TITLE: Frobenius homomorphism
QUESTION [5 upvotes]: Its easy to proof that any non-zero field homomorphism is injective:
Proof
Assume that $\exists a, b\in F: a\neq b~~and~~\psi(a)=\psi(b)$ then:
$$\psi(1)=\psi((a-b)^{-1}(a-b))=\psi((a-b)^{-1})\cdot 0,$$
$$\forall x\in F:~~\psi(x)=\psi(x\cdot 1)=\psi(x)\cdot\psi(1)=\psi(x)\cdot0=0.$$
So, $\psi\equiv 0.$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
But lets consider Frobenius homomorphism of algebraically closed field $F$:
$$F\ni x\stackrel{\Phi}{\longmapsto} x^p\in F.$$
Equation
$$x^p-a=0$$
have p different roots. So, $\Phi$ isn't injective. Where is mistake?
Thanks.

REPLY [9 votes]: Over a field of characteristic $p$ such an equation has either a single root or no roots at all. If $\xi$ is one solution of the equation $x^p-a=0$, then $\xi^p=a$. Consequently
$$
x^p-a=x^p-\xi^p=(x-\xi)^p,
$$
and $\xi$ is a root of multiplicity $p$.
The mistake was assuming that the $p$ roots would be distinct.