Base case

Assume \(R\) is Noetherian & show \(R[x]\) is Noetherian. Let \(J\subseteq R[x]\) be an ideal. We must show \(J\) is finitely generated.

Leading coefficient ideal

For each \(f(x)\in J\), let \(\mathrm{lc}(f)\) be its leading coefficient in \(R\). Define the leading-coefficient ideal \[ I=\{\,\mathrm{lc}(f)\mid f\in J\,\}\subseteq R \] Because \(R\) is Noetherian, \(I\) is finitely generated, say \(I=(a_1,\dots,a_n)\).

Generators with above leading coefficients

For each \(a_i\) pick \(f_i(x)\in J\) whose leading coefficient is \(a_i\) and whose degree is minimal among such elements. Let \(d_i=\deg f_i\) and set \(d=\max_i d_i\).

Generate all lower-degree elements

Let \[ J_{\lt d}=\{\,\text{coefficients of all }f\in J\text{ of degree }\lt d\,\}\subseteq R \] This is an ideal of \(R\), hence finitely generated: \(J_{\lt d}=(b_1,\dots,b_m)\). For each \(b_j\) pick \(g_j(x)\in J\) whose leading coefficient equals \(b_j\) and degree \( \lt d \).

Above elements generate \(J\), i.e. \((f_i,g_j) = J\)

Let \(f(x)\in J\). Write \(\mathrm{lc}(f)=\sum_i r_i a_i\) (since \(a_i\) generate \(I\)). Subtract \(\sum_i r_i x^{\deg f-d_i}f_i(x)\) from \(f(x)\); the result lies in \(J\) and has degree \( \lt \deg f \). Induction on degree show every element of \(J\) can be expressed as an \(R[x]\)-linear combination of the \(f_i,g_j\). Thus \(J\) is finitely generated.

Hence \(R[x]\) is Noetherian.

Inductive case

If \(R[x_1,\dots,x_n]\) is Noetherian, then by the same argument \(R[x_1,\dots,x_n][x_{n+1}]\) is Noetherian.
Therefore by induction every finitely generated polynomial extension of a Noetherian ring is Noetherian.

Summary