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The following is a theorem I am trying to understand.

If $n$ is a positive integer and $m=2*3^{n-1}$
We know that $t^m+t^{m/2}+1$ is irreducible over GF(2).

I am looking at the case when $n=2, m=6$: so $f(t)=t^6+t^3+1$

Q1. What are the basic features of the corresponding extension field $F[\alpha]$ of F?

Since we are working in $GF[2]$, $F[\alpha]$ has $(2^6)$ elements. But obviously $64-1$ ($F[\alpha ] \setminus \{0\}$) is not prime, so can't use Lagrange's theorem about primitive to calculate the order of $F[\alpha]$. How do I go about calculating that?

Q2. How do I expres $(1+\alpha)^{-1}$ as an F-linear combination of $1,\alpha,\alpha^2,...,\alpha^5$ ?

since $\alpha^6=\alpha^3+1$ here, solve $(1-\alpha)\sum(a_i\alpha^i)=1$

I get: $a_0 + a_5 = 1$
$a_0 + a_1 = 0$
$a_1 + a_2 = 0$
$a_2 + a_3 + a_5 = 0$
$a_3 + a_4 = 0$
$a_4 + a_5 = 0$

so $a_3=a_4=a_5=1$
and $a_0=a_1=a_2=0$

Can anyone go through my working and suggest things that can be improved, finished or missing? or even direct me to a book where i can read more about this ?

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    I am reasonably sure you want $F[\alpha] \setminus \{0\}$ and not $F[\alpha]\{0}$...Do you want the latter unrenderable version?2012-05-01
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    In case you are wondering about the theorem itself, [here's a proof](http://math.stackexchange.com/a/70159/11619) I wrote as an answer to another question on this site. That is an exercise in *Finite Fields* by Lidl & Niederreiter - the tome of finite fields.2012-05-01

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