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We define the Frobenius Homomorphism as that:

Let $F$ be a field of characteristic $p\gt 0$. Then we call the Frobenius homomorphism this map: $$\phi:F\to F, \phi(x)=x^p$$

I have the following questions:

  1. When this homomorphism is in fact an endomorphism or an automorphism?
  2. Why $x^p=0$ implies $x=0$ (I think it's true in a finite field $F$, but I don't know why)

I would appreciate it so much if someone could help me with this. I think it's a common doubt for beginners because I've already read some books without any clarification on this topic.

Thanks again

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1 Answer 1

up vote 1 down vote accepted

First, this homomorphism is always an endomorphism, because endomorphism means homomorphism from a space to itself. Indeed, $\phi$ maps $F$ to itself.

The frobenius map is an automorphism if and only if it is bijective and has a bijective inverse. What this says is that "$\phi$ is an automorphism" is equivalent with "$F$ contains a unique $p^{\mathrm th}$ root for every element $x\in F.$" In other words, $F$ is a perfect field of characteristic $p.$

Second, the fact that $x^p=0\Rightarrow x=0,$ is true because $F$ is a field, and thus has nilradical $(0).$ (Suppose $x\neq 0.$ Can you derive a contradiction?)

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In the second question, can I say also because $F$ is in particular an integral domain? Thank you for your answer :) –  user42912 Dec 12 '12 at 0:31
    
Dear @RafaelChavez, yes that certainly works. Or, since $F$ is a field, if we suppose $x\neq 0,$ then $x$ is invertible, and we cannot have $x^n=0.$ You're welcome! –  Andrew Dec 12 '12 at 0:34
    
Andrew, I didn't understand why you said that: "this homomorphism is always an endomorphism, because endomorphism means homomorphism from a space to itself". Endomorphisms are surjective homomorphisms and we have counterexamples of homomorphism from a space to itself, without be endomorphisms. –  user42912 Dec 15 '12 at 7:29
    
Dear @RafaelChavez, no endomorphisms need not be surjective. en.wikipedia.org/wiki/Endomorphism –  Andrew Dec 15 '12 at 16:29
    
yes, you're right, thank you –  user42912 Dec 16 '12 at 8:49

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