Hilbert's Basis Theorem - Polynomial of Minimal Degree

[Math Amateur]

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 $$I \ne 0$$.

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

It is clear that $$deg(f_0) \le deg(f_1) \le deg(f_2) \le$$ ... ... ... (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 $$f_0(x)$$, 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 $$(f_0(x))$$ includes all of these, so that $$I - (f_0(x))$$ contains no polynomials of degree 3 and so the degree of $$f_1$$ will be larger and so 1 holds.

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

Can someone clarify the above.

Peter

[This has.also been posted on MHF]

 

[jvicens]

Dear Peter,

Thank you for sharing your thoughts and questions about the Proof of Hilbert's Basis Theorem. I can understand your confusion and I am happy to help clarify the proof for you.

To answer your first question, yes, a polynomial of minimal degree simply means a polynomial of least or smallest degree in I. This is important because the proof relies on the fact that we can always find a polynomial of minimal degree in I, which is then used in the induction process.

Now, to address your second question, you are correct in assuming that the ideal (f_0) includes all the polynomials of the same degree as f_0(x). This is because, by definition, (f_0) is the ideal generated by f_0(x), which means it contains all polynomials that can be obtained by multiplying f_0(x) with other polynomials in R[x]. So, when we choose f_0(x) to be a polynomial of minimal degree in I, we can be sure that (f_0) includes all polynomials of the same degree as f_0(x).

I hope this helps clarify the proof for you. If you have any further questions, please do not hesitate to ask.