Exercise 11.16

Proof. (a) If $\alpha ,\beta \in F_s$, we know that $N_{F_s∕F}(\mathit{\alpha \beta })=N_{F_s∕F}(\alpha )N_{F_s∕F}(\beta )$ (Proposition 11.2.2). Therefore

This shows that ${\chi }^{\prime }$ is a character. (b) Assume that ${\chi }^{\prime }={\rho }^{\prime }$. Then for all $\alpha \in K^{\text{∗}}$, $\chi (N_{F_s∕F}(\alpha ))=\rho (N_{F_s∕F}(\alpha ))$. By Proposition 11.2.2 (d), the map

is surjective. Let $a$ be any element of $F^{\text{∗}}$. Since $\phi$ is surjective, there is some $\alpha \in F_s^{\text{∗}}$ such that $\phi (\alpha )=a$. Then $\chi (a)=\chi (N_{F_s∕F}(\alpha ))=\rho (N_{F_s∕F}(\alpha ))=\rho (a)$. Since this is true for every $a\in F^{\text{∗}}$, and $\chi (0)=0=\rho (0)$, this shows that $\chi =\rho$.

To conclude, ${\chi }^{\prime }={\rho }^{\prime }$ implies $\chi =\rho$, thus $\chi \ne \rho$ implies ${\chi }^{\prime }\ne {\rho }^{\prime }$. (c) If ${\chi }^m=𝜀$, then for all $\alpha \in K$, ${({\chi }^{\prime })}^m(\alpha )={\chi }^m(N_{F_s∕F}(\alpha ))=1,$ thus ${\chi {}^{\prime }}^m=𝜀$. (d) If $a\in F$, by Proposition 11.2.2(c), $N_{F_s∕F}(a)=a^s$, therefore

(e) Assume that $q\equiv 1(\mathit{mod}m)$. Write $C$ the group of character on $F$, $C^{\prime }$ the group of characters on $F_s$, $C_m$ the group of character on $F$ with order dividing $m$, and $C_m^{\prime }$ the group of character on $F_s$ with order dividing $m$. By the generalization of Proposition 8.1.3, $C$ is a cyclic group of order $q-1$, and $C^{\prime }$ a cyclic group of order $q^s-1$.

We know that if $m\mid q-1=|C|$, the subgroup $C_m=\{\chi \in C\mid {\chi }^m=𝜀\}$ of the cyclic group $C$ is cyclic of order $m$. Since $m\mid q-1\mid q^s-1$, it is the same for $C_m^{\prime }$:

Let $\psi$ be the map

Part (b) shows that $\psi$ is injective, and $|C_m|=|C_m^{\prime }|=m$, therefore $\psi$ is bijective. In other words, as $\chi$ varies over all characters of $F$ with order dividing $m$, ${\chi }^{\prime }$ varies over all characters of $F_s$ with order dividing $m$. □