Let $D$ be an integral domain $a,b \in D$. $m,n$ are relatively prime and $a^n=b^n$ and $a^m=b^m \implies a=b$.

Question

Let $D$ be an integral domain $a,b \in D$. $m,n$ are relatively prime and $a^n=b^n$ and $a^m=b^m \implies a=b$.

I quickly realized that since $m,n$ are relatively prime then $\exists p,q \in \mathbb{Z}$ s.t. $pm+qn=1$.

So we have $a^m=b^m \implies a^{pm}=b^{pm}$ and similarly $a^{qn}=b^{qn}$.

Then $a^{pm} a^{qn}=b^{pm} a^{qn} = b^{pm}b^{qn} \implies$ $a^{pm+qn}=b^{pm+qn} \implies a=b$.

Now my concern is I didn’t use any properties of integral domain. Am I jumping the gun here? What mistake did I make?

Answer 1

Here's the issue: how can you raise something to the $p^{th}$ power? Bezout's lemma only guarantees that such a $p, q \in \mathbb Z$ exist, not that they are positive. If these are negative, then expressions like $a^{pm}$ make no sense, unless $a^m$ was a unit. And now you can see that you want your ring to be a domain, as you can work in the quotient field. Here, you can raise to negative powers all you want and your same proof applies.