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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.

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    Once you've worked this out, you should try proving that Frobenius is an automorphism if and only if $F$ is perfect (so don't assume algebraically closed anymore). It is a neat little exercise and these sorts of ideas appear in it.2011-08-12

1 Answers 1

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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.