How would I find the minimum polynomial $m_4(x)$ of $\xi^4$, where $\xi= x \pmod{x^6+x+1}$ in $GF(2^6)$?
This is what I think I need to do so far:
Let $\alpha=\xi^4$. Then I would find the conjugacy class of $\alpha$, consisting of $\alpha^{2^0}, \alpha^{2^1}, \alpha^{2^2}, \ldots$. When If multiply these out using the log/antilog table for $\xi^n$, I get $\xi^4, \xi^8, \xi^{16}, \xi^{32}, \xi^{64}=\xi, \xi^{128} = \xi^2, \xi^{256}=\xi^4$. So the conjugacy class of $\alpha$ has 6 distinct elements because $\alpha^{2^6} = \alpha$.
Then $m_4(x)=(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)(x-\alpha^{16})(x-\alpha^{32})$.
Then to find the minimal polynomial, I should be able to multiply these out.
Or because I know that $m_4(x)$ is monic and of degree 6, and that $m_4(\alpha)=0$, then I should be able to write out:
$$m_4(x) = x^6 + b_5 x^5 + b_4 x^4 + b_3 x^3 + b_2 x^2 +b_1 x + b_0$$
And plug in $\alpha$ for $x$.
This gives me:
$$0 = \alpha^6 + b_5 \alpha^5 + b_4 \alpha^4 + b_3 \alpha^3 + b_2 \alpha^2 + b_1 \alpha + b_0$$
Or, equivalently, because working with $GF(2)$:
$$\alpha^6 = b_5 \alpha^5 + b_4 \alpha^4 + b_3 \alpha^3 + b_2 \alpha^2 + b_1 \alpha^1 + b_0$$
To solve this, I substitute $\xi^4$ back in place of $\alpha$:
$$\xi^{24} = b_5 \xi^{20} + b_4 \xi^{16} + b_3 \xi^{12} + b_2 \xi^8 + b_1 \xi^4 + b_0$$
And use the log/antilog table to find the values of $\xi^n$:
$$\begin{bmatrix}0\\1\\0\\0\\0\\1\end{bmatrix} = b_5\begin{bmatrix}1\\1\\1\\1\\0\\0\end{bmatrix} + b_4\begin{bmatrix}0\\1\\0\\0\\1\\1\end{bmatrix} + b_3\begin{bmatrix}0\\0\\0\\1\\0\\1\end{bmatrix} + b_2\begin{bmatrix}0\\0\\1\\1\\0\\0\end{bmatrix} + b_1\begin{bmatrix}0\\1\\0\\0\\0\\0\end{bmatrix} + b_0\begin{bmatrix}0\\0\\0\\0\\0\\1\end{bmatrix}$$
$$\begin{bmatrix}0\\1\\0\\0\\0\\1\end{bmatrix} = \begin{bmatrix} 1&0&0 & 0&0&0\\ 1&1&0 & 0&1&0\\ 1&0&0 & 1&0&0\\ 1&0&1 & 1&0&0\\ 0&1&0 & 0&0&0\\ 0&1&1 & 0&0&1\\ \end{bmatrix} \begin{bmatrix}b_5\\b_5\\b_3\\b_2\\b_1\\b_0\end{bmatrix}$$
Solving for this system of equations, I get the polynomial $m_4(x)=x^6+x+1$
I know this primitive because it's in Peterson's table as a primitive polynomial, but I'm not sure how to tell if it is the minimal polynomial of $\xi^4$
Is this correct/on track for how to find the minimal polynomial, or am I just completely lost?