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second degree irreducible polynomial over extension field

asked 2016-07-23 05:51:30 -0500

santoshi gravatar image

the given field is f_2^3 having third degree irreducible polynomial x^3+x+1 . if theta is a root of irreducible polynomial the field elements are {0,1,theta,theta^2,theta^3,theta^4,theta^5,theta^6}.

my question is i want second degree irreducible polynomial over f_{2^3}^2.

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I don't follow what you are asking for. Could you please clarify?

jaebond gravatar imagejaebond ( 2016-07-25 22:22:10 -0500 )edit

I presume the question is about how to build a quadratic extension of GF(8) explicitly.

Dima gravatar imageDima ( 2016-07-28 03:08:22 -0500 )edit

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answered 2017-04-06 15:02:16 -0500

dan_fulea gravatar image

A possible construction in the sens of the posting is as follows:

sage: R.<x> = PolynomialRing( GF(2) )
sage: F8.<t> = GF( 2**3, modulus=x^3+x+1 )
sage: F8
Finite Field in t of size 2^3
sage: RF8 = PolynomialRing( F8 )

We have so far $\mathbb{F}_8$, realized so that $t$ is a generator of the multiplicative group $\mathbb{F}_8^\times$ with $7=(8-1)$ elements. Starting from it, we can extend, by explicitly introducing also a root of order nine (but not $3$) of unity. So that we have one of order $63=(8-1)(8+1)=64-1$.

So we extend:

sage: RF8.<y> = PolynomialRing( F8 )
sage: factor( y^9-1 )
(y + 1) * (y^2 + y + 1) * (y^2 + t*y + 1) * (y^2 + t^2*y + 1) * (y^2 + (t^2 + t)*y + 1)
sage: F64.<u> = F8.extension( y^2 + t*y + 1 )  
sage: F64.is_field()
True
sage: F64.order()
64
sage: u^3, u^9
((t^2 + 1)*u + t, 1)
sage: g = u*t
sage: g^7, g^9, g^63
(t*u + t^2 + 1, t^2, 1)

So $u$ has multiplicative order $9$, and $g=ut$ correpsondingly $63$. The modulus for the constucted field, so that $g$ is a root of it is:

sage: for f,_ in factor( x^63-1 ):
    if f(g) == 0:
        print f
....:         
x^6 + x^4 + x^3 + x + 1
sage: g^6 + g^4 + g^3 + g + 1
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Asked: 2016-07-23 05:51:30 -0500

Seen: 84 times

Last updated: Apr 06