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In Washington's Cyclotomic Fields, he makes the following assertion (p. 233): For each positive integer $i$, let $\zeta_i$ be a primitive $i$-th root of unity, chosen such that $(\zeta_i)^{(i/j)}=\zeta_j$. Then

$\prod_{b=0}^{(i/j)-1} (\zeta_i^{a+bj}-1)=\zeta_j^a-1$ if $j | i$.

After playing around with this for a while and working a few small examples, I still can't see how to prove this generally, even though I'm sure it's very simple. Can someone please enlighten me? Thanks!

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Hint: Consider the cyclotomic polynomial $P(z)=\prod_{b=0}^{i/j-1}((\zeta_i^{j})^b-z)$ when $z=\zeta_i^{-a}$.

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  • $\begingroup$ Hi anon, thanks for your response. I've certainly been led already to consider this polynomial, since this is what pops out when you factor out the appropriate number of $\zeta_i^a$ (which helpfully leaves $\zeta_j^a$ on the outside), but I'm still unsure how to proceed. I'll keep working on it, though. $\endgroup$ – Jeff Aug 26 '11 at 16:33
  • $\begingroup$ Well, I feel dumb. Thanks, anon! $\endgroup$ – Jeff Aug 26 '11 at 16:55
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Let $i = jk$, $x = \zeta_i^a$, $z = \zeta_i^j$. Note that $z$ is a primitive $k$th root of unity. Rewriting your equation, $$ \prod_{b=0}^{k-1} (xz^b-1) = x^k - 1. $$ Multiplying both sides by $z^{k(k-1)/2}$, we obtain (noticing that the inverses $z^{-b}$ are the same as the original collection, but in a different order) $$ \prod_{c=0}^{k-1} (x-z^c) = z^{k(k-1)/2} (x^k - 1). $$ If $k$ is odd then $z^{k(k-1)/2}$ and we get an equation of polynomials, stating that all $k$th roots of unity are given by $1,z,\ldots,z^{k-1}$. If $k = 2l$ is even, then $z^{k(k-1)/2} = z^{l(2l-1)} = (-1)^{2l-1} = -1$, and the equation is actually wrong as stated.

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