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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 ?

MathPowerUser
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2 Answers2

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Any further factorization would give a root $\zeta \neq 1$ of $x^7 - 1$ in $\mathbb{F}_4$. This would be an element of order $7$ in the multiplicative group $\mathbb{F}_4^*$ of order $3$, which clearly cannot exist. So this is also the factorization over $\mathbb{F}_4$.

Arthur
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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 $\mathbb F_2(\alpha) \subseteq \mathbb F_4$.

Brent Kerby
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  • So the task is to enter the new elements a, a+1 to find new roots and to split as new linear factors ? Ist this correct ? – MathPowerUser Feb 25 '15 at 23:05
  • Well, if there were additional roots, then yes, that would be a way to find them. However, since the argument reached a contradiction, these roots cannot exist; i.e., the original factors remain irreducible, so the factorization over $\mathbb F_4$ is the same as over $\mathbb F_2$. – Brent Kerby Feb 25 '15 at 23:09