Let $R$ be a cyclic ring and $S$ be a subring of $R$. Then $R$ and $S$ are both cyclic rings. Let $r$ be a generator of the additive group of $R$ and $s$ be a generator of the additive group of $S$. Then $s\in R$. Thus, there exists $z\in\mathbb{Z}$ with $s=zr$.

Let $t\in R$ and $u\in S$. Then $u\in R$. Since multiplication is commutative in a cyclic ring, $tu=ut$. Since $t\in R$, there exists $a\in{\mathbb{Z}}$ with $t=ar$. Since $u\in S$, there exists $b\in{\mathbb{Z}}$ with $u=bs$.

Since $R$ is a ring, $r^{2}\in R$. Thus, there exists $k\in{\mathbb{Z}}$ with $r^{2}=kr$. Since $tu=(ar)(bs)=(ar)[b(zr)]=(abz)r^{2}=(abz)(kr)=(abkz)r=(abk)(zr)=(abk)s\in S$, it follows that $S$ is an ideal of $R$. ∎