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Thread: Extension

  1. #1
    Senior Member Sampras's Avatar
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    Extension

    Let $\displaystyle F(\alpha_1, \dots, \alpha_n) $ be an extension field of $\displaystyle F $. Show that any automorphism $\displaystyle \sigma $ of $\displaystyle F(\alpha_1, \dots, \alpha_n) $ leaving $\displaystyle F $ fixed is completely determined by the $\displaystyle n $ values of $\displaystyle \sigma(a_i) $.

    So suppose we have some isomorphism $\displaystyle \sigma_1: F \to F' $. We can extend this to isomorphism to $\displaystyle \sigma: F(\alpha_1, \dots, \alpha_n) \to L $ where $\displaystyle L $ is some subfield of $\displaystyle \bar{F'} $.

    So consider $\displaystyle \text{irr}(\alpha_i, F) $ and show that $\displaystyle \sigma(a_i) $ must be one of the zeros in $\displaystyle \bar{F'} $?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Sampras View Post
    Let $\displaystyle F(\alpha_1, \dots, \alpha_n) $ be an extension field of $\displaystyle F $. Show that any automorphism $\displaystyle \sigma $ of $\displaystyle F(\alpha_1, \dots, \alpha_n) $ leaving $\displaystyle F $ fixed is completely determined by the $\displaystyle n $ values of $\displaystyle \sigma(a_i) $.

    So suppose we have some isomorphism $\displaystyle \sigma_1: F \to F' $. We can extend this to isomorphism to $\displaystyle \sigma: F(\alpha_1, \dots, \alpha_n) \to L $ where $\displaystyle L $ is some subfield of $\displaystyle \bar{F'} $.

    So consider $\displaystyle \text{irr}(\alpha_i, F) $ and show that $\displaystyle \sigma(a_i) $ must be one of the zeros in $\displaystyle \bar{F'} $?
    I don't see why you need to look at "any subfield of F" or "$\displaystyle irrF(\alpha_i,F)$. Just use the definitions.

    Any number in the extension field, $\displaystyle F(\alpha_1, \dots, \alpha_n) $ can be written as $\displaystyle a_1\alpha_1+ a_2\alpha_2+ \cdot\cdot\cdot+ a_n\alpha_n$ where $\displaystyle a_1, a_2, \cdot\cdot\cdot, a_n$ are members of F.

    Since $\displaystyle \sigma$ is an "automorphism", it is, by definition, an isomorphism. In particular that means $\displaystyle \sigma(ax+ by)= \sigma(a)\sigma(x)+ \sigma(b)\sigma(y)$. Since it "leaves F fixed", $\displaystyle \sigma(a_i)= a_i$. Therefore $\displaystyle \sigma(x)= \sigma(a_1\alpha_1+ a_2\alpha_2+ \cdot\cdot\cdot+ a_n\alpha_n)= a_1\sigma(\alpha_1)+ a_2\sigma(\alpha_2)+ \cdot\cdot\cdot+ a_n\sigma(\alpha_n)$.
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