MHB Hilbert's Basis Theorem - Basic Question about proof

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The discussion revolves around a question regarding the proof of Hilbert's Basis Theorem, specifically how to confirm that the polynomial rx^ef - x^dg is in the ideal I of the polynomial ring R[x]. It is clarified that since I is an ideal and contains polynomials f and g, any linear combination of these polynomials, including rx^ef - x^dg, must also belong to I. The participant seeks assurance that the polynomial rx^e is in I, which is established by the properties of ideals. The conversation emphasizes the foundational aspects of ideal theory in the context of polynomial rings. Understanding these principles is crucial for grasping the proof of Hilbert's Basis Theorem.
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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

[This has also been posted on MHF]
 
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Peter said:
My problem: How do we know that the polynomial rx^ef - x^dg is in I?

We have $rx^e\in R[x]$, $x^d \in R[x]$, $f \in I$ and $g\in I$. Being $I$ an ideal and according to the definition of ideal, necessarily $rx^ef - x^dg \in I$.
 
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