cyclic rings and zero rings

Lemma 1.

Let $n$ be a positive integer and $R$ be a cyclic ring of order (http://planetmath.org/OrderRing) $R$. 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 (http://planetmath.org/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 $k\equiv nmodn$. By definition of behavior, $k$ divides (http://planetmath.org/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 $k{\mathbb{Z}}_{kn}$ (http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ_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 . Then the following are equivalent:

1. $R$ is a zero ring;

2. $R$ has behavior $0$;

3. $R$ is isomorphic (http://planetmath.org/Isomorphism7) to the subring $𝐁$ of ${𝐌}_{2\mathrm{x}2}(\mathbb{Z})$:

Proof.

To show that 1 implies 2, the contrapositive of the theorem that is stated and proven at cyclic rings that are isomorphic to $k\mathbb{Z}$ (http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ) 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 $\phi :R\to 𝐁$ defined by is a ring isomorphism.

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