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Math Help - Conjugation Isomorphism

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

    Theorem. Suppose we have the conjugation isomorphism  \psi_{\alpha, \beta}: F(\alpha) \mapsto F(\beta) defined by  \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.  \alpha and  \beta are conjugates). Then  \text{irr}(\alpha, F) = \text{irr}(\beta, F) .


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


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