Let $p_1,p_2,q_1,q_2$ be irreducible elements in integral domain $R$ such that none are associates to any of the others and $p_1p_2=q_1q_2$.

Prove that $p_1q_1q_2$ and $p_1p_2q_1$ do not have a greatest common divisor.

I have no idea to approach this problem.

It will be started with having gcd and show contradiction. But there are not much theorem to use show the question with integral domain. Help me!!

  • 2
    $\begingroup$ Well, $p_1q_1$ is obviously a common divisor. Using $p_1p_2 = q_1q_2$, you get that $p_1q_1q_2 = p_1^2p_2$, so $p_1p_2$ is a common divisor. You could try showing that neither of those are a divisor of the others. $\endgroup$
    – Arthur
    May 31, 2016 at 22:57

1 Answer 1


We have that $p_1q_1$ and $p_1p_2$ are both common divisors, so a greatest common divisor would be divisible by both of these. Suppose there are $a,b,c$ such that $$p_1q_1ac=p_1p_2bc=p_1q_1q_2=p_1^2p_2$$ with $p_1q_1a=p_1p_2b$ being the greatest common divisor. Then $ac=q_2$ and $bc=p_1$. Since these are irreducible, $c$ must be a unit, and it follows that the greatest common divisor is an associate of $p_1q_1q_2$. This implies that $p_1q_1q_2=p_1^2p_2$ divides $p_1p_2q_1$ so there is a $d$ with $p_1p_2q_1=p_1^2p_2d$. It follows that $q_1=p_1d$, which contradicts the fact that $q_1$ is irreducible.

  • $\begingroup$ @Pearl no problem. $\endgroup$ Jun 1, 2016 at 5:16

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