Exercise 10.21

Question

\def\imp{\Rightarrow}
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\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $\psi$ be a map from $\F_q$ to $\C^*$ such that $\psi(\alpha + \beta) = \psi(\alpha)\psi(\beta)$ for all $\alpha, \beta \in \F_q$. Show that there is a $\gamma \in \F_q$ such that $\psi(x) = \zeta^{\mathrm{tr}(\gamma x)}$ for all $x \in \F_q$, where $\zeta = {2i\pi/p}$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

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\begin{proof} Here $q = p^n$.

The map $\psi$ is a group homomorphism, from $(\F_q,+)$ to $(\C^*,	imes)$, thus $\psi(0) = 1$, and $\psi(a \alpha) = \psi(\alpha)^a$, where $\alpha \in \F_q$ and $a \in \Z$.

Let $(\omega_1,\ldots,\omega_n)$ be a basis for $\F_{p^n}$ over $\F_p$. For each $k \in \gcro 1,n\dcro$, since the characteristic of $\F_{q}$ is $p$,
$$\psi(\omega_k)^p = \psi( p \omega_i) = \psi(0) = 1.$$
Thus $\psi(\omega_k)$ is a $p$-th root of unity, of the form
$$\psi(\omega_k) = \zeta^{c_k}, \quad c_k \in \{0, \ldots, p-1\}.$$
Since $\zeta^{c_k} = \zeta^{c_k + lp} \ (l\in \Z)$, we can give a sense to $\psi(\omega_k) = \zeta^{c_k} = \zeta^{[c_k]}$, where $[c_k] \in \F_p$ is the class of $c_k$ modulo $p$.

Consider the map
$$
\varphi
\left\{
\begin{array}{ccl}
\F_q & 	o & (\F_p)^n\
\gamma & \mapsto & (\mathrm{tr}(\gamma \omega_1),\ldots, \mathrm{tr}(\gamma \omega_n)).
\end{array}
ight.
$$
We will show that the linear map $\varphi$ is bijective.

If $\gamma \in \ker(\varphi)$, then $\mathrm{tr}(\gamma \omega_1) = \ldots, \mathrm{tr}(\gamma \omega_n) = 0$. If $y$ is any element in $\F_q$, then $y = b_1 \omega_1+\cdots + b_n \omega_n$, where $b_1,\ldots, b_n \in \F_p$. Then  $\mathrm{tr}(\gamma y) = b_1 \mathrm{tr}(\gamma \omega_1) + b_n \mathrm{tr}(\gamma \omega_n) = 0$, which gives
$$\forall y \in \F_q,\ \mathrm{tr}(\gamma y ) = 0.$$ 
Reasoning by contradiction suppose that $\gamma 
e 0$. Since $\mathrm{tr}$ maps $\F_q$ onto $\F_p$( Proposition 10.3.1.(d)), there is some $\delta \in \F_q$ such that $\mathrm{tr}(\delta) = 1$. If $y = \delta \gamma^{-1}$, then $0 = \mathrm{tr}(\gamma y) = \mathrm{tr}(\delta) = 1$. This is a contradiction, so $\gamma = 0$, and this proves $\ker(\varphi) = \{0\}$.

Moreover $\mathrm{dim}_{\F_p} (\F_q) = \mathrm{dim}_{\F_p}(\F_p)^n = n$, thus $\varphi$ is a bijection.

Thus there exists $\gamma \in \F_q$ such that $$\mathrm{tr}(\gamma \omega_k) = [c_k],\qquad k=1,\ldots,n.$$

Then, if $x$ is any element in $\F_q$, we can write $x = a_1 \omega_1 + \cdots + a_n \omega_n$, where $a_1,\ldots,a_k \in \F_p$. Since $\psi(\omega_k)$ is a $p$-th root of unity,
\begin{align*}
\psi(x) &= \psi( a_1 \omega_1 + \cdots + a_n \omega_n)\
&=\psi(\omega_1)^{a_1} \cdots \psi(\omega_n)^{a_n}\
&=\zeta^{a_1 \mathrm{tr}(\gamma \omega_1) + \cdots + a_n \mathrm{tr}(\gamma\omega_n)}\
&=\zeta^{\mathrm{tr}(\gamma x)}.
\end{align*}

If $\psi$ is a group homomorphism from $\F_q$ to $\C^*$, then there is a $\gamma \in \F_q$ such that $\psi(x) = \zeta^{\mathrm{tr}(\gamma x)}$ for all $x \in \F_q$.
\end{proof}

Exercise 10.21

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Exercise 10.21

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