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If $n$ is divisible by $m$, why is it the case that the $m$th primitive roots of unity are also roots of $\binom{n}{k}_q$ if and only if $m$ does not divide $k$?

I'm viewing $\binom{n}{k}_q$ as a polynomial in $q$, but I don't see why the divisibility relations gives this condition of the primitive roots being roots of the polynomial. Thanks.

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The complex $m$th roots of unity are roots of $x^m-1$, and form a regular $m$-gon on the unit circle in the complex plane with one vertex at $1$ (for $0 < m \in \mathbb{N}$). Primitive $m$th roots of unity are roots of the irreducible polynomial factors of $x^m-1$, called cyclotomic polynomials $\phi_d(x)$ and satisfying $$ x^m-1 = \prod_{0$\phi(d)$ and, as stated, its roots are precisely the primitive $m$th roots of unity.

(The polynomial factorization formula above can also be seen as providing a partition of the full set of $m$th roots of unity, which forms a multiplicative subgroup of $\mathbb{C}$ of order $m$, into subsets based on their order -- for which minimal $d$ do they satisfy $x^d=1$ -- comprised of $\phi(d)$ generators $\{e^{2\pi ij/d}|(j,d)=1\}$ of the subgroup of order $d$ for each positive divisor $d$ of $n$. It is worthwhile once to draw some examples of these partitions.)

Now since $$ [m]_q=\frac{q^m-1}{q-1}=\prod_{11$ (all primitive $d$th roots of unity where $d>1$ and $d \mid m$). So a root of $$ \binom{n}{k}_q =\prod_{j=1}^{k}\frac{[n-j+1]_q}{[j]_q} =\prod_{1

In particular for $m=d \mid n$, we can see that the primitive $m$th roots of unity are roots of $\binom{n}{k}_q$ iff $$ \phi_m(q) \mid \binom{n}{k}_q \iff e_m = 1 \iff m \nmid{k} \iff m \nmid n-k \;. $$

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    Thanks bgins. Do you mind explaining further the case when $m\mid k$ if you have the time?2011-12-17
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    Wow, amazing! Thanks again bgins.2011-12-17