I am given a primitive element $\alpha$ in the Galoisfield $F_{2^6}$. The question is to find the mimimal polynomial of $\alpha^7$. How to I find this?
My thoughts so far: $$ \alpha^7 \rightarrow \alpha^2\\ \Rightarrow \alpha^7+ \alpha^2 = 0 $$ But it seems to simple to me...
$x^9-1 = (x^3-1)(x^6+x^3+1)=(x-1)(x^2+x+1)(x^6+x^3+1)$.
$\beta = \alpha^7$ is an element of order $9$ and thus a root of $x^9-1$. It cannot be a root of $x^3-1$ and so must be a root of $x^6+x^3+1$. So, the minimal polynomial of $\beta$ is $x^6+x^3+1$ or a divisor thereof. Now, the roots of the minimal polynomial of $\beta$ are $\beta$ and all its conjugates, that is, $\beta, \beta^2, \beta^{2^2}, \beta^{2^3}, \cdots$ are the roots of the minimal polynomial of $\beta$. (Also, the minimal polynomial of $\beta$ is irreducible over $\mathbb F_2$). How many (distinct) conjugates are there? What does that tell you about the degree of the minimal polynomial of $\beta$? Can you deduce the minimal polynomial of $\beta$ from all this information?