# Checking Finite Field Automorphism

My book contains the proof of the following claim:

Let $F$ be a finite field with characteristic p. Show that $F^p = F$

They construct a morphism $\phi: F \rightarrow F^p$ that sends $x \mapsto x^p$. The check this is a homomorphism, that it is one-to-one, and its onto. I followed most of the proof, except for checking surjection. They write one line:

Since F is finite, $\phi$ is onto.

Why is this true?

• This follows from the rank-nullity theorem, see here. Commented Dec 11, 2017 at 15:36
• @Dietrich $F$ is not a vector space, so how would this work? I mean I see Ker $\phi$ is zero so dim(Im($\phi$)) = dim F. Commented Dec 11, 2017 at 15:51
• @yoshi $F$ is a $1$-dimensional vector space over itself and $F^p$ is a subspace. Actually, I guess the fact that $F^p$ is closed under addition and scalar multiplication isn't clear... Commented Dec 11, 2017 at 15:56

Since $F^p \subseteq F$, then $\#(F^p) \leq \#F$. Since $\phi: F \to F^p$ is injective, then $\#F \leq \#(F^p)$, so $\#(F^p) = \#F$. Since these are finite sets, then $F^p = F$. An injective map from a finite set to itself must also be surjective.
• I have never seen the # or $\leq$ notation in relation to $F$. Could you define these or point me to a wiki? Commented Dec 11, 2017 at 15:53
• $\#F$ just means the number of elements in $F$, aka its cardinality. It's also sometimes written $|F|$. Commented Dec 11, 2017 at 15:54
• I'm almost there, why is $F^p \subseteq F$ true? Commented Dec 11, 2017 at 15:58
• $F^p$ is the set of all elements of $F$ that are $p^\text{th}$ powers: $F^p = \{a \in F : a = b^p \text{ for some } b \in F\}$. So $F^p$ is a subset by definition. Commented Dec 11, 2017 at 16:01
Here is a direct proof that $\phi$ is surjective.
Let $F$ have $q$ elements. Then $a^q = a$ for all $a \in F$ because of Lagrange's theorem applied to $F^{\times}$.
Since $q$ is a power of $p$, we have $q=tp$ and so $a=a^q=(a^t)^p \in F^p$.