**Proof plan.** Given any ideal $I \trianglelefteq R[X]$, extract leading coefficients at each degree to form a chain of ideals $I_0 \subseteq I_1 \subseteq \cdots$ in $R$. Since $R$ is Noetherian, this chain stabilises. A finite selection of polynomials realising the generators then spans all of $I$. **Step 1: Define the leading-coefficient ideals.** For each $n \geq 0$, let \begin{align*} I_n = \{r \in R : r = 0 \text{ or } r \text{ is the leading coefficient of some } f \in I \text{ with } \deg f = n\} \cup \{0\}. \end{align*} Each $I_n$ is an ideal of $R$ (using the ideal property of $I$ under multiplication by $X$ and ring elements). The chain $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ is ascending (multiplying a polynomial in $I$ by $X$ raises its degree and preserves its leading coefficient, so $I_n \subseteq I_{n+1}$). **Step 2: The chain stabilises.** [claim: Chain Stabilises] Since $R$ is Noetherian, there exists $N$ such that $I_N = I_{N+1} = I_{N+2} = \cdots$. [/claim] [proof] An ascending chain of ideals in a Noetherian ring eventually stabilises by definition. [/proof] **Step 3: Choose finitely many generators.** Since $R$ is Noetherian, each $I_n$ is finitely generated. For each $0 \leq n \leq N$, write $I_n = (r^{(n)}_1, \ldots, r^{(n)}_{k(n)})$, and for each generator $r^{(n)}_i$, choose a polynomial $f^{(n)}_i \in I$ with leading coefficient $r^{(n)}_i$ and degree $n$. **Step 4: These polynomials generate $I$.** [claim: Finite Generation] The polynomials $\{f^{(n)}_i : 0 \leq n \leq N,\, 1 \leq i \leq k(n)\}$ generate $I$. [/claim] [proof] Suppose not. Pick $g \in I$ of minimal degree not in the submodule generated by the $f^{(n)}_i$. *Case $\deg g = m \leq N$:* The leading coefficient $r$ of $g$ lies in $I_m = (r^{(m)}_1, \ldots, r^{(m)}_{k(m)})$, so $r = \sum_i \lambda_i r^{(m)}_i$. Then $g - \sum_i \lambda_i f^{(m)}_i$ has degree less than $m$, lies in $I$, and cannot be generated by the $f^{(n)}_i$ (since $g$ cannot), contradicting minimality. *Case $\deg g = m > N$:* The leading coefficient $r$ of $g$ lies in $I_m = I_N$ (since the chain has stabilised). So $r = \sum_i \lambda_i r^{(N)}_i$. Then $g - \sum_i \lambda_i X^{m-N} f^{(N)}_i$ has degree less than $m$, lies in $I$, and leads to the same contradiction. In both cases we obtain a contradiction, so the $f^{(n)}_i$ generate $I$. [/proof] Since every ideal of $R[X]$ is finitely generated, $R[X]$ is Noetherian. $\square$