Suppose R is an integral domain. Prove that $(a)=(b)$ if and only if $b = ua$ where $u$ is in $R^\times$

Question

I am lost on this one. I'm still new to ring theory, as we're only a couple weeks into the course, but it's already well over my head. I know that $R$ is an integral domain, so the additive and multiplicative identities are not equal, and if both $a$ and $b$ are nonzero, then their product will be nonzero. So $u$ can't be zero. I need to prove that $u$ is the multiplicative identity element, but I'm not sure how to do that without having knowledge of what kind of ring $R$ is.

Answer 1

We may assume wlog. that $a\ne 0$. From $b\in(a)$, there exists $u\in R$ with $b=ua$. From $a\in(b)$, there exists $v\in R$ with $a=vb$. Then $a=vua$, hence $(vu-1)a=0$. Since $r$ is a domain and $a\ne0$, we conclude $vu=1$.

Answer 2

Because $(a) = (b)$, $b \in (a)$. By the definition of $(a)$, there exists $u \in R$ such that $b = ua$. Similarly, there is a $v \in R$ such that $a = vb$. Combining these gives $a = vua$, and since this is an integral domain, we can cancel*, getting $vu = 1$. Thus, $u$ is a unit.

*We have to watch out for the case where $a = 0$. But if it is, then $(a) = \{ 0 \}$, so it's pretty clear that $b = 0$ as well, and so picking any unit as $u$ will suffice.

Answer 3

$\begin{eqnarray}{\bf Hint} & \quad\ \ \ aR &=&\, bR\\ \iff &b\in aR,\!\!\!&&\!\!\!\!\ a\in bR\\ \iff&\quad a\overset{\large \times\ u}\rightarrow &b& \overset{\large\times\ v}\rightarrow a\\ \iff&au = b\!\!\!\!\!\!&,&\!\! auv =a \\ \iff& au=b\!\!\!\!\!\!&,&\ uv = 1, \ \ \rm by\ \ cancel\ \ a\ [\ne 0\ \ wlog] \end{eqnarray}$