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.

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    $\begingroup$ isn't this about as hard as telling if an ideal is principal ? $\endgroup$
    – mercio
    Commented May 5, 2017 at 23:21
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    $\begingroup$ @mercio I think that's actually easier. If $\langle b, c \rangle$ is principal, then we can find $a$ such that $\langle a \rangle = \langle b, c \rangle$. But if a ring has three or more distinct ideal classes, it's not immediately obvious which ideal class a particular ideal belongs to. $\endgroup$ Commented May 6, 2017 at 18:14
  • $\begingroup$ Just to stir the pot further, I think $\mathbb{Z}[\sqrt{15}]$ would have made for a much better example. $\endgroup$ Commented May 8, 2017 at 20:04
  • $\begingroup$ No, scratch that, $\mathbb{Z}[\sqrt{79}]$. But I did mean to say 15 earlier. Mwahahaha! $\endgroup$ Commented May 8, 2017 at 20:13

4 Answers 4


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.


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.


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.

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    $\begingroup$ You are right, I wonder why it isn't mentioned there oeis.org/A094619 also $(3,1+\sqrt{10})(3,1-\sqrt{10})=(1+\sqrt{10})$ (that's what Sage says..) $\endgroup$
    – reuns
    Commented May 2, 2017 at 23:57
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    $\begingroup$ But if we don't know what the class number is in advance, we would be led astray if we assume the principal ideals are all together in one class and the non-principal ideals are all together in another class. The asker should have chosen an example of class number 3 or higher. $\endgroup$ Commented May 3, 2017 at 18:08
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    $\begingroup$ @RobertSoupe Well, thanks for you upvote, they are getting rarer these day. You may know that I wrote the first line because at the time there were comments claiming the opposite, and this seems the quickest way to settle the question. $\endgroup$ Commented May 3, 2017 at 19:26
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    $\begingroup$ Everything you've written is completely correct, but it is also quite easy to misunderstand. $\endgroup$ Commented May 5, 2017 at 21:38
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    $\begingroup$ The correspondence between forms and classes uses ideal classes in the strict sense. Which is not a big problem here since the fundamental unit has negative norm, but might lead to confusion in general. $\endgroup$
    – user23365
    Commented May 6, 2017 at 6:43

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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