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Homethe multiplicative identity of a cyclic ring must be a generator

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# the multiplicative identity of a cyclic ring must be a generator

###### Theorem.

Let $R$ be a cyclic ring with multiplicative identity $u$. Then $u$ generates the additive group of $R$.

###### Proof.

Let $k$ be the behavior of $R$. Then there exists a generator $r$ of the additive group of $R$ such that $r^{2}=kr$. Let $a\in\mathbb{Z}$ with $u=ar$. Then $r=ur=(ar)r=ar^{2}=a(kr)=(ak)r$. If $R$ is infinite, then $ak=1$, causing $a=k=1$ since $k$ is a nonnegative integer. If $R$ is finite, then $ak\equiv 1\operatorname{mod}|R|$. Thus, $\gcd(k,|R|)=1$. Since $k$ divides $|R|$, $k=1$. Therefore, $a\equiv 1\operatorname{mod}|R|$. In either case, $u=r$. ∎

Note that it was also proven that, if a cyclic ring has a multiplicative identity, then it has behavior one. Its converse is also true. See this theorem for more details.

## Mathematics Subject Classification

16U99*no label found*13F10

*no label found*13A99

*no label found*

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