Equivalence class of an ideal Let $K$ denote the number field $\mathbb{Q}(\sqrt{15}).$ According to standard lore, we have that $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{15}]$. Moreover,
$2\mathcal{O}_{K} = \langle 2, 1+\sqrt{15}\rangle^{2}$
and
$3\mathcal{O}_{K} = \langle 3, \sqrt{15}\rangle^{2}.$
Resorting to the law of quadratic reciprocity, I have proven that the ideal $\langle 3, \sqrt{15}\rangle$ is non-principal. Since $h(\mathbb{Q}(\sqrt{15}))=2$, it must be the case that 
$\langle 2, 1+\sqrt{15}\rangle \langle \alpha \rangle = \langle 3, \sqrt{15}\rangle \langle \beta \rangle$
for certain $\alpha, \beta \in \mathcal{O}_{K}\setminus \{0\}$.
Is there a quick way to determine such a pair $(\alpha, \beta)$?
Thanks!

Starting from $(2, 1+\sqrt{15})(\alpha) = (3, \sqrt{15})(\beta)$, we can multiply both sides by $(3,\sqrt{15})$ to get:
$$(2, 1+\sqrt{15})(3,\sqrt{15}) = (3\beta/\alpha)$$
Multiplying the ideals on the left together
$$(6, 2\sqrt{15}, 3 + 3\sqrt{15}, 15 + \sqrt{15}) = (3\beta/\alpha)$$
So the ideal on the left side is a principal ideal, and we just need to find a generator.  We repeatedly use the property $(x,y) = (x, y+zx)$ for ideals (adding a multiple of one generator to another doesn't change the ideal) to simplify the list of generators.
$$\begin{eqnarray}
 & (6, 2 \sqrt{15}, 3 + 3 \sqrt{15}, 15 + \sqrt{15}) \\
 = & (6, 2 \sqrt{15}, -42, 15 + \sqrt{15}) & \mbox{subtract 3 times 4th generator from third} \\
 =  & (6, 2 \sqrt{15}, 15 + \sqrt{15}) & \mbox { 42 is a multiple of 6 }\\
 = & (6, -30, 15 + \sqrt{15}) & \mbox { subtract 2 times 3rd generator from 2nd } \\
= & (6, 15 + \sqrt{15}) & \mbox { 30 is a multiple of 6 } \\
= & (6, 3 + \sqrt{15}) & \mbox { subtract 2 times 6 from 2nd generator } \\
= & (3 + \sqrt{15})\\
\end{eqnarray}
$$
For the last equality, note that $(3 + \sqrt{15})(3 - \sqrt{15}) = -6$.
So we can take $\alpha = 3$, $\beta = 3 + \sqrt{15}$.
