# Frobenius homomorphism

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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They're not all different; $a^\frac{1}{p}$ is a $p$-fold root of $x^p-a$ over an algebraically closed field of characteristic $p.$ – jspecter Aug 12 '11 at 6:22
One way to quickly recognize that the roots are not distinct is to note that the (formal) derivative is $px^{p - 1} = 0$. See the Wikipedia article on separable polynomials. – Dylan Moreland Aug 12 '11 at 15:30
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. – Matt Aug 12 '11 at 15:37

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.