I have to factorize the polynomial ($x^8-x$) in $\mathbb{F}_{2}$. I found the following factorization: ($x^8-x$) = x*($x+1$) * ($x^3$+x+1)* ($x^3$+$x^2$+1).
But now I change to the $\mathbb{F}_{4}$. Does the factorization change ?
I have to factorize the polynomial ($x^8-x$) in $\mathbb{F}_{2}$. I found the following factorization: ($x^8-x$) = x*($x+1$) * ($x^3$+x+1)* ($x^3$+$x^2$+1).
But now I change to the $\mathbb{F}_{4}$. Does the factorization change ?
Of course the factorization written is still valid; the question is whether the polynomials $x^3+x+1$ and $x^3+x^2+1$ remain irreducible over $\mathbb F_4$, or whether they split up into smaller factors.
So suppose $x^3+x+1$ factors over $\Bbb F_4$. Then it must have a linear factor, so there would be a root $\alpha\in\Bbb F_4$. On the other hand, since $x^3+x+1$ is irreducible over $\mathbb F_2$, adjoining a root will generate an extension field of degree 3, i.e. $\mathbb F_2(\alpha) \cong \mathbb F_8$, which is a contradiction since $\mathbb F_2(\alpha) \subseteq \mathbb F_4$.