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Thread: Frobenius Morphism

  1. #1
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    Frobenius Morphism

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

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

    I solved a) using Fermat's Little Theorem, but am unsure about b) and c). I think I solved c) by considering $\displaystyle Fr^7$, 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 $\displaystyle \mathbb{F}$ be a finite field of characteristic $\displaystyle p.$ As such it is a finite-dimensional vector space over $\displaystyle \mathbb{Z}_p$.

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


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

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



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


    2 is an eigenvalue iff $\displaystyle \exists\,0\neq a\in\mathbb{F}\,\,s.t.\,\,a^7=Fr(a)=2a\Longleftrig htarrow a^6=2$ , 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 $\displaystyle Fr^7$, 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 $\displaystyle a^6 = 2$ not possible?
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