simple additively idempotent semiring. Then just one of the following three cases takes place:

(α) S is isomorphic to one of the two-element semirings T3,T4,T5, table[I]. (β) The additive semilattice S(+) possesses ∞ (i.e., x + ∞ = ∞ for

every x ∈ S),S∞ = {∞},G = S\{∞} is an (abelian) subgroup of the

multiplicative semigroup S(.), and the semiring S is (isomorphic to)

the semiring V (G).

(γ) The semiring S is multiplicatively cancellative.

Proof. Let A = {x ∈ S; |Sx| = 1}. If a ∈ S\A and from (2.1.1) part(7) (ra) is a congruence relation of S, so |Sa| > 1. Then there exist at least two element c, d ∈ S such that ac 6= ad, so S × S 6= ra = ids, and the map x −→ ax is an injective 39

transformation of S. Now, if A = φ, then (γ) is true, hence assume that A 6= φ. Let a, b ∈ A, then |Sa| = 1, |Sb| = 1, let Sa = w, Sb = w1 and we have ab = w = w1, then SA = {Sx : x ∈ A} = {w}, and Sw = S(SA) = SS(A) = {w}, since SS ⊆ S.

And S + w is a bi-ideal of S . Now if S + w = S, then w = 0 is a neutral element of S(+), T × T ∪ ids is a congruence of S, where T = S\{0}, so ∃a 6= w such that

(a, w) ∈/ T × T ∪ ids, then T × T ∪ ids 6= S × S, so T × T ∪ ids = ids, then |T | = 1, and so |S| = 2, let S = {x, w}, so there is two cases for x ∈ S.

2 1. x ∈ A, so x = xw = w, let f : S −→ T4, such that f(x) = 0, f(w) = 1, look at ∼ the table (I) we have S = T4.

2 2. x∈ / A, then x = x, let h : S −→ T5, h(x) = 1, h(w) = 0 and look at the table ∼ (I) , so S = T5.

Hence, assume that S + w = {w} and w = ∞ is the absorbing element of S(+).

And from (3.1.1) part(7), the relation s on S by (x, y) ∈ s if and only if {z1 ∈ S : ax + z1 = ∞} = {z2 ∈ S : ay + z2 = ∞} for every a ∈ S is a congruence relation of S.

Now if s = S × S, we have (x, ∞) ∈ s for every x ∈ S, then {z1 ∈ S : ax + z1 = ∞} =

{z2 ∈ S : a∞ + z2 = ∞} = S, so ∀u, v, z ∈ S we have uv + z = ∞, in particular, let z = uv, we have uv + uv = ∞, and since S is additively idempotent, that means uv = uv + uv = ∞, ie., SS = {∞}, and A = S. Now If a ∈ S\{∞}, then S + a is a bi-ideal of S, and a, ∞ ∈ (S + a), since a + a = a, ∞ + a = ∞, hence S + a = S, a is a neutral element of S(+), let T = S\{∞}, so T × T ∪ ids is a congruence of S, ∞ ∈/ T , then (a, ∞) ∈/ T × T , then T × T ∪ ids 6= S × S, then T × T ∪ ids = ids, so |T | = 1, ∼ so |S| = 2, let f : S −→ T3, such that S = {∞, a}, f(∞) = 1, f(a) = 1, so S = T3.

Next assume that s = ids, so A 6= S, and if z ∈ S\ A, and v ∈ A, then zv = ∞ = z∞, implies v = ∞, this means that A = {∞}, and that G = S\{∞} is 40

a subsemigroup of S(.), G is a cancellative semigroup, because if xz = xy, and x, y, z ∈ G = S\{∞}, so x ∈ G = S\A, then (z, y) ∈ rx 6= S × S from above, then rx = ids, so z = y that means G is a cancellative semigroup. Now, let a, b ∈ S, a 6= b,we are going to show that a + b = ∞ proceeding by contradiction, assume that a + b = c 6= ∞, since a 6= b, we may also assume that c 6= a. Then (a, c) ∈/ s = ids, and there exist d, e ∈ S such that dc + e = ∞ 6= da + e, we have d 6= ∞, because if d = ∞, we have da + e = ∞a + e = ∞ + e = ∞, is a contradiction, and so dc 6= ∞.

Let B = {x ∈ S : x + vdc = x for some v ∈ S ∪ {1}} such that 1 is the multiplication identity, and since ∞ + vdc = ∞∀v ∈ S ∪ {1}, dc + 1.dc = dc + dc = dc, we have

∞, dc ∈ B, and since B is a bi-ideal of S, |B| ≥ 1, then B = S, da ∈ B and ﬁnally,

da + wdc = da, w ∈ S ∪ {1}, (3.1.7)

and since dc = d(a + b) = da + db, so

da + dc + wdc = da + da + db = da + db = dc, (3.1.8)

from eq(3.1.7) we have,

dc + da + wdc = dc + da (3.1.9)

,from eq(3.1.8), eq(3.1.9) we have dc+da = dc 6= da, and so w ∈ S. Quite similarly, da + u(da + e) = da for some u ∈ S ∪ {1}.

Now we have

da + uda = da + u(da + e) = uda = da + uda + ue = da + u(da + e) = da (3.1.10) 41

and since da + wdc = da, then

uda + wudc = uda, (3.1.11)

da + uda + wudc = da + uda, from eq(3.1.8) we have

da + wudc = da, (3.1.12)

adding eq(3.1.11), eq(3.1.12),we have, uda + da = wudc = da + uda + wudc, so da + wudc = da + u(da + wdc), since da = uda + da = da + wdc, then

da + wudc = da + uda = da, (3.1.13)

wdc + wda = w(da + db) + wda = wda + wdb + wda = wda + wdb = wdc

,wda = w(da + u(da + e)) = wda + wuda + wue

,da + wue = da + wdc + wue = wue + da + wda + wdc

= wue + da + wdc + wda + wuda + wue

= da + wdc + w(da + uda + ue) =

da + wdc + wda = da + wdc = da. (3.1.14)

From eq(3.1.13), eq(3.1.14), we have ∞ 6= da = da + da = da + wudc + da + wue

=da + wu(dc + e) = da + wu∞ = da + ∞ = ∞, this a contradiction.

We have proved that a + b = ∞ for all a, b ∈ S, a 6= b. From this it follows easily that Sa is a bi-ideal of S for every a ∈ S. If a ∈ G, then a, ∞ ∈ Sa, then |Sa| ≥ 2, and therefore Sa = S. Now we Show G(.) is a group, let a ∈ G, then Sa = S, but a ∈ S,then a = ta, then t is the identity for a, we show t is the identity ∀x ∈ G.

Now suppose that t.x 6= x, then there is two cases, 42

1. t.xa = x.a, and so tx = x, since G(.) is cancellative, this a contradiction for

assumption.

2. If txa 6= xa, then ax 6= ax a contradiction, then t is the identity in G(.).

Now, we show ∀x ∈ G, x has inverse, since t is the identity in G(.), t ∈ G, so t ∈ S, Sx = S ∀x ∈ G, then there exist a ∈ S\{∞} such that ax = t, then a is the inverse for x, then G(.) is a group, and so S ∼= V (G).

Theorem 3.1.4 [10]

Let S be a congruence-simple additively cancellative semiring. Then just one of the following three cases takes place:

(α) S is a zero-multiplication ring of prime order;

(β) S is aﬁeld;

(γ) The semiring S is multiplicatively cancellative and the additive semigroup

S(+) possesses no neutral element.

Proof. First, assume that there exist a neutral 0 ∈ S for the additive semigroup S(+), let M = {x ∈ S : x is an invertible element of S(+)}, and deﬁne the relation r on S by

(x, y) ∈ r if and only if y = x+u for some u ∈ M, x+0 = x, so x2 +0x = x2 = x2 +0, then 0x = 0 for all x ∈ S since S(+) is cancellative, so we have S0 = 0, SM ⊂ M, because xu+x(−u) = x(u−u) = 0, so xu is invertible ∀u ∈ M. Now r is a congruence relation of S, so if r = S × S, hence for all x ∈ S, u ∈ M, we have (u, x) ∈ r, then x = u + v for some v ∈ M, then x ∈ M, so M = S, S(+) is a group, then S is a ring,

S is an ideal-simple, and it (α) or (β) is true. 43

And if r = ids. Then M = 0 and we put q = ((S\{0}) × (S\{0})) ∪ ids. From

(3.1.1) part (5), we have the relation ha deﬁned by (x, y) ∈ ha if and only if ax = ay, ha is a congruence relation of S, if ra = S × S we have Sa = 0, if ha = ids, we have (S\{0})a 6= 0, from this we have q is a congruence relation of S, and q 6= S × S, so ∼ ∼ q = ids, then S\{0} has one element a, then |S| = 2, S = T7 or S = T8, such that T7 is a zero multiplication ring of order 2, T8 is a ﬁeld. Now, assume that 0 ∈/ S and put A = {x ∈ S : |Sx| = 1}. Proceeding in a manner similar to the proof (3.1.3), and so S(.) is cancellative.

Example 3.1.5 [10]

Let A be a non-zero subsemigroup of the additive group R(+) of real numbers, We shall consider the following commutative multiplicative cancellative additive idempotent semiring W = W (A) = W (⊕, ∗): W = A, a ⊕ b = min(a, b), and a ∗ b = a + b for all a, b ∈ A.

Theorem 3.1.6 [10]

Let S be a congruence-simple multiplicative cancellative additively idempotent semiring. Then there exists a subsemigroup A of the additive group R(+) of real number such that A∩ R+ 6= φ 6= A ∩ R− and the semirings S and W (A) are isomorphic.

Proof. See [3], [4], [5], [7], [10]

Main Theorem 3.1.8 [10]

Let S be a commutative, congruence-simple, ﬁnite semiring. Then one of the following holds: 44

1. S is isomorphic to one of ﬁve semirings T1,T2, ..., T5 of order two.

2. S is a ﬁnite ﬁeld.

3. S is a zero-multiplication ring (S2 = {o}) of prime order.

4. S is isomorphic to V (G), for some ﬁnite abelian group G.

Proof. Combine (3.1.2),(3.1.3),(3.1.4),(3.1.6).

3.2 Congruence-Simple Additively Commutative Semi- rings of Finite Order

Lemma 3.2.1 [1]

Let S be a ﬁnite additively commutative, congruence-simple semiring. If the multiplication table of S has two identical rows [columns], then one of the following holds.

1. There exists c ∈ S such that xy = c for all x, y ∈ S.

2. |S| = 2.

Proof. Observe that the relation ∼ deﬁned by x ∼ y if xz = yz for all z ∈ S, ∼ is a congruence relation of S.

Now by assumption, there exist r1 6= r2, such that r1z = r2z for all z ∈ S, so ∼ = S × S, since (S, ., +) is a congruence-simple. Thus

xz = yz ∀x, y, z ∈ S. (3.2.1) 45

Now suppose that (S,.) is not left cancellative, then ∃ a, b, c, d ∈ S, such that da = db = c, a 6= b, so from eq(3.2.1) we have xa = ya, xb = yb for all

x, y ∈ S, da = ya, db = yb. (3.2.2)

And so,

ya = yb = c, ∀y ∈ S. (3.2.3)

So observe that the relation ≈ deﬁned by x ≈ y if zx = zy for all z ∈ S, then ≈ is a congruence relation of S, since a 6= b, then ≈ = S × S, then

zx = zy ∀x, y ∈ S. (3.2.4)

Thus, from eq(3.2.2), eq(3.2.3), eq(3.2.4) we have xy = xa = c, for all x, y ∈ S.

Now suppose that (S, .) is left cancellative, and ﬁx x ∈ S, let z = x2. Then xz = x3 = x2x = zx, from (3.15), we have yz = yx = zx, then yz = yx, since xz = zx, then z = x because (S,.) is left cancellative, so x2 = x, then S is multiplicatively idempotent.

Furthermore, for all w ∈ S, we have (w+w)(w+w) = w+w = w2+w2 = (w+w)w, then w + w = w, since (S,.) is left cancellative so S is additively idempotent.

Now we will show |S| ≤ 2 by contradiction, suppose that |S| = n > 2. For each P P nonempty subset A ⊆ S, let σA= x for all x ∈ A, and let σ = σS = x, x ∈ S, and suppose that A ⊂ S with |A| = n−1. Consider the relation m = ids∪{(σA, σ), (σ, σA)} is an equivalence relation.

Now we show m is a congruence relation, so let x ∼ y, there are two cases: 46

1. If x = y, then xc m yc for all c ∈ S, cx m cy for all c ∈ S c + x m c + y for all

c ∈ S.

2. If x 6= y, then x = σA and y = σ, or x = σ and y = σA.

Since (S,.) is idempotent and from equation (3.15), implies that for each c ∈ S

we have,

cσA = σAσA = σA and cσ = σσ = σ

then (cσA, cσ) = (σA, σ) ∈ m, then cσA m cσ, and,

2 2 σAc = c = c and σc = c = c.

then (σAc, σc) = (c, c) ∈ ids, then (σAc, σc) ∈ m, then σA m σc P P Since (S, +) is idempotent, we have σ + c = ( x) + c = x = σ, and ( σA, if c ∈ A; σA + c = σ, otherwise.

Then (σ + c) m (σA + c), thus m is a congruence relation of S.

Then m = ids or m = S × S, but |S| = n > 2, m = ids ∪ {(σA, σ), (σ, σA)}, then |m| = n + 2, |m| = n + 2 < n2 = |S × S| for all n > 2, so m 6= S × S, thus

m = ids.

So σA = σ for all proper A ⊂ S, |A| = n − 1.

Now, by induction we will show σA = σ for any non-empty subset A ⊆ S. 47

Suppose this known to hold for all A with |A| = k ≥ 2. Now we show the statement is true ∀A ⊂ S, |A| = k − 1. And again consider the relation m = ids ∪ {(σA, σ), (σ, σA)}. As above m is a multiplicative equivalence relation. Furthermore c+σ = σ+c = σ, since (S,+) an idempotent, and

( σA, if c ∈ A; c + σA = σA + c = σA∪{c}, otherwise.

But if c∈ / A, implies that |A ∪ {c}| = k, so σA∪{c} = σ, by the inductive assumption, thus m is again congruence relation of S. Since m 6= S × S, it follows m = ids, so σA = σ, in particular, this shows that for each w ∈ S, w 6= σ, we have w = σw = σ, a contradiction.

Thus |S| ≤ 2, if |S| = 1 then (1) is hold, and so if |S| 6= 1, then |S| = 2, then (2) hold.

Now if S has two identical columns, consider the reciprocal semiring (S0, +, ⊗), deﬁned by (S0, +) = (S, +) and x⊗ y = yx.

This semiring is congruence-simple and has two identical rows, so the above argument applies.

Lemma 3.2.2 [1]

Let S be a ﬁnite, additively commutative, congruence-simple semiring. then one of the following holds

1. (S, +) is a group, hence (S, +,.) is a ring 48

2. S has an additively absorbing element α

Proof. Consider the relation ∼ deﬁned by x ∼ y if x + t = y + t for some t ∈ S, we have ∼ is a congruence relation.

and since S is a congruence-simple, so we have ,∼= ids or ∼ = S × S.

If ∼ =ids, if x + t = y + t, x, y, t ∈ S, then x = y, since ∼ = ids, so (S, +) is cancellative, and so (S, +) is a group from (1.1.8), and so (S, +,.) is a ring.

If ∼ = S × S, then for all x, y ∈ S there exist tx,y ∈ S such that x + tx,y = y + tx,y.

P 0 0 Set σ = x , α = σ + σ, for x, y ∈ S there exist σ ∈ S such that σ = tx,y + σ

0 P such that σ = x, x ∈ S\{tx,y}.

0 0 Then x + σ = x + tx,y + σ = y + tx,y + σ = y + σ∀x, y ∈ S. In particular let y = σ, then

x + σ = σ + σ ∀x ∈ S. (3.2.5)

Then x + α = x + σ + σ = (σ + σ) + σ = α + σ = σ + σ = α from eq(3.2.5), we have α is an additively absorbing in S.

Theorem 3.2.3 [1]

Let S be a ﬁnite, additively commutative, congruence simple. Then one of the following holds.

1. (S, +,.) is a ring.

2. S has an inﬁnity.

3. S is additively idempotent. 49

Proof. With respect to (3.2.2) one may assume that there is an additively absorbing element α ∈ S. Consider the relation T deﬁned by,

xT y if 2x = 2y.

Then T is a congruence relation, Whence T = ids or T = S × S. If T = S × S, Then for all x ∈ S, x + x = α + α = α. Thus, xα = x(α + α) = xα + xα = α.

Similarly, αx = α so α is an inﬁnity. Now if Suppose T = ids, Consider the congruence relation ∼ deﬁned by x ∼ y if there exist u, v ∈ S ∪ {0} and i ≥ 0 such that

2ix = y + u,

2iy = x + v.

Then 2(2x) = (x) + 3x and 2(x) = (2x) + 0, so x ∼ 2x for all x ∈ S. If ∼= ids, then x = 2x for all x ∈ S, whence (S, +) is idempotent. Suppose now that ∼ = S ×S and let x ∈ S.

Then xα ∼ α, so there exists v ∈ S ∪ {o} and i ≥ 0 such that 2ixα = α + v = α, then xα = x(2iα) = 2ixα = α, so xα = α. Similarly, αx = α, and so α is an inﬁnity.

Corollary 3.2.4 [1]

If S is a ﬁnite, additively commutative congruence-simple semiring with zero then one of the following holds.

∼ 1. S = Matn(Fq) for some n ≥ 1 and some finite field Fq, such that Matn(Fq) is n × n matrix over finite field.

2. S is a zero-multiplication ring (S2 = {o}) of prime order. 50

3. S is additively idempotent.

Proof. From the above theorem, we have no an inﬁnity element, because S has a zero, so (S, +,.) is a ring or S is additively idempotent. Now If (S, +,.) is a ring, so is an ideal simple ring, thus by The Wedderburn-Artin Theorem see [17], so either S2 = {o} ∼ or there exists a positive integer n and a ﬁnite Fq such that S = Matn(Fq). Further- more if S2 = {o}, it must be actually abelian group with no non-trivial subgroup,

Whence |S| = |(S, +)| = p, for some prime p.

lemma 3.2.5 [1]

Let S be a ﬁnite, additively commutative, congruence-simple semiring with ∞ and

|S| > 2. Then one of the following holds.

1. S is additively idempotent.

2. S + S = {∞} and (S, .) is a congruence-free semigroup.

Proof. Consider the congruence relation deﬁned by

xT y if 2x = 2y.

CaseI : T = ids. Then 2x = 2y iﬀ x = y. Set x ∼ y if there exists i ≥ 0 and u, v ∈ S ∪ {0} such that:

2ix = y + u

2iy = x + v. 51

Then ∼ is a congruence relation and x ∼ 2x for all x ∈ S. But x ∞ for x 6= ∞, so ∼ 6= S × S. Thus, ∼ = ids, and so S is additively idempotent. CaseII : T = S × S.

Then x + x = ∞ for all x ∈ S. For φ 6= A ⊆ S, let

X σA = x x∈A .

Let N = |S| and suppose that |A| = N − 1, Then for every c ∈ S, σA + c = ∞, since c ∈ A, c = ∞, or σA. Furthermore,

X cσA = cx x∈A ( ∞, if cx = cx for some distinct x , x ∈ A; = 1 2 1 2 σA, otherwise.

Similarly, σAc = ∞ or σAc = σA. Thus, B = {σA : A ⊂ S with |A| = N − 1} is a bi-ideal. Furthermore, ∞ ∈ A implies σA = ∞. Thus, |B| ≤ 2 and so B = S, then |S| = N − 1.

By induction, we will show that σA = ∞ for all A ⊂ S with |A| = 2.

Assume σA = ∞ for all A ⊂ S with |A| = K + 1 ≥ 2. Suppose now that A ⊂ S with |A| = k ≥ 2. Then for c ∈ S,

( ∞, if c ∈ A; σA + c = σA∪{c}, otherwise.

By assumption, if c∈ / A then σA∪{c} = ∞ for all c ∈ S. Also 52

X cσA = cx x∈A ( ∞, if cx = cx for some distinct x , x ∈ A,; = 1 2 1 2 σB, for some |B| = k otherwise.

The same is easily seen to hold for σAc. Observe that σX = ∞ for some X ⊂ S with |X| = k, so B = {σA : A ⊂ S with |A| = k} is a bi-ideal of S. CaseI : B = {∞}.

Then σ = ∞ for all A ⊂ S with |A| = k, so we may apply the induction and conclude that σA = ∞ for all A ⊂ with |A| = 2. Thus, x + y = ∞ for all x, y ∈ S. CaseII : B = S.

we will show directly that x + y = ∞ for all x, y ∈ S. By assumption this holds for x = y, so suppose x 6= y. Then there exist A1,A2 ⊂ S with |A1| = |A2| = k, and

σA1 = x, σA2 = y.

A1 ∩ A2 6= φ, then x + y = σA1 + σA2 = ∞.

A1 ∩ A2 = φ, then x + y = σA1 + σA2 = σA1∪A2

But |A1 ∪ A2| > k. In particular, either |A1∪ A2| = k + 1 or there exist φ 6=

B1,B2 ⊂ S with |B1| = k + 1, b1 ∩ B2 = φ and b1 ∪ B2 = A1∪ A2. By assumption,

σB1 = ∞ and we have x + y = σA1∪A2 = σB1∪B2 =σB1 + σB2 = ∞ + σB2 = ∞. Thus x + y = ∞ for all x, y ∈ S. Finally, note that since S + S = {∞}, any nontrivial congruence relation on (S, .) is also a nontrivial congruence relation on

(S, +,.), Whence (S, .) is a congruence-free semigroup. 53

Theorem 3.2.6 [9]

0 0 let I = {1, 2, ..., m},A = {1, 2, ..., n}, and p = (pij be an n × m matrix of 1 s and 0 s such that no row or column is identically zero, no two rows are identical, and no two columns are identical. Let S = (I × A) ∪ {∞} and deﬁne a binary relation on S by

( (i, µ), if p ; (i, λ)/(j, µ) = λj ∞, otherwise.

(i, λ).∞ = ∞.(i, λ) = ∞.∞.

Then S is a congruence-free semigroup of order mn + 1. Conversely, every ﬁnite congruence-free semigroup with an absorbing element is isomorphic to one of this kind.

Main Theorem 3.2.7

[1]

Let S be a ﬁnite, additively commutative, congruence-simple semiring. Then one of the following holds:

1. |S| ≤ 2

∼ 2. S = Matn(Fq) for some ﬁnite ﬁeld Fq and some n ≥ 1.

3. S is a zero multiplication ring of prime order.

4. S is additively idempotent.

5. (S, .) is a semigroup as in ( 3.2.6) with ∞ ∈ S and S + S = {∞}, so (S,.) is

congruence-free semiring. 54

Proof. Apply (3.2.3),(3.2.6),(3.2.5) and (3.2.4). Also notice that if (S, .) is a semigroup as in Theorem (2.3.2), and we deﬁne S + S = {∞}, then (S, +,.) is necessarily congruence-simple semiring. 55

TABLE I

ALL COMMUTATIVE SEMIRINGS OF ORDER TOW

(T1, +) 0 1 . 0 1 (T2, +) 0 1 . 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1

(T3, +) 0 1 . 0 1 (T4, +) 0 1 . 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 0 0 1 0 1 1 1 1

(T5, +) 0 1 . 0 1 (T6, +) 0 1 . 0 1 0 0 0 0 0 1 0 0 0 i 0 0 0 1 0 1 1 1 1 1 0 1 1 0 1

(T7, +) 0 1 . 0 1 (T8, +) 0 1 . 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 Chapter 4

Smaradache Semirings

4.1 Deﬁnition of Smarandache Semirings and Ex- amples

In this chapter we study Smarandache semirings, Smarandache zero divisors, and classiﬁcation zero divisors in a spacial semirings is a smarandache zero divisors or not. More information can be found in [14], [15], [16].

Deﬁnition 4.1.1

(Smarandache, Florentin):[15]

The Smarandach semiring S Which will be denoted from here on wards as S- semiring is deﬁned to be a semiring S such that a proper subset B of S is a semiﬁeld

(with respect to the same induced operation). that is φ 6= B ⊂ S.

Example 4.1.2

(Q, +,.) the ﬁeld of rational numbers, (H◦, +,.) the set of non-negative rational numbers, note that H◦ ⊂ Q. This means Q is S-semiring, since H◦ is a semiﬁeld.

56 57

Example 4.1.3

(P◦, +,.) the non-negative semiring of real numbers, let H◦ be the non-negative semiring of rational numbers, so P◦ is S-semiring, since H◦ ⊂ P◦ is a semiﬁeld.

Theorem 4.1.4 [15]

Every semiring in general need not be a S-semiring.

Proof. (Z◦, +,.) is the non-negative integer is a semiring, but not S-semiring.

Example 4.1.5

Let Mn×n = {(aij/aij ∈ P◦} be the set of all n×n matrices with entries from P◦,Mn×n is a semiring under matrix addition and matrix multiplication. clearly Mn×n is a non- commutative semiring, Mn×n is a S-semiring, since A = {(aij)/aij = 0 if i 6= j and aii ∈ P◦, so A ⊂ Mn×n is a semiﬁeld, so Mn×n is a S-semiring.

Deﬁnition 4.1.6 [15]

If the S-semiring has only finite number of elements, we say the S-semiring is finite, otherwise infinite.

Example 4.1.7

Let P the set of non-negative real number, P◦ is a semiring, and S-semiring, let

S = (P◦ × P◦ × P◦...... × P◦) (n times), let A = (P◦ × {0} × {0}...... × {0}) (n times), note that A ⊂ S, A is a semiﬁeld of S, then S is a S-semiring. 58

Deﬁnition 4.1.8 [15]

Let S be a semiring. Anon-empty proper subset A of S is said to be a Smarandache subsemiring (S-subsemiring) if A is S-semiring.

Theorem 4.1.9 [15]

Let S be a semiring having a S-subsemiring, then S is a S-semiring.

Proof. Given S is a semiring and A ⊂ S is a S-semiring of S, so A has a proper subset P ⊂ A . Now P ⊂ A ⊂ S, so P ⊂ S is a semiﬁeld Hence S is a S-semiring.

Theorem 4.1.10 [15]

Every subsemiring of a S-semiring need not in general be a S-subsemiring.

Proof. To prove this, let H◦ ⊂ Q, H◦ is a semiring, but H◦ is not a S-semiring.

Deﬁnition 4.1.11 [15]

Let S be a S-semiring. We say S is a Smarandache commutative semiring (S- commutative) if S has a S-subsemiring which is commutative. If the S-semiring has no commutative S-subsemiring then we say S is a Smarandache non-commutative semiring (S-non commutative semiring).

Example 4.1.12

Consider the semiring M3×3 = {(aij): aij ∈ H◦}. M3×3 is S-semiring. Let 59

  ½ a11 0 0 ¾     A =  0 a22 0  : a11, a22 ∈ H◦ . 0 0 0

A is a subsemiring of M3×3.A is a S-subsemiring of M3×3 for take

  ½ a11 0 0 ¾     P =  0 0 0  : a11, ∈ H◦ ,P is a semiﬁeld and P ⊂ A. So A is a S- 0 0 0   1 0 0     semiring, note that The unit of the semiﬁeld is  0 0 0  which is not the unit of 0 0 0 M3×3.   ½ a11 0 0 ¾     But the subsemiring A =  0 a22 0  : a11, a22 ∈ H◦ is a commutative 0 0 0 subsemiring.

Hence the semiring M3×3 is S-commutative semiring.

Deﬁnition 4.1.13 [15]

Let S be a semiring. A non-empty subset P of S is said to be Smarandache right(left) ideal [S-right(left)ideal] of S if the following conditions are satisﬁed.

1. P is a S-subsemiring.

2. For every p ∈ P and A ⊂ P where A is the semiﬁeld of P we have for all a ∈ A

and p ∈ P , ap(pa) is in A. 60

If P is simultaneously both a S-right ideal and a S-left ideal then we say P is a

Smarandache ideal (S-ideal) of S.

Example 4.1.14   ½ a11 0 0     Let M3×3 be the semiring given in (4.1.12). Clearly P =  0 a22 0  : 0 0 0 ¾ a11, a22 ∈ H◦ is a S-ideal of M3×3. It is easily veriﬁed that P is both a S-left ideal and a S-right ideal of M3×3.

Theorem 4.1.15 [15]

Let S be a S-semiring. Every S-ideal of S is a S-subsemiring of S but every S- subsemiring of S in general need not be a S-ideal of S.

Proof. Clearly by verifying the deﬁnition of the S-ideal we see every S-ideal is a

S-subsemiring. ½ ¾

Conversely, let M2×2 = (aij): aij ∈ M = {0, 1} ⊂ Z .,(M, +,.) be a semiring, such that a + b = max{a, b}, and a.b = min{a, b} for all a, b ∈ M. ½ Ã ! Ã ! 1 0 0 0 We have M2×2 is S-semiring with 16 element in it. For P= , , 0 1 0 0

Ã ! ¾ ½ Ã ! Ã ! ¾ 1 0 1 0 0 0 is S-subsemiring, take A= , 0 0 0 1 0 0

Ã ! Ã ! Ã ! 1 0 1 0 1 0 is a semiﬁeld and let ∈ P , let ∈ A, but, 0 0 0 1 0 0 61

Ã ! Ã ! 1 0 1 0 = ∈/ A, then P is S-subsemiring of M2×2, which is not a S- 0 1 0 0

ideal of M2×2.

4.2 Smarandache Special Elements in Semirings

Deﬁnition 4.2.1 [15]

Let S be any semiring. We say a and b ∈ S is a Smarandache zero divisor(S-zero divisor) if a.b = 0 and there exists x, y ∈ S\{a, b, 0}, x 6= y with

1. ax = 0 or xa = 0

2. by = 0 or yb = 0 and

3. xy 6= 0 or yx 6= 0.

Clearly if S is a semiﬁeld we will not have S-zero divisors.

Example 4.2.2

Let S = Z◦ × Z◦ × Z◦ × Z◦ be a semiring. Now a = (0, 0, 4, 2) and b = (5, 0, 0, 0) in S is such that a.b = (0, 0, 0, 0) i.e, they are zero divisors. Take x = (2, 8, 0, 0) and y = (0, 1, 0, 0) ∈ S\{a, b, (0, 0, 0, 0)}, and we have ax = 0, yb = 0 and by = 0, but xy = (0, 8, 0, 0) 6= (0, 0, 0, 0). so a = (0, 0, 4, 2) is a S-zero divisor in S. 62

Deﬁnition 4.2.3 [14]

Let S be a semiring. An element a ∈ S\{0} is aS-weak zero divisor if there exist b ∈ S\{0, a} such that a.b = 0 satisfying the following conditions, there exists x, y ∈

S\{0, a, b} such that

1. a.x = 0 or x.a = 0

2. b.y = 0 or y.b = 0

3. x.y = 0 or y.x = 0.

Example 4.2.4

Let S = Z◦ × Z◦ = {(x, y): x, y ∈ Z◦} then we have the only zero divisors in S has the form (x, 0), (0, y), x, y ∈ Z◦\{0}, take a = (x1, 0), b = (0, y1) ,we have a.b =

(x1, 0).(0, y1) = (0, 0) and let x = (0, c), y = (d, 0), such that c 6= y1 6= 0, x1 6= d 6= 0, then a.x = (0, 0) , b.y = (0, 0) and x.y = (0, 0), then a are S-weak zero divisors, not

S-zero divisor.

Theorem 4.2.5

Let S = Z◦ × Z◦ × ... × Z◦ (n-times) be a semiring, n ≥ 3. Let a = (x1, x2, ..., xn) =

n (xi)i=1 be a zero divisor, such that at least two components are zeros, and at most has (n − 1) components are zeros, then a is a S-zero divisor.

n Proof. Suppose (xi)i=1 has at least two components are zeros and at most n − 1 components are zeros, let j < s ≤ n, such that xj = 0, xs = 0,

let a = (x1, x2, .., 0j,xj+1, ..., 0s, xS+1, .., xn), b = (0, 0, .., xs, 0, .., 0), 63

then a.b = (0, 0, .., 0).

Now let x = (0, 0, .., bj, 0, ..., bs, 0, .., 0), and y = (x1, x2, .., xj, ..xs−1, 0s, xs+1, .., xn), and we have a.x = (0, 0, .., 0), b.y = (0, 0, ..., 0),

and x.y = (0, 0, .., bj, 0, ..., bs, 0, .., 0)(x1, x2, .., xj, .., xs−1, 0s, xs+1, .., xn)

= (0, 0, .., bjxj, .., bs.xs+1, 0, .., 0) 6= (0, 0, .., 0) (n times),

such that bj, bs, xj 6= 0 then a is S-zero divisor.

Theorem 4.2.6 [15]

Let S be a semiring if x ∈ S is S-zero divisor then x is a zero divisor.

Proof. By the verify deﬁnition of the S-zero divisor we see every S-zero divisor is a zero divisor.

The converse of this theorem is not true.

Deﬁnition 4.2.7 [15]

Let S be a semiring. An element x ∈ S is said to be a Smarandache anti-zero divisor

(S-anti-zero divisor) if we can ﬁnd a y such that xy 6= 0 and a, b ∈ S\{0, x, y} such that

1. ax 6= 0 or xa 6= 0

2. by 6= 0 or yb 6= 0

3. ab = 0 or ba = 0. 64

Example 4.2.8 ½ Ã ! ¾ Ã ! a b 1 0 Let M2×2 = : a, b, c, d ∈ Z◦ be the semiring. A = ∈ M2×2 is c d 0 1 Ã ! 0 1 an anti-zero divisor. For take B = is such that AB 6= 1 0

Ã ! Ã ! Ã ! 0 0 0 0 0 0 . Choose X = and Y = in M2×2. 0 0 1 0 0 1

Ã ! Ã ! Ã ! 0 0 0 0 0 0 AB 6= , AX 6= . But XY = 0 0 0 0 1 0

Ã ! Ã ! 0 0 0 0 = , thus A is a S-anti-zero divisor. 0 1 0 0

Theorem 4.2.9 [15]

Let S be a semiring. If x ∈ S is a S-anti zero divisor then x need not in general be a zero divisor. Ã ! 1 0 Proof. We see from (4.2.9). The element is the unit element of M2×2, so 0 1 Ã ! Ã ! 1 0 1 0 for non-zero element A in M2×2 we have A. = A, so is a S-anti 0 1 0 1 zero divisor and not zero divisor. 65

Theorem 4.2.10 [15]

If S is a semiring with unity and if S has zero divisors then the unity is an S-anti zero divisor.

Proof. Given S has zero divisors α, β ∈ S\{0} with αβ = 0. Let 1 be the identity of

S. Choose x=1, we have 1.x = 1 6= 0 such that α.1 6= 0, βx 6= 0 but αβ = 0, then 1 is S-anti zero divisor. conclusions

In this research we defined some algebraic structures on semirings, and illustrated them with examples, and analyzed some interesting theorems and results. Finally we defined a Smarandache zero-divisors, and Smarandache weak-zero-divisors. We have lot a spacial semiring, (S × S × ...... × S) (n-times), such that S a semiring with zero, we studied and classified all zero divisors in this semiring, may it a Smarandache zero- divisors or not. There are some research problems related to smarandache semirings, we hope we can find solutions for them, and we hope in the future there will be other wide researches on other algebraic structures that has important applications in the practical life, as the cryptography, and the coding.

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