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In the ring $\mathbb{Z}[\sqrt{6}]$, the ideal $(2,\sqrt{6})$ simplifies to $(\sqrt{6}-2)$, while in the ring $\mathbb{Z}[\sqrt{10}]$, the ideal $(2,\sqrt{10})$ is not principal (I think). Is there some simple way to tell if a finitely generated ideal is actually principal?

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    $\begingroup$ In which rings? From your tags I assume in integer rings of quadratic extensions. And I would assume those which are not PIDs, or would your question contain the question for an easy criterion whether such a ring is a PID? $\endgroup$ – Torsten Schoeneberg Jun 2 '18 at 2:39
  • $\begingroup$ @TorstenSchoeneberg I am chiefly concerned with quadratic integer rings which are not principal ideal domains, but if a criterion exists which can work for the ring of integers for any primitive extension of the rationals, that'd be great. $\endgroup$ – weux082690 Jun 2 '18 at 2:46
  • $\begingroup$ Related question: math.stackexchange.com/questions/1031546/… $\endgroup$ – Lisa May 31 at 21:09
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Especially with quadratic integer rings, but also for more general rings of integers, the norm is really useful. Note that if $I=(a)$, then $N(I)=|N_{K/\Bbb Q}(a)|$ where $K$ is the quotient field of your quadratic integer ring (this holds more generally in number fields.)

Thus if you can show that there exists no $a$ in your ring with $|N(a)|=N(I)$, it follows that the ideal is not principal. In the example, there exists no $a \in \Bbb Z[\sqrt{10}]$ such that $|N(a)|=2$, because the equation $x^2-10y^2=\pm 2$ has no solutions with $x,y \in \Bbb{Z}$.

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If you don't know whether the domain is a principal ideal domain or not, then I suppose there is no simple way to know whether any ideal generated by two numbers might be a "disguised" principal ideal.

You could try computing the GCD of the generators using the Euclidean algorithm, probably using the absolute value of the norm as the Euclidean function. In quadratic integer rings, that would be $|N(a + b \sqrt{d})| = |a^2 - db^2|$.

For example, in the case of $\langle 2, \sqrt{6} \rangle$, we see that $N(2) = 4$ and $N(\sqrt{6}) = -6$. Then we solve $\sqrt{6} = 2m + r$ such that $|N(r)| < 4$. The fundamental unit is $5 + 2 \sqrt{6}$. Since $\sqrt{6}$ ensures $db^2$ is even, $a^2$ must be odd, and so must $a$ be. But multiplying by $2$ ensures the $a$ and $b$ of $2m$ are both even, so $N(r) = \pm 1$ is impossible.

Is $N(r) = \pm 2$ possible? It is. Thus we find that $\sqrt{6} = (2 \times 1) + (2 + \sqrt{6})$. And from there it readily follows that $2 = (2 + \sqrt{6})(2 - \sqrt{6}) + 0$. Furthermore, $\langle 2 - \sqrt{6} \rangle = \langle 2 + \sqrt{6} \rangle$. And since $\langle 2 \rangle \in \langle 2 + \sqrt{6} \rangle$ and $\langle \sqrt{6} \rangle \in \langle 2 + \sqrt{6} \rangle$, we conclude $\langle 2, \sqrt{6} \rangle$ is a principal ideal, despite not having been presented as such.

Now try to do the same thing with $\langle 2, \sqrt{10} \rangle$, to solve $\sqrt{10} = 2m + r$ such that $|N(r)| < 4$. Once again we see that $r$ with unit norm is impossible. And no number in this domain has norm $\pm 2$, since no perfect square in $\mathbb{Z}$ has $2$ nor $8$ for a least significant digit. Likewise, no number in this domain has norm $\pm 3$, since no perfect square ends $3$ or $7$.

That our Euclidean algorithm fails for our chosen Euclidean function suggests $\langle 2, \sqrt{10} \rangle$ is not a principal ideal. But even if it was principal, that would mean there is some number in this domain with the same norm as our ideal, and we already saw there isn't. We must therefore conclude $\langle 2, \sqrt{10} \rangle$ is not a principal ideal after all.

Is this simple? No, I don't suppose that it is.

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