1. ## 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

2. ## Re: Hilbert's Basis Theorem - Fundamental Question about R[x]

Originally Posted by Bernhard
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.