Splitting field of $f=x^6+x^5+1$, orders of elements, and minimal polynomials. 
Let $F$ be the splitting field over $\Bbb{Z}_2$ of $f=x^6+x^5+1$, an irreducible polynomial over $\Bbb{Z}_2$.
We know that:
$\bullet$ The Galois group of $F$ over $\Bbb{Z}_2$ is cyclic and is generated by the Frobenius autormorphism.
$\bullet$ The lattice of the subgroups of $Gal(F/\Bbb{Z}_2)$ is isomorphic to that of $\Bbb{Z}_6$. Let's denote the one of degree 2 over $\Bbb{Z}_2$ as $\Bbb{Z}_2(\omega)$ and the one of degree 3 over $\Bbb{Z}_2$ as $\Bbb{Z}_(\beta)$.
$\bullet$ There are four intermediate fields  with orders $|F|=64$, $|\Bbb{Z}_2(\beta)|=8$, $|\Bbb{Z}_2(\omega)|=4$, and $|\Bbb{Z}_2|=2$ as additive groups.
a) For each intermediate field $E$ (including $F$), what are the possible multiplicative orders of the units $\gamma$ for which $E=\Bbb{Z}_2(\gamma)$?
b) What is the minimal polynomial of an element of order $9$?

a) Since fields are also additive subgroups, the orders of the multiplicative groups is just Euler's totient function. So $|F^{\times}|=\phi(63)=(7-1)(3-1)(3)=36$, $|\Bbb{Z}_2(\beta)|=2^2=4$, $|\Bbb{Z}_2(\omega)|=2$, and $|\Bbb{Z}_2|=2$.
Orders of the units must divide the orders of the multiplicative groups, so for the possibilities, we have (let $F=\Bbb{Z}(\alpha)$)
$|\alpha|=2,3,6,4,9$
$|\beta|=2,4$
$|\omega|=2$
$1$ is not an option, because otherwise we would have $\Bbb{Z}_2(\gamma)=\Bbb{Z}_2$.
b) We know that the minimal polynomial of that element (call it $\epsilon$), must divide $x^9-1$.
We also know that $x^9-1=\Psi_1(x)\Psi_3(x)\Psi_9(x)$, where $\Psi_k(x)$ is the $k$th cyclotomic polynomial.
Now we need to find $\Psi_9(x)$ in order to use this.
$x^9-1=(x-1)(x^2+x+1)\Psi_9(x)$
$\implies \Psi_9(x) = \frac{x^9-1}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{(x-1)(x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1)}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1}{x^2+x+1}$.
However, when I tried to do long division, I kept getting a remainder of -2...and I'm not sure why. Doesn't $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$ for sure need to be divisible by $x^2+x+1$? Because we know for sure that $x^9-1$ equals $\Psi_1(x)\Psi_3(x)\Psi_9(x)$, right?
Thank you in advance
 A: $(x^9-1)/(x-1)$ is not $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$, but $x^8+x^7+\cdots +x+1$.
Indeed, 
$$
x^9-1=(x^6+x^3+1)(x^2+x+1)(x-1),\quad \frac{x^9-1}{x-1}=(x^6+x^3+1)(x^2+x+1).
$$
A: For starters, if $E$ is a finite field, then $|E^{\times}|=|E \setminus \{0\}|=|E|-1$. So $|F^{\times}|=63, |\Bbb Z_2(\beta )^{\times}|=7$ and $|\Bbb Z_2(\omega )^{\times}|= 3$. Taking the Euler function of these numbers gives you the number of generators of the multiplicative group (which is cyclic). Now since the multiplicative group of a finite field is cyclic, every divisor (proper or not) occurs as the order of some element: Let $n=|F^{\times}|$, $d$ a divisor of $n$ and $g$ a generator of $F^{\times}$, write $m= \frac nd$. Then $g^m$ has order $d$. Finally, $\gamma$ generates $E$ as a field iff it is not contained in any proper subfield or equivalenly iff its order in $E^{\times}$ doesn't divide the order of the multiplicative group of a proper subfield.  This gives you: $|\omega |=3, |\beta |=7, |\alpha |= 9, 21, 63$.
A: Adding a few points about the minimal polynomial of a ninth root of unity $\epsilon$.
Part A) revealed that the fields of four or eight elements don't have elements of order nine. On the other the field of 64 elements does have ninth roots of unity. As none of the proper subfields can contain $\epsilon$ we know that $F=\mathbb{Z}_2(\epsilon)$. Therefore the minimal polynomial has degree six.
As $\epsilon^3-1\neq0$, it is a zero of the polynomial in Dietrich Burde's answer
$$
\frac{x^9-1}{x^3-1}=x^6+x^3+1.
$$
As this has the correct degree, it must be the minimal polynomial of $\epsilon$.
This is also just the usual (char zero) ninth cyclotomic polynomial that happens to remain irreducible when reduced modulo two.
As an auxiliary check we verify that $\epsilon$ has six conjugates in $F$. These are gotten by iterating the Frobenius automorphism. It permutes the ninth roots of unity as follows:
$$
\epsilon\mapsto\epsilon^2\mapsto\epsilon^4\mapsto\epsilon^8\mapsto\epsilon^{16}=\epsilon^7\mapsto\epsilon^{14}=\epsilon^5\mapsto\epsilon^{10}=\epsilon.
$$
It is worth observing that the above calculation also shows that the residue class of $2$ generates the group of units $\mathbb{Z}_9^*$ of the ring $\mathbb{Z}_9$. All the powers $\epsilon^m$, $\gcd(m,9)=1$ appear in the above list of conjugates. 
For extra credit: Show that the characteristic zero cyclotomic polynomial $\psi_k(x)$, $k$ odd, is irreducible modulo two, if and only if $2$ generates the group $\mathbb{Z}_k^*$. For this to happen it is necessary (but not sufficient) that $k$ is a power of an odd prime.
