hilberts basis theorem | Arithmetic variety

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.