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Thread: automorphism

  1. #1
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    automorphism

    If $\displaystyle \sigma_{1}, \sigma_2, \ldots, \sigma_{n} $ is a group of automorphisms of a field $\displaystyle E $ and if $\displaystyle F $ is a fixed field of $\displaystyle \sigma_{1}, \sigma_{2}, \ldots, \sigma_{n}, $ then $\displaystyle (E/F) = n $.

    How would I prove this?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    If $\displaystyle \sigma_{1}, \sigma_2, \ldots, \sigma_{n} $ is a group of automorphisms of a field $\displaystyle E $ and if $\displaystyle F $ is a fixed field of $\displaystyle \sigma_{1}, \sigma_{2}, \ldots, \sigma_{n}, $ then $\displaystyle (E/F) = n $.
    There is a result due to Artin . Let $\displaystyle G$ be a finite group of automorphism of $\displaystyle E$, let $\displaystyle F=E^G$ be the fixed subfield, then $\displaystyle E/F$ is finite and $\displaystyle [E:F] \leq |G|$. Note, you need the $\displaystyle \leq $ sign.
    Last edited by ThePerfectHacker; Mar 1st 2008 at 07:35 PM.
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