Number of ideals of a given norm












5












$begingroup$


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







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jun 4 '13 at 7:33









      user29253user29253

      1463




      1463






















          1 Answer
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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          4












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          4












          4








          4





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 9:27

























          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


















          • $begingroup$
            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


















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