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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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The proof of uniqueness almost writes itself. Suppose that $a+b\zeta=c+d \zeta$. Then $a-c=(d-b)\zeta$. But $\dots$. – André Nicolas Jun 29 '12 at 18:07
André. I already did Part 1. How get the part two an three? – Andres Jun 29 '12 at 18:10
Dear Andres, Then you should say that you have done Part 1 in the body of the question, so that André does not spend his time telling you what you already know. For (b), can you show that those six elements are in fact units? That's a start. – Dylan Moreland Jun 29 '12 at 18:13
Inthe part 1. I suposse that $a+\zeta b=a'+\zeta b'$ and prove that $a=a'$ and $b=b'$ – Andres Jun 29 '12 at 18:16
Are you familiar with "accepting answers"? Check the FAQ. You've asked 10 questions, but have not accepted a single answer; have all answers been unsatisfactory? – Arturo Magidin Jun 29 '12 at 18:46

1 Answer 1

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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Arturo and for the part 3? – Andres Jun 29 '12 at 18:38
@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. – Dylan Moreland Jun 29 '12 at 18:47

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