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.
 A: In general, use the definition of equivalence and show that there exist elements $\alpha$ and $\beta$ with $\alpha {\mathfrak a} = \beta{\mathfrak b}$.
In your example we have $(2,\sqrt{10})(3,1-\sqrt{10}) = (2 + \sqrt{10})$ and $(2,\sqrt{10})(3,1+\sqrt{10}) = (2 - \sqrt{10})$, from which it follows that the two ideals in question belong to the same ideal class. If you divide the first equation by the second and clear denominators, then you get
$$ (2 - \sqrt{10})(3,1-\sqrt{10}) = (2 + \sqrt{10})(3,1+\sqrt{10}) , $$
which gives you the elements $\alpha$ and $\beta$ mentioned above. The ideals in the displayed equation both equal $3(2,\sqrt{10})$, by the way (look at their prime ideal factorization).
How do we find these equations? Well, we have ideals of norms $2$ and $3$, so to find whether they are in the same class we look for elements whose norms are $\pm 6$, $\pm 12$, $\pm 18$ etc.
A: If we want to determine if two nonzero ideals $I$ and $J$ of $\mathcal{O}_K$ are in the same ideal class, first find another integral ideal $I'$ so that $II'$ is principal—i.e. $I'$ is a representative of the inverse ideal class, in the quadratic case the conjugate ideal will always do the job for this—and so the question amounts to determining if $I'J$ is principal.
In other words, rather than work with $2$ ideals and seek two elements $\alpha$ and $\beta$, the real task is to consider a single nonzero ideal $I$ of $\mathcal{O}_K$ and to determine if there exists an element $a$ in $\mathcal{O}_k$ such that $I = (a)$. By computing the norm $\text{N}(I) = \#(\mathcal{O}_K/I)$, at least we know what the value of $\text{N}_{K/\mathbb{Q}}(a)$ has to be, but that is very limited information in general—much more constraining in the quadratic case. Perhaps the "right" thing to do for a general method—as being sought here—is to let $S$ be the set of prime ideals over the primes dividing $\text{N}(I)$, so any possible element $a$ that could work must at least be an $S$-unit.
That brings us to the task of trying to make the $S$-unit theorem effective—i.e. find generators for the $S$-units, a well-studied task in lattice techniques in computational number theory—and somehow bounding where to look within that lattice. Which should not be too hard: we can bound the possible exponents of prime ideals factors of $a$, and $\mathcal{O}^*_K$-ambiguity does not matter, so it ought to come down to a finite search.
Please take a look at Henri Cohen's "A course in computational number theory". You will undoubtedly find good information about computing class groups and generators of $S$-unit groups, plus much much more on this subject.
A: The class number of $\mathbb{Q}(\sqrt{10})$ is $2$. Thus these ideals, which are not principal must be equivalent.
Another (maybe more systematic) method is to exploit the connection with quadratic forms. Here $(a, b, c)$ (shorthand for $ax^2 + bxy + cy^2$) is associated with the ideal $\left(a, \frac{b - \sqrt{D}}{2}\right)$ where $D$ is the discriminant, in this case $40$. So we are looking at the forms $(3, 2, 3)$ and $(3, -2, 3)$. Now the associated ideals are equivalent if and only if their quadratic forms are equivalent (you'll have to look this concept up, it's easy but too much to explain here). In any case these forms are clearly equivalent, in general $(a, b, c)$ and $(c, -b, a)$ are always equivalent.
A: 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
$$\frac{\alpha}{k} = \frac{k}{\beta} \ \textrm{so}\ \frac{\alpha}{k} (k, \beta)= (\alpha, k)$$
