Is $\mathbb{Z}[√13] $ a Unique factorization domain? I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.
Hint for the title question: We have $${\displaystyle (1+{\sqrt {13}})^{2}=2(7+{\sqrt {13}})}.$$
Since $2$ is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ but does not divide ${\displaystyle 1+{\sqrt {13}}}$, we see that ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ is not a UFD.
$\mathbb{Z}[\sqrt{13}]$ is not even integrally closed (consider for instance $\displaystyle \frac{3+\sqrt{13}}{2}$), Hence it can't be a principal ideal domain. It also can't be a unique factorization domain because a Dedekind domain is a PID if and only if it is a UFD, but $\mathbb{Z}[\sqrt{13}]$ is not even integrally closed thus not Dedekind. At any rate: if you're not integrally closed then you're not a UFD.