Prove that every extension of a finite field is normal In book by Roman 'Field Theory' it is written that it is straightforward that every extension of a finite field is normal. However I just cannot see it. Can you help me with this problem?
Thank
 A: An (algebraic) field extension is normal if and only if it is the splitting field of a family of polynomials, i.e. if the extension $K/k$ contains an element $\alpha$, it also contains all conjugates of $\alpha$ in an algebraic closure. Thus, we can prove the extension is normal by proving this property for every element!
Now, consider an element $\alpha\in K$. Since $K$ is algebraic over $k$, there is a finite extension $k\subset L \subset K$ such that $\alpha \in L$. Now we'd like to prove that every finite extension of a finite field is normal. Why is that straightforward? For one, we can list all the finite extensions of finite fields, since we know all the finite fields...
Can you continue from here?
A: If the field $F$ is a subfield of $K$ with $K$ finite of order $q$, then the elements of $K$ are all roots of $x^q-x \in F[x]$, so this equation has $q$ (distinct) roots in $K$, and so  $K$ is its splitting field over $F$. So $K$ is a Galois extension of $F$.
A: I haven't thought about what this would mean for infinite and possibly non-algebraic "extensions", so I'll pretend they don't exist.
From a finite extension of fields $E/F$ we get inequalities
$$
|\operatorname{Aut}(E/F)| \leq |\operatorname{Hom}_{F\text{-alg}}(E, \overline{F})| \leq [E:F].
$$
Normality means that the first inequality is actually an equality. The trick is that in this case it's easy to write $[E:F]$ elements of the automorphism group down.
