Hilbert’s Basis Theorem

Here is a proof of Hilbert’s Basis Theorem I thought of last night.

Let R be a noetherian ring. Consider an ideal J in R[X]. Let I_i be the ideal in R generated by the leading coefficients of the polynomials of degree i in I. Notice that I_i \subseteq I_{i+1}, since if P \in I_i, xP \in I_{i+1}, and it has the same leading coefficient. Thus we have an ascending chain \dots \subseteq I_{i-1} \subseteq I_i \subseteq I_{i+1} \subseteq \dots, which must terminate, since R is noetherian. Suppose it terminates at i = n, so I_n = I_{n+1} = \dots.

Now for each I_i choose a finite set of generators (which we can do since R is noetherian). For each generator, choose a representative polynomial in J with that leading coefficient. This gives us a finite collection polynomials: define J_i to be the ideal of R[x] generated by these polynomials.

Let J' = J_0 + J_1 + \dots + J_n. I claim J = J'. Assume for the sake of contradiction that there is a polynomial P of minimal degree (say i) which is in J but J'. If i \leq n, there is an element of P' \in J' with the same leading coefficient, so P - P' is not in J' but has degree smaller than i: contradiction. If P is of degree i > n, then there is an element of P' of J_n which has the same leading coefficient as P. Thus P - x^{i-n}P' is of degree smaller than i but is not in J': contradiction.

Thus J = J'. Since J is therefore finitely generated, this proves R[x] is noetherian.

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