If $x^3 + x + 1$ is reducible over $F$, then why does it follow that $E\subseteq F$? I'm trying to understand a solution to the following problem: Show that $K = \mathbb Z_2[y,z]/\langle y^4 + y^3 + 1,z^3 + z + 1\rangle$ is a field.
A solution is as follows (the part I don't understand is in bold):
Consider $x^3 + x + 1\in \mathbb Z_2[x]$. This has no zeros over $\mathbb Z_2$ and is thus irreducible. Write $E = \mathbb Z_2[x]/\langle x^3 + x + 1\rangle$.
Now consider $x^4 + x^3 + 1\in \mathbb Z_2[x]$. This has no zeros over $\mathbb Z_2$ and is also not the square of the unique irreducible polynomial of degree $2$ in $\mathbb Z_2[x]$, so it is irreducible over $\mathbb Z_2$. Write $F = \mathbb Z_2[x]/\langle x^4 + x^3 + 1\rangle$.
In order to show that $K$ is a field, we endeavor to show that $x^3 + x + 1$ is irreducible over $F$. If it were reducible over $F$, it would have a root over $F$, and thus we have a field isomorphic to $E$ contained in $F$. Then $4 = [F:\mathbb Z_2] = [F:E][E:\mathbb Z_2] = [F:E] \cdot 3$, which is impossible. Thus, $x^3 + x + 1$ is irreducible over $F$, and $K = F[x]/\langle x^3 + x + 1\rangle$ is a field.
Why do we know that a copy of $E$ is contained in $F$? I know $E = \mathbb Z_2(\alpha)$ where $\alpha = [x]\in \mathbb Z_2[x]/\langle x^3 + x + 1\rangle$, but how do we know that $\alpha\in F$? (Which would imply that $E = \mathbb Z_2(\alpha)\subseteq F$.)
 A: If any cubic polynomial is reducible, then it must split into at least two factors. By a comparison of degrees, it must have a linear factor. 
Hence if $x^3+x+1$ is reducible over $F$, then it has a linear factor, and hence a root in $F$ - say $\alpha$. Then $\mathbb Z_2[\alpha]$ is a subring of $F$ isomorphic to $E$ via the surjective homomorphism (the evaluation homomorphism)
$$\mathbb Z_2[x]\to \mathbb Z_2[\alpha]\\x\mapsto \alpha$$
which has kernel $\langle x^3+x+1\rangle$. So $ \mathbb Z_2[\alpha]\cong E$, and hence is a field. 
A: If $f(x)$ is irreducible over a field, say $\mathbb{Z}_2$ in your case, then $\mathbb{Z}_2(\alpha)$ is isomorphic to $\mathbb{Z}_2(\beta)$ for any roots $\alpha,\beta$ of $f(x)$ in any extension of $\mathbb{Z}_2$. (This fact holds over any field, not just $\mathbb{Z}_2$.) 
So, in the part that you aren't clear on, the polynomial $x^3+x+1$, if assumed to be reducible over $F$, would have to have at least one factor that is linear; i.e. it has a root, say $\beta \in F$. This means that $F$ contains a subfield, namely $\mathbb{Z}_2(\beta) \simeq \mathbb{Z}_2(\alpha) = E.$
