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Thread: Hilbert's Basis Theorem - Fundamental Question about R[x]

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    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, $\displaystyle 0 \in L $.

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

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

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

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

    Can someone clarify this situation for me?

    Peter
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    Last edited by Bernhard; Jun 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 $\displaystyle rx^ef - x^dg $ is in I?
    Since $\displaystyle f\in I$ and $\displaystyle rx^e \in R[x]$,$\displaystyle rx^ef \in I $ by definition of ideal. Similarly, $\displaystyle g\in I,-x^d\in R[x]$ means $\displaystyle -x^dg\in I$. Hence $\displaystyle rx^ef - x^dg \in I $

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