Exercise 10.22

Proof. Here $\psi :{𝔽}_q\to ℂ$ is defined by $\psi (\alpha )={\zeta }_p^{\mathrm{tr}(\alpha )}$, and the Gauss sum for a character $\chi$ of ${𝔽}_q$ by

First we generalize Proposition 8.1.2, with the same proof. If $\chi \ne 𝜀$, there is an $a\in {𝔽}_q^{\text{∗}}$ such that $\chi (a)\ne 1$. Then, if $T={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{t\in {𝔽}_q}\chi (t)$, then

Since $\chi (a)T=T$ and $\chi (a)\ne 1$, it follows that $T=0$. This proves, for a non trivial character $\chi$,

(a)

If $\alpha \in {𝔽}_q^{\text{∗}}$, $$\begin{array}{llll}\hfill \chi (\alpha )g_{\alpha }(\chi )& =\underset{t\in {𝔽}_q}{\sum }\chi (\alpha )\chi (t)\psi (\mathit{\alpha t})& \hfill & \\ \hfill & =\underset{t\in {𝔽}_q}{\sum }\chi (\mathit{\alpha t})\psi (\mathit{\alpha t})& \hfill & \\ \hfill & =\underset{s\in {𝔽}_q}{\sum }\chi (s)\psi (s)(s=\mathit{\alpha t})& \hfill & \\ \hfill & =g(\chi ).& \hfill & \end{array}$$

Since $|\chi (\alpha )|=1$, $\chi {(\alpha )}^{-1}=\overline{\chi (\alpha )}$, thus

$$g_{\alpha }(\chi )=\overline{\chi (\alpha )}g(\chi ).$$

(b)

Since ${(-1)}^2=1$, ${(\chi (-1))}^2=1$, thus $\chi (-1)=\pm 1$ is real, therefore $\overline{\chi (-1)}=\chi (-1)$. This gives $$\begin{array}{llll}\hfill \overline{g(\chi )}& =\underset{t\in {𝔽}_q}{\sum }\overline{\chi (t)}{\zeta }_p^{-\mathrm{tr}(t)}& \hfill & \\ \hfill & =\underset{t\in {𝔽}_q}{\sum }\overline{\chi (-1)\chi (-t)}{\zeta }_p^{-\mathrm{tr}(t)}& \hfill & \\ \hfill & =\chi (-1)\underset{t\in {𝔽}_q}{\sum }\overline{\chi (-t)}{\zeta }_p^{\mathrm{tr}(-t)}& \hfill & \\ \hfill & =\chi (-1)\underset{s\in {𝔽}_q}{\sum }\overline{\chi (s)}{\zeta }_p^{\mathrm{tr}(s)}(s=-t)& \hfill & \\ \hfill & =\chi (-1)g(\bar{\chi })& \hfill & \\ \hfill & & \hfill \end{array}$$

We have seen in part (a) that ${\chi }^{-1}=\bar{\chi }$. This gives

$$g({\chi }^{-1})=g(\bar{\chi })=\chi (-1)\overline{g(\chi )}.$$

(c)

Here we assume that $\chi \ne 𝜀$. By part (a), $|g_{\alpha }(\chi )|=|g(\chi )|$, so it it sufficient to verify $|g(\chi )|=q^{1∕2}$.

We evaluate the sum $S={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in {𝔽}_q}{g}_a(\chi )\overline{g_a(\chi )}$ in two ways.

$\text{∙}$

We have proved in the introduction that $g_0(\chi )=0$. If $a\in {𝔽}_q^{\text{∗}}$, then $g_a(\chi )=\chi (a^{-1})g(\chi )$, and $\overline{g_a(\chi )}=\overline{\chi (a^{-1})}\overline{g(\chi )}=\chi (a)\overline{g(\chi )}$. It follows that $$\begin{array}{llll}\hfill S& =\underset{a\in {𝔽}_q^{\text{∗}}}{\sum }\chi (a^{-1})g(\chi )\chi (a)\overline{g(\chi )}& \hfill & \\ \hfill & =\underset{a\in {𝔽}_q^{\text{∗}}}{\sum }|g(\chi ){\text{|}}^2& \hfill & \\ \hfill & =(q-1)|g(\chi ){\text{|}}^2& \hfill & \end{array}$$

$\text{∙}$

Furthermore $$g_a(\chi )\overline{g_a(\chi )}=\underset{x\in {𝔽}_q}{\sum }\underset{y\in {𝔽}_q}{\sum }\chi (x)\overline{\chi (y)}\psi (a(x-y)).$$

Therefore,

$$\begin{array}{llll}\hfill S& =\underset{a\in {𝔽}_q}{\sum }\underset{x\in {𝔽}_q}{\sum }\underset{y\in {𝔽}_q}{\sum }\chi (x)\overline{\chi (y)}\psi (a(x-y))& \hfill & \\ \hfill & =\underset{x\in {𝔽}_q}{\sum }\underset{y\in {𝔽}_q}{\sum }\chi (x)\overline{\chi (y)}\left(\underset{a\in {𝔽}_q}{\sum }\psi (a(x-y))\right)& \hfill & \end{array}$$

By Proposition 10.3.3,

$$\begin{array}{llll}\hfill \underset{a\in {𝔽}_q}{\sum }\psi (a(x-y))& =\mathit{q\delta }(x,y)& \hfill & \end{array}$$

Therefore,

$$\begin{array}{llll}\hfill S& =q\underset{x\in {𝔽}_q}{\sum }\underset{y\in {𝔽}_q}{\sum }\chi (x)\overline{\chi (y)}\delta (x,y)& \hfill & \\ \hfill & =q\underset{x\in {𝔽}_q}{\sum }\chi (x)\overline{\chi (x)}& \hfill & \\ \hfill & & \hfill \end{array}$$

Since $\chi (x)\overline{\chi (x)}=1$ if $x\ne 0$, and $\chi (x)\overline{\chi (x)}=0$ if $x=0$, we obtain

$$S=q(q-1).$$

The comparison of these two results gives

$$(q-1)|g(\chi ){\text{|}}^2=(q-1)q,$$

thus

$$|g_{\alpha }(\chi )|=|g(\chi )|=\sqrt{q}.$$

(d)

Here $\chi \ne 𝜀$. Then, by parts (b) and (c), $$\begin{array}{llll}\hfill g(\chi )g({\chi }^{-1})& =\chi (-1)g(\chi )\overline{g(\chi )}& \hfill & \\ \hfill & =\chi (-1)|g(\chi ){\text{|}}^2& \hfill & \\ \hfill & =\chi (-1)q.& \hfill & \end{array}$$