I was wondering if there is a classification for this:
For which $d$ is $D=\mathbb{Z}[\sqrt{d}]$ a UFD, with $d > 1$?
For $d \equiv 1 $ (mod $ 4$), $D$ is not a UFD (proof here).
algebraic-number-theoryring-theoryunique-factorization-domains
I was wondering if there is a classification for this:
For which $d$ is $D=\mathbb{Z}[\sqrt{d}]$ a UFD, with $d > 1$?
For $d \equiv 1 $ (mod $ 4$), $D$ is not a UFD (proof here).
Best Answer
When $d \not\equiv 1\; (\textrm{mod}\; 4)$, the ring of integers of $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$. $\mathbb{Z}[\sqrt{d}]$ is a UFD if and only if it has trivial class group (i.e., the class number of $\mathbb{Q}(\sqrt{d})$ is $1$).
However, it's an open question as to whether or not there are infinitely many $d>0$ with $\mathbb{Q}(\sqrt{d})$ having class number 1, so the answer is not known.