If the product of two relatively prime elements is a cube then the factors are also cubes .

Question

I am not really sure if it is correct. My idea is:

Let $ab=x^3 $ and $(a, b) = 1$ and $a, b, x$ are some elements in an arbitrary ring $R$.

Then $a|x^3 $ and $b|x^3$ which means there exists some elements $c,d \in R$ that satisfies $ac = x^3$ and $bd = x^3$. We then know that $a|d$ because $a, b$ are coprime and $a\space|\space x^3 = bd$.

For example: Consider the ring $R=Z[i]$

I want to find all solutions $x,y \in \mathbb{Z}$ with $y^2=x^3-1$

Now the factored form in $R$ is $$x^3=(y+i)(y-i) $$

I would like to know why the factors have to be cubes .

Answer 1

If the $a, b$ are relatively prime, it means they contain no common prime factors. Since we know that,

$$ab = x^3$$

We also know that all exponents of the primes $ab$ contain, must be a multiple of $3$. Because $a$ and $b$ have no common prime factor, their all prime factors have to be exponentiated to a multiple of $3$. So they are cubes. QED.