Prove that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain and $\mathbb{Z}[\sqrt{-10}]$ is not 
Prove that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain and $\mathbb{Z}[\sqrt{-10}]$ is not.

I know that in general, to prove something is a Euclidean domain, I must prove the existence of a division algorithm involving a norm. In the case of $\mathbb{Z}[\sqrt{-2}],$ the norm is $a^2 + 2b^2.$ I know how to prove that the Gaussian integers $\mathbb{Z}[\sqrt{-1}]$ is a Euclidean domain, but I'm not sure if the proof for that relates to this proof.
Also, proving that $\mathbb{Z}[\sqrt{-10}]$ is not a Euclidean domain involves determining which ideals are not principal, but I'm not sure how to find a non-principal ideal. I think it should be generated by at least two elements of $\mathbb{Z}[\sqrt{-10}]$ though.

I'm new to abstract algebra, so I'd like more than just a simple hint, if possible.


If $\mathbb{Z}[\sqrt{-10}]$ were an Euclidean domain, then it would be a UFD.
By considering norms, we see that $2$, $5$, and $\sqrt{-10}$ are irreducible in $\mathbb{Z}[\sqrt{-10}]$. 
Since $10 = (-1)(\sqrt{-10})^2 = 2 \times 5$ are two distinct factorizations into irreducibles, $\mathbb{Z}[\sqrt{-10}]$ is not a UFD and so cannot be an Euclidean domain.
