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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.

Bill Dubuque
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Angel Fernando Aviles Lopez
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    Using the UFD-property is surely the way to go. For fun, you can also use Bezout's identity (or the PID-property): there exist elements $u,v\in\Bbb{E}$ such that $$1=ua+vb.\qquad()$$ Raising $()$ to power $n$, use binomial formula on the r.h.s., shows that there exist elements $U,V\in\Bbb{E}$ such that $$1= Ua^n+Vb.$$ – Jyrki Lahtonen Mar 22 '24 at 06:33

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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.

Asigan
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