# 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?

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No, it is not an UFD – Blah Mar 3 '12 at 12:36

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 $1+\sqrt{-3}$.
@mal Proving that the factors are irreducible in $\rm\:R = \mathbb Z[\sqrt{-3}]\:$ is not enough to prove that the factorization is nonunique. It is essential to also prove that $2$ is not associate to $1\pm \sqrt{3}$ in $\rm\:R\:$ (which it is in its UFD integral closure $\rm\:R' = \mathbb Z[(1+\sqrt{-3})/2]\:$) – Bill Dubuque Mar 3 '12 at 16:17
No need to delete, just say "nonassociate irreducibles" (or say it shows that $2$ is irreducible but not prime, since $\:2\nmid 1\pm\sqrt{3}\:$ in $\mathbb Z[\sqrt{-3}]\:).\:$ I emphasized that because many students overlook it. – Bill Dubuque Mar 3 '12 at 17:17
The elements of $\mathbb{Q}(\sqrt{69})$ which are algebraic integers are an example There is a largish literature, fair number of examples, some open problems. – André Nicolas Mar 3 '12 at 18:21