## You are here

Homebehavior exists uniquely (infinite case)

## Primary tabs

# behavior exists uniquely (infinite case)

The following is a proof that behavior exists uniquely for any infinite cyclic ring $R$.

###### Proof.

Let $r$ be a generator of the additive group of $R$. Then there exists $z\in\mathbb{Z}$ with $r^{2}=zr$. If $z\geq 0$, then $z$ is a behavior of $R$. Assume $z<0$. Note that $-z>0$ and $-r$ is also a generator of the additive group of $R$. Since $(-r)^{2}=(-1)^{2}r^{2}=(-1)^{2}(zr)=(-z)(-r)$, it follows that $-z$ is a behavior of $R$. Thus, existence of behavior has been proven.

Let $a$ and $b$ be behaviors of $R$. Then there exist generators $s$ and $t$ of the additive group of $R$ such that $s^{2}=as$ and $t^{2}=bt$. If $s=t$, then $as=s^{2}=t^{2}=bt=bs$, causing $a=b$. If $s\neq t$, then it must be the case that $t=-s$. (This follows from the fact that 1 and -1 are the only generators of $\mathbb{Z}$.) Thus, $as=s^{2}=(-1)^{2}s^{2}=(-s)^{2}=t^{2}=bt=b(-s)=-bs$, causing $a=-b$. Since $a$ and $b$ are nonnegative, it follows that $a=b=0$. Thus, uniqueness of behavior has been proven. ∎

## Mathematics Subject Classification

13A99*no label found*16U99

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections