# Is $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$ is a UFD

Is it true that the quadratic ring of integers $$O :=\mathcal{O}_{\mathbb{Q}(\sqrt{5})} =\mathbb{Z}(\frac{\sqrt{5}+1}{2})$$ is a UFD? How to show it?

My guess: I know that if a quadratic ring of integers is a PID, then it is a UFD. If $$O$$ has a Dedekine-Hasse norm, then it is a PID (Dummit & Foote p.281), then a UFD. Possibly one may imitate Dummit & Foote p.282 where they show that $$\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$$ is a PID, right? Also, this table indicates $$O$$ has class number 1, I guess it means it is a UFD.

Possibly related: Quadratic rings of integers which are UFD, Real quadratic fields which are UFD (P.s. On algebra, I have only exposure to Dummit & Foote.)

• yes, class number $1$ means it's a UFD – J. W. Tanner Jan 30 at 11:51
• Ah, thanks, but I also want to show it (just added) – Violapterin Jan 30 at 11:55
• Some more interesting facts about this ring can be found here: math.stackexchange.com/q/18589/101420. Unfortunately the answers don't seem to give a proof of the fact that it is a UFD, but just use it. However one answer states that the ring is 'norm Euclidean' and therefore a UFD. This latter implication works exactly the same as in the case of $\mathbb{Z}$. So in order to get a nice elementary proof, we just need to figure out which norm they are talking about and check that it does indeed make the ring Euclidean. Perhaps the most obvious choice already works? I didn't check it yet – Vincent Jan 30 at 12:41

The ring $$\Bbb Z[\sqrt{5}]$$ is not a UFD, but the ring of integers $$\Bbb Z[\frac{1+\sqrt{5}}{2}]$$ of the number field $$\Bbb Q(\sqrt{5})$$ is a Dedekind ring, which is a PID and hence is factorial.

For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?
Edit: The proof that $$K=\Bbb Q(\sqrt{5})$$ has class number $$1$$ is very easy, because the Minkowski bound satisfies $$B_K<2$$, so that the class number is $$1$$. More precisely, $$B_K=\frac{\sqrt{5}}{2}=1.11803398874989\cdots$$ The formula in general is $$B_K=\frac{n!}{n^n}\left(\frac{4}{\pi} \right)^s\sqrt{|d_K|}$$ where $$n$$ is the degree of the extension (here $$n=2$$), $$d_K$$ the discriminant.
• $d_K=5?$ and what is $s$? – J. W. Tanner Jan 31 at 3:00
• @J.W.Tanner We have $s=0$. Hence for $\Bbb Q(\sqrt{d})$ with squarefree $d>0$ and $d\equiv 1\bmod 4$ we have $B_K=\frac{1}{2}\sqrt{|d|}$. In this case also $|d|=|d_K|$. I have edited it, thank you. – Dietrich Burde Jan 31 at 9:18
• thanks for editing. It looks good now. +1. I couldn't figure out how you got that other number for $B_K$ – J. W. Tanner Jan 31 at 13:57