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Let $\zeta$ be the cube root of 1 given by $\zeta=\frac{-1}{2}+i\frac{\sqrt{3}}{2}$ and let $\mathbb{Z}[\zeta]=\{a+\zeta b: a, b\in \mathbb{Z}\}$, called the "Eisenstein integers".

How prove the following exercises?

(a) Every element of $\mathbb{Z}[\zeta]$ can be uniquenly written in the form $a+\zeta b$ for some $a, b\in\mathbb{Z}$.

(b) Let $N(a+\zeta b)=(a+\zeta b)(a+\overline{\zeta} b)=a^2-ab+b^2$. Show that the units of $\mathbb{Z}[\zeta]$ are $\{\pm 1,\pm\zeta, \pm \zeta^2 \}$.

(c) Show $\mathbb{Z}[\zeta]$ is an Euclidean domain with size function $N$.

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    Inthe part 1. I suposse that $a+\zeta b=a'+\zeta b'$ and prove that $a=a'$ and $b=b'$2012-06-29

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For part 1, note that $\zeta^2$ can be written as $a+b\zeta$, so that any element of $\mathbb{Z}[\zeta]$ can be written at least one way as $a+b\zeta$; uniqueness then follows because $\zeta$ is not rational (since there are no rational roots of $x^2+x+1$, the polynomial that $\zeta$ satisfies).

For part 2, show that $N$ is multiplicative: for any $\alpha,\beta\in\mathbb{Z}[\zeta]$, $N(\alpha\beta) = N(\alpha)N(\beta)$. Conclude that $\alpha$ is a unit if and only if $N(\alpha) = \pm 1$. Now rewrite $a^2-ab+b^2$ as $a^2-ab+b^2 = \left( a - \frac{1}{2}b\right)^2 + \frac{3}{4}b^2$ so that it is clear that the norm is always positive. Consider what you need for this sum of squares to be equal to $1$.

For part 3, there is a beautiful geometric argument given by Klein; I've posted about it before.

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    @Andres Do you know how to prove that $\mathbb Z[i]$ is a Euclidean domain? This is quite similar. It helps me to draw a picture of the lattice of Eisenstein integers in the complex plane.2012-06-29