Corollary   A finite cyclic ring of order $n$ with behavior $k$ is isomorphic to $k{\mathbb{Z}}_{kn}$

Proof. Note that $k{\mathbb{Z}}_{kn}$ is a cyclic ring and that $k$ is a generator of its additive group. As groups, $k{\mathbb{Z}}_{kn}$ and $\mathbb{Z}_n$ are isomorphic. Thus, $k{\mathbb{Z}}_{kn}$ has order $n$ Since $k^2=k(k)$ then $k\mathbb{Z}$ has behavior $k$