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Math Help - Field Theory - The Degree of alpha over F

  1. #1
    Super Member Bernhard's Avatar
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    Field Theory - The Degree of alpha over F

    I am trying to clarify my understanding of Proposition 11 of Dummit and Foote Ch13 Field Theory concerning the degree of  \alpha over F.

    Proposition 11 reads as follows:

    ---------------------------------------------------------------------------------------------------------------------------------------------------------

    Proposition 11. Let  \alpha be algebraic over the field F and let  F(\alpha) be the field generated by  \alpha over F.

    Then   F(\alpha) \cong F[x]/(m_{\alpha}(x))

    so that in particular  [F(\alpha) \ : \ F] = deg  \  m_{\alpha}(x) = deg \ \alpha

    i.e. the degree of  \alpha over F is the degree of the extension it generates over F

    --------------------------------------------------------------------------------------------------------------------------------------------------------------


    However Corollary 7 (Dummit and Foote page 518) states the following:


    -----------------------------------------------------------------------------------------------------------------------------------------------------------------

    Corollary 7. Suppose in Theorem 6 that p(x) is of degree n. Then

      F(\alpha) = \{ a_0 + a_1 {\alpha} + a_2 {\alpha}^2 + ... ... a_{n-1} {\alpha}^{n-1} \  | \ a_0, a_1, ... ... a_{n-1} \in F \} \subseteq K

    -------------------------------------------------------------------------------------------------------------------------------------------------------------------

    Given that  F(\alpha)  consists of polynomials of degree (n-1) should not the degree of  [F(\alpha) \ : \ F] = deg  \  m_{\alpha}(x) - 1 = deg \  \alpha -1  - that is the degree of  \alpha over F be one less than the degree of the minimal polynomial?


    Can someone please clarify this situation for me

    Peter
    Last edited by Bernhard; September 27th 2013 at 01:43 AM.
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  2. #2
    Super Member Bernhard's Avatar
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    Re: Field Theory - The Degree of alpha over F

    I have just been reflecting about my own question above.

    It is possible that  F(\alpha) viewed as a vector space of deg (n-1) polynomials would have the following as a basis:

     1, {\alpha}, {\alpha}^2, {\alpha}^3, ... ... {\alpha}^{n-1}

    Then the degree of the space would be n which would equal the degree of the minimal polynomial involved.

    If this thought is actually correct, could someone please confirm this as the case.

    Peter
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  3. #3
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    Re: Field Theory - The Degree of alpha over F

    That is correct. Any higher degree polynomials would contain a linear dependency (some linear combination with nonzero coefficients would sum to zero).
    Thanks from Bernhard
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  4. #4
    Super Member Bernhard's Avatar
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    Re: Field Theory - The Degree of alpha over F

    Thanks SlipEternal

    Sorry my thank you was so late - my day job intervened! :-(
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