Cubes in $\mathbb{Z}[\sqrt{-5}]$ If $\alpha\beta$ is a cube in $\mathbb{Z}[\sqrt{-5}]$, and $\alpha$ and $\beta$ are relatively prime, are $\alpha$ and $\beta$ necessarily cubes as well? Intuitively I think that the answer should be yes, since the class number is 2, but I am not quite sure how to proceed. Can someone help?
Edit: new to this site, but curious why the downvote? Should I clarify something?
 A: The class number ignores units, so this sort of argument isn't sufficient. For example, in the ring of Eisenstein integers $\mathbb{Z}[\omega]$, where $\omega = e^{2\pi i/3}$ is a cube root of unity, we have $1^3 = 1 = \omega \omega^2$, and $\omega$ is a unit and therefore relatively prime to everything, but $\omega$ isn't a cube in $\mathbb{Z}[\omega]$.
However, you can at least get a general result "up to units": in any Dedekind domain $A$, if $\alpha \beta = \gamma^3$, then by uniqueness of prime factorization of ideals of Dedekind domains, we have $(\alpha) = I^3$ and $(\beta) = J^3$ for some ideals $I, J$.
Let $C$ be the ideal class group of $A$. Then $I$ and $J$ represent $3$-torsion classes in $C$, so if $C$ has trivial $3$-torsion (for example, if $C$ is finite and of order coprime to $3$), then $I$ and $J$ are principal. Writing $I = (r)$ and $J = (s)$, we then have $\alpha = ur^3$ and $\beta = vs^3$ for some units $u, v \in A$. Moreover,
$$\gamma^3 = \alpha \beta = uvr^3 s^3,$$
so $uv$ must be a cube in the field of fractions of $A$ (and therefore a cube in $A$, since $A$ is integrally closed).
Returning to $\mathbb{Z}[\sqrt{-5}]$, we can now reduce to the following question: are there two units whose product is a cube in $\mathbb{Q}(\sqrt{-5})$? Fortunately, the group of units of $\mathbb{Z}[\sqrt{-5}]$ is just $\{\pm 1\}$, both of which are cubes. So in $\mathbb{Z}[\sqrt{-5}]$, it's indeed true that if $\alpha \beta$ is a cube, then $\alpha$ and $\beta$ are cubes.
More generally, by Dirichlet's unit theorem, the roots of unity are the only units in the ring of integers of an imaginary quadratic field. With the sole exceptions of the Gaussian and Eisenstein integers, this is just $\{\pm 1\}$, so the same applies as long as the class number is coprime to $3$. In the Eisenstein integers, we also have examples like the one I gave at the start. For the Gaussian integers $\mathbb{Z}[i]$, we also have $\pm i$, but these are cubes as well. In number fields beyond the imaginary quadratic case, there are also units of infinite order, so more possibilities emerge.
A: An example showing that we need the right definition of relatively prime, that is $(\alpha, \beta) = 1$.
Take $\alpha$, $\beta$ with prime decomposition
$$(\alpha) = \mathfrak{p}\cdot \mathfrak{q^3}\\
(\beta) = \mathfrak{p}^2$$
where $\mathfrak{p}$,  $\mathfrak{q}$ are distinct non-principal prime ideals.  Then $\alpha \cdot \beta$  is a cube, but neither $\alpha$, nor $\beta$ is a cube.
Example
$$\mathfrak{p}= (2,  1+\sqrt{-5}), \\ 
\mathfrak{p}^2 = (2, 1+ \sqrt{-5}) \cdot (2, 1- \sqrt{-5}) = (2)$$
and
$$\mathfrak{q} = (3, 1+\sqrt{-5}), \\ 
\mathfrak{p}\cdot \mathfrak{q} = (1+\sqrt{-5}), \ \ \mathfrak{q}^2 = (-2+ \sqrt{-5}) $$
Take
$$\alpha = (1+ \sqrt{-5})(-2 + \sqrt{-5}) = - 7 - \sqrt{-5} \\
\beta = 2$$
and check that
$$\alpha \cdot \beta = (1+ \sqrt{-5})^3$$
Note the only elements in $\mathbb{Z}[-\sqrt{-5}]$ that divide both $\alpha$ and $\beta$ are the units ($\pm 1$). However, $\alpha$, $\beta$ are not relatively prime since the ideal generated by them is $\mathfrak{p}\ne 1$.
