Is PID with one divisible element a field?

Question

Let $R$ be a PID and $r\in R$ be the divisible element i.e. for any $a\in R$ there exists $y\in R$ s.t. $r=ay$. Can we imply that $R$ is a field?

Answer 1

Here is a hint: Let $x\in R$, and $a=rx$. Then we can find $y$ such that $r=ay=rxy$. What does this tell us?

Answer 2

For any non-zero $a\in R$, by definition $r\in(a)$. But $r$ is divisible by $r^2$ too, i.e. $r=xr^2$ for an $x\in R$; hence (assuming $r\neq 0$) $1=xr$, as $R$ is a domain.

If the ideal $(a)$ contains a unit, $a$ is a unit.