If $a$ and $b$ are coprimes in an euclidean domain, then $a^n$ and $b$ are coprimes.

Question

Let $a,b\in\mathbb{E}$ where $\mathbb{E}$ is an euclidean domain. Prove that if $a$ and $b$ are coprimes, then $a^n$ and $b$ are coprimes for all $n\in\mathbb{N}$. I think the proof is by induction on $n$ and that's how I started it, but any suggestions are appreciated.

Answer 1

Denote $Ra$, the principal ideal generated by $a$, by $(a)$.

Let us recall the following statement (some textbooks state it as part of the CRT (Chinese Remainder Theorem)):

Let $R$ be a commutative ring with $1$. If ideals $I,J$ are coprime ($A,B$ are coprime ideals provided $A+B=R$), then so are $I^n,J^m$ for all $m,n\in \mathbb Z_+$. (Hint for the proof: If we have proved ${\color {skyblue} I}, {\color{violet}{J^m}}$ are coprime, then using this statement again implies so are ${\color {skyblue} I}^n, {\color{violet}{J^m}}$)

By Bezout's identity, two elements in a PID are coprime iff their correponding principal ideals are coprime. $a,b$ are coprime$\implies$ the ideal $(a),(b)$ are coprime ideals$\implies$ $(a)^n=(a^n),(b)$ are coprime ideals$\implies$ $a^n,b$ are coprime.