criteria for cyclic rings to be isomorphic


Theorem.

Two cyclic rings are isomorphicPlanetmathPlanetmathPlanetmath if and only if they have the same order and the same behavior.

Proof.

Let R be a cyclic ring with behavior k and r be a generatorPlanetmathPlanetmathPlanetmath (http://planetmath.org/Generator) of the additive groupMathworldPlanetmath of R with r2=k⁢r. Also, let S be a cyclic ring.

If R and S have the same order and the same behavior, then let s be a generator of the additive group of S with s2=k⁢s. Define φ:R→S by φ⁢(c⁢r)=c⁢s for every c∈ℤ. This map is clearly well defined and surjectivePlanetmathPlanetmath. Since R and S have the same order, φ is injectivePlanetmathPlanetmath. Since, for every a,b∈ℤ, φ⁢(a⁢r)+φ⁢(b⁢r)=a⁢s+b⁢s=(a+b)⁢s=φ⁢((a+b)⁢r)=φ⁢(a⁢r+b⁢r) and

φ⁢(a⁢r)⁢φ⁢(b⁢r)=(a⁢s)⁢(b⁢s)=(a⁢b)⁢s2=(a⁢b)⁢(k⁢s)=(a⁢b⁢k)⁢s=φ⁢((a⁢b⁢k)⁢r)=φ⁢((a⁢b)⁢(k⁢r))=φ⁢((a⁢b)⁢r2)=φ⁢((a⁢r)⁢(b⁢r)),

it follows that φ is an isomorphismPlanetmathPlanetmathPlanetmathPlanetmath.

Conversely, let ψ:R→S be an isomorphism. Then R and S must have the same order. If R is infiniteMathworldPlanetmath, then S is infinite, and k is a nonnegative integer. If R is finite, then k divides (http://planetmath.org/Divisibility) |R|, which equals |S|. In either case, k is a candidate for the behavior of S. Since r is a generator of the additive group of R and ψ is an isomorphism, ψ⁢(r) is a generator of the additive group of S. Since (ψ⁢(r))2=ψ⁢(r2)=ψ⁢(k⁢r)=k⁢ψ⁢(r), it follows that S has behavior k. ∎

Title criteria for cyclic rings to be isomorphic
Canonical name CriteriaForCyclicRingsToBeIsomorphic
Date of creation 2013-03-22 16:02:39
Last modified on 2013-03-22 16:02:39
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 14
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 13A99
Classification msc 16U99