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Math Help - Grobner Bases

  1. #1
    Super Member Bernhard's Avatar
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    Grobner Bases

    I am reading Dummit and Foote Section 9.6 Polynomials in Several Variables Over a Field and Grobner Bases.

    I have a problem understanding a step in the proof of Proposition 24, Page 322 of D&F

    Proposition 24 reads as follows:

    Proposition 24. Fix a monomial ordering on  R= F[x_1, ... , x_n] and let I be a non-zero ideal in R

    (1) If  g_1, ... , g_m are any elements of I such that  LT(I) = (LT(g_1), ... ... LT(g_m) )

    then  \{ g_1, ... , g_m \} is a Grobner Basis for I

    (2) The ideal I has a Grobner Basis


    The proof of Proposition 24 begins as follows:


    Proof: Suppose  g_1, ... , g_m \in I with  LT(I) = (LT(g_1), ... ... LT(g_m) ) .

    We need to see that  g_1, ... , g_m generate the ideal I.

    If  f \in I use general polynomial division to write  f = \sum q_i g_i + r where no non-zero term in the remainder r is divisible by any  LT(g_i)

    Since  f \in I , also  r \in I ... ... etc etc



    My question: How do we know  f \in I \Longrightarrow r \in I ?

    Would appreciate some help

    Peter
    Last edited by Bernhard; June 29th 2013 at 09:25 PM.
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Re: Grobner Bases

    Quote Originally Posted by Bernhard View Post
    /snip/

    If  f \in I use general polynomial division to write  f = \sum q_i g_i + r where no non-zero term in the remainder r is divisible by any  LT(g_i)
    Since  f \in I , also  r \in I ... ... etc etc
    My question: How do we know  f \in I \Longrightarrow r \in I ?
    Since g_i \in I, using the definition of an ideal, we have  q_i g_i \in I. Since an ideal is an additive group, and f \in I we conclude f - \sum q_i g_i \in I.
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Grobner Bases

    Thanks for the help

    Peter
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