Q:

Algebra Splitting fields

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 $3$. 

A:

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 $1,\theta,\theta^2$: 
$$\theta^4 = 1+\theta+\theta^2$$

A:

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. 

