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:
$\mathfrak{p}=(2,\sqrt{82})_R$
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.
Best Answer
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 $n=0, 1$. But you can easily check $2$ and $2\varepsilon$ are not squares 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.