Is weight sum of Dirichlet character always an algebraic integer (up to power of $2$ and $3$)?
$begingroup$
Let $chi: (mathbb Z/Nmathbb Z)^{times} rightarrow mathbb C^{times}$ be a character, consider
$a=frac{1}{N}sum_{i=1}^N chi(i)i$
where $chi(n)=0$ if $n$ is not coprime to $N$.
If $chi$ is the Legendre symbol assigned to an imaginary quadratic field, then $a$ is an algebraic integer up to power of $2$ and $3$, because $a$ is essentially the class number by the class number formula.
What about general case?
number-theory algebraic-number-theory
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
$endgroup$
add a comment |
$begingroup$
Let $chi: (mathbb Z/Nmathbb Z)^{times} rightarrow mathbb C^{times}$ be a character, consider
$a=frac{1}{N}sum_{i=1}^N chi(i)i$
where $chi(n)=0$ if $n$ is not coprime to $N$.
If $chi$ is the Legendre symbol assigned to an imaginary quadratic field, then $a$ is an algebraic integer up to power of $2$ and $3$, because $a$ is essentially the class number by the class number formula.
What about general case?
number-theory algebraic-number-theory
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
$endgroup$
add a comment |
$begingroup$
Let $chi: (mathbb Z/Nmathbb Z)^{times} rightarrow mathbb C^{times}$ be a character, consider
$a=frac{1}{N}sum_{i=1}^N chi(i)i$
where $chi(n)=0$ if $n$ is not coprime to $N$.
If $chi$ is the Legendre symbol assigned to an imaginary quadratic field, then $a$ is an algebraic integer up to power of $2$ and $3$, because $a$ is essentially the class number by the class number formula.
What about general case?
number-theory algebraic-number-theory
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
$endgroup$
Let $chi: (mathbb Z/Nmathbb Z)^{times} rightarrow mathbb C^{times}$ be a character, consider
$a=frac{1}{N}sum_{i=1}^N chi(i)i$
where $chi(n)=0$ if $n$ is not coprime to $N$.
If $chi$ is the Legendre symbol assigned to an imaginary quadratic field, then $a$ is an algebraic integer up to power of $2$ and $3$, because $a$ is essentially the class number by the class number formula.
What about general case?
number-theory algebraic-number-theory
number-theory algebraic-number-theory
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
edited Dec 24 '18 at 17:07
zzy
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
asked Dec 23 '18 at 16:52
zzyzzy
2,6331420
2,6331420
add a comment |
add a comment |
1 Answer
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$begingroup$
Presumably the question is whether there exists natural numbers $ell,m$ such that $2^ellcdot3^ma$ is an algebraic integer.
A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $Bbb{Z}_N^*$. Let's try a quartic character defined by $chi(2^t)=i^t$.
I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$
The sum becomes
$$
begin{aligned}
sum_{t=0}^{19}i^ts(t)
&=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\
&quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\
&=-15-5i.
end{aligned}
$$
Implying that $a=-(3+i)/5$ which is not of the required form.
If the numerator were $phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
$endgroup$
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
Presumably the question is whether there exists natural numbers $ell,m$ such that $2^ellcdot3^ma$ is an algebraic integer.
A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $Bbb{Z}_N^*$. Let's try a quartic character defined by $chi(2^t)=i^t$.
I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$
The sum becomes
$$
begin{aligned}
sum_{t=0}^{19}i^ts(t)
&=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\
&quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\
&=-15-5i.
end{aligned}
$$
Implying that $a=-(3+i)/5$ which is not of the required form.
If the numerator were $phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
$endgroup$
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
add a comment |
$begingroup$
Presumably the question is whether there exists natural numbers $ell,m$ such that $2^ellcdot3^ma$ is an algebraic integer.
A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $Bbb{Z}_N^*$. Let's try a quartic character defined by $chi(2^t)=i^t$.
I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$
The sum becomes
$$
begin{aligned}
sum_{t=0}^{19}i^ts(t)
&=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\
&quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\
&=-15-5i.
end{aligned}
$$
Implying that $a=-(3+i)/5$ which is not of the required form.
If the numerator were $phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
$endgroup$
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
add a comment |
$begingroup$
Presumably the question is whether there exists natural numbers $ell,m$ such that $2^ellcdot3^ma$ is an algebraic integer.
A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $Bbb{Z}_N^*$. Let's try a quartic character defined by $chi(2^t)=i^t$.
I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$
The sum becomes
$$
begin{aligned}
sum_{t=0}^{19}i^ts(t)
&=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\
&quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\
&=-15-5i.
end{aligned}
$$
Implying that $a=-(3+i)/5$ which is not of the required form.
If the numerator were $phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
$endgroup$
Presumably the question is whether there exists natural numbers $ell,m$ such that $2^ellcdot3^ma$ is an algebraic integer.
A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $Bbb{Z}_N^*$. Let's try a quartic character defined by $chi(2^t)=i^t$.
I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$
The sum becomes
$$
begin{aligned}
sum_{t=0}^{19}i^ts(t)
&=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\
&quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\
&=-15-5i.
end{aligned}
$$
Implying that $a=-(3+i)/5$ which is not of the required form.
If the numerator were $phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
answered Dec 24 '18 at 17:58
Jyrki LahtonenJyrki Lahtonen
110k13171380
110k13171380
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
add a comment |
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
Thank you! Because the class number formula claim it's such integer in special cases.
$endgroup$
– zzy
Dec 24 '18 at 20:54
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
$begingroup$
If $N$ is a prime number, maybe that's true.
$endgroup$
– zzy
Dec 24 '18 at 20:55
add a comment |
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