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Math Help - Extension

  1. #1
    Senior Member Sampras's Avatar
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    Extension

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

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

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

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

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

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

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

    Since \sigma is an "automorphism", it is, by definition, an isomorphism. In particular that means \sigma(ax+ by)= \sigma(a)\sigma(x)+ \sigma(b)\sigma(y). Since it "leaves F fixed", \sigma(a_i)= a_i. Therefore \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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