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!

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You don't need quadratic reciprocity to see that the ideal $(3,\sqrt{15})$ is non-principal. If it were principal, say with generator $\gamma$, then $(\gamma)^2 = (3)$, so $\gamma^2 = 3u$ for some unit $u$ in ${\mathbf Z}[\sqrt{15}]$. Taking the norm of both sides, ${\rm N}(\gamma)^2 = \pm 9$. The norm of $\gamma$ is an integer, so we must have 9 on the right side. Then $\gamma$ has norm $\pm 3$. Writing $\gamma = a + b\sqrt{15}$, $a^2 - 15b^2 = \pm 3$, so $\pm 3 \equiv a^2 \bmod 5$. But neither 3 nor $-3$ is a square mod 5, so we have a contradiction. – KCd Dec 19 '11 at 9:33

It is easier to look for a principal ideal (of norm $6$) that factors into the product of your ideals :

$\langle 2, 1+\sqrt{15}\rangle \langle 3, \sqrt{15}\rangle = \langle 6, 2\sqrt{15}, 3+3 \sqrt{15},15+\sqrt{15}\rangle = \langle 3+\sqrt{15}, \ldots \rangle = \langle 3 + \sqrt{15} \rangle$, because $3+\sqrt{15}$ is of norm $6$.

Therefore, $\langle 2, 1+\sqrt{15}\rangle \langle 3+ \sqrt{15} \rangle = \langle 3,\sqrt{15} \rangle \langle 2 \rangle$

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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 $\alpha = 3$, $\beta = 3 + \sqrt{15}$.

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