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A month ago I've asked two questions about rationality of the zeta function. The pages that belongs to my question are (linked here) Unfortunately I'm still clueless, but some steps are clear now.

We start with:

$$ \sum_{\chi_0^{(s)},...,\chi_n^{(s)}} \chi_0^{(s)}(a_o^{-1}) \cdots \chi_n^{(s)}(a_n^{-1})g(\chi_0^{(s)}) \cdots g(\chi_n^{(s)}) \mbox{ } \mbox{ } \mbox{ }\mbox{ } \mbox{ } $$

$\chi_i^{(s)}$ are multiplicative characters of $F_s$, such that $\chi_i^{(s)m} = \varepsilon$, $\chi_i^{(s)} \neq \varepsilon$ und $\chi_0^{(s)} \cdots \chi_n^{(s)} = \varepsilon$ .

We showed that $\chi^{'} = \chi \circ N_{F_s/F}$ is a character of $F_s$ with:

(a) $ \chi \neq \rho $ implies that $\chi^{'} \neq \rho^{'} $

(b) $ \chi^{m} = \varepsilon $ implies that $\chi^{'m} = \varepsilon $

(c) $ \chi^{'}(a) = \chi(a)^s $ for all $a \in F$.

with the help of the information above I can almost replace $$ \sum_{\chi_0^{(s)},...,\chi_n^{(s)}} \chi_0^{(s)}(a_o^{-1}) \cdots \chi_n^{(s)}(a_n^{-1})g(\chi_0^{(s)}) \cdots g(\chi_n^{(s)}) \mbox{ } \mbox{ } \mbox{ }\mbox{ } \mbox{ } $$ with $$ \sum_{\chi_0,...,\chi_n} \chi_0(a_o^{-1})^s \cdots \chi_n(a_n^{-1})^sg(\chi_0^{'}) \cdots g(\chi_n^{'})$$ why almost? : I only need to show the part " it follows that as $\chi$ varies over characters of $F$ of order dividing $m$, $\chi^{'}$ varies over characters of $F_s$ of order dividing $m$ ". Please help me here.

If you need more information. Please let me know.

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  • $\begingroup$ I'm not sure I got everything, but the statement it follows that as χ varies over characters of F of order dividing m, χ′ varies over characters of F of order dividing m. follows easily from the cyclicity of the character group. $\endgroup$ – Jyrki Lahtonen Dec 25 '18 at 20:24
  • $\begingroup$ Presumably $m$ is a factor of $q-1$, $F$ is the field with $q$ elements, and $F_s$ is the field with $q^s$ elements. Whenever $G$ is a cyclic group of order $\ell$, the same holds for its dual group $\hat G$. It follows that (assuming $m\mid\ell$) $G$ has exactly $m$ distinct characters $\chi\in\hat G$ with the property $\chi^m=\varepsilon$. In view of this, both $F^*$ and $F_s^*$ have exactly $m$ characters with that property. Because the norm map $N:F_s\to F$ is surjective, composing it with distinct characters of $F$ gives rise to distinct characters of $F_s$. Done. $\endgroup$ – Jyrki Lahtonen Dec 25 '18 at 20:29
  • $\begingroup$ I did see your earlier questions, but I was busy with other duties, and didn't have the time to look at the details. Sorry about that. $\endgroup$ – Jyrki Lahtonen Dec 25 '18 at 20:31
  • $\begingroup$ Of course you are supposed to start with $n=1$.. As I said in the first question everything is about the function $1_{x \in (F_{q^s})^m}$ which can be defined as a sum over the characters of $F_{q^s}$ whose order $|m$. Thus, if $m$ divides $q−1$, letting $r=\frac{q^s−1}{q−1}$, assuming $gcd(m,r)=1$, as a sum over the characters $\chi \circ N_{F_{q^s}/F_q}(a) = \chi(a^r)$ with $\chi$ in the characters of $F_q$ whose order $|m$. If $x \in F_q$ then $\chi(a^r) = \chi(a)^r$ by definition of characters. $\endgroup$ – reuns Dec 25 '18 at 22:07
  • $\begingroup$ Thank you Jyrki Lahtonen and reuns for your answers. (: $\endgroup$ – RukiaKuchiki Jan 3 at 19:09

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