CYCLIC RINGS 1. Introduction the Goal of This Work Is to Explore Cyclic Rings, Derive Some Results About These Structures, and I

CYCLIC RINGS 1. Introduction the Goal of This Work Is to Explore Cyclic Rings, Derive Some Results About These Structures, and I

CYCLIC RINGS 1. Introduction The goal of this work is to explore cyclic rings, derive some results about these structures, and investigate how these results influence ring classification. Some basic facts from abstract algebra will be used without a proof supplied in this work. Also, although the focus of this work is ring theory, some results from number theory are needed, and one of these results will be used without a proof supplied in this work. Many different definitions are commonly used for a ring. The following will be the working definition for a ring. Definition. Let R be a set with well-defined addition and multiplication operations. Then R is a ring if R is an abelian group under addition, the multiplication is associative, and multiplication distributes over addition. Note that, by the definition, a ring need not have a multiplicative identity. In some sections of this work, however, alternative definitions of a ring will be con- sidered to see what proven facts would hold true even if a different definition were used for a ring. If R is a ring, then R+ will be used to refer to the additive group of R. The order of the additive group of R+ will be referred to as both |R| and the order of R. Also, 0R will be used to denote the additive identity of R. Some basic facts from group theory and number theory will be used in this work without proof. Readers who wish to see proofs of these results can refer to [3]. By the definition of a ring, R+ must be abelian. The focus of this work will be on rings whose additive groups are cyclic. This is the motivation for the following definition. Definition. A ring R is a cyclic ring if R+ is a cyclic group. 2. Basic Facts about Cyclic Rings One reason why the study of cyclic rings is important is that many familiar structures are cyclic rings. Among these are Zn and nZ for any positive integer n. Cyclic rings are relatively easy to study because their additive structure forces them to have certain properties. For instance, if R is an infinite cyclic ring, r is a generator of R+, and a, b ∈ Z, then ar = br if and only if a = b. If R is a finite cyclic ring of order n, r is a generator of R+, and a, b ∈ Z, then ar = br if and only if a ≡ b mod n. Also, subrings of cyclic rings are cyclic rings. All three of these results follow immediately from elementary results about cyclic groups. Proofs shall be supplied for other basic properties of cyclic rings. 1 2 CYCLIC RINGS 2.1. Cyclic Rings and the Commutative Property. The following fact is well known and is stated in [4]. Lemma 1. Cyclic rings are commutative under multiplication. Proof: Let R be a cyclic ring, r be a generator of R+, and s, t ∈ R. Then there exist a, b ∈ Z with s = ar and t = br. Since st = (ar)(br) = (ab)r2 = (ba)r2 = (br)(ar) = ts, it follows that R is commutative. 2.2. Subrings and Ideals of Cyclic Rings. The next lemma, which is proven in [2], is analogous to the theorem from group theory which states that every subgroup of an abelian group is normal. Lemma 2. Every subring of a cyclic ring is an ideal. Proof: Let R be a cyclic ring and r be a generator of R+. Since R is a ring, then r2 ∈ R. Let k ∈ Z with r2 = kr. Let S be a subring of R. Then S is a cyclic ring. Let s be a generator of S+. Since S is a subring of R, then s ∈ R. Thus, there exists z ∈ Z with s = zr. Let t ∈ R and u ∈ S. Then there exist a, b ∈ Z with t = ar and u = bs. Since tu = (ar)(bs) = (ar)[b(zr)] = (ar)[(bz)r] = (abz)r2 = (abz)(kr) = (abkz)r = (abk)(zr) = (abz)s ∈ S and multiplication is commutative in R, it follows that S is an ideal of R. 2.3. Generators of Cyclic Rings. Given a ring R and a generator r of R+, it is desirable to find all of the generators of R+ in terms of r. For instance, the next lemma follows immediately from the fact that 1 and −1 are the only generators of Z+. Lemma 3. Let R be an infinite cyclic ring and r be a generator of R+. Then r and −r are the only generators of R+. The next result immediately follows from the fact that, if k ∈ Zn, then k is a + generator of Zn if and only if gcd(k, n) = 1. Lemma 4. Let R be a finite cyclic ring of order n, r be a generator of R+, and + k ∈ Zn. Then kr is a generator of R if and only if gcd(k, n) = 1. 3. Constructing Cyclic Rings A simple method exists for constructing a cyclic ring from a cyclic group. In order to define the multiplication for a cyclic group R, a generator r should be chosen. After that, let r2 be any element of R. Finally, define multiplication for all elements as follows: If a, b ∈ Z, then (ar)(br) = (ab)r2. It is routine to check that, for any cyclic group, this procedure defines a multiplication such that the associative property of multiplication and the distributive property hold. Thus, for any cyclic group, this procedure does produce a cyclic ring. This method is demonstrated for R = {0R, r, 2r, 3r}, a cyclic group of order four. Following are possible multiplication tables for R. CYCLIC RINGS 3 Before defining r2: · 0R r 2r 3r 0R 0R 0R 0R 0R 2 2 2 r 0R r 2r 3r 2 2 2r 0R 2r 0R 2r 2 2 2 3r 0R 3r 2r r 2 r = 0R: · 0R r 2r 3r 0R 0R 0R 0R 0R r 0R 0R 0R 0R 2r 0R 0R 0R 0R 3r 0R 0R 0R 0R r2 = r: · 0R r 2r 3r 0R 0R 0R 0R 0R r 0R r 2r 3r 2r 0R 2r 0R 2r 3r 0R 3r 2r r r2 = 2r: · 0R r 2r 3r 0R 0R 0R 0R 0R r 0R 2r 0R 2r 2r 0R 0R 0R 0R 3r 0R 2r 0R 2r r2 = 3r: · 0R r 2r 3r 0R 0R 0R 0R 0R r 0R 3r 2r r 2r 0R 2r 0R 2r 3r 0R r 2r 3r 4 CYCLIC RINGS 4. Some Problems The only other generator of R+ is 3r. What if it had been chosen instead of r to 2 2 determine the multiplicative nature of R? When r = 0R = 0r, (3r) = 0R = 0(3r), and, when r2 = 2r, (3r)2 = 2r = 6r = 2(3r). Thus, in these cases, the relationship between a generator and its square is always the same regardless of which generator is chosen to define the multiplication. Unfortunately, this is not always true. For example, when r2 = r, (3r)2 = r = 9r = 3(3r). Thus, the square of one generator is itself, but the square of the other generator is its triple. In this case, which generator should be chosen to define the multiplication of R? Another question is, given a cyclic group with generator r, which values of r2 produce isomorphic rings? For the previous example, it can easily be verified that the only two cases that produce isomorphic rings are r2 = r and r2 = 3r. This may yield a hint of how to determine in general which cyclic rings are isomorphic given a generator r of the additive group and the value of r2. These problems will be remedied by investigating a certain concept. 5. The Solution: Behavior 5.1. Definitions. Definition. Let R be an infinite cyclic ring and k be a nonnegative integer. R has behavior k if R+ has a generator r such that r2 = kr. Definition. Let R be a finite cyclic ring of order n and k be a positive divisor of n. R has behavior k if R+ has a generator r such that r2 = kr. 5.2. Theorems Regarding Existence and Uniqueness of Behavior. It will be proven that, for any cyclic ring, behavior exists uniquely. Thus, behavior makes the choice of a generator to determine multiplication quite natural. Theorem 1. Any infinite cyclic ring has a unique behavior. Proof: Let R be an infinite cyclic ring and r be a generator of R+. Then there exists z ∈ Z with r2 = zr. If z ≥ 0, the z is a behavior of R. If z < 0, then −z > 0. Note that −r is a generator of R+. Since (−r)2 = (−1)2r2 = (−1)2(zr) = (−z)(−r), then −z is a behavior of R. Thus, existence of behavior has been proven. Let a and b be behaviors of R. Then R+ has generators s and t such that s2 = as and t2 = bt. If s = t, then as = s2 = t2 = bt = bs, causing a = b. If s 6= t, then t = −s. In this case, as = s2 = (−1)2s2 = (−s)2 = t2 = bt = b(−s) = −bs, causing a = −b. Since a and b are nonnegative, then a = b = 0. Thus, uniqueness of behavior has been proven. CYCLIC RINGS 5 Theorem 2. Any finite cyclic ring has a unique behavior. Proof: Let R be a cyclic ring of order n and r be a generator of R+ with r2 = ar for some a ∈ Z. Let k = gcd(a, n) and b ∈ Z with a = bk.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    16 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us