Computing the zeta function of curves over finite fields.












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$begingroup$


Given a nonsingular curve $X/mathbb{F}_q$, how can one efficiently compute its zeta function $Z(X/mathbb{F},T)$?



My current strategy is to determine the genus $g$ of $X$, then count the points $N_1,...N_g$, then use that these determine $Z(X/mathbb{F}_q,T)$.



This is doable by hand for small primes/genus, but quickly gets out of hand, and I am wondering if there are more intelligent ways to get $Z(X/mathbb{F}_q,T)$.



For instance, how could one efficiently find $Z(X/mathbb{F}_q,T)$ by hand where $X$ is the nonsingular curve associated to $y^2+y+x^5+x^3+1$ over $mathbb{F_2}$?



Presumably computer algebra packages can solve this problem, but I'm not really familiar with any, so any reference would be much appreciated.






finite-fields zeta-functions computer-algebra-systems function-fields







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    1












    $begingroup$


    Given a nonsingular curve $X/mathbb{F}_q$, how can one efficiently compute its zeta function $Z(X/mathbb{F},T)$?



    My current strategy is to determine the genus $g$ of $X$, then count the points $N_1,...N_g$, then use that these determine $Z(X/mathbb{F}_q,T)$.



    This is doable by hand for small primes/genus, but quickly gets out of hand, and I am wondering if there are more intelligent ways to get $Z(X/mathbb{F}_q,T)$.



    For instance, how could one efficiently find $Z(X/mathbb{F}_q,T)$ by hand where $X$ is the nonsingular curve associated to $y^2+y+x^5+x^3+1$ over $mathbb{F_2}$?



    Presumably computer algebra packages can solve this problem, but I'm not really familiar with any, so any reference would be much appreciated.






    finite-fields zeta-functions computer-algebra-systems function-fields







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Given a nonsingular curve $X/mathbb{F}_q$, how can one efficiently compute its zeta function $Z(X/mathbb{F},T)$?



      My current strategy is to determine the genus $g$ of $X$, then count the points $N_1,...N_g$, then use that these determine $Z(X/mathbb{F}_q,T)$.



      This is doable by hand for small primes/genus, but quickly gets out of hand, and I am wondering if there are more intelligent ways to get $Z(X/mathbb{F}_q,T)$.



      For instance, how could one efficiently find $Z(X/mathbb{F}_q,T)$ by hand where $X$ is the nonsingular curve associated to $y^2+y+x^5+x^3+1$ over $mathbb{F_2}$?



      Presumably computer algebra packages can solve this problem, but I'm not really familiar with any, so any reference would be much appreciated.






      finite-fields zeta-functions computer-algebra-systems function-fields







      share|cite|improve this question









      $endgroup$




      Given a nonsingular curve $X/mathbb{F}_q$, how can one efficiently compute its zeta function $Z(X/mathbb{F},T)$?



      My current strategy is to determine the genus $g$ of $X$, then count the points $N_1,...N_g$, then use that these determine $Z(X/mathbb{F}_q,T)$.



      This is doable by hand for small primes/genus, but quickly gets out of hand, and I am wondering if there are more intelligent ways to get $Z(X/mathbb{F}_q,T)$.



      For instance, how could one efficiently find $Z(X/mathbb{F}_q,T)$ by hand where $X$ is the nonsingular curve associated to $y^2+y+x^5+x^3+1$ over $mathbb{F_2}$?



      Presumably computer algebra packages can solve this problem, but I'm not really familiar with any, so any reference would be much appreciated.






      finite-fields zeta-functions computer-algebra-systems function-fields




      finite-fields zeta-functions computer-algebra-systems function-fields






      share|cite|improve this question













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      asked Jan 16 at 12:26









      user277182user277182

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          $begingroup$

          On magma http://magma.maths.usyd.edu.au/calc/



          F<b> := FiniteField(5^2);
          
 A<x,y> := AffineSpace(F,2);
          
 f := y^2+2*x^3+x-b-1;
          
 C := ProjectiveClosure(Curve(A,f));
          
 IsSingular(C); // SingularPoints(C);
          
 g:= Genus(C);
          
 for n in [1..2*g] do
          
   Cn := BaseChange(C,FiniteField((5^2)^n));
          
   cn := #Points(Cn);
          
   {n,cn};
          
 end for;


          Then (for non-singular projective curve) you want to identify the $alpha_j$ such that $$c_n = q^{n}+1-sum_{j=1}^{2g} alpha_j^n$$



          Let $$zeta(T) = frac{prod_{j=1}^{2g}(1-alpha_j T)}{(1-T)(1-q T)}= exp(sum_{n=1}^infty frac{c_n}{n} T^n), qquad F(T) = sum_{n=1}^{2g} frac{c_n}{n} T^n $$



          Then $$zeta(T) = exp(F(T)+O(T^{2g+1})) =exp( F(T))(1+O(T^{2g+1}))$$



          $$prod_{j=1}^{2g}(1-alpha_j T) = (1-T)(1-qT)exp( F(T))(1+O(T^{2g+1}))$$
          Let $$(1-T)(1-qT) exp(F(T)) = sum_{l=0}^{2g}b_l T^l +O(T^{2g+1})$$
          then
          $$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$






          share|cite|improve this answer











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            1 Answer
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            active

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            1












            $begingroup$

            On magma http://magma.maths.usyd.edu.au/calc/



            F<b> := FiniteField(5^2);
            
 A<x,y> := AffineSpace(F,2);
            
 f := y^2+2*x^3+x-b-1;
            
 C := ProjectiveClosure(Curve(A,f));
            
 IsSingular(C); // SingularPoints(C);
            
 g:= Genus(C);
            
 for n in [1..2*g] do
            
   Cn := BaseChange(C,FiniteField((5^2)^n));
            
   cn := #Points(Cn);
            
   {n,cn};
            
 end for;


            Then (for non-singular projective curve) you want to identify the $alpha_j$ such that $$c_n = q^{n}+1-sum_{j=1}^{2g} alpha_j^n$$



            Let $$zeta(T) = frac{prod_{j=1}^{2g}(1-alpha_j T)}{(1-T)(1-q T)}= exp(sum_{n=1}^infty frac{c_n}{n} T^n), qquad F(T) = sum_{n=1}^{2g} frac{c_n}{n} T^n $$



            Then $$zeta(T) = exp(F(T)+O(T^{2g+1})) =exp( F(T))(1+O(T^{2g+1}))$$



            $$prod_{j=1}^{2g}(1-alpha_j T) = (1-T)(1-qT)exp( F(T))(1+O(T^{2g+1}))$$
            Let $$(1-T)(1-qT) exp(F(T)) = sum_{l=0}^{2g}b_l T^l +O(T^{2g+1})$$
            then
            $$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              On magma http://magma.maths.usyd.edu.au/calc/



              F<b> := FiniteField(5^2);
              
 A<x,y> := AffineSpace(F,2);
              
 f := y^2+2*x^3+x-b-1;
              
 C := ProjectiveClosure(Curve(A,f));
              
 IsSingular(C); // SingularPoints(C);
              
 g:= Genus(C);
              
 for n in [1..2*g] do
              
   Cn := BaseChange(C,FiniteField((5^2)^n));
              
   cn := #Points(Cn);
              
   {n,cn};
              
 end for;


              Then (for non-singular projective curve) you want to identify the $alpha_j$ such that $$c_n = q^{n}+1-sum_{j=1}^{2g} alpha_j^n$$



              Let $$zeta(T) = frac{prod_{j=1}^{2g}(1-alpha_j T)}{(1-T)(1-q T)}= exp(sum_{n=1}^infty frac{c_n}{n} T^n), qquad F(T) = sum_{n=1}^{2g} frac{c_n}{n} T^n $$



              Then $$zeta(T) = exp(F(T)+O(T^{2g+1})) =exp( F(T))(1+O(T^{2g+1}))$$



              $$prod_{j=1}^{2g}(1-alpha_j T) = (1-T)(1-qT)exp( F(T))(1+O(T^{2g+1}))$$
              Let $$(1-T)(1-qT) exp(F(T)) = sum_{l=0}^{2g}b_l T^l +O(T^{2g+1})$$
              then
              $$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                On magma http://magma.maths.usyd.edu.au/calc/



                F<b> := FiniteField(5^2);
                
 A<x,y> := AffineSpace(F,2);
                
 f := y^2+2*x^3+x-b-1;
                
 C := ProjectiveClosure(Curve(A,f));
                
 IsSingular(C); // SingularPoints(C);
                
 g:= Genus(C);
                
 for n in [1..2*g] do
                
   Cn := BaseChange(C,FiniteField((5^2)^n));
                
   cn := #Points(Cn);
                
   {n,cn};
                
 end for;


                Then (for non-singular projective curve) you want to identify the $alpha_j$ such that $$c_n = q^{n}+1-sum_{j=1}^{2g} alpha_j^n$$



                Let $$zeta(T) = frac{prod_{j=1}^{2g}(1-alpha_j T)}{(1-T)(1-q T)}= exp(sum_{n=1}^infty frac{c_n}{n} T^n), qquad F(T) = sum_{n=1}^{2g} frac{c_n}{n} T^n $$



                Then $$zeta(T) = exp(F(T)+O(T^{2g+1})) =exp( F(T))(1+O(T^{2g+1}))$$



                $$prod_{j=1}^{2g}(1-alpha_j T) = (1-T)(1-qT)exp( F(T))(1+O(T^{2g+1}))$$
                Let $$(1-T)(1-qT) exp(F(T)) = sum_{l=0}^{2g}b_l T^l +O(T^{2g+1})$$
                then
                $$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$






                share|cite|improve this answer











                $endgroup$



                On magma http://magma.maths.usyd.edu.au/calc/



                F<b> := FiniteField(5^2);
                
 A<x,y> := AffineSpace(F,2);
                
 f := y^2+2*x^3+x-b-1;
                
 C := ProjectiveClosure(Curve(A,f));
                
 IsSingular(C); // SingularPoints(C);
                
 g:= Genus(C);
                
 for n in [1..2*g] do
                
   Cn := BaseChange(C,FiniteField((5^2)^n));
                
   cn := #Points(Cn);
                
   {n,cn};
                
 end for;


                Then (for non-singular projective curve) you want to identify the $alpha_j$ such that $$c_n = q^{n}+1-sum_{j=1}^{2g} alpha_j^n$$



                Let $$zeta(T) = frac{prod_{j=1}^{2g}(1-alpha_j T)}{(1-T)(1-q T)}= exp(sum_{n=1}^infty frac{c_n}{n} T^n), qquad F(T) = sum_{n=1}^{2g} frac{c_n}{n} T^n $$



                Then $$zeta(T) = exp(F(T)+O(T^{2g+1})) =exp( F(T))(1+O(T^{2g+1}))$$



                $$prod_{j=1}^{2g}(1-alpha_j T) = (1-T)(1-qT)exp( F(T))(1+O(T^{2g+1}))$$
                Let $$(1-T)(1-qT) exp(F(T)) = sum_{l=0}^{2g}b_l T^l +O(T^{2g+1})$$
                then
                $$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Jan 16 at 13:16

























                answered Jan 16 at 13:00









                reunsreuns

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