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Simple extensions

  1. Feb 12, 2008 #1
    [SOLVED] simple extensions

    1. The problem statement, all variables and given/known data

    My book (Farleigh) says the following (I am paraphrasing):

    Let E be an extension field of a field F, and let [itex] \alpha \in E[/itex]. Suppose [itex]\alpha[/itex] is algebraic over a field F. Let [itex]\phi_{\alpha}[/itex] be the evaluation homomorphism of F[x] into E with [itex]\phi_{\alpha}(a)=a[/itex] for [itex]a \in F[/itex] and [itex]\phi_{\alpha}(x)=\alpha[/itex] .

    Suppose \alpha is algebraic over F. Then [itex]F[x]/<irr(\alpha,F)>[/itex] is isomorphic to the image of [itex]\phi_{\alpha}[F[x]][/itex] in E.

    What I do not understand is how this can be true when [itex]\alpha[/itex] is in F. When [itex]\alpha[/itex] is in F, [itex]\phi_{\alpha}[F[x]][/itex] = F, but I think it is pretty clear that [itex]F[x]/<irr(\alpha,F)>=F[x]/<x-\alpha>[/itex] will not equal F because x - \alpha will generate a nontrivial ideal. What is wrong here!!!!

    Also, I do not understand why the stuff in bold is necessary. I thought that was obvious from the definition of an evaluation map.

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2
    I am starting to kind of see why F[x]/<x-alpha> is isomorphic to F, but can someone explain/prove that please? It is probably really easy.
  4. Feb 13, 2008 #3
  5. Feb 13, 2008 #4


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    If alpha is in F, then the evaluation at alpha homomorphism from F[x] to F is surjective and its kernel is <x-alpha>. So by the first isomorphism theorem, F[x]/<x-alpha> =~ F.

    And about the bold stuff: it's probably there because that's the definition of phi_alpha.
  6. Feb 13, 2008 #5
    That makes sense.
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