# criteria for cyclic rings to be isomorphic

###### 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 $\varphi\colon R\to S$ by $\varphi(cr)=cs$ for every $c\in\mathbb{Z}$. This map is clearly well defined and surjective  . Since $R$ and $S$ have the same order, $\varphi$ is injective  . Since, for every $a,b\in\mathbb{Z}$, $\varphi(ar)+\varphi(br)=as+bs=(a+b)s=\varphi((a+b)r)=\varphi(ar+br)$ and

$\begin{array}[]{rl}\varphi(ar)\varphi(br)&=(as)(bs)\\ &=(ab)s^{2}\\ &=(ab)(ks)\\ &=(abk)s\\ &=\varphi((abk)r)\\ &=\varphi((ab)(kr))\\ &=\varphi((ab)r^{2})\\ &=\varphi((ar)(br)),\end{array}$

Conversely, let $\psi\colon R\to S$ 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 $k$. ∎

Title criteria for cyclic rings to be isomorphic CriteriaForCyclicRingsToBeIsomorphic 2013-03-22 16:02:39 2013-03-22 16:02:39 Wkbj79 (1863) Wkbj79 (1863) 14 Wkbj79 (1863) Theorem msc 13A99 msc 16U99