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?
Is $\mathbb Z[\sqrt{-3}]$ Euclidean under some other norm?
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ring-theory
algebraic-number-theory
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0Related: https://math.stackexchange.com/questions/70976/ – 2018-11-28
1 Answers
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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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7The 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