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Take the irreducible polynomial $x^3 + x^2 + 1$ over $F_2$ (field of order $2$). Find the splitting field and its roots in that field.


Where I am:

I understand what splitting fields are, and I also know that the splitting field is $F_8$ because $8=2^3$ and we can work in the field $F_2[\theta]/(\theta^3 + \theta^2 + 1)$ in which the polynomial $x^3 + x^2 + 1$ will have linear factors.

What I want to know is: is there are a quick method for finding such roots? Should I simply plug in each element of the splitting field every time given such a question in order to find a linear factor, and work from there? However, I understand that this only works for polynomials up to deg $3$.

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in $F_2[Y]/(Y^3+Y^2+1)$, an obvious root of $P(X)=X^3+X^2+1$ is $Y$.

Another one will be $Y^2$ because in $F_2[Y]$, for any polynomial $Q$, $Q(Y^2)=Q(Y)^2$

Hence $$X^3+X^2+1=(X-Y)(X-Y^2)(X-z)$$

But looking at the coefficient of $X^2$, you get $z=Y^2+Y+1$

Or as $Y^2$ is a root $(Y^2)^2=Y^4$ is too. But $Y^4=Y^2+Y+1$.

$$X^3+X^2+1=(X-Y)(X-Y^2)(X-(Y^2+Y+1))$$

Note that $(Y^2+Y+1)^2=Y^4+Y^2+1=Y$, so there are only 3 different roots.

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  • $\begingroup$ So what is $\theta$ or $Y$ at you? Weare just working with some abstract numbers. We actualy don't know what is $Y$? $\endgroup$ – Aqua Apr 5 '18 at 11:10
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    $\begingroup$ @ChristianF $Y$ is defined by its property that $Y^3+Y^2+1=0$ exactly like in $\mathbb C$, $i$ is defined by $i^2+1=0$. $\endgroup$ – Xoff Apr 5 '18 at 11:50
  • $\begingroup$ Yes, but with $i$ I have at least some geometric insight. What about $Y$? Thanks. $\endgroup$ – Aqua Apr 5 '18 at 11:53
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    $\begingroup$ @ChristianF We are in a finite field $F_{2 ^3}$ here. You can probably see the answer to math.stackexchange.com/questions/124695/… $\endgroup$ – Xoff Apr 5 '18 at 12:08
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Clearly $\theta$ (or rather, its image in $F_2[\theta]/(\theta^3+\theta^2+1)$) will be one such root, and you can find the others by taking the Galois conjugates of $\theta$: since the Galois group is cyclic of order 3, generated by the Frobenius automorphism $\phi : y \mapsto y^2$, we get that the roots are $\theta, \theta^2$, and $(\theta^2)^2=\theta^4$. So, over $F_8$, the polynomial $x^3+x^2+1$ will factor as $$x^3+x^2+1=(x-\theta)(x-\theta^2)(x-\theta^4)$$ And of course here we could rewrite $\theta^4$ as an $F_2$-linear combination of the basis elements $1,\theta,\theta^2$: $$\theta^4 = 1+\theta+\theta^2$$

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