How to index all elements of a finite field?

finite-field

MKS
37 ●1 ●3 ●8

I would like to assign an integer corresponding to element of a finite field $GF(p^m)$ , where $p^m\in[ {13^2,3^5, 131,137,139,251}]$
MWE:

 F.<x> = GF(3^5, impl='givaro')

THe elements of GF(3^5) are 0,1,2,x,x^2 etc, we would like to indexing each element such as $0-->0, 1-->1,2-->2,x-->3, x^2--4$ etc. Not only that, if I call any element for example if I call x^2 it should rerun 4 and conversely.

This process should work for the field order prime $p^m, m=1$ also. How can I do this?

Comments

What is the logic behind this indexing? The more usual way is to associate $f(x)$ to $f(p) \in \mathbb{N}$ , so $x^2$ corresponds to $9$ and $x+1$ corresponds to $4$ .

rburing ( 4 years ago )

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answered 4 years ago

eric_g
8564 ●10 ●62 ●168 https://luth.obspm.fr/...

updated 4 years ago

If you don't have any privileged choice for the ordering of the elements, simply use list(F) to generate a list containing all elements of the finite field. For instance:

sage: F.<x> = GF(3^5, impl='givaro')                                                                
sage: LF = list(F)

The first 10 elements of the list LF are

sage: LF[:10]                                                                                       
[0,
 x,
 x^2,
 x^3,
 x^4,
 x + 2,
 x^2 + 2*x,
 x^3 + 2*x^2,
 x^4 + 2*x^3,
 2*x^4 + x + 2]

Accessing to element of index 5:

sage: LF[5]                                                                                         
x + 2

The reverse operation:

sage: LF.index(x + 2)                                                                               
5

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answered 4 years ago

rburing

10964 ●4 ●80 ●219 https://www.rburing.nl/

Using the more standard indexing:

sage: F.<x> = GF(3^5, impl='givaro')
sage: F.fetch_int(4)
x + 1
sage: (x+1).integer_representation()
4

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