# Integral Domain

• Apr 15th 2009, 06:01 PM
chiph588@
Integral Domain
If $\displaystyle R$ is an integral domain and $\displaystyle a, b \in R$, show if $\displaystyle Ra = Rb$ then $\displaystyle a=ub$ where $\displaystyle u$ is a unit.
• Apr 15th 2009, 06:20 PM
ThePerfectHacker
Quote:

Originally Posted by chiph588@
If $\displaystyle R$ is an integral domain and $\displaystyle a, b \in R$, show if $\displaystyle Ra = Rb$ then $\displaystyle a=ub$ where $\displaystyle u$ is a unit.

Since $\displaystyle (a) = (b)$ it means $\displaystyle (a)\subseteq (b) \text{ and }(b) \subseteq (a)$.
Thus, $\displaystyle b|a \text{ and } a|b$ so $\displaystyle a,b$ are associates and so $\displaystyle a=ub$.
• Apr 15th 2009, 07:33 PM
NonCommAlg
Quote:

Originally Posted by chiph588@

If $\displaystyle R$ is an integral domain and $\displaystyle a, b \in R$, show if $\displaystyle Ra = Rb$ then $\displaystyle a=ub$ where $\displaystyle u$ is a unit.

to see why we need R to be an integral domain:

if $\displaystyle a=0,$ then $\displaystyle Rb=0$ and so $\displaystyle b=0.$ in this case choose $\displaystyle u=1.$ so we may assume that $\displaystyle a \neq 0.$ now we have $\displaystyle a \in Ra = Rb.$ so $\displaystyle a=rb,$ for some $\displaystyle r \in R.$ similarly $\displaystyle b=sa,$ for some $\displaystyle s \in R.$

thus $\displaystyle a=rb=rsa.$ hence $\displaystyle (1-rs)a=0.$ since $\displaystyle R$ is an integral domain and $\displaystyle a \neq 0,$ we must have $\displaystyle 1-rs=0,$ i.e. $\displaystyle rs=1.$ so $\displaystyle r$ is a unit and we choose $\displaystyle u=r.$