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How to compute function field of a curve
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What does $barF_q(C)$ (function field of a curve) mean? And, how can I compute it when I know C and $F_q$ (or $barF_q$)? (especially for q=2)
algebraic-geometry function-fields
asked Jul 23 at 19:18
ssss1
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up vote
1
down vote
favorite
What does $barF_q(C)$ (function field of a curve) mean? And, how can I compute it when I know C and $F_q$ (or $barF_q$)? (especially for q=2)
algebraic-geometry function-fields
asked Jul 23 at 19:18
ssss1
61
2
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What does $barF_q(C)$ (function field of a curve) mean? And, how can I compute it when I know C and $F_q$ (or $barF_q$)? (especially for q=2)
algebraic-geometry function-fields
asked Jul 23 at 19:18
ssss1
61
What does $barF_q(C)$ (function field of a curve) mean? And, how can I compute it when I know C and $F_q$ (or $barF_q$)? (especially for q=2)
algebraic-geometry function-fields
asked Jul 23 at 19:18
ssss1
61
asked Jul 23 at 19:18
ssss1
61
asked Jul 23 at 19:18
ssss1
61
asked Jul 23 at 19:18
ssss1
61
61
2
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17
 |Â
show 1 more comment
2
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17
2
2
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17
 |Â
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2
The function field is basically defined by the equation(s) of the curve. If $C$ is a plane curve, given by, say the equation $y^2+xy=x^3+1$, then the function field is $BbbF_2(x)(y)$ where $BbbF_2(x)$ is the field of rational functions of $x$, and we then extend it by adjoining $y$, a solution of the equation $y^2+xy=x^3+1$. The equation is quadratic in $y$, so $BbbF_2(x,y)$ is a degree two extension of $BbbF_2(x)$.
– Jyrki Lahtonen
Jul 23 at 19:31
You need to explain how is your $C$ determined. If it is described just by the zero set of an equation $f(x,y)=0$, the previous comment gives the answer.
– xarles
Jul 23 at 20:42
@ Lahtonen: What is $F_2(x)(y)$ ? Does it mean I just add y to $F_2(x)$ ?
– ssss1
Jul 24 at 6:50
@ xarles:I suppose yes, it is represented by some polynomial $F[X, Y] in F_q[X, Y]$. So, I must be able to compute its zero set. $C$ is a smooth, projective, absolutely irreducible curve defined over $F_q$.
– ssss1
Jul 24 at 8:06
@ Lahtonen: Sorry if my question was somehow odd. My major is computer science and I am familiar with basic definitions of algebra. If I am correct, $F_q$ is made of polynomials; so, how can I add adjoin the "point" y to "a field" to make its extension? It seems weird to me. Must I make a polynomial which y is its solution, and then add this polynomial to $F_q$?
– ssss1
Jul 24 at 8:17