$A$ is the ring of integers generated by the cube roots of unity. How will we prove that it is an Euclidean ring?

We have following results and theorem:

1) Any integral quadratic extension $A$ of $\Bbb Z$ is obtained as the ring of algebraic integers in $\mathbb Q(\sqrt d)$, [where d is a square-free integer different from 1]. The elements of $A$ are of the form $a + \sqrt d$ here $(a, b\in Z)$ if $d \equiv 2$ or $3 \pmod 4$ , and $a + b(1 + \sqrt d) / 2$ if $d \equiv 1\pmod 4$

2) The ring of integers in $\Bbb Q(\sqrt d)$ is Euclidean for $d=2,3,5,-1,-2,-3, -7, -11.$

  • 1
    $\begingroup$ $x^3-1=0$ $\implies (x-1)(x^2+x+1)=0$ if$ x=1$ then $\Bbb Z=<1>$ therefore euclidean. But if $x=\frac {-1\pm\sqrt{-3}}2$ .... $\endgroup$ Nov 26, 2013 at 6:15

1 Answer 1


The idea is to mimic the proof of $\mathbb{Z}[i]$ : Let $\omega = \frac{-1}{2} + \frac{\sqrt{3}}{2}i = e^{2\pi i/3}$, and define $d : \mathbb{Z}[\omega]\setminus\{0\} \to \mathbb{N}$ by $$ d(z) = |z|^2 $$ where you think of $\mathbb{Z}[\omega] \subset \mathbb{C}$.

It is easily checked that $d(z_1z_2) \leq d(z_2)$, so we check Euclidean division. If $z_1, z_2\in \mathbb{Z}[w]\setminus\{0\}$, consider $$ t = \frac{z_1}{z_2} = x+iy \in \mathbb{C}. $$ Now $\mathbb{Z}[\omega]$ is a "lattice" in $\mathbb{C}$ whose fundamental region is a parellelogram. $t$ lies in one such parallelogram, so choose the corner $(a,b)$ of the parallelogram that is closest to $t$.

Check that $|a-x| < 1/2$ and $|b-y| < 1/2$, hence if $q = a+bi$, then $q\in \mathbb{Z}[\omega]$ (since it is on the lattice), and $$ d(z_1 - qz_2) \leq \frac{1}{2}|z_2|^2 < d(z_2). $$

  • $\begingroup$ Thank you Prahlad. Can we use above theorem and result instead of defining this norm? Here our $d=-3$. $\endgroup$ Nov 26, 2013 at 7:13

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