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Math Help - Frobenius Morphism

  1. #1
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    Frobenius Morphism

    Let  \mathbb{F} be a finite field of characteristic p. As such it is a finite-dimensional vector space over \mathbb{Z}_p.

    a) Prove that the Frobenius morphism Fr:\mathbb{F} \rightarrow \mathbb{F}, Fr(a) = a^p is a linear map over \mathbb{Z}_p.
    b) Prove that the geometric multiplicity of 1 as an eigenvalue of  Fr is 1.
    c) Let \mathbb{F} have dimension 2 over \mathbb{Z}_7. Prove that 2 is not an eigenvalue of Fr.

    I solved a) using Fermat's Little Theorem, but am unsure about b) and c). I think I solved c) by considering Fr^7, but I'm not sure if there's a simpler proof.

    Thanks!
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  2. #2
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    Quote Originally Posted by h2osprey View Post
    Let  \mathbb{F} be a finite field of characteristic p. As such it is a finite-dimensional vector space over \mathbb{Z}_p.

    a) Prove that the Frobenius morphism Fr:\mathbb{F} \rightarrow \mathbb{F}, Fr(a) = a^p is a linear map over \mathbb{Z}_p.
    b) Prove that the geometric multiplicity of 1 as an eigenvalue of  Fr is 1.


    This means you have to prove the dimension of the eigenspace corresponding to 1 is

    1, but a=:Fr(a)=a^p\Longleftrightarrow a=a^p\Longleftrightarrow a\in \mathbb{Z}_p , so...



    c) Let \mathbb{F} have dimension 2 over \mathbb{Z}_7. Prove that 2 is not an eigenvalue of Fr.


    2 is an eigenvalue iff \exists\,0\neq a\in\mathbb{F}\,\,s.t.\,\,a^7=Fr(a)=2a\Longleftrig  htarrow a^6=2 , which

    of course is impossible.

    Tonio



    I solved a) using Fermat's Little Theorem, but am unsure about b) and c). I think I solved c) by considering Fr^7, but I'm not sure if there's a simpler proof.

    Thanks!
    .
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  3. #3
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    Quote Originally Posted by tonio View Post
    .
    I get b) now, but for c) why is  a^6 = 2 not possible?
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