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.