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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?

Dylan Moreland
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mal
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2 Answers2

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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 $1+\sqrt{-3}$.

André Nicolas
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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 $\rm:R' = \mathbb Z[(1+\sqrt{-3})/2]:$) – Bill Dubuque Mar 03 '12 at 16:17
  • @Bill Dubuque: The system will not let m delete. How does one take care of that problem? – André Nicolas Mar 03 '12 at 17:06
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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. – Bill Dubuque Mar 03 '12 at 17:17
  • @Bill Dubuque: Thanks! – André Nicolas Mar 03 '12 at 17:23
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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. – Dylan Moreland Mar 03 '12 at 18:07
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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. – André Nicolas Mar 03 '12 at 18:21
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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.

Arthur Conmy
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