Lemma 1   Let $n$ be a positive integer and $R$ be a cyclic ring of order $R$ . Then the following are equivalent:

1. $R$ is a zero ring;
2. $R$ has behavior $n$ ;
3. $R \cong n\mathbb{Z}_{n^2}$ .

Proof. To show that 1 implies 2, let $R$ have behavior $k$ . Then there exists a generator $r$ of the additive group of $R$ such that $r^2=kr$ . Since $R$ is a zero ring, $r^2=0_R$ . Since $kr=r^2=0_R=nr$ , it must be the case that $k \equiv n \mod n$ . By definition of behavior, $k$ divides $n$ . Hence, $k=n$ .

The fact that 2 implies 3 follows immediately from the theorem that is stated and proven at cyclic rings that are isomorphic to $k\mathbb{Z}_{kn}$ .

The fact that 3 implies 1 follows immediately since $n\mathbb{Z}_{n^2}$ is a zero ring.

Lemma 2   Let $R$ be an infinite cyclic ring. Then the following are equivalent:

1. $R$ is a zero ring;
2. $R$ has behavior $0$ ;
3. $R$ is isomorphic to the subring $\mathbf{B}$ of $\mathbf{M}_{2\operatorname{x}2}(\mathbb{Z})$ :
   

Proof. To show that 1 implies 2, the contrapositive of the theorem that is stated and proven at cyclic rings that are isomorphic to $k\mathbb{Z}$ can be used. If $R$ does not have behavior $0$ , then its behavior $k$ must be positive by definition, in which case $R \cong k\mathbb{Z}$ . It is clear that $k\mathbb{Z}$ is not a zero ring.

To show that 2 implies 3, let $r$ be a generator of the additive group of $R$ . It can be easily verified that $\varphi \colon R \to \mathbf{B}$ defined by is a ring isomorphism.

The fact that 3 implies 1 follows immediately since $\mathbf{B}$ is a zero ring.