# criteria for cyclic rings to be isomorphic

###### Theorem.

Two cyclic rings are isomorphic^{} 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 generator^{} (http://planetmath.org/Generator) of the additive group^{} of $R$ with ${r}^{2}=kr$. 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 ${s}^{2}=ks$. Define $\phi :R\to S$ by $\phi (cr)=cs$ for every $c\in \mathbb{Z}$. This map is clearly well defined and surjective^{}. Since $R$ and $S$ have the same order, $\phi $ is injective^{}. Since, for every $a,b\in \mathbb{Z}$, $\phi (ar)+\phi (br)=as+bs=(a+b)s=\phi ((a+b)r)=\phi (ar+br)$ and

$\begin{array}{cc}\hfill \phi (ar)\phi (br)& =(as)(bs)\hfill \\ & =(ab){s}^{2}\hfill \\ & =(ab)(ks)\hfill \\ & =(abk)s\hfill \\ & =\phi ((abk)r)\hfill \\ & =\phi ((ab)(kr))\hfill \\ & =\phi ((ab){r}^{2})\hfill \\ & =\phi ((ar)(br)),\hfill \end{array}$

it follows that $\phi $ is an isomorphism^{}.

Conversely, let $\psi :R\to S$ be an isomorphism. Then $R$ and $S$ must have the same order. If $R$ is infinite^{}, 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 $\psi $ is an isomorphism, $\psi (r)$ is a generator of the additive group of $S$. Since ${(\psi (r))}^{2}=\psi ({r}^{2})=\psi (kr)=k\psi (r)$, it follows that $S$ has behavior $k$.
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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 |