# Cyclotomic extensions over $\mathbb{Q}$

Let $q(n)$ denote the primitive $n$th roots of unity and let $K=\mathbb{Q}(q(n))$ be the associated cyclotomic field. Let $a$ denote the trace of $q(n)$ from $K$ to $\mathbb{Q}.$

How to

prove that $a=1$ if $n=1$, $a=0$ if $n$ is divisible by the square of a prime, and $a=(-1)^r$ if $n$ is the product of $r$ distinct primes?

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There is no reason to ever use all capital letters, even for a title. It is considered an extremely rude practice on the internet. Please don't do it again. – Zev Chonoles Mar 25 '12 at 3:56
Also, please explain what your thoughts are about your problem so far - are you stuck somewhere in particular? To do so is polite, and helps others better help you. – Zev Chonoles Mar 25 '12 at 3:58

You are asked to compute the sum of all primitive $n$th roots of unity. Hints:
1. What would you get if you computed the sum of all (not necessarily primitive) $n$th roots of unity?
2. If an $n$th root of unity is not primitive, then it is a primitive $d$th root of unity for some divisor $d$ of $n$.