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Hilbert's Basis Theorem - Polynomial of Minimal Degree

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading the Proof of Hilbert's Basis Theorem in Rotman's Advanced Modern Algebra ( See attachment for details of the proof in Rotman).

Hilbert's Basis Theorem is stated as follows: (see attachment)

Theorem 6.42 (Hilbert's Basis Theorem) If R is a commutative noetherian ring, the R[x] is also noetherian.

The proof begins as follows: (see attachment)

Proof: Assume that I is an ideal in R[x] that is not finitely generated; of course [TEX] I \ne 0 [/TEX].

Define [TEX] f_0(x) [/TEX] to be a polynomial in I of minimal degree, and define, inductively [TEX] f_{n+1}(x) [/TEX] to be a polynomial of minimal degree in [TEX] I - (f_0, f_1, f_2, ... ... f_n) [/TEX].

It is clear that [TEX] deg(f_0) \le deg(f_1) \le deg(f_2) \le [/TEX] ... ... ... (1)

Question: Is polynomial of minimal degree simply any polynomial of least or smallest degree in I. If so how can we be sure (1) holds.

If for the stage of choosing [TEX] f_0(x) [/TEX], for example the minimum degree is 3 and there are a number (possibly infinite) of such polynomials in I, how can we be sure that [TEX] (f_0(x)) [/TEX] includes all of these, so that [TEX] I - (f_0(x)) [/TEX] contains no polynomials of degree 3 and so the degree of [TEX] f_1 [/TEX] will be larger and so 1 holds.

To try to answer my own question ... ... I am assuming that the ideal [TEX] (f_0) [/TEX] includes all the polynomials of the same degree as [TEX] f_0(x) [/TEX]. Is that correct?

Can someone clarify the above.

Peter

[This has.also been posted on MHF]
 
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