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Hilbert's Basis Theorem - Basic Question about proof


Well-known member
MHB Site Helper
Jun 22, 2012
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, [TEX] 0 \in L [/TEX].

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

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

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

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

Can someone clarify this situation for me?


[This has also been posted on MHF]

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
My problem: How do we know that the polynomial [TEX] rx^ef - x^dg [/TEX] 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 $rx^ef - x^dg \in I$.