$chi$ varies over characters of $F$ of order dividing $m$, $chi^{'}$ varies over characters of $F_s$ of order...
$begingroup$
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.
number-theory finite-fields characters zeta-functions
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
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.
number-theory finite-fields characters zeta-functions
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
$begingroup$
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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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
add a comment |
$begingroup$
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.
number-theory finite-fields characters zeta-functions
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
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.
number-theory finite-fields characters zeta-functions
number-theory finite-fields characters zeta-functions
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
edited Dec 25 '18 at 20:46
RukiaKuchiki
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
asked Dec 25 '18 at 19:36
RukiaKuchikiRukiaKuchiki
337211
337211
$begingroup$
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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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
add a comment |
$begingroup$
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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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
$begingroup$
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
$begingroup$
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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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
$begingroup$
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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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
$begingroup$
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
$begingroup$
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
$begingroup$
Thank you Jyrki Lahtonen and reuns for your answers. (:
$endgroup$
– RukiaKuchiki
Jan 3 at 19:09
add a comment |
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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 $mmidell$) $G$ has exactly $m$ distinct characters $chiinhat G$ 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 $N:F_sto F$ 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