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# cyclic rings and zero rings

###### 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 $\mathbf{M}_{{2\operatorname{x}2}}(\mathbb{Z})$:

$\mathbf{B}=\left\{\left.\left(\begin{array}[]{cc}n&-n\\ n&-n\end{array}\right)\right|n\in\mathbb{Z}\right\}.$

###### 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 $\displaystyle\varphi(nr)=\left(\begin{array}[]{cc}n&-n\\ n&-n\end{array}\right)$ is a ring isomorphism.

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

## Mathematics Subject Classification

13M05*no label found*13A99

*no label found*16U99

*no label found*

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