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Math Help - Hilbert's Basis Theorem - Fundamental Question about R[x]

  1. #1
    Super Member Bernhard's Avatar
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    Hilbert's Basis Theorem - Fundamental Question about R[x]

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

    I have a very basic question regarding the beginning of the proof of Hilbert's Basis Theorem (see attachment for a statement of the Theorem and details of the proof)

    Theorem 21 (Hilbert's Basis Theorem) If R is a Noetherian ring then so is the polynomial ring R[x]

    The proof begins as follows:

    Proof: Let I be an ideal in R[x] and let L be the set of all leading coefficients of the elements in I. We will first show that L is an ideal of R, as follows. Since I contains the zero polynomial,  0 \in L .

    Let  f = ax^d + ... and  g = bx^e + ... be polynomials in I of degrees d, e and leading coefficients a, b \in R .

    Then for any  r \in R either ra - b is zero or it is the leading coefficient of the polynomial  rx^ef - x^dg . Since the latter polynomial is in I ... ...???

    My problem: How do we know that the polynomial  rx^ef - x^dg is in I?

    For  rx^ef  to belong to I we need  rx^e \in I . Now it seems to me that  rx^e \in I if  x^e \in I (right?) but how do we know that or be sure that  x^e \in I ?

    Can someone clarify this situation for me?

    Peter
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    Last edited by Bernhard; June 7th 2013 at 05:48 PM.
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    Re: Hilbert's Basis Theorem - Fundamental Question about R[x]

    Quote Originally Posted by Bernhard View Post
    My problem: How do we know that the polynomial  rx^ef - x^dg is in I?
    Since f\in I and rx^e \in R[x],  rx^ef \in I by definition of ideal. Similarly, g\in I,-x^d\in R[x] means -x^dg\in I. Hence  rx^ef - x^dg \in I

    For  rx^ef  to belong to I we need  rx^e \in I . Now it seems to me that  rx^e \in I if  x^e \in I (right?) but how do we know that or be sure that  x^e \in I ?
    It is not true in general that ab\in I \implies a \in I. If this were the case, taking a=1 and b\in I, we obtain 1\in I so I=R. This would mean that there are no non-trivial ideals.
    Thanks from Bernhard
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