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# criterion for cyclic rings to be principal ideal rings

###### Theorem.

A cyclic ring is a principal ideal ring if and only if it has a multiplicative identity.

###### Proof.

Let $R$ be a cyclic ring. If $R$ has a multiplicative identity $u$, then $u$ generates the additive group of $R$. Let $I$ be an ideal of $R$. Since $\{0_{R}\}$ is principal, it may be assumed that $I$ contains a nonzero element. Let $n$ be the smallest natural number such that $nu\in I$. The inclusion $\langle nu\rangle\subseteq I$ is trivial. Let $t\in I$. Since $t\in R$, there exists $a\in\mathbb{Z}$ with $t=au$. By the division algorithm, there exists $q,r\in\mathbb{Z}$ with $0\leq r<n$ such that $a=qn+r$. Thus, $t=au=(qn+r)u=(qn)u+ru=q(nu)+ru$. Since $ru=t-q(nu)\in I$, by choice of $n$, it must be the case that $r=0$. Thus, $t=q(nu)$. Hence, $\langle nu\rangle=I$, and $R$ is a principal ideal ring.

Conversely, if $R$ is a principal ideal ring, then $R$ is a principal ideal. Let $k$ be the behavior of $R$ and $r$ be a generator of the additive group of $R$ such that $r^{2}=kr$. Since $R$ is principal, there exists $s\in R$ such that $\langle s\rangle=R$. Let $a\in\mathbb{Z}$ such that $s=ar$. Since $r\in R=\langle s\rangle$, there exists $t\in R$ with $st=r$. Let $b\in\mathbb{Z}$ such that $t=br$. Then $r=st=(ar)(br)=(ab)r^{2}=(ab)(kr)=(abk)r$. If $R$ is infinite, then $abk=1$, in which case $k=1$ since $k$ is nonnegative. If $R$ is finite, then $abk\equiv 1\operatorname{mod}|R|$, in which case $k=1$ since $k$ is a positive divisor of $|R|$. In either case, $R$ has behavior one, and it follows that $R$ has a multiplicative identity. ∎

## Mathematics Subject Classification

16U99*no label found*13A99

*no label found*13F10

*no label found*

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