Let $R$ be a UFD and $p,q,r \in R$. $pq=r^3$ and $\gcd(p,q)=1$ then $p,q$ are cubes up to associates.

Question

I'm not too sure how to prove this statement.

This seems like a relatively small problem however I can't for the life of me figure out how to start this so I don't really have any working to show. I imagine a small push in the right direction is all I need to find a solution.

Answer 1

Let $$r = r_{1}r_{2}\cdots r_{s},~ p = p_{1}p_{2}\cdots p_{t},~ q = q_{1}q_{2}\cdots q_{u}$$ be the factorizations of $r,p,q \in R$ respectively for some $s,t,u\in \mathbb{N}$. Because $\mathrm{gcd}(p,q) = 1$, it's true that $p_i \ne q_j$ for every $i\le t,j\le u$.

Now since $r^3 = r^{3}_{1}r^{3}_{2}\cdots r^{3}_{s} = p_{1}p_{2}\cdots p_{t}q_{1}q_{2}\cdots q_{u}$ and you are in the UFD, the factorizations must be the same up to the associates (ie. every $p_i$ and $q_j$ is associated with some $r^3_k$). Now use definition of "to be associated"...