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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 UFD2012-03-03
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    Related: https://math.stackexchange.com/questions/70976/2018-11-28

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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 $1+\sqrt{-3}$.

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    @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 $\rm\:R' = \mathbb Z[(1+\sqrt{-3})/2]\:$)2012-03-03
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    @Bill Dubuque: The system will not let m delete. How does one take care of that problem?2012-03-03
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    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.2012-03-03
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    @Bill Dubuque: Thanks!2012-03-03
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    Is there an example of a ring (of integers) which isn't Euclidean under its standard norm, but which becomes Euclidean under some other norm? I've never thought about this.2012-03-03
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    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.2012-03-03