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Suppose I have equation of curve C:

curve 
# => y^2 + (x^2 + x)*y + x
curve.parent()
# => Univariate Polynomial Ring in y over Rational function field in x over Rational Field

and I know that x that is root of another equation:

equation
# => x^3 + x^2 - 2*x - 9/2
equation.parent()
# => Univariate Polynomial Ring in x over Rational Field

is x-coordinate of some point of C.

How to properly get y-coordinate of C in x?

It is not as easy as it seems because of conversion problems.

My workaround is as following:

FF.<z> = NumberField(equation)
P.<x,y> = QQ[]

u = P(curve).subs(x=z)

P.<y> = FF[]
return z, P(u).roots()[0][0]

and it doesn't seem right.

Are there more elegant way of doing it?

Suppose I have equation of curve C:

curve 
# => y^2 + (x^2 + x)*y + x
curve.parent()
# => Univariate Polynomial Ring in y over Rational function field in x over Rational Field

and I know that there is a point with x that is value is a root of another equation: equation (i.e. element of corresponding Number Field):

equation
# => x^3 + x^2 - 2*x - 9/2
equation.parent()
# => Univariate Polynomial Ring in x over Rational Field

is x-coordinate of some point of C.

How to properly get y-coordinate of C in x?

It is not as easy as it seems because of conversion problems.

My workaround is as following:

FF.<z> = NumberField(equation)
P.<x,y> = QQ[]

u = P(curve).subs(x=z)

P.<y> = FF[]
return z, P(u).roots()[0][0]

and it doesn't seem right.

Are there more elegant way of doing it?

Suppose I have equation of curve C:

curve 
# => y^2 + (x^2 + x)*y + x
curve.parent()
# => Univariate Polynomial Ring in y over Rational function field in x over Rational Field

and I know that there is a point with x value is a root of another equation (i.e. element of corresponding Number Field):

equation
# => x^3 + x^2 - 2*x - 9/2
equation.parent()
# => Univariate Polynomial Ring in x over Rational Field

is x-coordinate of some point of C.

How to properly get y-coordinate of C in x?

It is not as easy as it seems because of conversion problems.

My workaround is as following:

FF.<z> = NumberField(equation)
P.<x,y> = QQ[]

u = P(curve).subs(x=z)

P.<y> = FF[]
return z, P(u).roots()[0][0]

and it doesn't seem right.

Are there more elegant way of doing it?

  curve = y^2 + (x^2 + x)*y + x;

Suppose I have equation of curve C:

curve 
# => y^2 + (x^2 + x)*y + x
curve.parent()
# => Univariate Polynomial Ring in y over Rational function field in x over Rational Field

and I know that there is a point with x value is a root of another equation (i.e. element of corresponding Number Field):

equation
# => x^3 + x^2 - 2*x - 9/2
equation.parent()
# => Univariate Polynomial Ring in x over Rational Field

is x-coordinate of some point of C.

How to properly get y-coordinate of C in x?

It is not as easy as it seems because of conversion problems.

My workaround is as following:

FF.<z> = NumberField(equation)
P.<x,y> = QQ[]

u = P(curve).subs(x=z)

P.<y> = FF[]
return z, P(u).roots()[0][0]

and it doesn't seem right.

Are there more elegant way of doing it?

P.S. curve is constructed as follows:

  F = FunctionField(QQ, 'x')
   x = F.gen()
   R.<y> = F[]
 
F[]

  curve = y^2 + (x^2 + x)*y + x;
 
``` 
But it was done in another place so I have no direct access to x and y from above code.