1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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?)

$\endgroup$
  • $\begingroup$ In the second question, can I say also because $F$ is in particular an integral domain? Thank you for your answer :) $\endgroup$ – user42912 Dec 12 '12 at 0:31
  • $\begingroup$ 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! $\endgroup$ – Andrew Dec 12 '12 at 0:34
  • $\begingroup$ 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. $\endgroup$ – user42912 Dec 15 '12 at 7:29
  • $\begingroup$ Dear @RafaelChavez, no endomorphisms need not be surjective. en.wikipedia.org/wiki/Endomorphism $\endgroup$ – Andrew Dec 15 '12 at 16:29
  • $\begingroup$ yes, you're right, thank you $\endgroup$ – user42912 Dec 16 '12 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.