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Let $w$ be a primitive $p^{th}$ root of unity, where $p$ is prime . Let $I,J \subset F_{p}$ where $F_{p}$ is a field of $p$ elements. Chebotarev's theorem states that if $I$ and $J$ have equal cardinality then the matrix $(w^{ij})_{i \in I, j\in J}$ has non-zero determinant.

What is the intuition behind this result? Note that it does not hold in general if $p$ is not prime. For example if $p=4$ then $I=\left\{0,2\right\}$ and $J=\left\{0,2\right\}$ gives a counterexample. Intuitively why doesn't it hold in general when $w$ is a primitive $n^{th}$ root of unity, $n$ being composite?

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I only knew Chebotarev's density theorem but now I see there's a Chebotarev's theorem on roots of unity. Thanks for educating me. – lhf May 27 '11 at 16:38
Why do you ask why it doesn't hold? Do you have an actual counterexample? From a very quick google search, the paper… gives an analogue for composite n. – KCd May 27 '11 at 16:53
See the 4th page of the paper on Chebotarev by Lenstra and Stevenhagen for a description of Chebotarev's proof for roots of unity with prime order p, which involves p-adic completions. They give some intuition. – KCd May 27 '11 at 17:02
@lhf: I was also going to comment the same :). Good to see our minds thinking along the same lines. – user9413 May 27 '11 at 17:03
Myke: please put the counterexample into the question itself so readers will know there is something different in the case of prime vs. non-prime n. – KCd May 27 '11 at 17:49

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