MHB Hilbert's Basis Theorem - Basic Question about proof

[Math Amateur]

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]

 

[Fernando Revilla]

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 $rx^ef - x^dg \in I$.