1. ## Find trace

Let p be prime, and let $\displaystyle K = \mathbb {Q} [w]$, w is a primitive pth root of unity. Compute $\displaystyle Tr_{K}(w)$ and $\displaystyle Tr_{K} (1-w)$

Also, show that $\displaystyle (1-w)(1-w^2)... (1-w^{p-1}) = p$

proof.

So I have $\displaystyle w^p = 1$ with $\displaystyle w^{k} \neq 1 \ \ \ \forall k < p$

Let p be prime, and let $\displaystyle K = \mathbb {Q} [w]$, w is a primitive pth root of unity. Compute $\displaystyle Tr_{K}(w)$ and $\displaystyle Tr_{K} (1-w)$
If $\displaystyle E/F$ is a finite Galois extension then we define $\displaystyle \text{Tr}(\alpha) = \sum_{\theta \in \text{Gal}(E/F)}\theta (\alpha)$ for $\displaystyle \alpha \in E$. Now if $\displaystyle \omega$ is a primitive $\displaystyle p$-th root of unity then $\displaystyle \mathbb{Q}(\omega)$ is finite and Galois over $\displaystyle \mathbb{Q}$, furthermore, since $\displaystyle \{ 1,\omega, ... , \omega^{p-1}\}$ for a basis for this extension and $\displaystyle \mathbb{Q}$ is a field with $\displaystyle \text{Char}(\mathbb{Q}) = 0$ it means $\displaystyle G=\text{Gal}(K/\mathbb{Q})$ is a group of order $\displaystyle p$, so it must be a cyclic group. Thus, there is an automorphism $\displaystyle \theta$ of $\displaystyle K$ leaving $\displaystyle \mathbb{Q}$ fixed so that $\displaystyle \left< \theta \right> = G$. Suppose that $\displaystyle \theta (\omega) = \omega^k$ where $\displaystyle 0< k \leq p-1$. Then it means $\displaystyle \{ \text{Id},\theta, \theta^2, ... ,\theta^{p-1} \} = G$. And so $\displaystyle \text{Tr}(\omega) = \text{Id}(\omega) + \theta(\omega) + \theta^2 (\omega) + ... + \theta^{p-1} (\omega)$. Thus, $\displaystyle \mbox{Tr}(\omega) = \omega + \omega^k + \omega^{2k} + ... + \omega^{(p-1)k}= \omega - 1$.