Show ideal generated by unit times element is the same as ideal generated by element

Question

Let $R$ be a UFD and let $a \in R$ be nonzero and $u \in R$ be a unit.
I need to show that $(a)=(ua)$, where $(x)$ is the ideal generated by $x$.
To show $(a) \subseteq (ua)$, take $x \in (a)$, so that $x=ar$ for some $r \in R$.
Then can I write $x=uu^{-1}ar=ua(u^{-1}r) \in (ua)$?
For the inclusion $(ua) \subseteq (a)$, let $y \in (ua)$, so that $y=uar$ for some $r \in R$.
Then can I say that $y=uar=a(ur) \in (a)$?

Answer 1

Hint: $\ b\in (a)\!\iff\! \overbrace{a\mid b\!\iff\! \color{#c00}{ua}\mid b}^{\textstyle \color{#c00}u\,(\color{#c00}a\mid u^{-1}b)}\!\iff\! b\in (ua)\,\ \ $ QED

The overbracee means $\ \color{#c00}{ua}\mid uu^{-1}b = b\ $ i.e. scale the divisibility by $\,u.\ $ More abstractly:

Lemma $\ $ If $\ a\mid a'\mid a\ $ then $\ a\mid b \iff a'\mid b,\, $ i.e. associates have equal sets of multiples.

Proof $\ \ (\Rightarrow)\ \ a'\mid a\mid b.\ $ $\ (\Leftarrow)\ \ a\mid a'\mid b$.

Beware $ $ Only in domains are associates always unit multiples - see here.