Why $\mathbb{Z}[\sqrt{-3}]$ is not a Euclidean domain?

Question

I need to prove that $\mathbb{Z}[\sqrt{-3}]$ is not a Euclidean domain. I tried to show that $\mathbb{Z}[\sqrt{-3}]$ is not a P.I.D. but all ideals that I generate by two elements, turn out to be principal.

I already appreciate your help in advance.

$\mathbf{Z}[(1 + \sqrt{-3})/2]$ is a Euclidean domain. Try focusing on the difference between the two rings. — , Nov 12 '14 at 17:30
You can try to show it is not a UFD. The solution is here after you try it: http://math.stackexchange.com/questions/70976/why-is-mathbbz-sqrt-n-not-a-ufd — Aman, Nov 12 '14 at 17:34

score 7 · Accepted Answer · answered Nov 12 '14 at 18:30

7

Hint:

$$\left(1-\sqrt3\,i\right)\left(1+\sqrt3\,i\right)=4=2\cdot 2$$

Now just show $\;2\;$ is not associate with $\;1\pm\sqrt3\,i\;$ and you get your ring is not a UFD.

answered Nov 12 '14 at 18:30

Timbuc

34,191

Why $\mathbb{Z}[\sqrt{-3}]$ is not a Euclidean domain?

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