1

Construct a finite field of order $2^6$.

My solution: We just need to find a degree six polynomial over $\mathbb{F}_2$. The cyclotomic polynomial $\Phi_7(x)=1+x+x^2+x^3+x^4+x^5+x^6$ is monic and irreducible. Hence the quotient $\mathbb{F}_2[x]/\langle \Phi_7(x)\rangle$ is a field containing $2^6$ elements. Moreover, the elements satisfy $a\alpha+b:\Phi_7(\alpha)=0$. Is this right?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
cho221
  • 87
  • 2
    What is your $a$ and $b$? If your $a,b$ are in $\Bbb{F}2$, then your conclusion is wrong, since $\Bbb{F}{16}$ is a $4$-dimensional vector space over $\Bbb{F}_2$. – Zerox Jan 17 '23 at 14:21
  • 4
    You need to find an irreducible polynomial $\in \Bbb{F}_2[x]$ of degree $6$. No $\Phi_7$ doesn't work as its splits in $\Bbb{F}_8$. – reuns Jan 17 '23 at 14:27
  • @reuns: I see the factorization in the answer below, but I'm wondering how you knew that $\Phi_7$ splits in $\mathbb F_8$ -- is there a general rule that you used? – J. W. Tanner Jan 17 '23 at 15:40
  • 3
    @J.W.Tanner There is a rule, see for example this post. – Dietrich Burde Jan 17 '23 at 15:44

1 Answers1

4

Is this right?

Nope.

$1+x+x^2+x^3+x^4+x^5+x^6=(1 + x + x^3) (1 + x^2 + x^3)$ so it is not irreducible over $F_2$. You'll need to find an actual irreducible. One which would work is $x^6 + x + 1$.

If you're up for the challenge of fully constructing it by finding a primitive element and working out the multiplication, it might be worthwhile. 64 is big enough for a nontrivial challenge, but not impossible.

rschwieb
  • 160,592
  • This is the point, +1. If memory serves, $x^6+x+1$ is actually a primitive polynomial. Meaning that its roots have multiplicative order $2^6-1$. – Jyrki Lahtonen Jan 18 '23 at 05:59