Question about characters ( Section : The Rationality of the Zeta Function associated to...
I am currently reading " A Classic Introduction to Modern Number Theory " by Kenneth Ireland and Michael Rosen. In the 11th chapter they consider the zeta function. In the third section of this chapter they want to show that the Zeta Function is associated to $a_0x_0^m+a_1x_1^m+...+a_nx_n^m$. And I have a lot of questions to this chapter. In this question I will start with an easy question. Let us start with:
My question : So I don't understand (a) and (b). Like you can see we need Proposition 11.2.2. So here:
Like I said: can you proof (a) and (b)? ( not the (a) and (b) from the Proposition. I mean the (a) and (b) of the first picture. ) It is not clear to me. I solved (c) by my own, I think:
$ chi^{'}(a) = chi(N_{F_s/F}(a)) = chi(a^s * N_{F_s/F}(1)) = chi(a^s) = (chi(a))^s $ I used Proposition 11.2.2 (c) for the second equation.
So if you need more information let me know.
number-theory finite-fields characters zeta-functions
asked Nov 23 at 21:22
RukiaKuchiki
299211
add a comment |
I am currently reading " A Classic Introduction to Modern Number Theory " by Kenneth Ireland and Michael Rosen. In the 11th chapter they consider the zeta function. In the third section of this chapter they want to show that the Zeta Function is associated to $a_0x_0^m+a_1x_1^m+...+a_nx_n^m$. And I have a lot of questions to this chapter. In this question I will start with an easy question. Let us start with:
My question : So I don't understand (a) and (b). Like you can see we need Proposition 11.2.2. So here:
Like I said: can you proof (a) and (b)? ( not the (a) and (b) from the Proposition. I mean the (a) and (b) of the first picture. ) It is not clear to me. I solved (c) by my own, I think:
$ chi^{'}(a) = chi(N_{F_s/F}(a)) = chi(a^s * N_{F_s/F}(1)) = chi(a^s) = (chi(a))^s $ I used Proposition 11.2.2 (c) for the second equation.
So if you need more information let me know.
number-theory finite-fields characters zeta-functions
asked Nov 23 at 21:22
RukiaKuchiki
299211
For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51
add a comment |
I am currently reading " A Classic Introduction to Modern Number Theory " by Kenneth Ireland and Michael Rosen. In the 11th chapter they consider the zeta function. In the third section of this chapter they want to show that the Zeta Function is associated to $a_0x_0^m+a_1x_1^m+...+a_nx_n^m$. And I have a lot of questions to this chapter. In this question I will start with an easy question. Let us start with:
My question : So I don't understand (a) and (b). Like you can see we need Proposition 11.2.2. So here:
Like I said: can you proof (a) and (b)? ( not the (a) and (b) from the Proposition. I mean the (a) and (b) of the first picture. ) It is not clear to me. I solved (c) by my own, I think:
$ chi^{'}(a) = chi(N_{F_s/F}(a)) = chi(a^s * N_{F_s/F}(1)) = chi(a^s) = (chi(a))^s $ I used Proposition 11.2.2 (c) for the second equation.
So if you need more information let me know.
number-theory finite-fields characters zeta-functions
asked Nov 23 at 21:22
RukiaKuchiki
299211
I am currently reading " A Classic Introduction to Modern Number Theory " by Kenneth Ireland and Michael Rosen. In the 11th chapter they consider the zeta function. In the third section of this chapter they want to show that the Zeta Function is associated to $a_0x_0^m+a_1x_1^m+...+a_nx_n^m$. And I have a lot of questions to this chapter. In this question I will start with an easy question. Let us start with:
My question : So I don't understand (a) and (b). Like you can see we need Proposition 11.2.2. So here:
Like I said: can you proof (a) and (b)? ( not the (a) and (b) from the Proposition. I mean the (a) and (b) of the first picture. ) It is not clear to me. I solved (c) by my own, I think:
$ chi^{'}(a) = chi(N_{F_s/F}(a)) = chi(a^s * N_{F_s/F}(1)) = chi(a^s) = (chi(a))^s $ I used Proposition 11.2.2 (c) for the second equation.
So if you need more information let me know.
number-theory finite-fields characters zeta-functions
number-theory finite-fields characters zeta-functions
asked Nov 23 at 21:22
RukiaKuchiki
299211
asked Nov 23 at 21:22
RukiaKuchiki
299211
edited Nov 25 at 14:25
asked Nov 23 at 21:22
RukiaKuchiki
299211
asked Nov 23 at 21:22
RukiaKuchiki
299211
asked Nov 23 at 21:22
RukiaKuchiki
299211
299211
For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51
add a comment |
For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51
For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51
add a comment |
1 Answer
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Let us start with (a): So you know that $ chi neq rho $. That means that there is an element $ a in F^*$ with $chi(a) neq rho(a)$.
We use Proposition 11.2.2 (d). That means that there is an element $ beta in F_s^* $ with $N_{F_s/F}(beta)= a$
We get with the help of Proposition 11.2.2(a): $chi^{'}(beta) = chi(N_{F_s/F}(beta)) neq rho(N_{F_s/F}(beta))=rho^{'}(beta). $ So $ chi^{'} neq rho^{'}$.
Now let us solve (b) :
Remember $(chi^{'})^m = varepsilon$ <=> $(chi^{'})^m(alpha) = 1$.
We want to use Proposition 11.2.2(a) again. $chi^{'}(alpha)^m = chi(N_{F_s/F}(alpha))^m = 1 mbox{ } mbox{ } forall alpha in F_s $
answered Dec 8 at 19:24
Mugumble
455212
add a comment |
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Let us start with (a): So you know that $ chi neq rho $. That means that there is an element $ a in F^*$ with $chi(a) neq rho(a)$.
We use Proposition 11.2.2 (d). That means that there is an element $ beta in F_s^* $ with $N_{F_s/F}(beta)= a$
We get with the help of Proposition 11.2.2(a): $chi^{'}(beta) = chi(N_{F_s/F}(beta)) neq rho(N_{F_s/F}(beta))=rho^{'}(beta). $ So $ chi^{'} neq rho^{'}$.
Now let us solve (b) :
Remember $(chi^{'})^m = varepsilon$ <=> $(chi^{'})^m(alpha) = 1$.
We want to use Proposition 11.2.2(a) again. $chi^{'}(alpha)^m = chi(N_{F_s/F}(alpha))^m = 1 mbox{ } mbox{ } forall alpha in F_s $
answered Dec 8 at 19:24
Mugumble
455212
add a comment |
Let us start with (a): So you know that $ chi neq rho $. That means that there is an element $ a in F^*$ with $chi(a) neq rho(a)$.
We use Proposition 11.2.2 (d). That means that there is an element $ beta in F_s^* $ with $N_{F_s/F}(beta)= a$
We get with the help of Proposition 11.2.2(a): $chi^{'}(beta) = chi(N_{F_s/F}(beta)) neq rho(N_{F_s/F}(beta))=rho^{'}(beta). $ So $ chi^{'} neq rho^{'}$.
Now let us solve (b) :
Remember $(chi^{'})^m = varepsilon$ <=> $(chi^{'})^m(alpha) = 1$.
We want to use Proposition 11.2.2(a) again. $chi^{'}(alpha)^m = chi(N_{F_s/F}(alpha))^m = 1 mbox{ } mbox{ } forall alpha in F_s $
answered Dec 8 at 19:24
Mugumble
455212
add a comment |
Let us start with (a): So you know that $ chi neq rho $. That means that there is an element $ a in F^*$ with $chi(a) neq rho(a)$.
We use Proposition 11.2.2 (d). That means that there is an element $ beta in F_s^* $ with $N_{F_s/F}(beta)= a$
We get with the help of Proposition 11.2.2(a): $chi^{'}(beta) = chi(N_{F_s/F}(beta)) neq rho(N_{F_s/F}(beta))=rho^{'}(beta). $ So $ chi^{'} neq rho^{'}$.
Now let us solve (b) :
Remember $(chi^{'})^m = varepsilon$ <=> $(chi^{'})^m(alpha) = 1$.
We want to use Proposition 11.2.2(a) again. $chi^{'}(alpha)^m = chi(N_{F_s/F}(alpha))^m = 1 mbox{ } mbox{ } forall alpha in F_s $
answered Dec 8 at 19:24
Mugumble
455212
Let us start with (a): So you know that $ chi neq rho $. That means that there is an element $ a in F^*$ with $chi(a) neq rho(a)$.
We use Proposition 11.2.2 (d). That means that there is an element $ beta in F_s^* $ with $N_{F_s/F}(beta)= a$
We get with the help of Proposition 11.2.2(a): $chi^{'}(beta) = chi(N_{F_s/F}(beta)) neq rho(N_{F_s/F}(beta))=rho^{'}(beta). $ So $ chi^{'} neq rho^{'}$.
Now let us solve (b) :
Remember $(chi^{'})^m = varepsilon$ <=> $(chi^{'})^m(alpha) = 1$.
We want to use Proposition 11.2.2(a) again. $chi^{'}(alpha)^m = chi(N_{F_s/F}(alpha))^m = 1 mbox{ } mbox{ } forall alpha in F_s $
answered Dec 8 at 19:24
Mugumble
455212
answered Dec 8 at 19:24
Mugumble
455212
answered Dec 8 at 19:24
Mugumble
455212
answered Dec 8 at 19:24
Mugumble
455212
455212
add a comment |
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For $F = mathbf{F}_q,E= mathbf{F}_{q^s}$ then $N_{E/F}(a) =prod_{l=1}^s a^{q^l}= a^{(q^s-1)/(q-1)} $ is a root of $y^q-y$ so is in $F$
– reuns
Nov 23 at 21:30
Then for $m | q-1$ pick some $g in F^times$ of order $q-1$. Then $chi(0) = 0,chi(g^n) = e^{2i pi n /m}$ is a character of order $m$ and for $gcd(s,m)=1$, $chi_E(a) = chi(N_{E/F}(a))$ is a character of order $m$ of $E^times$ and $exists b in E^times, a = b^m$ iff $chi_E(a) = 1$ iff $sum_{l=1}^m chi(a)^l = m$, the last sum being over the characters of order $| m $.
– reuns
Nov 23 at 21:45
Pardon. Am I wrong or do you proof some parts of Proposition 11.2.2? My question was (a) $chi neq rho $ implies $chi^{'} neq rho^{'} $ and (b) $chi^m = varepsilon$ implies $chi^{'m} = varepsilon$. If I misunderstood something, I'm sorry.
– RukiaKuchiki
Nov 23 at 21:51