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Math Help - Hilbert's Basis Theorem - Polynomial of Minimal Degree

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    Super Member Bernhard's Avatar
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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  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
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    MHF Contributor Drexel28's Avatar
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    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
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