Show that $S^p = \sum\limits_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}.$

Question

Let $a \in \Bbb Z$ and $p,q$ be primes. Define $\left (\frac a p \right )$ as follows $:$

$$\left(\frac{a}{p}\right) = \begin{cases}\;\;\,0&\text{ if }p \text { divides } a\\+1&\text{ if } a \operatorname{R} p \text{ and }p \text { does not divide } a\\-1&\text{ if }a \operatorname{N} p \text{ and }p \text{ does not divide } a\end{cases}$$

Let $\omega$ be a primitive $q$-th root of unity. Let $$S = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^x.$$ Show that $$S^p = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}.$$

How can I show that? Please help me in this regard.

Thank you very much.

Answer 1

That is not true: when $p \neq q$, one can show that $S^p$ has modulus $q^{p/2}$, while the RHS has modulus $q^{1/2}$.

You must have misunderstood something. What is true, is that $S^p$ is congruent to $\sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^{xp}$ modulo the ideal $(p)$, inside $\mathbb Z[\omega]$.