How to find one element of residue field as a vector over base field

[chandra chowdhury]

Dear all,

   I have written the following code:

K.<x> = FunctionField(GF(2)); 

R.<y> = K[]

L.<y> = K.extension(y^2 +y+x+1/x)

print L.places(2)

p = L.places(2)[1]

print p

G=p.divisor()

LG=G.basis_function_space()

print LG

Output is as follows:

[Place (x^2 + x + 1, x*y + 1), Place (x^2 + x + 1, x*y + x + 1)]
Place (x^2 + x + 1, x*y + x + 1)
[1, (x/(x^2 + x + 1))*y + 1/(x^2 + x + 1)]

What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by

(x^2 + x + 1, x*y + 1). 

What is the value of $\frac{xy}{(x^2 + x + 1) } + 
\frac{1}{x^2 + x + 1}+$ Place $(x^2 + x + 1, x y + 1)$? 

It is an element of residue field which is isomorphic to 
$\mathbb{F}_{2^2}$. Since $\mathbb{F}_{2^2}$ is isomorphic 
to $\mathbb{F}^2_{2}$ as a vector space, 
I want value in $\mathbb{F}^2_{2}$.

[Kwankyu]

Hi Chandra,

What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by

(x^2 + x + 1, x*y + 1). 

No. Place (x^2 + x + 1, x*y + 1) is the unique place of the function field 

at which both functions x^2 + x +1, x*y + 1 vanish.

 

What is the value of $\frac{xy}{(x^2 + x + 1) } + 
\frac{1}{x^2 + x + 1}+$ Place $(x^2 + x + 1, x y + 1)$? 

You cannot add an element of the function field with a place.

 

It is an element of residue field which is isomorphic to 
$\mathbb{F}_{2^2}$. Since $\mathbb{F}_{2^2}$ is isomorphic 
to $\mathbb{F}^2_{2}$ as a vector space, 
I want value in $\mathbb{F}^2_{2}$.

vector(a)

or you can use the maps returned by

k.vector_space(map=True)

if k is the residue field.

 

[Santanu Sarkar]

On Wed, 15 May 2019 at 17:03, Kwankyu <ekwa...@gmail.com> wrote:

Hi Chandra,

What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by

(x^2 + x + 1, x*y + 1). 

No. Place (x^2 + x + 1, x*y + 1) is the unique place of the function field 

at which both functions x^2 + x +1, x*y + 1 vanish.

Thank you for your response. We know that a place is the unique maximal ideal of a local (valuation) ring obtained from the valuation map, which is well known to be a principle ideal.

 So, there will be a single generator for a place. But here it is represented by two polynomials. We didn't get what it means. Can we find the corresponding valuation ring, valuation map

ant the generator for the place?

 

 

What is the value of $\frac{xy}{(x^2 + x + 1) } + 
\frac{1}{x^2 + x + 1}+$ Place $(x^2 + x + 1, x y + 1)$? 

You cannot add an element of the function field with a place.

Actually by this we meant the element modulo the place ( a maximum ideal).

 

 

It is an element of residue field which is isomorphic to 
$\mathbb{F}_{2^2}$. Since $\mathbb{F}_{2^2}$ is isomorphic 
to $\mathbb{F}^2_{2}$ as a vector space, 
I want value in $\mathbb{F}^2_{2}$.

vector(a)

or you can use the maps returned by

k.vector_space(map=True)

if k is the residue field.

 

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[Kwankyu]

On Wednesday, May 15, 2019 at 10:08:05 PM UTC+9, Santanu wrote:

On Wed, 15 May 2019 at 17:03, Kwankyu <ekwa...@gmail.com> wrote:

Hi Chandra,

What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by

(x^2 + x + 1, x*y + 1). 

No. Place (x^2 + x + 1, x*y + 1) is the unique place of the function field 

at which both functions x^2 + x +1, x*y + 1 vanish.

Thank you for your response. We know that a place is the unique maximal ideal of a local (valuation) ring obtained from the valuation map, which is well known to be a principle ideal. So, there will be a single generator for a place. But here it is represented by two polynomials. We didn't get what it means.

A place generally cannot be represented by a single polynomial. The generator of the principal ideal does not specify the local ring itself. 

Can we find the corresponding valuation ring, valuation map

ant the generator for the place?

Use p.valuation_ring() for the valuation ring, which is only a facade for the ring. Use p.local_uniformizer() for the generator. 

Look at this example:

sage: p1,p2 = L(x^2+x+1).zeros()

sage: p1.local_uniformizer()

x^2 + x + 1

sage: p2.local_uniformizer()

[Kwankyu]

On Wednesday, May 15, 2019 at 10:08:05 PM UTC+9, Santanu wrote:

What is the value of $\frac{xy}{(x^2 + x + 1) } + 
\frac{1}{x^2 + x + 1}+$ Place $(x^2 + x + 1, x y + 1)$? 

You cannot add an element of the function field with a place.

Actually by this we meant the element modulo the place ( a maximum ideal).

Use R.residue_field() for the valuation ring R.