I have an exam coming up on coding theory stuff and I'm stumped on how to find the minimum polynomial. A study guide I was given is here: http://www.csee.umbc.edu/~lomonaco/f14/653/handouts/StudyProbs653.pdf
Problem 3. on that page is what I'm looking at. So we have a primitive element of $GF(2^6)$ $ξ$ such that: $1 + ξ + ξ^6$ = 0. They are trying to find the minimum polynomial $m_{36}(x)$ of $ξ^{36}$
They say the roots of $m_{36}(x)$ consist of $ξ^{36}$ and all its conjugates and lists the roots as $ξ^{36}$, $ξ^{9}$, $ξ^{18}$.
How exactly did they get these roots? They then say $ξ^{36}$ is of degree 3, i.e. $m_{36}(x)$ = $x^3 + a_2x^2 + a_1x + a_0$. Not sure how they got this polynomial either. I checked the log/antilog table at the bottom of the page and it doesn't match up for me. Any tips on understanding this would be great, thanks!
For any primitive element $\xi\in GF(p^n)$, its $n$ conjugates are $\xi^{p^0}, \xi^{p^1}, \dots, \xi^{p^{n-1}}$. Moreover, the minimal polynomial of $\xi$ is just the product $$ (x - \xi^{p^0})(x - \xi^{p^1}) \cdots (x -\xi^{p^{n-1}}) = \prod_{i=0}^{n-1} (x - \xi^{p^i}). $$ Any primitive element has (multiplicative) order $p^n-1$, that is, $\xi^{p^n-1} = 1$, and the nonzero elements of $GF(p^n)$ form a cyclic multiplicative group $\{1 = \xi^{0}, \xi^1, \xi^2, \dots, \xi^{p^n-2}\}$. Hence we can do exponent arithmetic modulo $p^n-1$.
These facts should allow you to do the desired calculations.