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.

Title the multiplicative identity of a cyclic ring must be a generator TheMultiplicativeIdentityOfACyclicRingMustBeAGenerator 2013-03-22 15:56:59 2013-03-22 15:56:59 Wkbj79 (1863) Wkbj79 (1863) 16 Wkbj79 (1863) Theorem msc 16U99 msc 13F10 msc 13A99 CyclicRing3 CriterionForCyclicRingsToBePrincipalIdealRings CyclicRingsOfBehaviorOne