# Thread: Hilbert's Basis Theorem - Polynomial of Minimal Degree

1. ## Hilbert's Basis Theorem - Polynomial of Minimal Degree

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

2. ## Re: Hilbert's Basis Theorem - Polynomial of Minimal Degree

Hello,

I don't seem to see the problem. There is only the claim of inequality, not strict inequality? For example, in $I=(x^2+1,x^2+2)\subseteq \mathbb{Q}[x]$ one can see that $x^2+1$ is of minimal degree, but so is $x^2+2$, but that they are coprime.

Best,
Alex