$\mathbb{F_2}(\alpha)/\mathbb{F_2}$ is a splitting field of $f(x)$ over $\mathbb{F_2}$ Let $\alpha$ be a zero of $f(x)=x^3+x+1 \in \mathbb{F_2}[x]$. 

Show that $\mathbb{F_2}(\alpha)/\mathbb{F_2}$ is a splitting field of $f(x)$ over $\mathbb{F_2}$

So we need to show that  $\mathbb{F_2}(\alpha)/\mathbb{F_2}$ is the smallest field over which $f(x)$ is reducible.
Does this mean that there are no field extensions between $\mathbb{F_2}(\alpha)$ and $\mathbb{F_2}$... i.e. $\alpha$ is the only root?
 A: Let $E$ be a splitting field of $f$ over $\mathbb{F}_2$ containing $\alpha$. Clearly, $\mathbb{F}_2 (\alpha) \subseteq E$. But $$X^3 + X + 1 = (X-\alpha)(X-\alpha ^2)(X-(\alpha  + \alpha ^2))$$
Then $E \subseteq \mathbb{F}_2(\alpha)$ and so $\mathbb{F}_2 (\alpha)= E$.
A: It does not mean $\alpha$ is the only root, since this is a finite field, irreducible polynomials are separable, so they have all distinct roots. In fact, there are indeed $3$ roots to the polynomial, all of them are in $\Bbb F_2(\alpha)$. The abstract way to see this is to note that since $[\Bbb F_2(\alpha):\Bbb F_2]=3$ we have that the multiplicative group of the extension has size $2^3-1=7$ is cyclic (even if you don't know the theorem that all such extensions have cyclic unit groups, it's a finite group of prime order).
But then all elements of $\Bbb F_2(\alpha)$ satisfy $x^8-x=0$. This polynomial has distinct roots, since it's derivative is $-1$ which is coprime to $x^8-x$. But since $F[x]$ is a UFD when $F$ is a field, and $\Bbb F_2(\alpha)$ is certainly a field, it must be that all roots of this polynomial are in $\Bbb F_2(\alpha)$. Now note that $(x^3 + x + 1)(x^3+x^2+1)(x)(x+1) = x^8-x\mod 2$ and a fortiori all the roots of $x^3+x+1$ are in $\Bbb F_2(\alpha)$.
A: It means all other roots are of the form 
$$a+b\alpha+c\alpha^2$$
where $a, b, c \in \mathbb{F}_2$.
