Coprime elements in a PID satisfy that any of their powers are coprime

Question

I have recently met this problem in my abstract algebra dealing with PID rings and coprimes stating:

Let D be a PID ring $ a,b \in D $ two coprime elements. We are to show that for all $ m,n \in \mathbb{N} $ we have $ a^m,b^n $ are also coprime.

Now to be honest I really have no idea on this one as I know intuitively this holds for integers but for an arbitrary PID, this I cannot show. I certainly appreciate the help, thanks.

Answer 1

I'm not entirely sure what you mean by coprime.

If you mean that $(a,b) = (1)$. Then for any $m,n\in\mathbb{N}$, $(a,b)^{n+m}$ is the ideal generated by $a^{n+m},a^{n+m-1}b,\ldots,ab^{n+m-1},b^{n+m}$. Each of these generators is divisible by either $a^m$ or $b^n$, and thus $(a,b)^{n+m}\subset (a^m,b^n)$, but $(a,b) = (1)$, so $(a,b)^{n+m} = (1)^{n+m} = (1)$, so $(1)\subset (a^m,b^n)$, so $(a^m,b^n) = (1)$

If you mean that in the unique factorization of $a,b$, any prime that divides $a$ cannot divide $b$, and vice versa, then the same is obviously true for $a^m,b^n$ (by the factorization of $a,b$ into primes, you have an obvious factorization of $a^m,b^n$ into primes. Now use uniqueness of this factorization).

Answer 2

If $a$ and $b$ are coprime, the ideal generated by $a$ and $b$ is $1$ so you can write $ca+db=1, (ca+db)^{n+m}=Ua^m+Vb^m=1$. done