Prove that, if $(a)=(a')$, then $a'=ua$

Question

Let $R$ be integral domain. Show that if $2$ principal ideals $(a)$ and $(a')$ are equal (where $a,a'\in R$) then there exists $u\in R^{\times}$ such that, $a'=ua$

Now if $(a)=(a')$ then $a\in(a')$ and $a'\in(a)$ therefore there exists some elements $u,v$ in $R$, such that $a=ua'$ and $a'=va$, which implies that $a=uva$.

Hence $uv=1$ and $u$ is a unit, $u\in R^{\times}(=R\setminus\{0\})$

but I didn't make use of the property that, $R$ is an integral domain, so $R$ can be any ring or not ?

Answer 1

You stated that $a=uva$ implies $uv = 1$. This is the cancellation law which holds in integral domains (but not for any ring).

To prove this for integral domains, note that if $ca=cb$ for $c \neq 0$, then $c(a-b)=0$ so $a=b$.

A counter-example for a ring which is not an integral domain is $\mathbb{Z}/4\mathbb{Z}$, where $2 \cdot 1 = 2 \cdot 3$.

Answer 2

The implication $a=uva\implies uv=1$ is not valid in just any ring $R$. It needs to be a ring $R$ with no zero divisors. An integral domain does not have any zero divisors....

So your proof is almoast complete it just needs one extra line:

Now if $(a)=(a')$ then $a\in(a')$ and $a'\in(a)$. Therefore there exist some elements $u,v$ in $R$, such that $a=ua'$ and $a'=va$, which implies that $a=uva$.

Since $R$ is an integral domain, this implies $uv=1$, etc...