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Given an algebraic alpha be of degree n over F, show at most . . .

  1. Jul 8, 2007 #1
    let alpha be algebraic over F of degree n, show that there exists at most n isomorphisms mapping F(alpha) onto a subfield of bar F (this means the algebraic closure).

  2. jcsd
  3. Jul 8, 2007 #2


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    Let phi be one of the isomorphisms, and apply phi to the minimal polynomial for alpha. This should give you a condition on phi(alpha).
  4. Jul 8, 2007 #3
    Please, reread your questions. This is not the first time you made an error. I was lucky to understand what you asked.

    I presume you want to show there are at most n isomorphisms mapping F(a) onto a subfield of bar(F) leaving F fixed.

    If [tex]\phi:F(a) \mapsto E[/tex] is an isomorphism for [tex]E\leq \bar F[/tex] then [tex]a\mapsto b[/tex] where [tex]a\mbox{ and }b[/tex] are conjugates. Conversely, if [tex]a\mapsto b[/tex] and their are conjugates then by the Conjugation Isomorphism Theorem there is exactly one such isomorphism leaving [tex]F[/tex] fixed. Let the irreducible monic polynomial for [tex]a[/tex] be [tex]c_0+c_1x+...+c_nx^n[/tex] then there are at most [tex]n[/tex] zeros, and hence at most [tex]n[/tex] conjugate elements.
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