of the order (p − 1). Finally, we note that every ring (Z

(1)

Lidia Obojska, Paul O’Hara

Some Remarks on Rings

Abstract. Algebraic structures such as Rings, Fields, Boolean Algebras (Set Theory) and σ−Fields are well known and much has been written about them. In this paper we explore some properties of rings related to the distribution law. Specifically, we shall show that for rings there exists only one distribution law. Moreover, for the ring (Z

_p(p−1)n

+, ·), where (p, n) = 1 there exist isomorphic groups (G, +), (H, ·), G, H ⊆ Z

_p(p−1)n

of the order (p − 1). Finally, we note that every ring (Z

pn

, +, ·) contains subfields mod(pn).

2000 Mathematics Subject Classification: 06E20, 20E34.

Key words and phrases: Distribution law, Rings, Isomorphism.

In this short paper we investigate properties of Rings, putting special emphasis on the relationship between the two operations “+” and “·”. We begin with a few examples and then formulate some theorems.

Example 1 Consider the field: Ϝ(R, +, ·) where A = R; B = R ⁺ .

Let φ : R −→ R ⁺ such that φ : e ^x+y −→ e ^x · e ^y on the two groups: (R, +) and (R ⁺ , ·). Clearly, φ defines an isomorphism.

Note that ∀x, y, z ∈ R : there exists only one distribution law dist(·/+) given by:

x · (y + z) = x · y + x · z.

Example 2 Let S be a set with powerset P (S).

Consider the Boolean Algebra (P (S), ∩, ∪), then ∀A, B, C ∈ P (S) there exist two distribution laws:

dist 1 : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) dist 2 : A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Also, we can define a homomorphism ^c (the complement) by

c : (P (S), ∩) −→ (P (S), ∪) : (A ∩ B) ^c = A ^c ∪ B ^c .

c is an automorphism on P (S).

(2)

Example 3 Consider a ring: (Z 20 , +, ·) such that:

(A, +) = ({0, 5, 10, 15}, + ²⁰ ) (B, ·) = ({4, 8, 12, 16}, · ²⁰ )

Then ({0, 5, 10, 15}, + 20 ) ∼ = (Z 4 , + 4 ), and ({4, 8, 12, 16}, · 20 ) ∼ = (Z ₅ ^∗ , · 5 ).

An isomporphism φ : (A, +) −→ (B, ·) can be defined as follows:

φ(0) = 16 φ(5) = 12 φ(10) = 4 φ(15) = 8.

Note φ(a + b) = φ(a) · φ(b) and dist(·/+) is defined as in Example 1.

Lemma 4 If ∃o ∈ Ω such that x ∗ o = o ∗ x = o ∀x ∈ Ω

and there exists a homomorphism φ : (Ω, ∗) −→ (Ω, ◇) then φ(o) is a zero for ◇.

Proof x ∗ o = o by definition.

φ(x ∗ o) = φ(o)

=⇒ φ(x)◇φ(o) = φ(x ∗ o) = φ(o).

Example 5 Consider the structure hP (S), ∩, ∪, ^c , dist 1 : A ∩ (B ∪ C)i of Example 2.

Then ∀A ∈ P (S) : A ∩ ∅ = ∅.

Also φ − automorphism =⇒ φ(A ∩ B) = φ(A) ∪ φ(B) = A ^c ∪ B ^c such that φ(A ^c ∩ ∅) = φ(A ^c ) ∪ φ(∅) = A ∪ S = S and A ∪ S = S.

Therefore, S is zero for ∪ iff ∅ is zero for ∩.

Lemma 6 Consider a ring (Ω, ∗, ◇).

If e ∈ (Ω, ∗) is identity for ∗ then e is zero for ◇.

Proof Let a, b, c ∈ Ω

Let dist(◇/∗) be a distribution law.

a◇(e ∗ b) = a◇b by identity By dist(◇/∗)

a◇(e ∗ b) = (a◇e) ∗ (a◇b) = a◇b

Now since (Ω, ∗) is a group, there exists an inverse.

(a◇e) ∗ (a◇b) ∗ (a◇b) ⁻¹ = (a◇b) ∗ (a◇b) ⁻¹

=⇒ a◇e = e, i.e. e is zero for ◇.

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Comment: Lemma 6 is just to state that x · 0 = 0, ∀x ∈ A, where (A, +, ·, e) is an algebra describing a ring.

Theorem 7 For every “non zero” ring (Ω, ∗, ◇) there exist only one distribution law.

Proof Consider the ring (Ω, ∗, ◇) with group (G, ∗) and d 1 = dist(◇/∗) : a◇(b ∗ c) = (a◇b) ∗ (a◇c).

Let e be identity for ∗ .

Assume there exists a second distribution law d 2 : a ∗ (b◇c) = (a ∗ b)◇(a ∗ c).

Let b = e and c = a ⁻¹ under ∗.

Lemma 6 ⇒ e is a zero for ◇. Therefore,

d 2 : a ∗ (e◇c) = (a ∗ e)◇(a ∗ a ⁻¹ ).

⇒ a ∗ e = a◇e

⇒ a = 0 by zero property _■

This is a contradiction.

The result follows.

Comment: A close review of the above indicates that the proof depends upon the existence of inverse elements resulting from the group structure (G, ∗). Interestingly if the inverse axiom of the group structure is relaxed then as Example 2 indicates we can have two distribution properties which are related by a homomorphism:

φ(A ∩ (B ∪ C)) = φ(A) ∪ φ(B ∪ C) = A ^c ∪ (B ^c ∩ C ^c )

Also φ(A∩(B∪C)) = φ(A∩B)∩φ(A∩C) = (A∩B) ^c ∩(A∩C) ^c = (A ^c ∪B ^c )∩(A ^c ∪C ^c ) then (A ∪ (B ∩ C)) = (A ∪ B) ∩ (A ∪ C) and vice versa.

Theorem 8 Consider a ring (Ω, ∗, ◇) where (Ω, ∗) is a cyclic group. For every finite (Ω, ∗, ◇) such that:

o(Ω) = p(p − 1)n, where (p, n) = 1, there exist isomorphic groups (G, ∗) and (H, ◇), G, H ⊆ Ω of the order (p − 1).

Proof Since (Ω, ∗) is cyclic group then it is isomorphic to Z ^p (p−1)n . Now let

(G, ∗) = (G, +) = ({0, pn, 2pn, · · · , (p − 2)pn}, + p (p−1)n ),

(H, ◇) = (H, ·) = ({(p − 1)n, 2(p − 1)n, · · · , (p − 1) ² n }, · p (p−1)n ). _■

Then both (G, +) ∼ = Z p −1 and (H, ·) ∼ = Z p −1 and consequently (G, +) ∼ = (H, ·) Note

that (p, n) = 1 implies (H, ·) is closed under · and contains an identity and inverse.

(4)

Note that the presented Therem is only a sufficient condition and not necessary.

Example 9 (Z ₁₈ , + 18 , · 18 )

h{0, 9}, + 18 , · 18 , dist( ·/+)i is a subfield of Z 18 . ({0, 3, 6, 9, 12, 15}, + ¹⁸ ) ∼ = ({2, 4, 8, 10, 14, 16}, · ¹⁸ ) φ(0) = 10

φ(3) = 2 φ(6) = 4 φ(9) = 8 φ(12) = 16 φ(15) = 14

In fact there is another variation on Theorem 8 based on the use of the Euler function. Let n = o(Ω). Let H - the Euler group such that ϕ(n) = o(H). If there exists a prime number p such that p|n and p|o(H) then there exists isomorphic groups (G, + n ) and (H, · ⁿ ) of order p. This follows from Cauchy’s theorem. Note that if n and ϕ(n) are relatively prime then there is only the trivial isomorphism.

Example 10 Consider the Euler group of Z 18 and call it H.

Then H = {1, 5, 7, 11, 13, 17} . o(H) = 6 and 3 |6 and 3|18.

So by Cauchy’s theorem there has to be a cyclic subgroup of H of order 3.

In fact K = {1, 7, 13} is such a group under multiplication · ¹⁸ and K is isomorphic to ({0, 6, 12}, + ¹⁸ ).

Lemma 11 Let (p, n) = 1. For every ring (Z pn , +, ·) there exists a subset F ⊆ Z ^pn such that (F, + n , · ⁿ ) is a field.

Proof Let Ϝ = {0, n, 2n, ..., (p − 1)n} then (Ϝ, + ⁿ , · ⁿ , dist( ·/+)) is clearly a finite field.

Definition 12 We will say that F is a subfield of a ring (Ω, ∗, ◇) if F ⊆ Ω and (F, ∗,◇) is a field.

Example 13 Let (Ω, ∗, ◇) be (Z 6 , +, ·).

H 2 = ({0, 3}, + ⁶ ) − subgroup K ₃ ^x = ({2, 4}, · ⁶ ) − subgroup

φ : (H 2 , + 6 ) −→ (K 3 ^x , · ⁶ ) − isomorphism

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φ(0) = 4 φ(3) = 2

Example 14 (Z 20 , + 20 , · ²⁰ )

h{0, 4, 8, 12, 16}, + ²⁰ , · ²⁰ , dist( ·/+)i is a subfield of Z ²⁰ . ({0, 5, 10, 15}, + ²⁰ ) ∼ = ({4, 8, 12, 16}, · ²⁰ ) (see Ex. 3) and ({0, 10}, + ²⁰ ) ∼ = ({5, 15}, · 20 ) (φ(0) = 5, φ(10) = 15)

Example 15 (Z 30 , + 30 , · ³⁰ )

({0, 6, 12, 18, 24}, + ³⁰ , · ³⁰ , dist( ·/+)). is a subfield of Z ³⁰ .

Example 16 (G = {0, 3, 6}, + ⁹ , · ⁹ , dist( ·/+)) is not a subfield. (G, + ⁹ ) is a group but (G ^∗ , · ⁹ ) = ({3, 6}, · ⁹ ) is not closed. Specifically: 6 · 6 = 36 = 0 /∈ G ^∗ .

Example 17 In the case of (Z 8 , +, ·) note that ({1, 3, 5, 7}, · ⁸ ) is isomorphic to the Klein-Gordan group of order 4, but is not isomorphic to Z ₄ . In this case (2, 4) 6= 1 and the conditions of the theorem are violated.

References

[1] Mostowski, A. and Stark, M., Algebra wyższa, v.1-3, PWN 1953.

[2] Scott, W. R., Group Theory, Prentice-Hall, Inc., Englewood Cliffs, N.J. 1964.

[3] Sierpiński, W., Elementary Theory of Numbers, PWN 1964.

Lidia Obojska

University of Podlasie, Faculty of Sciences, Dept. of Mathematics and Physics ul. 3 Maja 54, 08-110 Siedlce, Poland

Istituto Universitario Sophia

San Vito 28, 50-064 Incisa Val D’Arno (FI),Italy

E-mail: obojskal@ap.siedlce.pl; lidia.obojska@iu-sophia.org Paul O’Hara

Dept. of Mathematics, Northeastern Illinois University Chicago, IL 60625

E-mail: pohara@neiu.edu

(Received: 13.08.2009)

of the order (p − 1). Finally, we note that every ring (Z

Full text

Lidia Obojska, Paul O’Hara

Some Remarks on Rings

+, ·), where (p, n) = 1 there exist isomorphic groups (G, +), (H, ·), G, H ⊆ Z

of the order (p − 1). Finally, we note that every ring (Z

, +, ·) contains subfields mod(pn).

2000 Mathematics Subject Classification: 06E20, 20E34.

Key words and phrases: Distribution law, Rings, Isomorphism.

In this short paper we investigate properties of Rings, putting special emphasis on the relationship between the two operations “+” and “·”. We begin with a few examples and then formulate some theorems.

Example 1 Consider the field: Ϝ(R, +, ·) where A = R; B = R + .

Let φ : R −→ R + such that φ : e x+y −→ e x · e y on the two groups: (R, +) and (R + , ·). Clearly, φ defines an isomorphism.

Note that ∀x, y, z ∈ R : there exists only one distribution law dist(·/+) given by:

x · (y + z) = x · y + x · z.

Example 2 Let S be a set with powerset P (S).

Consider the Boolean Algebra (P (S), ∩, ∪), then ∀A, B, C ∈ P (S) there exist two distribution laws:

dist 1 : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) dist 2 : A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Also, we can define a homomorphism c (the complement) by

c : (P (S), ∩) −→ (P (S), ∪) : (A ∩ B) c = A c ∪ B c .

c is an automorphism on P (S).

Example 3 Consider a ring: (Z 20 , +, ·) such that:

(A, +) = ({0, 5, 10, 15}, + 20 ) (B, ·) = ({4, 8, 12, 16}, · 20 )

Then ({0, 5, 10, 15}, + 20 ) ∼ = (Z 4 , + 4 ), and ({4, 8, 12, 16}, · 20 ) ∼ = (Z 5 ∗ , · 5 ).

An isomporphism φ : (A, +) −→ (B, ·) can be defined as follows:

φ(0) = 16 φ(5) = 12 φ(10) = 4 φ(15) = 8.

Note φ(a + b) = φ(a) · φ(b) and dist(·/+) is defined as in Example 1.

Lemma 4 If ∃o ∈ Ω such that x ∗ o = o ∗ x = o ∀x ∈ Ω

and there exists a homomorphism φ : (Ω, ∗) −→ (Ω, ◇) then φ(o) is a zero for ◇.

Proof x ∗ o = o by definition.

φ(x ∗ o) = φ(o)

=⇒ φ(x)◇φ(o) = φ(x ∗ o) = φ(o).

Example 5 Consider the structure hP (S), ∩, ∪, c , dist 1 : A ∩ (B ∪ C)i of Example 2.

Then ∀A ∈ P (S) : A ∩ ∅ = ∅.

Also φ − automorphism =⇒ φ(A ∩ B) = φ(A) ∪ φ(B) = A c ∪ B c such that φ(A c ∩ ∅) = φ(A c ) ∪ φ(∅) = A ∪ S = S and A ∪ S = S.

Therefore, S is zero for ∪ iff ∅ is zero for ∩.

Lemma 6 Consider a ring (Ω, ∗, ◇).

If e ∈ (Ω, ∗) is identity for ∗ then e is zero for ◇.

Proof Let a, b, c ∈ Ω

Let dist(◇/∗) be a distribution law.

a◇(e ∗ b) = a◇b by identity By dist(◇/∗)

a◇(e ∗ b) = (a◇e) ∗ (a◇b) = a◇b

Now since (Ω, ∗) is a group, there exists an inverse.

(a◇e) ∗ (a◇b) ∗ (a◇b) −1 = (a◇b) ∗ (a◇b) −1

=⇒ a◇e = e, i.e. e is zero for ◇.

Comment: Lemma 6 is just to state that x · 0 = 0, ∀x ∈ A, where (A, +, ·, e) is an algebra describing a ring.

Theorem 7 For every “non zero” ring (Ω, ∗, ◇) there exist only one distribution law.

Proof Consider the ring (Ω, ∗, ◇) with group (G, ∗) and d 1 = dist(◇/∗) : a◇(b ∗ c) = (a◇b) ∗ (a◇c).

Let e be identity for ∗ .

Assume there exists a second distribution law d 2 : a ∗ (b◇c) = (a ∗ b)◇(a ∗ c).

Let b = e and c = a −1 under ∗.

Lemma 6 ⇒ e is a zero for ◇. Therefore,

d 2 : a ∗ (e◇c) = (a ∗ e)◇(a ∗ a −1 ).

⇒ a ∗ e = a◇e

⇒ a = 0 by zero property ■

This is a contradiction.

The result follows.

φ(A ∩ (B ∪ C)) = φ(A) ∪ φ(B ∪ C) = A c ∪ (B c ∩ C c )

Also φ(A∩(B∪C)) = φ(A∩B)∩φ(A∩C) = (A∩B) c ∩(A∩C) c = (A c ∪B c )∩(A c ∪C c ) then (A ∪ (B ∩ C)) = (A ∪ B) ∩ (A ∪ C) and vice versa.

Theorem 8 Consider a ring (Ω, ∗, ◇) where (Ω, ∗) is a cyclic group. For every finite (Ω, ∗, ◇) such that:

o(Ω) = p(p − 1)n, where (p, n) = 1, there exist isomorphic groups (G, ∗) and (H, ◇), G, H ⊆ Ω of the order (p − 1).

Proof Since (Ω, ∗) is cyclic group then it is isomorphic to Z p (p−1)n . Now let

(G, ∗) = (G, +) = ({0, pn, 2pn, · · · , (p − 2)pn}, + p (p−1)n ),

(H, ◇) = (H, ·) = ({(p − 1)n, 2(p − 1)n, · · · , (p − 1) 2 n }, · p (p−1)n ). ■

Then both (G, +) ∼ = Z p −1 and (H, ·) ∼ = Z p −1 and consequently (G, +) ∼ = (H, ·) Note

that (p, n) = 1 implies (H, ·) is closed under · and contains an identity and inverse.

Note that the presented Therem is only a sufficient condition and not necessary.

Example 9 (Z 18 , + 18 , · 18 )

h{0, 9}, + 18 , · 18 , dist( ·/+)i is a subfield of Z 18 . ({0, 3, 6, 9, 12, 15}, + 18 ) ∼ = ({2, 4, 8, 10, 14, 16}, · 18 ) φ(0) = 10

φ(3) = 2 φ(6) = 4 φ(9) = 8 φ(12) = 16 φ(15) = 14

Example 10 Consider the Euler group of Z 18 and call it H.

Then H = {1, 5, 7, 11, 13, 17} . o(H) = 6 and 3 |6 and 3|18.

So by Cauchy’s theorem there has to be a cyclic subgroup of H of order 3.

In fact K = {1, 7, 13} is such a group under multiplication · 18 and K is isomorphic to ({0, 6, 12}, + 18 ).

Lemma 11 Let (p, n) = 1. For every ring (Z pn , +, ·) there exists a subset F ⊆ Z pn such that (F, + n , · n ) is a field.

Proof Let Ϝ = {0, n, 2n, ..., (p − 1)n} then (Ϝ, + n , · n , dist( ·/+)) is clearly a finite field.

Definition 12 We will say that F is a subfield of a ring (Ω, ∗, ◇) if F ⊆ Ω and (F, ∗,◇) is a field.

Example 13 Let (Ω, ∗, ◇) be (Z 6 , +, ·).

H 2 = ({0, 3}, + 6 ) − subgroup K 3 x = ({2, 4}, · 6 ) − subgroup

φ : (H 2 , + 6 ) −→ (K 3 x , · 6 ) − isomorphism

φ(0) = 4 φ(3) = 2

Example 1 Consider the field: Ϝ(R, +, ·) where A = R; B = R ⁺ .

Let φ : R −→ R ⁺ such that φ : e ^x+y −→ e ^x · e ^y on the two groups: (R, +) and (R ⁺ , ·). Clearly, φ defines an isomorphism.

Also, we can define a homomorphism ^c (the complement) by

c : (P (S), ∩) −→ (P (S), ∪) : (A ∩ B) ^c = A ^c ∪ B ^c .

(A, +) = ({0, 5, 10, 15}, + ²⁰ ) (B, ·) = ({4, 8, 12, 16}, · ²⁰ )

Then ({0, 5, 10, 15}, + 20 ) ∼ = (Z 4 , + 4 ), and ({4, 8, 12, 16}, · 20 ) ∼ = (Z ₅ ^∗ , · 5 ).

Example 5 Consider the structure hP (S), ∩, ∪, ^c , dist 1 : A ∩ (B ∪ C)i of Example 2.

Also φ − automorphism =⇒ φ(A ∩ B) = φ(A) ∪ φ(B) = A ^c ∪ B ^c such that φ(A ^c ∩ ∅) = φ(A ^c ) ∪ φ(∅) = A ∪ S = S and A ∪ S = S.

(a◇e) ∗ (a◇b) ∗ (a◇b) ⁻¹ = (a◇b) ∗ (a◇b) ⁻¹

Let b = e and c = a ⁻¹ under ∗.

d 2 : a ∗ (e◇c) = (a ∗ e)◇(a ∗ a ⁻¹ ).

⇒ a = 0 by zero property _■

φ(A ∩ (B ∪ C)) = φ(A) ∪ φ(B ∪ C) = A ^c ∪ (B ^c ∩ C ^c )

Also φ(A∩(B∪C)) = φ(A∩B)∩φ(A∩C) = (A∩B) ^c ∩(A∩C) ^c = (A ^c ∪B ^c )∩(A ^c ∪C ^c ) then (A ∪ (B ∩ C)) = (A ∪ B) ∩ (A ∪ C) and vice versa.

Proof Since (Ω, ∗) is cyclic group then it is isomorphic to Z ^p (p−1)n . Now let

(H, ◇) = (H, ·) = ({(p − 1)n, 2(p − 1)n, · · · , (p − 1) ² n }, · p (p−1)n ). _■

Example 9 (Z ₁₈ , + 18 , · 18 )

h{0, 9}, + 18 , · 18 , dist( ·/+)i is a subfield of Z 18 . ({0, 3, 6, 9, 12, 15}, + ¹⁸ ) ∼ = ({2, 4, 8, 10, 14, 16}, · ¹⁸ ) φ(0) = 10

In fact K = {1, 7, 13} is such a group under multiplication · ¹⁸ and K is isomorphic to ({0, 6, 12}, + ¹⁸ ).

Lemma 11 Let (p, n) = 1. For every ring (Z pn , +, ·) there exists a subset F ⊆ Z ^pn such that (F, + n , · ⁿ ) is a field.

Proof Let Ϝ = {0, n, 2n, ..., (p − 1)n} then (Ϝ, + ⁿ , · ⁿ , dist( ·/+)) is clearly a finite field.

H 2 = ({0, 3}, + ⁶ ) − subgroup K ₃ ^x = ({2, 4}, · ⁶ ) − subgroup

φ : (H 2 , + 6 ) −→ (K 3 ^x , · ⁶ ) − isomorphism

Example 14 (Z 20 , + 20 , · ²⁰ )

h{0, 4, 8, 12, 16}, + ²⁰ , · ²⁰ , dist( ·/+)i is a subfield of Z ²⁰ . ({0, 5, 10, 15}, + ²⁰ ) ∼ = ({4, 8, 12, 16}, · ²⁰ ) (see Ex. 3) and ({0, 10}, + ²⁰ ) ∼ = ({5, 15}, · 20 ) (φ(0) = 5, φ(10) = 15)

Example 15 (Z 30 , + 30 , · ³⁰ )

({0, 6, 12, 18, 24}, + ³⁰ , · ³⁰ , dist( ·/+)). is a subfield of Z ³⁰ .

Example 16 (G = {0, 3, 6}, + ⁹ , · ⁹ , dist( ·/+)) is not a subfield. (G, + ⁹ ) is a group but (G ^∗ , · ⁹ ) = ({3, 6}, · ⁹ ) is not closed. Specifically: 6 · 6 = 36 = 0 /∈ G ^∗ .

Example 17 In the case of (Z 8 , +, ·) note that ({1, 3, 5, 7}, · ⁸ ) is isomorphic to the Klein-Gordan group of order 4, but is not isomorphic to Z ₄ . In this case (2, 4) 6= 1 and the conditions of the theorem are violated.