### 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.*

^{x+y}

^{x}

^{y}

*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}

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

^{c}

^{c}

^{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.*

^{c}

*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.*

^{c}

^{c}

^{c}

^{c}

*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}

^{−1}

^{−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 ∗.*

^{−1}

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

*d* 2 *: a ∗ (e◇c) = (a ∗ e)◇(a ∗ a* ^{−1} *).*

^{−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} )

^{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.*

^{c}

^{c}

^{c}

^{c}

^{c}

^{c}

*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

^{p}

*(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*

### 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.*

^{n}

*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.

^{pn}

^{n}

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

^{n}

^{n}

*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*

^{x}

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

^{x}

*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* ^{∗} *.*

^{∗}

^{∗}