Exercise 11.17

Proof. In Theorem 2,

where $A$ is the set of $(n+1)$-tuples $({\chi }_0,\dots ,{\chi }_n)$ of characters on $F$ such that ${\chi }_i^m=𝜀,{\chi }_i\ne 𝜀(i=0,\dots ,n)$ and ${\chi }_0\cdots {\chi }_n=𝜀$. In each factor, the coefficient of $u$ is not zero, thus each factor has degree $1$. Therefore the degree $d$ of $P$ is $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(P)=|A|$. Write $C_m$ the subgroup of characters on F such that ${\chi }^m=𝜀$.

Since $q\equiv 1(\mathit{mod}m)$ is an hypothesis of Theorem 2, $C_m$ is a subgroup of order $m$: $|C_m|=m$ and $|C_m-\{𝜀\}|=m-1$. We count the number $d$ of $(n+1)$-tuples $({\chi }_0,\dots ,{\chi }_n)\in {(C_m-\{𝜀\})}^{n+1}$ such that ${\chi }_0\cdots {\chi }_n=𝜀$, that is ${\chi }_n={\chi }_0^{-1}\cdots {\chi }_{n-1}^{-1}$, and ${\chi }_n\ne 𝜀$. Let $\chi$ be a character of order $m$ (such a character exists because $C_m$ is cyclic). Write ${\chi }_i={\chi }^{k_i}$, where $1\le k_i\le m-1$. Then $d$ is the number of $n$-tuples $(k_0,\dots ,k_{n-1})\in \text{[[}1,m-1{\text{]]}}^n$ such that

In other words, $d$ is the number of $n$-tuples $(a_0,\dots ,a_{n-1})\in {({(\mathbb{Z}∕\mathit{m\mathbb{Z}})}^{\text{∗}})}^n$ such that

To begin an induction, fix the integer $m\in {\mathbb{N}}^{\text{∗}}$, and write

For $n\ge 2$, if $(a_0,\dots ,a_{n-2})\in {({(\mathbb{Z}∕\mathit{m\mathbb{Z}})}^{\text{∗}})}^{n-1}$ is given, we count the number of $a_{n-1}\in {(\mathbb{Z}∕\mathit{m\mathbb{Z}})}^{\text{∗}}$ such that $a_{n-1}\ne -a_0-a_1-\cdots -a_{n-2}$.

There are two cases.

If $a_0+\cdots +a_{n-2}\ne 0$, there are $m-2$ choices for $a_{n-1}\in \mathbb{Z}∕\mathit{m\mathbb{Z}}\setminus \{0,-a_0-\cdots -a_{n-2}\}$, and if $a_0+\cdots +a_{n-2}=0$, there are $m-1$ choices for $a_{n-1}\in \mathbb{Z}∕\mathit{m\mathbb{Z}}\setminus \{0\}$. This gives the relation

Since $d_1=\mathrm{Card}\{a\in {(\mathbb{Z}∕\mathit{m\mathbb{Z}})}^{\text{∗}}\mid a\ne 0\}=m-1$, we obtain by immediate induction

Then

This is the waited answer,

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