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 $\displaystyle I \ne 0 $.

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

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

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

Can someone clarify the above.

Peter