Let \(R\) be a Noetherian ring. Prove that \(R[x]\) and \(R[[x]]\) are both Noetherian. (The first part of the question is asking you to prove the Hilbert Basis Theorem, not to use it!)
Solution
We prove the two assertions separately.
1. The polynomial ring \(R[x]\). Let \(I\subseteq R[x]\) be an ideal. For \(d\ge0\), let \[ L_d=\{a\in R:\text{there is }f\in I\text{ of degree at most }d\text{ whose }x^d\text{-coefficient is }a\}. \] Then \(L_d\) is an ideal of \(R\), and \(L_d\subseteq L_{d+1}\), since multiplying a polynomial by \(x\) preserves its leading coefficient. Because \(R\) is Noetherian, the ascending chain \[ L_0\subseteq L_1\subseteq L_2\subseteq\cdots \] stabilizes, say at \(L_N\).
For each \(0\le d\le N\), choose finitely many polynomials \[ f_{d1},\ldots,f_{dr_d}\in I \] of degree at most \(d\) whose \(x^d\)-coefficients generate \(L_d\). We claim these finitely many polynomials generate \(I\).
Take \(f\in I\), \(f\ne0\), of degree \(m\). If \(m\le N\), its leading coefficient lies in \(L_m\), so subtracting a suitable \(R\)-linear combination of the \(f_{mj}\) lowers the degree. If \(m>N\), stabilization gives \(L_m=L_N\), so the leading coefficient of \(f\) is an \(R\)-linear combination of the \(x^N\)-coefficients of the \(f_{Nj}\); subtracting the corresponding combination of \(x^{m-N}f_{Nj}\) lowers the degree. Induction on degree expresses \(f\) in the ideal generated by the chosen finite set. Hence every ideal of \(R[x]\) is finitely generated, so \(R[x]\) is Noetherian.
2. The formal power-series ring \(R[[x]]\). Let \(I\subseteq R[[x]]\) be an ideal. For each \(n\ge0\), define \[ J_n=\{a\in R:\text{there is }f\in I\cap x^nR[[x]]\text{ whose coefficient of }x^n\text{ is }a\}. \] Each \(J_n\) is an ideal, and \(J_n\subseteq J_{n+1}\), because multiplication by \(x\) sends a witness for \(a\in J_n\) to one for \(a\in J_{n+1}\). Since \(R\) is Noetherian, choose \(N\) with \[ J_N=J_{N+1}=J_{N+2}=\cdots. \] For every \(0\le n\le N\), choose finitely many series \[ g_{n1},\ldots,g_{nr_n}\in I\cap x^nR[[x]] \] whose coefficients of \(x^n\) generate \(J_n\). Let \(I_0\) be the ideal generated by all these finitely many \(g_{nj}\). We show \(I=I_0\).
Take \(f\in I\). We cancel its coefficients successively. At stage \(m<N\), after the coefficients of degrees \(<m\) have been killed, the current remainder lies in \(I\cap x^mR[[x]]\); its \(x^m\)-coefficient belongs to \(J_m\), so an \(R\)-linear combination of the \(g_{mj}\) cancels it without changing lower coefficients. At stage \(m\ge N\), the current \(x^m\)-coefficient lies in \(J_m=J_N\), so an \(R\)-linear combination of the shifted series \(x^{m-N}g_{Nj}\) cancels it.
Thus for each degree \(m\) we obtain correction coefficients. For a fixed generator \(g_{nj}\), collect the corrections multiplying its shifts into a formal series \(h_{nj}(x)\in R[[x]]\). Coefficientwise, the recursive cancellation says \[ f=\sum_{n=0}^{N}\sum_{j=1}^{r_n}h_{nj}(x)g_{nj}(x). \] (The coefficient of any fixed power of \(x\) involves only finitely many terms, so the identity is well-defined in \(R[[x]]\).) Hence \(f\in I_0\). Therefore \(I=I_0\) is finitely generated.
So both \(R[x]\) and \(R[[x]]\) are Noetherian.