In a finite field $F_q$, I've read that one can get the minimum polynomial $f(z)$ of an element $\beta \in F_q$ using this formula:
$$f(z) = (z-\beta)(z-\beta^2)(z-\beta^4)(z-\beta^8)...$$
I'm trying to prove this to myself in the case of the finite field $F_8 = F_2[x]/(x^3 + x + 1)$
The elements of this field are:
$$0, 1, x, x^2, x^3\equiv (x+1), x^4\equiv (x^2+x), x^5\equiv (x^2+x+1), x^6\equiv (x^2 +1)$$
Let's say I want to find the minimal polynomial of $x$, which should also be the minimal polynomial for its conjugates $x^2$ and $x^4 \equiv x^2+x$. Since $x^8\equiv x$, I can ignore the higher powers.
Since coefficients are in $F_2$, $+1$ and $-1$ are considered equivalent:
$f(z) = (z+x)(z+x^2)(z+x^4)$
$f(z) = z^3 + z^2(x^4 + x^2 + 1) + z(x^6 + x^5 + x^3) + x^7$
Using the conversions I've listed above, I can replace all the powers of $x^3$ or higher with lower-degree polynomials:
$f(z) = z^3 + z^2((x^2+x) + x^2 + 1) + z((x^2+1) + (x^2+x+1) + (x+1)) + 1$
$f(z) = z^3 + z^2(x + 1) + z + 1$
This isn't what I'm expecting... I would have expected the $z^2$ term to go to zero. Am I making an algebra mistake here?