Computing the zeta function of curves over finite fields.
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
asked Jan 16 at 12:26
user277182user277182
518212
518212
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1 Answer
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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
$$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$
$endgroup$
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1 Answer
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1 Answer
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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
$$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$
$endgroup$
add a comment |
$begingroup$
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
$$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$
$endgroup$
add a comment |
$begingroup$
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
$$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$
$endgroup$
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
$$ prod_{j=1}^{2g}(1-alpha_j T) = sum_{l=0}^{2g}b_l T^l$$
edited Jan 16 at 13:16
answered Jan 16 at 13:00
reunsreuns
20.5k21353
20.5k21353
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