Proof. If $R$ is a cyclic ring and $r$ is a generator of the additive group of $R$ , then $|r|=|R|$ . Since, for every $s \in R$ , $|s|$ divides $|R|$ , then it follows that $\operatorname{char}~R=|R|$ . Conversely, if $R$ is a finite ring such that $\operatorname{char}~R=|R|$ , then the exponent of the additive group of $R$ is also equal to $|R|$ . Thus, there exists $t \in R$ such that $|t|=|R|$ . Since $\langle t \rangle$ is a subgroup of the additive group of $R$ and $|\langle t \rangle |=|t|=|R|$ , it follows that $R$ is a cyclic ring.