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

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.

1 Answer

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 $$\mathbb Z[\omega]$$.

• Would you please suggest me some text where I can able to find it and can able to understand it on my own? I came to know this fact from the lecture notes provided by our instructor where the things are not too explicit. Please try to provide me some text if you can. Thank you very much. – Dbchatto67 Feb 8 at 17:57
• You can look at Ireland and Rosen, A classical introduction to modern number theory. The chapter about quadratic Gauss sums contains everything you are looking for. Especially Proposition 6.1.6 and 6.3.2. – punctured dusk Feb 9 at 10:34
• Thank you so much @nbarto for your valuable suggestion. I will definitely check this book. Is it available online for free as pdf or djvu formats? – Dbchatto67 Feb 9 at 15:27
• It is available online. You can try libgen. – punctured dusk Feb 9 at 17:07