TITLE: Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
QUESTION [9 upvotes]: Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$.
Ok so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots.  So then I tried looking at the quotient field $\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't.  Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\mathbb{Z}_2$?
Edit: Ok it turns out I miscalculated in my quotient field, $\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots.

REPLY [5 votes]: Two good answers already here, but I wanted to emphasize the usefulness of the Frobenius endomorphism. Let $f(X) = X^3 + X + 1$. If I let $\alpha$ denote the image of $X$ in $k = \mathbb F_2[X]/(f(X))$ then applying Frobenius to $0 = f(\alpha)$ gives
\[
0 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1.
\]
Hence $\alpha^2$ is also a root of $f$, and $\alpha^2 \neq \alpha$ because $\alpha \neq 0, 1$. Since $k$ contains two roots of the cubic $f$, it contains three.
To bring in slightly more technology: extensions of finite fields are Galois and in particular normal, and this implies that if you adjoin one root of an irreducible polynomial over a finite field then you've actually adjoined all of them. This is, of course, not true in general!