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The "integers" of quadratic field $\mathbb Q[\sqrt{d}]$ , for a squarefree integer $d$ , forms an integral domain . I know that for $d<0$ , the quadratic integers of the quadratic number fields satisfy Euclidean algorithm only for $d=-1,-2,-3,-7,-11$ . I want to know , for which $d<0$ , it is true that every ideal of the integral domain of quadratic integers of $\mathbb Q[\sqrt{d}]$ is a principal ideal , that is when is the subring of quadratic integers of imaginary quadratic number fields is a PID ? I want to know this as PID do not imply ED but does imply UFD . Please help . Thanks in advance

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With $d<0$, the ring of integers of $\mathbb Q(\sqrt d)$ is a PID exactly when $$d= −1, −2, −3, −7, −11, −19, −43, −67, −163$$ Checking that these rings are PIDs is not too hard, but checking that no other values give an integer ring that is a PID requires some heavy machinery - see the class number 1 problem.

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  • $\begingroup$ I know that these are UFD 's , but are they also PID ? $\endgroup$ – user217921 Mar 30 '15 at 11:57
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    $\begingroup$ @SaunDev The ring of integers of a number field is a Dedekind Domain. It is a general theorem that a Dedekind domain is a UFD if and only if it is a PID. The key fact is that a UFD is a PID in which every non-zero prime ideal is maximal. $\endgroup$ – Mathmo123 Mar 30 '15 at 20:32
  • $\begingroup$ I'm looking specifically at the example Q(sqrt(-67)) and trying to show it is a PID So I should be looking at the ring of integers first and then determining whether it is a dedekind domain, then using the theorem you metioned above showing it is a UFD and the hence a PID. Is this correct or am I looking at this the wrong way? $\endgroup$ – Padraic May 27 '15 at 13:13
  • $\begingroup$ @Padraic all rings of integers of number fields are Dedekind domains. It might be possible to manually show that this is a UFD, but the best way will be to try to compute the class number $\endgroup$ – Mathmo123 May 27 '15 at 13:16
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The $d < 0$ such that $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is a principal ideal domain are: $$-1, -2, -3, -7, -11, -19, -43, -67, -163.$$ Multiplied by $-1$, these are the Heegner numbers (see Sloane's A003173).

If $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is a Euclidean domain, then it is also a principal ideal domain, and if it is a principal ideal domain, it is also a unique factorization domain. But it can be non-Euclidean and still be a principal ideal domain. For example, $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is not Euclidean: try for example to compute $$\gcd\left(\frac{3}{2} + \frac{\sqrt{-19}}{2}, 10\right)$$ using the Euclidean algorithm. Frustrating, right? But it is a principal ideal domain.

Sloane tells us Gauss discovered the nine $d$ listed above (and also listed in other answers to your question) but "Heegner proved the list is complete." Well, he had a little help from Tony Stark. The proof exists, has been published and has been verified. But it's too long to include here.

I can at least show you why $-2$ is the only even value. Suppose $n$ is positive, odd and squarefree. Then, in $\mathbb{Z}[\sqrt{-2n}]$, we can be sure the following numbers are irreducible: 2, $n$, $\sqrt{-2n}$. So $2n$ has at least two factorizations: $2n = (-1)(\sqrt{-2n})^2$. Therefore the prime ideal that properly contains $\langle 2 \rangle$ and $\langle n \rangle$ can't be a principal ideal.

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