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

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 $k$.∎