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Thread: Conjugation Isomorphism

  1. #1
    Senior Member Sampras's Avatar
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    Conjugation Isomorphism

    Theorem. Suppose we have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \mapsto F(\beta) $ defined by $\displaystyle \psi_{\alpha, \beta}(a_0+a_{1} \alpha + \cdots + a_{n-1} \alpha^{n-1}) = a_0+a_{1} \beta + \cdots + a_{n-1} \beta^{n-1} $ (e.g. $\displaystyle \alpha $ and $\displaystyle \beta $ are conjugates). Then $\displaystyle \text{irr}(\alpha, F) = \text{irr}(\beta, F) $.


    The notion is that we have to show that $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $ and $\displaystyle \text{irr}(\beta, F) $ divides $\displaystyle \text{irr}(\alpha, F) $. But why can't we just stop and say $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $? Because, by definition, if $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $, wouldn't that imply that $\displaystyle \text{irr}(\alpha, F) = \text{irr}(\beta, F) $?
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    Quote Originally Posted by Sampras View Post
    Theorem. Suppose we have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \mapsto F(\beta) $ defined by $\displaystyle \psi_{\alpha, \beta}(a_0+a_{1} \alpha + \cdots + a_{n-1} \alpha^{n-1}) = a_0+a_{1} \beta + \cdots + a_{n-1} \beta^{n-1} $ (e.g. $\displaystyle \alpha $ and $\displaystyle \beta $ are conjugates). Then $\displaystyle \text{irr}(\alpha, F) = \text{irr}(\beta, F) $.


    The notion is that we have to show that $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $ and $\displaystyle \text{irr}(\beta, F) $ divides $\displaystyle \text{irr}(\alpha, F) $. But why can't we just stop and say $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $? Because, by definition, if $\displaystyle \text{irr}(\alpha, F) $ divides $\displaystyle \text{irr}(\beta, F) $, wouldn't that imply that $\displaystyle \text{irr}(\alpha, F) = \text{irr}(\beta, F) $?

    If you already know that the corresponding irreducible polynomials are (monic, of course) and of the same degree, then yes: it is enough to show that one divides the other. Perhaps in this problem though they do not assume this...

    Tonio
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