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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?2012-04-30
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    and it's $\mathbb{Z}[\omega]$.2012-04-30
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    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".2012-04-30
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    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)!2012-04-30
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    -1 Question ill-posed, without due definitions and without answering back requests for clarification.2012-06-23

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