Is $\Bbb Z[\sqrt{6}]$ a UFD?
Solution 1:
$\mathbb Z[\sqrt 6]$ is indeed a unique factorization domain. Notice, for one thing, that $\langle 2 \rangle$ and $\langle 4 + \sqrt 6 \rangle$ are both contained in $\langle 2 + \sqrt 6 \rangle$. As for $\langle 5, 4 + \sqrt 6 \rangle$, that turns out to be the whole ring.
Maybe you meant $\mathbb Z[\sqrt{-6}]$ is not UFD? In which case the ideal $\langle 2, \sqrt{-6} \rangle$ would be far more pertinent than either $\langle 2, 4 + \sqrt{-6} \rangle$ or $\langle 5, 4 + \sqrt{-6} \rangle$.