Theorem   Let $R$ be a cyclic ring with multiplicative identity $u$ . Then $u$ generates the additive group of $R$ .

Proof. Let $k$ be the behavior of $R$ . Then there exists a generator $r$ of the additive group of $R$ such that $r^2=kr$ . Let $a \in \mathbb{Z}$ with $u=ar$ . Then $r=ur=(ar)r=ar^2=a(kr)=(ak)r$ . If $R$ is infinite, then $ak=1$ , causing $a=k=1$ since $k$ is a nonnegative integer. If $R$ is finite, then $ak \equiv 1 \operatorname{mod} |R|$ . Thus, $\gcd(k,|R|)=1$ . Since $k$ divides $|R|$ , $k=1$ . Therefore, $a \equiv 1 \operatorname{mod} |R|$ . In either case, $u=r$ .