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Incorrect result for x in ideal. Sage reports false, but I provide a correct factorization

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

Incorrect result for x in ideal. Sage reports false, but I provide a correct factorization

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but when I get errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

Incorrect result for x in ideal. Sage reports false, but I provide a correct factorization

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but when I get errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

Incorrect result for x in ideal. Sage reports false, but I provide a correct factorizationHow to test element is in multivariate function field's ideal

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but when I get errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

How to test element is in multivariate function field's ideal

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but I get singular errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

How to test element is in multivariate function field's ideal

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[x, y] K[V] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but I get singular errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks

How to test element is in multivariate function field's ideal

I'm trying to calculate the valuation of a function in a the function field of a coordinate ring $K(V) = { f / g : f, g \in K[V] }$.

My first attempt is to construct the coordinate ring $K[V] = K[x, y] / \langle C(x, y) \rangle$

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: S
Quotient of Multivariate Polynomial Ring in x, y over Ring of integers modulo 11 by the ideal (10*x^3 + y^2 + 7*x)

Now I want to see if a function $f = y - 2x$ lies in the ideal $I = \langle u \rangle$ where $u = x - 2$.

sage: I = S.ideal(x - 2)
sage: S(y - 2*x) in I
False

But sage is wrong. $y - 2x \in \langle x - 2 \rangle$ as shown by the following code:

sage: f1 = S(y - 2*x)
sage: f1
9*xbar + ybar
sage: f2 = S(
....:     (x - 2) * (
....:         (x - 2)^2*(y + 4) - 5*(x - 2)*(y + 4) - 2*((x - 2)^3 - 5*(x - 2)^2 + 5*(x - 2)
....:     ))) / S((y + 4)^2)
sage: f2
9*xbar + ybar
sage: bool(f1 == f2)
True

As plainly visible, f2 has the factor $(x - 2)$.

How can I calculate this in sage without having to factor the polynomial myself?

I assume this is because we need an actual function field in sage, but I get singular errors when I attempt to turn S into a fraction field.

sage: K.<x, y> = Integers(11)[]
sage: S = K.quotient(y^2 - x^3 - 4*x)
sage: R = FractionField(S)
# ...    
RuntimeError: error in Singular function call 'primdecSY':
ASSUME failed:   ASSUME(0, hasFieldCoefficient(basering) );
error occurred in or before primdec.lib::primdecSY_i line 5983: `  return (attrib(rng,"ring_cf")==0);`
leaving primdec.lib::primdecSY_i (5983)

Is there a way to construct local rings and maximal ideals? Or a way to calculate valuations (order of vanishing for poles and zeros) on elliptic curves?

Thanks