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Math Help - prime subfield

  1. #1
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    prime subfield

    Let F be a field and let E be the prime subfield of F. Prove that every automorphism of F fixes E.
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    Quote Originally Posted by dori1123 View Post
    Let F be a field and let E be the prime subfield of F. Prove that every automorphism of F fixes E.
    If \sigma: F\to F is an automorphism then \sigma (1) = 1.
    Let n\cdot 1 = \underbrace{1 + 1 + ... 1}_{n\text{ times }} for n>0.
    If n=0 then let n\cdot 1 = 0 and if n<0 then n\cdot 1 = - [(-n)\cdot 1].

    Now let D = \{ n\cdot 1 | n \in \mathbb{Z} \} then \sigma (x) = x for all x\in D.
    To see this notice \sigma (1+...+1) = \sigma (1) + ... + \sigma(1) = n\cdot \sigma(1) = n\cdot 1.
    This is because \sigma (a+b) = \sigma(a) + \sigma(b) and \sigma (-a) = -\sigma(a) since \sigma is automorphism.

    The prime subfield is E = \{ \alpha/\beta | \alpha,\beta \in D , \beta \not = 0 \}.
    Notice that \sigma (x) = x for all x\in E.
    To see this notice, \sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta.
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    Quote Originally Posted by ThePerfectHacker View Post
    The prime subfield is E = \{ \alpha/\beta | \alpha,\beta \in D , \beta \not = 0 \}.
    Notice that \sigma (x) = x for all x\in E.
    To see this notice, \sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta.
    I understand the first two cases. But why is that the third case? How about \{(n\cdot 1)^{-1}| n \in \mathbb{Z}\}?
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    Quote Originally Posted by dori1123 View Post
    I understand the first two cases. But why is that the third case? How about \{(n\cdot 1)^{-1}| n \in \mathbb{Z}\}?
    I have no idea what you are trying to ask. What are you asking? Are you asking why E is the prime subfield in the way I defined it?
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    Yes.
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  6. #6
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    Quote Originally Posted by dori1123 View Post
    Yes.
    The prime subfield is the intersection of all subfields i.e. it is the smallest subfield. If E is a subfield then it means 1\in E but then n\cdot 1\in E because E is closed under addition and it has 0 and it has all additive inverses. Therefore D must be contained in E. However, E is a field and so \alpha/\beta \in E for all \alpha,\beta \in D,\beta\not =0. Therefore, the set of fraction is contained in E, but this set of fractions is itself a field and so E is in fact the set of all these fractions that we can form.
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