I must find the irreducible factors of $f(x) = x^8 - x$ in $Z/2Z[x]$ and that's what I did:
$f(x) = x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$
but of course the correct solution is:
$f(x) = x(x-1)(x^3 + x^2 + 1)(x^3 + x + 1)$
I do not understand how I can reach the correct solution by logical reasoning (without going on attempts). The solution of the exercise says:
1) "The irreducible factors are all the irreducible polynomials of degree 1 and 3 alone" Why?
2) "You find exactly $6/3 = 2$ irreducible polynomials of degree 3". Does this happens because the 6th degree polynomial I found is not reducible (since the only ones have degree 1 or 3) and it has no zeros?
I note here $GF(p^n)$ the finite field of characteristic $p$ prime having $p^n$ elements.
The elements of $GF(2^3)$ are the roots of $f$ in a splitting field. Also note that the multiplicative group of $GF(2^3)$ is of order $7$ which is a prime. Hence all elements of the multiplicative group of $GF(2^3)$ except the identity $1$ are of order $7$.
From this it follows that an irreducible factor of $g(x) = x^7-1$ can only be of degree $1$, which is the case for the identity or of degree $3$. If an irreducible polynomial of $g$ would be of degree $2$, $GF(2^3)$ would have a subfield of order $4$ and at least an element of order $3$ which is not the case as we've seen above.
So we're left to find the irreducible polynomials (over $GF(2)$) of degree $3$. This is quite easy. The only polynomials of degree $3$ are: $$x^3, x^3+1, x^3+x, x^3+x+1, x^3+x^2+1$$
Only the two last ones are irreducible.
We're done.