# When is the Frobenius endomorphism an automorphism?

I was thinking about some problem:

Lef $F$ be a finite field with the characteristic $p$. Then a map $x\mapsto x^p$ is an automorphism.

I figured out that this map is called the Frobenius endomorphism, and that it can be a hint, if we show that any non-zero field homomorphism to itself is automatically an automorphism, but I don't know why it has to work.

Any hints?

• ok, but any details? :) – Yelon Oct 22 '17 at 21:22

In a field $F$ with characteristic $p$, the map $x\mapsto x^p$ is a field endomorphism. Indeed, $(xy)^p=x^py^p$ and $$(x+y)^p=\sum_{k=0}^p \binom{p}{k}x^ky^{p-k} =x^p+y^p$$ because, when $p$ is prime, $$p\mid \binom{p}{k}$$ for $0<k<p$ (prove it). Finally, $1^p=1$.
A field homomorphism $\alpha\colon F\to F'$ is necessarily injective, because $\ker\alpha$ is a proper ideal of $F$, hence it is $\{0\}$.
If a set $X$ is finite, an injective map $f\colon X\to X$ is also surjective.
Note that, in general, an endomorphism of a (nonfinite) field is not surjective. For instance, if $F$ is any field of characteristic $p$, then $t\mapsto t^p$ is also an endomorphism of the field $F(X)$ of rational functions on $F$, but it's not surjective because $X$ is not in the image.
By the way, if $F$ is a finite field of characteristic $p$, with $|F|=p^n$, then every endomorphism (automorphism) of $F$ is of the form $x\mapsto x^{p^k}$, for some $k$ with $0\le k<n$.