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I'm having trouble proving that $\mathbb{Z}[\sqrt{6}]$ is a PID. I thought that the way to go would be to prove that it's a Euclidean Domain, but I got stuck there as well. Any suggestions on how to proceed?

I'm looking for a way to prove that it's a PID without having to prove that it's a Euclidean Domain

Naweed G. Seldon
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    I'm looking for a way to prove that it's a PID without having to prove that it's a Euclidean Domain. Because the proof there tries adopting the proof for $\mathbb{Z}[\sqrt{2}]$ being a ED, but that doesn't work here. – Naweed G. Seldon Nov 01 '17 at 17:49
  • I think this ring in the ring of integers of $\mathbb{Q}(\sqrt{6})$. So you show that the class number is one. – Rene Schipperus Nov 01 '17 at 17:54
  • Since this ring is a Dedekind ring (it is the ring of integers of the above field), it is PID iff it is UFD. So prove this using prime and irreducible. Of course, it is easier to show that the ring is Euclidean. – Dietrich Burde Nov 01 '17 at 19:57
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    As @DietrichBurde suggests, the easiest way would be to succeed in your attempt to prove Euclideanness. Next would be to use the Minkowski Bound, which requires some fancy knowledge. A third, and most tedious way, would be to make heavy use of Quadratic Reciprocity, verifying that if $6$ is a square modulo $p$ and thus $(p)$ splits in $\Bbb Z[\sqrt6,]$, then $p$ may be written $(m+n\sqrt6,)(m-n\sqrt6,)$. This may require yet more technique, I haven’t carried out the details. – Lubin Nov 01 '17 at 22:05

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