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 (http://planetmath.org/Generator) the additive group of $R$ .

Proof.

Let $k$ be the behavior of $R$ . Then there exists a generator (http://planetmath.org/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 (http://planetmath.org/CyclicRingsOfBehaviorOne) for more details.

单词	TheMultiplicativeIdentityOfACyclicRingMustBeAGenerator
释义	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 (http://planetmath.org/Generator) the additive group of $R$ . Proof. Let $k$ be the behavior of $R$ . Then there exists a generator (http://planetmath.org/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 (http://planetmath.org/CyclicRingsOfBehaviorOne) for more details.
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