<s>[INST] How to tell two ideals belong to the same ideal class group Potentially dumb question here: how can you tell if two ideals belong to the same ideal class group?
Let's say we're looking at $\textbf{Z}[\sqrt{10}]$. It has infinitely many ideals, but by the Minkowski bound we need only concern ourselves with ideals with a norm of $2$ or $3$.
My gut tells me $\langle 3, 1 - \sqrt{10} \rangle$ and $\langle 3, 1 + \sqrt{10} \rangle$ are in the same ideal class, but that could be altogether wrong.
 [/INST] Let's show that the ideal $(3, 1+ \sqrt{10})\cdot (3, 1-\sqrt{10})^{-1}$ is principal. We have
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = \frac{(3, 1+ \sqrt{10})\cdot(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})\cdot (3, 1+\sqrt{10})}$$
Now, for fractional ideals of number fields we have
$$(\alpha_1, \alpha_2)\cdot( \beta_1, \beta_2) = (\alpha_1 \beta_1, \alpha_1 \beta_2 + \alpha_2 \cdot \beta_1, \alpha_2 \cdot \beta_2)$$
( Gauss lemma for Dedekind rings, see Herman Weyl- Algebraic Number Theory)
Therefore, the denominator in the fraction $(*)$ is
$$(9, 6, 1-\sqrt{10}^2) = (9, 6, -9)= (3)$$
while the numerator is
$$(9, 6(1+ \sqrt{10}), (1+\sqrt{10})^2)=(9, 6(1+ \sqrt{10}), 11 + 2 \sqrt{10})=\\ =(9, 6(1+ \sqrt{10}), 2 + 2\sqrt{10}) =((1-\sqrt{10})(1+\sqrt{10}), 2(1+\sqrt{10})) \\=(1+\sqrt{10})\cdot (1-\sqrt{10}, 2)$$
Now $(1-\sqrt{10}, 2) = (1)$, since $((1-\sqrt{10})(1+\sqrt{10}), 2) = (9,2) = (1)$
We conclude:
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = (\frac{1+\sqrt{10}}{3})$$
or  $\frac{1+\sqrt{10}}{3}\cdot (3, 1-\sqrt{10})= (1+\sqrt{10}, 3)$.
This is obvious, once we know the factor...
$\bf{Added:}$ If $\alpha \cdot \beta =  k^2$ then
$$\frac{\alpha}{k} = \frac{k}{\beta} \ \textrm{so}\ \frac{\alpha}{k} (k, \beta)= (\alpha, k)$$
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