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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?

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The $z^2$ term is $x^4 + x^2 + x$, not $x^4+x^2+1$ as you have written.

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