Surjectivity of the Frobenius map I wish to prove the following claim:

Let $\mathbb{F}$ be a finite field of characteristic $p$ then the map $a\to a^p$ is surjective

Dummit and Foote's Abstract Algebra says that this map is injective hence surjective, but isn't this is just an application of Fermat's little theorem ?
 A: No, Fermat's little theorem is about $\mathbb F_p$ and you need to it for $\mathbb F_q$, where $q$ is a power of $p$.
But you don't need to know that $\mathbb F$ has order a power of $p$, just that it is finite and that every injective map is surjective, as the book says.
A: If $\mathbb{F} = \mathbb{Z}/p\mathbb{Z}$, then yes, this is Fermat's Little Theorem.  In general there's a little more to it than that.
For that matter, the claim of injectivity requires some justification as well.  Here's one way to do this: in fact here $F$ can be any field of characteristic $p > 0$.  Suppose that $x,y \in F$ are such that $x^p = y^p$.  Then 
$0 = x^p - y^p = x^p + (-y)^p = (x-y)^p$, 
due to the fact that all the intermediate binomial coefficients are divisible by $p$ and thus vanish in characteristic $p$. But since $F$ is a field, if $z * z * ... * z = 0$, then we must have $z = 0$, i.e., $x =y$.  
(For the second equality: if $p$ is odd, we have an odd number of minus signs.  If $p = 2$, then $-1 = +1$ anyway, so no matter!)
Added: On the other hand, you can adapt the method of proof of FlT.  In general, this gives something which I have come to call Lagrange's Little Theorem: let $G$ be a finite commutative group of order $N$.  Then for every $x \in G$, $x^N = 1$.  
To apply this to the finite field $\mathbb{F}$, we need to know that its order, say $q$, is a prime power $p^a$.  (This is true, for instance, because it is a finite-dimensional vector space over $\mathbb{F}_p$, where $p$ is its characteristic.)  It then follows that for all $x \in \mathbb{F}$, $x^q = x$.  (If $x = 0$, no problem.  Otherwise, $x$ lies in the multiplicative group of $\mathbb{F}$, of order $q-1$, so 
by LlT $x^{q-1} = 1$ and thus $x^q = x$.) 
Now let $f_p$ be the map $x \mapsto x^p$ and $f_q$ be the map $x \mapsto x^q = x^{p^a}$.  Then $f_q = f_p \circ \cdots \circ f_p = f_p \circ g$, say.  In general, if I have $h = f \circ g$ and $h$ is surjective, then so is $f$, so this shows that $f_p$ is surjective because $f_q$ is the identity and hence surjective.  
