1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?