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I've got a comprehension question:

Be the polynomial: $f(x) = x^3 + x + 1$ over $\mathbb{F}_2[X]$

I know, that it's splitting field is $\mathbb{F}_8$, but that means, $f$ splits into linear factors in $\mathbb{F}_8$, but I don't know the way how to determine these linear factors. It seems, that $\alpha = x$ is a root of $f$ in $\mathbb{F}_8$ but don't see how that helps me there, since the linear factor would be $(x+\alpha) = (x+x) = 0$ so division is no option. I'm sure I'm missing some understanding of polynomials over finite fields here. Anyone got any clues, how these linear factors could be found or how I'm supposed to determine the correct splitting field of a polynomial over $F_{p^n}$?

Andreas H.
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    A. Pongrácz gave you the solution (comes with the Galois theory of finite fields). If you want to gain a bit of familiarity with the arithmetic of $\Bbb{F}8$ and $\Bbb{F}{16}$ you can take a peek at this Q&A I prepared for referrals like this. – Jyrki Lahtonen Jul 21 '18 at 20:19
  • @JyrkiLahtonen: Thanks for that post. That way I was able to follow A.Pongrácz's answer, sort of. – Andreas H. Jul 22 '18 at 13:55

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Let $\alpha$ be a root. Then the splitting field is $\mathbb{F}_2(\alpha)\cong \mathbb{F}_8$. It is well-known that given a finite field with characteristic $p$ (a prime), the mapping $x\mapsto x^p$ is an automorphism. This is called the Frobenius automorphism. Clearly, the automorphic image of a root is again a root. As we are in characteristic 2, the polynomial must be $(x-\alpha)(x-\alpha^2)(x-\alpha^4)$. Note that $\alpha^8=\alpha$, so the Frobenius automorphism does not find another root. This way, you can find all roots of an irreducible polynomial over the prime field.

It is easy to represent powers of $\alpha$ in the form $a+b\alpha+c\alpha^2$ with $a,b,c\in \mathbb{F}_2$, is you prefer that form.

A. Pongrácz
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