Let $R$ be a principal ideal domain containing distinct prime elements $p, q$. Let $n, m \geq 1$ be integers. Is there a simple proof for the claim that the elements $p^n$ and $q^m$ of $R$ are relatively prime, i.e. that $(p^n) + (q^m) = R$, or equivalently that the greatest common divisor of $p^n$ and $q^m$ is $1$?
Powers of distinct primes in PID are relatively prime
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abstract-algebra
ring-theory
principal-ideal-domains
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0Do you know that PIDs are UFDs? – 2017-01-31
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0@BillDubuque Yes. – 2017-01-31
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2Then $\,(p^m,q^m) = (d)\,\Rightarrow\, d\,$ divides the coprimes $\,p^n,q^m\,$ so $\,d\,$ is a unit, by unique prime factorization. – 2017-01-31
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1Most all of the common proofs are ln [this thread](http://math.stackexchange.com/questions/166839/if-gcda-b-1-then-gcdan-bn-1/166856#166856) will work here. If for some reason those those proofs don't work for you then please refine your question to say why, and we can reopen. – 2017-01-31