Suppose that R is a commutative ring without zero-divisors. Show that all the nonzero elements of R have the same additive order.

Consider p and q be any nonzero element of R.
therefore,
$pq+pq+pq+...n\times =n\cdot (pq)$
$=(n\times p)\times q$
$=p(n\times q)$
Since,
$n\times p=0$
$(n\times p)q=0$
$p(n\times q)=0$
With no zero divisiors,
$n\times q=0$, therefore,
and if the $pq+pq+pq+...m\times =m\cdot (pq)$
$=(m\cdot p)\cdot q$
$=p(m\cdot q)$
and, $m\cdot q=0$
$(m\cdot p)\cdot q=0$
$p(m\cdot q)=0$
with no zero divisors, , $m\cdot q=0$
Therefore, ${\displaystyle n\le m\text{and}m\le n}$