Is $\mathbb Z[\sqrt{-3}]$ Euclidean under some other norm? I know that $\mathbb Z[\sqrt{-3}]$ is not a Euclidean domain under the usual norm $N(x + y\sqrt{-3}) = x^2 + 3y^2$, but that does not necessarily mean that it can't be a Euclidean domain. Is it possible to define some norm that could make it into a Euclidean domain?
 A: It isn't possible. If it were, then $\mathbb{Z}[\sqrt{-3}]$ would be a Unique Factorization Domain. But 
$$4=(2)(2)=(1-\sqrt{-3})(1+\sqrt{-3}),$$
and $2$ and $1\pm\sqrt{-3}$ are non-associate irreducibles.   
Alternately, $2$ is irreducible in our ring. But $2$ is not prime, since $2$ divides the product $(1-\sqrt{-3})(1+\sqrt{-3})$, but $2$ divides neither $1-\sqrt{-3}$ nor $1+\sqrt{-3}$.
A: It's also possible to show this by showing that $\mathbb{Z}[\sqrt{-3}]$ is not a principal ideal domain.
If $(2,1+\sqrt{3})$ is principal (this particular ideal is natural to consider after working out why the proof that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain breaks down), then if must be either $(1)$ or $(\sqrt{-3})$, because of the modulus of $2$ being small. But if
$1 = 2(a+b\sqrt{-3}) + (1+\sqrt{-3})(c+d\sqrt{-3}) = (2a+c-3d) + (2b+c+d)\sqrt{-3}$
then we can reach a contradiction via a partity argument: $2b+c+d=0$ so $c+d$ is even, so $c-3d$ is even, so the real term in the above $2a+c-3d$ is even. But it's $1$, contradiction.
An almost identical parity argument also shows that $(2,1+\sqrt{-3})\neq(\sqrt{-3})$. So there's a non-principal ideal in $\mathbb{Z}[\sqrt{3}]$, so it can't be a Euclidean domain.
