State and prove the Hilbert Basis Theorem: if \(R\) is a Noetherian ring, then the polynomial ring \(R[x]\) is also Noetherian.
Solution
Let \(R\) be Noetherian and let \(I\subseteq R[x]\) be an ideal. For \(d\ge0\), let \(L_d\subseteq R\) be the ideal consisting of \(0\) together with the leading coefficients of degree-\(d\) elements of \(I\). Then \[ L_0\subseteq L_1\subseteq\cdots \] because multiplying by \(x\) preserves the leading coefficient. Since \(R\) is Noetherian, this chain stabilizes: choose \(N\) with \(L_d=L_N\) for all \(d\ge N\).
For each \(0\le d\le N\), choose generators \[ L_d=(a_{d1},\dots,a_{dr_d}) \] and choose \(f_{di}\in I\) of degree \(d\) with leading coefficient \(a_{di}\). Let \(J\) be the ideal of \(R[x]\) generated by these finitely many \(f_{di}\).
We prove \(I=J\) by degree reduction. Let \(g\in I\) have degree \(m\) and leading coefficient \(c\). If \(m<N\), write \(c\) as an \(R\)-linear combination of the \(a_{mi}\) and subtract the corresponding combination of the \(f_{mi}\). If \(m\ge N\), write \(c\) as a combination of the \(a_{Ni}\) and subtract the same combination of \(x^{m-N}f_{Ni}\). In either case the degree strictly decreases. Induction gives \(g\in J\).
Thus every ideal of \(R[x]\) is finitely generated, so \(R[x]\) is Noetherian.