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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 .

Define to be a polynomial in I of minimal degree, and define, inductively to be a polynomial of minimal degree in .

It is clear that ... ... ... (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 , 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 includes all of these, so that contains no polynomials of degree 3 and so the degree of will be larger and so 1 holds.

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

Can someone clarify the above.

Peter

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 one can see that is of minimal degree, but so is , but that they are coprime.

Best,

Alex