# If $\alpha \in \mathbb{Z}(\omega)$, show that $\alpha$ is congruent to either $0, 1$ or $-1$ modulo $1-\omega$.

If $\alpha \in \mathbb{Z}(\omega)$, show that $\alpha$ is congruent to either $0, 1$ or $-1$ modulo $1-\omega$.

Exercise 1 page 134 in the book 'A Classical Introduction to Modern Number Theory' of K. Ireland and M. Rosen.

Thanks a lot.

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$\omega$ being a primitive cubic root of unit? – lhf Apr 30 '12 at 14:05
and it's $\mathbb{Z}[\omega]$. – lhf Apr 30 '12 at 14:07
For the OP: using $\mathbb{Z}[\omega]$ instead of $\mathbb{Z}(\omega)$ is important since the latter suggests a field. Specifically when using a field, $\mathbb{F}[x]$ is standard notation for the polynomial ring, and $\mathbb{F}(x)$ is used for the field of "rational functions". – rschwieb Apr 30 '12 at 15:06
Try and draw the 2D grid on the plane with the numbers of $\mathbb{Z}[\omega]$ marked with dots. Circle the ones that are congruent to zero modulo $1-\omega$. All will be clear (and this is a useful exercise in its own right)! – Jyrki Lahtonen Apr 30 '12 at 21:11
-1 Question ill-posed, without due definitions and without answering back requests for clarification. – DonAntonio Jun 23 '12 at 22:03

Hint $\rm\: \omega\equiv 1\:\Rightarrow\: 0 = \omega^2\!+\!\omega\!+\!1\equiv 3\$ so $\rm\ \omega\equiv 1,\ 3\equiv 0\:\Rightarrow\: m\!+\!n\omega \equiv (m\!+\!n)\:\! mod\ 3\equiv 0,\pm1$