I just found out about Heegner numbers but there is something I quite don't understand about them : As I understand it, they are numbers D, such that $\mathbb{Z}[\sqrt{-D}]$ is still a UFD, correct? For instance, 2 is a Heegner number. But why is $\mathbb{Z}[\sqrt{-2}]$ an UFD?
Don't I have $3 = (1-\sqrt{-2})(1+\sqrt{-2})$ for example (and these elements are prime?)?
Edit
A better example to show 2 decomposition would have been : $2 \times 3 = 6 = (2-\sqrt{-2})(2+\sqrt{-2})$
$R=\Bbb Z[\sqrt{-2}]$ is a Euclidean domain (and so a UFD). The Euclidean function is $N(a+b\sqrt{-2})=|a+b\sqrt{-2}|^2=a^2+2b^2$.
To confirm this is a Euclidean function, one has to prove that if $\alpha$, $\beta\in R$ with $\beta\ne0$ there is $\gamma\in R$ with $N(\alpha-\gamma\beta) <N(\beta)$. To prove this, write $\alpha/\beta=x+y\sqrt{-2}$ with $x$, $y$ real, and take $\gamma=r+s\sqrt{-2}$ with $r$, $s$ nearest integers to $x$, $y$.