PID question in Ireland and Rosen

Question

Context: In Ireland and Rosen's 'A classic introduction to number theory' on page 11, the proof that in a PID$=R$, there is an integer $n$ such that, for a prime $p$ and any $b\in R$, $p^n \mid b , p^{n+1}\not \mid b.$

The proof is by contradiction, stating that if for each $m>0$ there is a $b_m$ such that $b=p^mb_m$. They then use that $p^mb_m=p^mp b_{m+1}\Rightarrow b_m=pb_{m+1}$

Question: Why is it true that in a PID, $ab=cb\Rightarrow a=c$, given that $b^{-1}$ doesn't necessarily exist? Or does the inverse of a prime always exist in a PID?

Excuse the elementary nature of the question, I can't find an answer anywhere.

Answer 1

Cancellation is always possible, if the ring is integral. Because then: $ab=cb \Rightarrow (a-c)b=0$ and if $b \neq 0$ it follows that $a-c=0$, so $a=c$. And all PIDs are integral.

Answer 2

A Domain by definition has no zero-divisors, i.e. $\,a\ne 0,\, ax = 0\,\Rightarrow\, x=0,\,$ i.e. $\,\ker(ax) = 0.\,$This is equivalent to: $ $ elements $\,a\ne 0\,$ are cancellable, i.e. $\,ax=ay\,\Rightarrow\,x= y,\,$ i.e. $\,ax\,$ is $1$-$1,\,$ since

$\qquad a\,$ non-zero-divisor $\iff \ker(ax) = 0\iff ax\,$ is $1$-$1\iff a\,$ is cancellable