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 $r=ur=(ar)r=a{r}^2=a(kr)=(ak)r$. If $R$ is infinite, then $ak=1$, causing $a=k=1$ since $k$ is a nonnegative integer. If $R$ is finite, then $ak\equiv 1\mathrm{mod}|R|$. Thus, $\gcd (k,|R|)=1$. Since $k$ divides $|R|$, $k=1$. Therefore, $a\equiv 1\mathrm{mod}|R|$. 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
Canonical name	TheMultiplicativeIdentityOfACyclicRingMustBeAGenerator
Date of creation	2013-03-22 15:56:59
Last modified on	2013-03-22 15:56:59
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	16
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 16U99
Classification	msc 13F10
Classification	msc 13A99
Related topic	CyclicRing3
Related topic	CriterionForCyclicRingsToBePrincipalIdealRings
Related topic	CyclicRingsOfBehaviorOne