# 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 ?

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There is (surely?) more than one book with the title Abstract Algebra. I wonder which one you are using? – Pete L. Clark May 4 '12 at 16:53
@PeteL.Clark - amazon.com/Abstract-Algebra-Edition-David-Dummit/dp/0471433349 – Belgi May 4 '12 at 16:54

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.

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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.

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I like the name Lagrange's Little Theorem! It's my pet theorem. – lhf May 4 '12 at 16:57
Why do you need the last paragraph ? I thought that you proved the claim already...also, are you saying in the last paragraph that $f_p$ is surjective since $f_q$ is surjective ? (because this seems like the special case where $a=1$. – Belgi May 4 '12 at 17:14
@lhf: Sure, okay. I added what you suggested. – Pete L. Clark May 4 '12 at 18:55
@Belgi: Yes, I am saying in the last paragraph is that $f_p$ is surjective since $f_q$ is surjective. But I am making this claim about a map on a finite field of order $p^a$ so (unless I'm mistaken) this doesn't follow immediately from the case of $\# \mathbb{F} = p$. – Pete L. Clark May 4 '12 at 18:57