The number field in question is $K=\mathbb{Q}(\sqrt{82})$, and the ideal considered is in its ring of integers $R=\mathcal{O}_K=\mathbb{Z}[\sqrt{82}]$. The ideal is:


It is a prime ideal (this follows from a theorem on the splitting behaviour of ideals generated by prime integers), and it appears quite naturally in the unique factorisation $\mathfrak{p}^2=(2)_R$, so we know that $N(\mathfrak{p})=\pm2$.

I want to show that the ideal is not principal. My attempts so far have been to consider the equation $x^2-82y^2=\pm2$ reduced modulo small integers and then arriving at a contradiction, but this has been to no avail. I have therefore resorted to asking the community the following:

(1) Is there any general method to proving that an ideal in the ring of integers of a number field is non-principal, or are there general methods for separate cases $\textit{e.g.}$ in the case of a quadratic number field? The case of $K=\mathbb{Q}(\sqrt{d})$ for $d<0$ square-free of course admits fairly simple solution, as it can be easily observed that $x^2-dy^2=c$ for some $c\in K$ has no solutions when this is the case.

(2) If the answer to (1) is either that there is no known general method or that any known general method is impracticable, and that in the case of an ideal in the ring of integers of a quadratic number field $K=\mathbb{Q}(\sqrt{d})$ the best one can do is consider equations of the form $x^2-dy^2=c$ modulo small integers, is there any general algorithm or technique that applies here other than some wit and a keen eye?

All help would, as always, be highly appreciated.

  • 3
    $\begingroup$ Did you really just stumble into this example at random? It is an example of a Diophantine equation that has a solution modulo $m$ for all $m > 1$ but no integer solution. To handle this, see Lemma 8.6 on p. 215 of Cassels's "Local Fields" and its application to $u^2 - 82v^2 = \pm 2$ on pp. 216-217. $\endgroup$ – KCd Jan 31 at 1:38
  • 1
    $\begingroup$ See also Theorems 3.3 and 5.1 of kconrad.math.uconn.edu/blurbs/ugradnumthy/pelleqn2.pdf for methods that involve units and continued fractions. $\endgroup$ – KCd Jan 31 at 1:44

Here is an ad hoc way to prove $\mathfrak{p}=(2,\sqrt{82})$ is not principal. The fundamental unit of $\mathbb{Q}(\sqrt{82})$ is $\varepsilon = 9+\sqrt{82}$.

The idea is to use information at the Archimedean place. You already know $\mathfrak{p}^2 = (2)$. Assume $\mathfrak{p}=(a)$ for some $a\in \mathbb{Z}[\sqrt{82}]$, we can assume $1<a<\varepsilon$. Then $a^2 = 2\varepsilon^n$ for some $n\in \mathbb{Z}$. That is, $$\frac{1}{2} < \varepsilon^n < \frac{\varepsilon^2}{2}$$ the only case is $n=1$. But you can easily check $2\varepsilon$ is not a square in $\mathbb{Q}(\sqrt{82})$.

Your follow up question: for general number fields, we have algorithm to determine whether a given ideal is principal, but the complexity increases exponentially as the degree, and is as hard as finding a system of fundamental units. See chapters 4,5,6 of A Course in Computational Algebraic Number Theory by Henri Cohen.

For real quadratic field, which the problem amounts to solve a Pell equation, more specialized algorithm is known, like continued fraction.


If $\langle 2, \sqrt{82} \rangle$ was a principal ideal, that would mean that the $a$ in $\langle a \rangle = \langle 2, \sqrt{82} \rangle$ is a common divisor of 2 and $\sqrt{82}$.

I believe, but have not been able to prove, that such a divisor, if it exists, can be found by the Euclidean algorithm, even in a non-Euclidean domain. I suppose the worst that can happen from my going ahead with this answer is that I lose a few points.

Consider for example $\langle 5, \sqrt{95} \rangle$. Since $$\frac{\sqrt{95}}{5} \approx 2,$$ we see that $\sqrt{95} = (2 \times 5) + (-10 + \sqrt{95})$ and it just so happens that $N(-10 + \sqrt{95}) = 5$. Then we readily see that $5 = (-1)(-10 + \sqrt{95})$ $(10 + \sqrt{95}) + 0$. So $\langle 5, \sqrt{95} \rangle = \langle 10 + \sqrt{95} \rangle$.

Now, modular arithmetic can still be very useful in some cases. For example, $\langle 2, \sqrt{95} \rangle$ must also be a ramifying ideal. If it's also a principal ideal, it would mean $x^2 - 95y^2 = \pm 2$ has solutions in integers. However, we can quickly check that $x^2 \equiv \pm 2 \pmod 5$ is impossible in integers.CORRECTION: I should have written $\langle 2, 1 + \sqrt{95} \rangle$, not $\langle 2, \sqrt{95} \rangle$.

As it turns out, $\mathbb Z[\sqrt{82}]$ is in some ways a very thorny example. Examples with really small and neat numbers seem kinda hard to find. Maybe you've come across $\langle 103 \rangle = \langle 103, 44 - \sqrt{82} \rangle \langle 103, 44 + \sqrt{82} \rangle$.

To be perfectly clear, neither $\langle 103, 44 - \sqrt{82} \rangle$ nor $\langle 103, 44 + \sqrt{82} \rangle$ is a ramifying ideal. But if either of them is a principal ideal, I still assert, without proof, that we can discover $\gcd(103, 44 \pm \sqrt{82})$ by the Euclidean algorithm.

I'm not going to bore you with the calculations. Suffice it to say that $\gcd(103, 44 + \sqrt{82}) = 15 - 2 \sqrt{82}$, which has a norm of $-103$.

Returning to your question of $\langle 2, \sqrt{82} \rangle$, trying to find $\gcd(2, \sqrt{82})$ by the Euclidean algorithm is likely to be as frustrating as trying to find a number $m$ that $x^2 - 82y^2 \equiv 2$ or $80 \pmod m$ is impossible.

A program I got off GitHub answers this query with $-2$, which is its way of telling me it could not carry out the Euclidean algorithm to a satisfactory conclusion in this instance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.