Showing that a given field is the splitting field of a given polynomial 
Let $F = Z_2$; show that the splitting field of $f(x) = x^3 + x^2 + 1 \in F[x]$ is a finite field with $8$ elements.

As $f$ has degree $3$, it is reducible if it has root in $F = Z_2$ but by substituting shows that neither of $0, 1 \in Z_2$ is root of $f$, hence $f$ is an irreducible polynomial of degree $3$ over the prime field $Z_2$, and hence
$Z_2[x]/ (f(x))$ is a field of order $2^3 = 8$ in which $f(x)$ has a root. 
Now, is $Z_2[x]/ (f(x))$ the splitting field of $f$? How can we check that $f(x)$ splits in this field? Or in any other spitting field (if this is not a spitting field)?
 A: Yes, $F / (f(x))$ is the splitting field of $f \in F[x]$, and we can see this directly: If we let $\alpha$ denote a root of $f$ in $F[x]$, then computing directly gives that $\alpha^2$ and $\alpha^4  = \alpha^2 + \alpha + 1$ are also roots of $f$, and all three of these are distinct, so $f$ factors over $F / (f(x)) = F[\alpha]$ as
$$f(x) = (x - \alpha) (x - \alpha^2) (x - \alpha^4).$$
A: Let $f$ be an irreducible polynomial of degree $n$ with coefficients in $\Bbb Z/p\Bbb Z=\Bbb F_p$, and let $z$ be a root, in a field $k$ with $p^n$ elements. Then the full set of roots of $f$ is $\{z, z^p,z^{p^2},\cdots,z^{p^{n-1}}\}$. All of these are in $k$, of course. To see why $f(z)=0\Rightarrow f(z^p)=0$, just take the equation $f(z)=0$ and go from that to $\bigl(f(z)\bigr)^p=0$. But $\bigl(f(X)\bigr)^p=f(X^p)$, as you see directly, using the fact that $a^p=a$ when $a\in\Bbb F_p$.
More abstractly, the Galois group of $\overline{\Bbb F_p}$ over $\Bbb F_p$ is abelian, so all algebraic extensions of a finite field are abelian and normal. Either way, adjoin one root and get all for free.
