I was going through the proof given in A Classical Introduction to Modern Number Theory by Michael Rosen and Kenneth Ireland (Relevant portion attached) for the claim that in an Euclidian domain $R$, if $a \in R$ and $a \ne 0$ and $p$ is a prime in $R$, $\exists n \in \mathbb{Z}$ such that $p^n | a$ and $p^{n+1} \nmid a$
According to my understanding $R$ is an integral domain, i.e., a commutative ring with no zero divisors and multiplicative inverse need not exist.
Thus I am unable to figure out a proof for the fact that $pb_{m+1} = b_{m}$. How does this statement follow from $a = p^{m}b_{m}$ and $a = p^{m+1}b_{m+1}$?
It's quite simple I think. If $a=p^mb_m$ and $a=p^{m+1}b_{m+1}$, then $p^{m+1}b_{m+1}=p^mb_m$, so $p^m(pb_{m+1}-b_{m})=0$. Now since we're working in an integral domain we must have $p^m=0$ or $pb_{m+1}=b_m$, but $p^m=0$ is impossible (why?), hence $pb_{m+1}=b_m$.