Understanding Output in SageMath Regarding Dirichlet Characters
1
p=7
G = DirichletGroup(p); G
m=3; n=ZZ((p-1)/m); print m,n
c=G[1]
c1=c^n;c1
The output is:
Dirichlet character modulo 7 of conductor 7 mapping 3 |--> zeta6 - 1
Can anyone explain what zeta6 is? Is this the Riemann-Zeta function? Is this the whole group of units? Is there a relation to the Eisenstein primes? I'm still a bit weak in this material and am having trouble grasping some of these sage outputs. Thank you in advance!
EDIT: If I take the log and then exponentiate I get, for p=7;
$$ -frac{3i}{2}sqrt{3} +frac{1}{2}$$
Not sure what this is exactly.
abstract-algebra number-theory sagemath
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
add a comment |
1
p=7
G = DirichletGroup(p); G
m=3; n=ZZ((p-1)/m); print m,n
c=G[1]
c1=c^n;c1
The output is:
Dirichlet character modulo 7 of conductor 7 mapping 3 |--> zeta6 - 1
Can anyone explain what zeta6 is? Is this the Riemann-Zeta function? Is this the whole group of units? Is there a relation to the Eisenstein primes? I'm still a bit weak in this material and am having trouble grasping some of these sage outputs. Thank you in advance!
EDIT: If I take the log and then exponentiate I get, for p=7;
$$ -frac{3i}{2}sqrt{3} +frac{1}{2}$$
Not sure what this is exactly.
abstract-algebra number-theory sagemath
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
1
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
2
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
1
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42
add a comment |
1
1
1
p=7
G = DirichletGroup(p); G
m=3; n=ZZ((p-1)/m); print m,n
c=G[1]
c1=c^n;c1
The output is:
Dirichlet character modulo 7 of conductor 7 mapping 3 |--> zeta6 - 1
Can anyone explain what zeta6 is? Is this the Riemann-Zeta function? Is this the whole group of units? Is there a relation to the Eisenstein primes? I'm still a bit weak in this material and am having trouble grasping some of these sage outputs. Thank you in advance!
EDIT: If I take the log and then exponentiate I get, for p=7;
$$ -frac{3i}{2}sqrt{3} +frac{1}{2}$$
Not sure what this is exactly.
abstract-algebra number-theory sagemath
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
p=7
G = DirichletGroup(p); G
m=3; n=ZZ((p-1)/m); print m,n
c=G[1]
c1=c^n;c1
The output is:
Dirichlet character modulo 7 of conductor 7 mapping 3 |--> zeta6 - 1
Can anyone explain what zeta6 is? Is this the Riemann-Zeta function? Is this the whole group of units? Is there a relation to the Eisenstein primes? I'm still a bit weak in this material and am having trouble grasping some of these sage outputs. Thank you in advance!
EDIT: If I take the log and then exponentiate I get, for p=7;
$$ -frac{3i}{2}sqrt{3} +frac{1}{2}$$
Not sure what this is exactly.
abstract-algebra number-theory sagemath
abstract-algebra number-theory sagemath
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
edited Dec 5 '18 at 17:29
Nicklovn
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
asked Dec 5 '18 at 17:09
NicklovnNicklovn
343110
343110
1
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
2
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
1
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42
add a comment |
1
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
2
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
1
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42
1
1
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
2
2
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
1
1
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42
add a comment |
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1
the best place to ask this question is here
– Masacroso
Dec 5 '18 at 17:12
2
$3$ is a generator of $(mathbb{Z}/7mathbb{Z})^times$ and the Dirichlet character is defined by "$chi(3) = zeta_6,chi(n+7) = chi(n)$" which implies $chi(7n) = 0, chi(3^l+7n) = zeta_6^l$ where $zeta_6 = e^{2i pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(mathbb{Z}/7mathbb{Z})^times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $chi$ is a generator of it.
– reuns
Dec 5 '18 at 18:12
1
Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/…
– kcrisman
Dec 7 '18 at 18:14
I know. I posted the question :)
– Nicklovn
Dec 7 '18 at 23:20
I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though.
– kcrisman
Dec 20 '18 at 12:42