# Why $\mathbb{Z}[\mu_6]$ is a PID?

Let $\mu_6$ be the 6th unit root of $1$. Why $\mathbb{Z}[\mu_6]$ is PID?

I tried to prove it by the method proving $\mathbb{Z}[i]$ is a PID but failed.

Anyone can help?

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Hmm. IIRC the same method should work. Which step did you have problems with? – Jyrki Lahtonen Jan 8 '13 at 9:56
Note that $\mathbb{Z}[\mu_6] = \mathbb{Z}[\mu_3]$. To prove that it's a PID, prove that it's a Euclidean domain by exhibiting a Euclidean norm function. – Alex B. Jan 8 '13 at 9:57
@JyrkiLahtonen. Yeah I see, I did not realize that $\mu_6^2=-\mu_6-1$..Now I'm clear. – hxhxhx88 Jan 8 '13 at 10:58
@AlexB....I worked out. Thank you very much! – hxhxhx88 Jan 8 '13 at 10:58

For every complex number $z$ there is at least one element $x\in\Bbb Z[\mu_6]$ such that $|z-x|<1$ (in fact the maximal necessary distance is $1/\sqrt3$, which is less than the maximal distance $1/\sqrt2$ for the Gaussian integers). This means that one can define a Euclidean division on $\Bbb Z[\mu_6]$ by rounding the exact quotient $z$ towards such an element $x$. It is an easy exercise to show the remainder is now less in (complex) absolute value than then the divisor in the Euclidean division, as it should. So $\Bbb Z[\mu_6]$ is norm-Euclidean, and therefore a PID. See Eisenstein integer for more details.