Let R be a domain with 1. Show that if aR = bR, then au = b for some unit u ∈ R.

Question

Let $R$ be a domain with $1$. Show that if $aR = bR$, then $au = b$ for some unit $u ∈ R$.

Any hint on how to start this proof?

Answer 1

There exist $g,h$ such that $$a=bg$$ and $$b=ah$$ Thus $$a=ahg$$ It follows that $hg=1$, hence $g$ and $h$ are units.

Answer 2

If

$aR = bR \tag 1$

for

$a, b \in R, \tag 2$

a domain, then since

$\exists 1_R \in R, \; \forall r \in R, 1_Rr = r1_R = r, \tag 3$

we have

$a = a1_R \in aR \Longrightarrow a \in bR \Longrightarrow a = br, \; r \in R; \tag 4$

likewise,

$\exists s\in S, \; b = as; \tag 5$

thus,

$a = br = asr \Longrightarrow a(1_R - sr) = a - asr = 0; \tag 6$

then since $R$ is a domain,

$a \ne 0 \Longrightarrow 1_R = sr = rs, \tag 7$

showing that $r$ and $s$ are units as required. In the event that $a = 0$, then

$b \in aR = \{ 0 \} \Longrightarrow b = 0, \tag 8$

so

$a = bc, \forall c \in R; \tag 9$

in particular, (9) binds for all units $c \in R^\times$; thus, we are done.