Number of ideals of a given norm
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Is there any analytic expression(those involving exp, sin, cos etc...) which gives the number of ideals of a given norm?
of course we are lying in an algebraic number field.
algebraic-number-theory
asked Jun 4 '13 at 7:33
user29253user29253
1463
$endgroup$
add a comment |
$begingroup$
Is there any analytic expression(those involving exp, sin, cos etc...) which gives the number of ideals of a given norm?
of course we are lying in an algebraic number field.
algebraic-number-theory
asked Jun 4 '13 at 7:33
user29253user29253
1463
$endgroup$
add a comment |
$begingroup$
Is there any analytic expression(those involving exp, sin, cos etc...) which gives the number of ideals of a given norm?
of course we are lying in an algebraic number field.
algebraic-number-theory
asked Jun 4 '13 at 7:33
user29253user29253
1463
$endgroup$
Is there any analytic expression(those involving exp, sin, cos etc...) which gives the number of ideals of a given norm?
of course we are lying in an algebraic number field.
algebraic-number-theory
algebraic-number-theory
asked Jun 4 '13 at 7:33
user29253user29253
1463
asked Jun 4 '13 at 7:33
user29253user29253
1463
asked Jun 4 '13 at 7:33
user29253user29253
1463
asked Jun 4 '13 at 7:33
user29253user29253
1463
asked Jun 4 '13 at 7:33
user29253user29253
1463
1463
add a comment |
add a comment |
1 Answer
1
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Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $mathfrak{p}lhdmathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[mathfrak{p}]$, then we can find an inverse say $[mathfrak{q}]$ in the class group, and hence $mathfrak{p}mathfrak{q}=(alpha)$ is principal with $alphainmathfrak{q}$ and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$. And if $alphainmathfrak{q}$, and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$, then $mathfrak{p}=(alpha)mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(alpha)$, where $alphainmathfrak{q}$ (and $[mathfrak{q}]=X^{-1}$ in the class group), and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$.
Let $x_1,dots,x_d$ be an integral basis for $mathcal{O}_K$, and let $alpha=r_1x_1+cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(alpha)$ into one of counting the lattice points $(r_1,dots,r_d)inmathbb{Z}^d$. This is explained in considerably more detail here.
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
$endgroup$
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $mathfrak{p}lhdmathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[mathfrak{p}]$, then we can find an inverse say $[mathfrak{q}]$ in the class group, and hence $mathfrak{p}mathfrak{q}=(alpha)$ is principal with $alphainmathfrak{q}$ and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$. And if $alphainmathfrak{q}$, and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$, then $mathfrak{p}=(alpha)mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(alpha)$, where $alphainmathfrak{q}$ (and $[mathfrak{q}]=X^{-1}$ in the class group), and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$.
Let $x_1,dots,x_d$ be an integral basis for $mathcal{O}_K$, and let $alpha=r_1x_1+cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(alpha)$ into one of counting the lattice points $(r_1,dots,r_d)inmathbb{Z}^d$. This is explained in considerably more detail here.
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
$endgroup$
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
add a comment |
$begingroup$
Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $mathfrak{p}lhdmathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[mathfrak{p}]$, then we can find an inverse say $[mathfrak{q}]$ in the class group, and hence $mathfrak{p}mathfrak{q}=(alpha)$ is principal with $alphainmathfrak{q}$ and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$. And if $alphainmathfrak{q}$, and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$, then $mathfrak{p}=(alpha)mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(alpha)$, where $alphainmathfrak{q}$ (and $[mathfrak{q}]=X^{-1}$ in the class group), and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$.
Let $x_1,dots,x_d$ be an integral basis for $mathcal{O}_K$, and let $alpha=r_1x_1+cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(alpha)$ into one of counting the lattice points $(r_1,dots,r_d)inmathbb{Z}^d$. This is explained in considerably more detail here.
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
$endgroup$
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
add a comment |
$begingroup$
Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $mathfrak{p}lhdmathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[mathfrak{p}]$, then we can find an inverse say $[mathfrak{q}]$ in the class group, and hence $mathfrak{p}mathfrak{q}=(alpha)$ is principal with $alphainmathfrak{q}$ and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$. And if $alphainmathfrak{q}$, and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$, then $mathfrak{p}=(alpha)mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(alpha)$, where $alphainmathfrak{q}$ (and $[mathfrak{q}]=X^{-1}$ in the class group), and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$.
Let $x_1,dots,x_d$ be an integral basis for $mathcal{O}_K$, and let $alpha=r_1x_1+cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(alpha)$ into one of counting the lattice points $(r_1,dots,r_d)inmathbb{Z}^d$. This is explained in considerably more detail here.
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
$endgroup$
Not really. There are upper bounds on the number of ideals of a given norm in a given ideal class, but I don't believe that the expression you're looking for exists.
If you take an ideal $mathfrak{p}lhdmathcal{O}_K$ of norm bounded by $n$ and consider its image in the class group $[mathfrak{p}]$, then we can find an inverse say $[mathfrak{q}]$ in the class group, and hence $mathfrak{p}mathfrak{q}=(alpha)$ is principal with $alphainmathfrak{q}$ and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$. And if $alphainmathfrak{q}$, and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$, then $mathfrak{p}=(alpha)mathfrak{q}^{-1}$ is an ideal of norm bounded by $n$.
So the number of ideals of norm bounded by $n$ in a given class, say $X$, is equal to the number of principal ideals $(alpha)$, where $alphainmathfrak{q}$ (and $[mathfrak{q}]=X^{-1}$ in the class group), and $|operatorname{Nm}(alpha)|le noperatorname{Nm}(mathfrak{q})$.
Let $x_1,dots,x_d$ be an integral basis for $mathcal{O}_K$, and let $alpha=r_1x_1+cdots+r_dx_d$. Then we can try and turn the problem of finding principal ideals $(alpha)$ into one of counting the lattice points $(r_1,dots,r_d)inmathbb{Z}^d$. This is explained in considerably more detail here.
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
edited Dec 18 '18 at 9:27
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
answered Jun 4 '13 at 10:26
Warren MooreWarren Moore
3,2731128
3,2731128
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
add a comment |
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
The link appears broken.
$endgroup$
– rabota
Dec 17 '18 at 20:42
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
$begingroup$
@barto thank you for pointing that out, I’ve updated the link to one that works.
$endgroup$
– Warren Moore
Dec 18 '18 at 9:29
add a comment |
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