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.