[proofplan] We reduce to the single-variable case $R[T]$ by induction on $n$, using $R[T_1, \ldots, T_n] \cong R[T_1, \ldots, T_{n-1}][T_n]$. To prove $R[T]$ is noetherian, we show every ideal $\mathfrak{a} \trianglelefteq R[T]$ is finitely generated. The strategy is to extract from $\mathfrak{a}$ the leading-coefficient ideals $\mathfrak{a}(i) \trianglelefteq R$ at each degree $i$. These form an ascending chain in $R$ that stabilises by the noetherian hypothesis. We then build a finitely generated ideal $\mathfrak{b} \subseteq \mathfrak{a}$ whose leading-coefficient ideals match those of $\mathfrak{a}$ at every degree, and use a minimal-degree argument to show $\mathfrak{b} = \mathfrak{a}$. [/proofplan]

[step:Reduce to the single-variable case by induction on $n$] We prove that if $R$ is noetherian, then $R[T_1, \ldots, T_n]$ is noetherian for every $n \geq 1$, by induction on $n$. The base case $n = 1$ is the content of the remaining steps. For the inductive step, suppose $R[T_1, \ldots, T_{n-1}]$ is noetherian. Since $R[T_1, \ldots, T_n] \cong R[T_1, \ldots, T_{n-1}][T_n]$ as $R$-algebras (by the [Universal Property of Polynomial Algebra](/theorems/2820) applied iteratively), and $R[T_1, \ldots, T_{n-1}]$ is noetherian by the inductive hypothesis, the base case applied to the noetherian ring $R[T_1, \ldots, T_{n-1}]$ shows $R[T_1, \ldots, T_n]$ is noetherian. For the "finitely generated algebra" formulation: if $A$ is a finitely generated $R$-algebra with generators $a_1, \ldots, a_n$, then by the [Universal Property of Polynomial Algebra](/theorems/2820) there is a surjective $R$-algebra homomorphism $R[T_1, \ldots, T_n] \twoheadrightarrow A$. Since $R[T_1, \ldots, T_n]$ is noetherian, and the image of a noetherian ring under a surjective ring homomorphism is noetherian (any ascending chain of ideals in $A$ lifts to an ascending chain in $R[T_1, \ldots, T_n]$ via preimages), $A$ is noetherian. [/step]

[step:Define the leading-coefficient ideals $\mathfrak{a}(i) \trianglelefteq R$ and show they form an ascending chain]Assume $R$ is noetherian. Let $\mathfrak{a} \trianglelefteq R[T]$ be an ideal. For each $i \geq 0$, define \begin{align*} \mathfrak{a}(i) = \{c \in R : \text{there exists } f \in \mathfrak{a} \text{ with } \deg(f) = i \text{ and leading coefficient } c\} \cup \{0\}. \end{align*} We verify that $\mathfrak{a}(i) \trianglelefteq R$ is an ideal. Let $c, c' \in \mathfrak{a}(i)$ with corresponding polynomials $f, f' \in \mathfrak{a}$ of degree $i$ and leading coefficients $c, c'$ respectively. If $c + c' \neq 0$, then $f + f'$ has degree $i$ and leading coefficient $c + c'$, so $c + c' \in \mathfrak{a}(i)$. If $c + c' = 0$, then $c + c' = 0 \in \mathfrak{a}(i)$ by definition. For $r \in R$, the polynomial $rf \in \mathfrak{a}$ has degree $i$ (when $rc \neq 0$) with leading coefficient $rc$, so $rc \in \mathfrak{a}(i)$; when $rc = 0$, we have $rc = 0 \in \mathfrak{a}(i)$. The chain $\mathfrak{a}(0) \subseteq \mathfrak{a}(1) \subseteq \mathfrak{a}(2) \subseteq \cdots$ is ascending: if $c \in \mathfrak{a}(i)$ via a polynomial $f \in \mathfrak{a}$ of degree $i$ with leading coefficient $c$, then $Tf \in \mathfrak{a}$ has degree $i + 1$ and leading coefficient $c$, so $c \in \mathfrak{a}(i+1)$.[/step]

[guided]The idea behind the leading-coefficient ideals is to "project" the information in $\mathfrak{a}$ down to $R$. A polynomial in $R[T]$ of degree $i$ is determined by its coefficients in $R$, and the leading coefficient carries the most information for the purpose of generating $\mathfrak{a}$. For each $i \geq 0$, define \begin{align*} \mathfrak{a}(i) = \{c \in R : \text{there exists } f \in \mathfrak{a} \text{ with } \deg(f) = i \text{ and leading coefficient } c\} \cup \{0\}. \end{align*} Why include $\{0\}$ separately? Because the zero polynomial has no well-defined degree or leading coefficient, yet $0$ must belong to every ideal. We verify $\mathfrak{a}(i) \trianglelefteq R$. Let $c, c' \in \mathfrak{a}(i)$ with witnessing polynomials $f, f' \in \mathfrak{a}$ of degree $i$. Adding: $f + f' \in \mathfrak{a}$. If $c + c' \neq 0$, then $f + f'$ has degree $i$ and leading coefficient $c + c'$, so $c + c' \in \mathfrak{a}(i)$. If $c + c' = 0$, the leading terms cancel, but $0 \in \mathfrak{a}(i)$ by definition. For $r \in R$: $rf \in \mathfrak{a}$ has leading coefficient $rc$ (when nonzero), so $rc \in \mathfrak{a}(i)$; when $rc = 0$, we use $0 \in \mathfrak{a}(i)$. Why is the chain ascending? If $f \in \mathfrak{a}$ has degree $i$ and leading coefficient $c$, then $T \cdot f \in \mathfrak{a}$ (since $\mathfrak{a}$ is an ideal of $R[T]$) has degree $i + 1$ and the same leading coefficient $c$. So every element of $\mathfrak{a}(i)$ belongs to $\mathfrak{a}(i+1)$, giving $\mathfrak{a}(i) \subseteq \mathfrak{a}(i+1)$.[/guided]

[step:Use the noetherian hypothesis on $R$ to stabilise the chain and select polynomial witnesses]Since $R$ is noetherian, the ascending chain $\mathfrak{a}(0) \subseteq \mathfrak{a}(1) \subseteq \cdots$ of ideals of $R$ stabilises: there exists $m \geq 0$ such that $\mathfrak{a}(i) = \mathfrak{a}(m)$ for all $i \geq m$. For each $0 \leq i \leq m$, the ideal $\mathfrak{a}(i)$ is finitely generated (since $R$ is noetherian). Write $\mathfrak{a}(i) = (b_{i,1}, \ldots, b_{i,n_i})$ with generators $b_{i,j} \in R$. For each generator $b_{i,j}$, choose a polynomial $f_{i,j} \in \mathfrak{a}$ of degree $i$ with leading coefficient $b_{i,j}$. Define the ideal \begin{align*} \mathfrak{b} = (f_{i,j} : 0 \leq i \leq m, \; 1 \leq j \leq n_i) \trianglelefteq R[T]. \end{align*} This is a finitely generated ideal of $R[T]$ contained in $\mathfrak{a}$ (since each $f_{i,j} \in \mathfrak{a}$). The total number of generators is $n_0 + n_1 + \cdots + n_m < \infty$.[/step]

[guided]The noetherian property of $R$ enters in two ways. First, the ascending chain $\mathfrak{a}(0) \subseteq \mathfrak{a}(1) \subseteq \cdots$ must stabilise at some index $m$, because an infinite strictly ascending chain of ideals would violate the noetherian condition. Second, each ideal $\mathfrak{a}(i)$ is finitely generated — this is equivalent to the noetherian property by the [characterisation of noetherian modules via finitely generated submodules](/theorems/2810). So fix $m$ with $\mathfrak{a}(i) = \mathfrak{a}(m)$ for all $i \geq m$. For each $0 \leq i \leq m$, write $\mathfrak{a}(i) = (b_{i,1}, \ldots, b_{i,n_i})$ and for each $b_{i,j}$, pick a witnessing polynomial $f_{i,j} \in \mathfrak{a}$ of degree $i$ with leading coefficient $b_{i,j}$. Define $\mathfrak{b} = (f_{i,j} : 0 \leq i \leq m, \; 1 \leq j \leq n_i)$. This is a finitely generated ideal of $R[T]$ with $\mathfrak{b} \subseteq \mathfrak{a}$. The role of $\mathfrak{b}$ is to be a "finite approximation" to $\mathfrak{a}$ that captures all the leading-coefficient data.[/guided]

[step:Show $\mathfrak{b} = \mathfrak{a}$ via a minimal-degree argument]We show $\mathfrak{a} \subseteq \mathfrak{b}$ (the reverse inclusion is immediate). Suppose for contradiction that $\mathfrak{a} \setminus \mathfrak{b} \neq \varnothing$. Among all polynomials in $\mathfrak{a} \setminus \mathfrak{b}$, choose $f$ of minimal degree $d$. **Case 1: $d \leq m$.** The leading coefficient $c$ of $f$ lies in $\mathfrak{a}(d) = (b_{d,1}, \ldots, b_{d,n_d})$. Write $c = \sum_{j=1}^{n_d} r_j b_{d,j}$ for some $r_j \in R$. Define $g = \sum_{j=1}^{n_d} r_j f_{d,j} \in \mathfrak{b}$. Then $g$ has degree $d$ and leading coefficient $\sum_j r_j b_{d,j} = c$, the same as $f$. Hence $f - g$ has degree strictly less than $d$, and $f - g \in \mathfrak{a}$ (since $f \in \mathfrak{a}$ and $g \in \mathfrak{b} \subseteq \mathfrak{a}$). By the minimality of $d$, we have $f - g \in \mathfrak{b}$, so $f = (f - g) + g \in \mathfrak{b}$, contradicting $f \notin \mathfrak{b}$. **Case 2: $d > m$.** The leading coefficient $c$ of $f$ lies in $\mathfrak{a}(d) = \mathfrak{a}(m) = (b_{m,1}, \ldots, b_{m,n_m})$. Write $c = \sum_{j=1}^{n_m} r_j b_{m,j}$. Define $g = \sum_{j=1}^{n_m} r_j T^{d-m} f_{m,j} \in \mathfrak{b}$. Then $g$ has degree $d$ and leading coefficient $\sum_j r_j b_{m,j} = c$. Again $f - g$ has degree less than $d$, $f - g \in \mathfrak{a}$, so $f - g \in \mathfrak{b}$ by minimality, giving $f \in \mathfrak{b}$, a contradiction. In both cases we reach a contradiction, so $\mathfrak{a} \setminus \mathfrak{b} = \varnothing$ and $\mathfrak{a} = \mathfrak{b}$. Since $\mathfrak{b}$ is finitely generated, so is $\mathfrak{a}$. As $\mathfrak{a}$ was an arbitrary ideal of $R[T]$, the ring $R[T]$ is noetherian.[/step]

[guided]The heart of the proof is this minimal-degree argument. We want to show that the finitely generated ideal $\mathfrak{b}$ captures all of $\mathfrak{a}$. The strategy: assume some $f \in \mathfrak{a}$ escapes $\mathfrak{b}$, pick such an $f$ of smallest possible degree $d$, then use the leading-coefficient data to "shave off" the leading term and reduce to a polynomial of smaller degree — which must already lie in $\mathfrak{b}$ by the minimality assumption. Suppose $f \in \mathfrak{a} \setminus \mathfrak{b}$ has minimal degree $d$ and leading coefficient $c$. If $d \leq m$, then $c \in \mathfrak{a}(d)$ and the generators $b_{d,1}, \ldots, b_{d,n_d}$ of $\mathfrak{a}(d)$ are the leading coefficients of the polynomials $f_{d,1}, \ldots, f_{d,n_d} \in \mathfrak{b}$, each of degree $d$. Express $c = \sum_j r_j b_{d,j}$ and form $g = \sum_j r_j f_{d,j} \in \mathfrak{b}$. This $g$ has degree $d$ and the same leading coefficient as $f$, so $f - g$ has $\deg(f - g) < d$. Since $f - g \in \mathfrak{a}$ and has degree strictly less than $d$, the minimality of $d$ forces $f - g \in \mathfrak{b}$, hence $f = (f-g) + g \in \mathfrak{b}$ — contradiction. If $d > m$, we use the stabilisation $\mathfrak{a}(d) = \mathfrak{a}(m)$. The leading coefficient $c$ is still expressible as $c = \sum_j r_j b_{m,j}$ using generators of $\mathfrak{a}(m)$. But the polynomials $f_{m,j}$ have degree $m < d$, so we multiply by $T^{d-m}$ to boost them to degree $d$: define $g = \sum_j r_j T^{d-m} f_{m,j}$. This $g$ lies in $\mathfrak{b}$ (since $T^{d-m} f_{m,j}$ is an $R[T]$-multiple of $f_{m,j} \in \mathfrak{b}$), has degree $d$, and has leading coefficient $c$. Again $\deg(f - g) < d$ and $f - g \in \mathfrak{a}$, so $f - g \in \mathfrak{b}$ by minimality, giving $f \in \mathfrak{b}$. Both cases yield contradictions, so no such $f$ exists: $\mathfrak{a} = \mathfrak{b}$. Since $\mathfrak{b}$ is finitely generated, every ideal of $R[T]$ is finitely generated, which means $R[T]$ is noetherian.[/guided]