[proofplan] We reduce to showing that if $R$ is noetherian then $R[T]$ is noetherian, since $R[T_1, \dots, T_n] \cong R[T_1, \dots, T_{n-1}][T_n]$ and the result follows by induction on the number of indeterminates. Given an arbitrary ideal $\mathfrak{a} \trianglelefteq R[T]$, we extract the leading-coefficient ideals $\mathfrak{a}(i) \trianglelefteq R$ for each degree $i$, observe they form an ascending chain that stabilises by the noetherian hypothesis on $R$, and use chosen polynomial witnesses to build a finitely generated ideal $\mathfrak{b} \subseteq \mathfrak{a}$ whose leading-coefficient ideals match those of $\mathfrak{a}$ at every degree. A minimal-degree argument then forces $\mathfrak{b} = \mathfrak{a}$. [/proofplan] [step:Reduce to the single-variable case $R[T]$] It suffices to prove: if $R$ is a noetherian ring, then $R[T]$ is noetherian. Indeed, $R[T_1, \dots, T_n] \cong R[T_1, \dots, T_{n-1}][T_n]$, so by induction on $n$, the ring $R[T_1, \dots, T_n]$ is noetherian whenever $R$ is. For a finitely generated $R$-algebra $A$, there exist $n \geq 0$ and a surjective $R$-algebra homomorphism $\varphi: R[T_1, \dots, T_n] \to A$ (by the [Finitely Generated Algebras and Surjections](/theorems/2902) characterisation). Since $R[T_1, \dots, T_n]$ is noetherian, $A \cong R[T_1, \dots, T_n] / \ker \varphi$ is a quotient of a noetherian ring and hence noetherian. [guided] The statement "every finitely generated $R$-algebra is noetherian" involves an algebra with arbitrarily many generators. Why does it suffice to handle a single indeterminate? Because $R[T_1, \dots, T_n]$ is built iteratively: $R[T_1, \dots, T_n] = R[T_1, \dots, T_{n-1}][T_n]$. If we know the single-variable result "$S$ noetherian implies $S[T]$ noetherian," we apply it with $S = R$ to get $R[T_1]$ noetherian, then with $S = R[T_1]$ to get $R[T_1, T_2]$ noetherian, and so on. After $n$ applications, $R[T_1, \dots, T_n]$ is noetherian. For an arbitrary finitely generated $R$-algebra $A$, the [Finitely Generated Algebras and Surjections](/theorems/2902) characterisation provides a surjection $\varphi: R[T_1, \dots, T_n] \twoheadrightarrow A$. Thus $A \cong R[T_1, \dots, T_n] / \ker \varphi$. A quotient of a noetherian ring is noetherian: any ideal of $A$ corresponds to an ideal of $R[T_1, \dots, T_n]$ containing $\ker \varphi$, and the ascending chain condition passes to quotients. So $A$ is noetherian. [/guided] [/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 arbitrary ideal. For each integer $i \geq 0$, define the set of leading coefficients at degree $i$: \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*} [claim:Each $\mathfrak{a}(i)$ is an ideal of $R$] Let $c, c' \in \mathfrak{a}(i)$ and $r \in R$. If $c = 0$ or $c' = 0$, the claims $c + c' \in \mathfrak{a}(i)$ and $rc \in \mathfrak{a}(i)$ are immediate. Otherwise, choose $f, f' \in \mathfrak{a}$ of degree $i$ with leading coefficients $c$ and $c'$ respectively. Then $f + f' \in \mathfrak{a}$ has degree at most $i$ with leading coefficient $c + c'$ if $c + c' \neq 0$ (and $c + c' = 0 \in \mathfrak{a}(i)$ otherwise). Similarly, $rf \in \mathfrak{a}$ has degree $i$ with leading coefficient $rc$ when $rc \neq 0$ (and $rc = 0 \in \mathfrak{a}(i)$ otherwise). More precisely, if $\deg(f + f') < i$, then the leading coefficient $c + c'$ of degree $i$ is zero, which is in $\mathfrak{a}(i)$ by definition. Hence $\mathfrak{a}(i)$ is closed under addition and scalar multiplication by elements of $R$, so $\mathfrak{a}(i) \trianglelefteq R$. [/claim] [proof] Contained in the claim statement above. [/proof] [claim:The chain $\mathfrak{a}(0) \subseteq \mathfrak{a}(1) \subseteq \mathfrak{a}(2) \subseteq \cdots$ is ascending] Let $c \in \mathfrak{a}(i)$ with $c \neq 0$, witnessed by $f \in \mathfrak{a}$ with $\deg f = i$ and leading coefficient $c$. Then $Tf \in \mathfrak{a}$ (since $\mathfrak{a}$ is an ideal) has degree $i + 1$ and leading coefficient $c$. Hence $c \in \mathfrak{a}(i+1)$. [/claim] [proof] Contained in the claim statement above. [/proof] 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 $i \geq m$. [guided] The idea behind the leading-coefficient ideals is to reduce the problem of understanding $\mathfrak{a}$ (an ideal in the polynomial ring $R[T]$, which we do not yet know is noetherian) to understanding ideals in $R$ (which is noetherian by hypothesis). By collecting the leading coefficients at each degree, we obtain a filtration $\mathfrak{a}(0) \subseteq \mathfrak{a}(1) \subseteq \cdots$ of ideals in $R$. Why is the chain ascending? If a polynomial $f \in \mathfrak{a}$ has degree $i$ and leading coefficient $c$, then $Tf$ has degree $i+1$ and the same leading coefficient $c$. Multiplying by $T$ is the mechanism that pushes leading coefficients upward in degree. Since $R$ is noetherian, every ascending chain of ideals in $R$ stabilises (this is the ascending chain condition). Hence there exists $m \geq 0$ with $\mathfrak{a}(i) = \mathfrak{a}(m)$ for all $i \geq m$. This stabilisation index $m$ will be the degree bound controlling the finitely generated ideal we build next. [/guided] [/step] [step:Choose polynomial witnesses and build the finitely generated ideal $\mathfrak{b}$] For each $0 \leq i \leq m$, write $\mathfrak{a}(i) = (b_{i,1}, \dots, b_{i,n_i})$ for finitely many generators $b_{i,j} \in R$. This is possible because $R$ is noetherian, so every ideal of $R$ is finitely generated. For each pair $(i, j)$ with $0 \leq i \leq m$ and $1 \leq j \leq n_i$, choose a polynomial $f_{i,j} \in \mathfrak{a}$ of degree $i$ with leading coefficient $b_{i,j}$. Such a polynomial exists by definition of $\mathfrak{a}(i)$. 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 $R[T]$ (it has $\sum_{i=0}^{m} n_i$ generators) and $\mathfrak{b} \subseteq \mathfrak{a}$ since each $f_{i,j} \in \mathfrak{a}$. [guided] We are building a finitely generated ideal $\mathfrak{b} \subseteq \mathfrak{a}$ that we will show equals $\mathfrak{a}$. The generators of $\mathfrak{b}$ are polynomials $f_{i,j}$, one for each generator $b_{i,j}$ of the leading-coefficient ideal $\mathfrak{a}(i)$, for degrees $i = 0, 1, \dots, m$. Since $R$ is noetherian, each $\mathfrak{a}(i)$ is finitely generated, and there are finitely many degrees to consider (only $0$ through $m$, since the chain stabilises at $m$). So $\mathfrak{b}$ has finitely many generators in total. The key property we need is that $\mathfrak{b}$ "captures" all the leading coefficients that $\mathfrak{a}$ has at every degree. This is what we verify next. [/guided] [/step] [step:Verify that $\mathfrak{b}(i) = \mathfrak{a}(i)$ for all $i \geq 0$] Since $\mathfrak{b} \subseteq \mathfrak{a}$, we have $\mathfrak{b}(i) \subseteq \mathfrak{a}(i)$ for all $i \geq 0$. We show the reverse inclusion. For $0 \leq i \leq m$: each generator $b_{i,j}$ of $\mathfrak{a}(i)$ is the leading coefficient of $f_{i,j} \in \mathfrak{b}$, so $b_{i,j} \in \mathfrak{b}(i)$. Since $\mathfrak{b}(i)$ is an ideal of $R$ containing all the generators of $\mathfrak{a}(i)$, we have $\mathfrak{a}(i) \subseteq \mathfrak{b}(i)$. For $i > m$: $\mathfrak{a}(i) = \mathfrak{a}(m) \subseteq \mathfrak{b}(m) \subseteq \mathfrak{b}(i)$. The first equality is the stabilisation at $m$. The second inclusion follows from the case $i = m$ established above. The third inclusion holds because $\mathfrak{b}(m) \subseteq \mathfrak{b}(i)$ by the same multiplication-by-$T$ argument (if $g \in \mathfrak{b}$ has degree $m$ and leading coefficient $c$, then $T^{i-m} g \in \mathfrak{b}$ has degree $i$ and leading coefficient $c$). Hence $\mathfrak{b}(i) = \mathfrak{a}(i)$ for all $i \geq 0$. [guided] Why do we need $\mathfrak{b}(i) = \mathfrak{a}(i)$? Because the leading-coefficient ideals control membership in the ideal: if a polynomial $f \in \mathfrak{a}$ has the same leading coefficient as some polynomial in $\mathfrak{b}$ of the same degree, we can subtract to reduce the degree. The equality of leading-coefficient ideals ensures this subtraction is always possible. For degrees $i \leq m$, the equality is immediate: we built $\mathfrak{b}$ to contain polynomial witnesses $f_{i,j}$ whose leading coefficients generate $\mathfrak{a}(i)$. For degrees $i > m$, we use the stabilisation: $\mathfrak{a}(i) = \mathfrak{a}(m)$. We already know $\mathfrak{a}(m) = \mathfrak{b}(m)$, and $\mathfrak{b}(m) \subseteq \mathfrak{b}(i)$ because multiplying any degree-$m$ polynomial in $\mathfrak{b}$ by $T^{i-m}$ produces a degree-$i$ polynomial in $\mathfrak{b}$ with the same leading coefficient. Thus $\mathfrak{a}(i) \subseteq \mathfrak{b}(i)$, and the reverse inclusion $\mathfrak{b}(i) \subseteq \mathfrak{a}(i)$ holds because $\mathfrak{b} \subseteq \mathfrak{a}$. [/guided] [/step] [step:Conclude $\mathfrak{b} = \mathfrak{a}$ by a minimal-degree argument] Suppose for contradiction that $\mathfrak{a} \not\subseteq \mathfrak{b}$. Among all elements of $\mathfrak{a} \setminus \mathfrak{b}$, choose $f$ of minimal degree $d$. Let $c$ denote the leading coefficient of $f$, so $c \in \mathfrak{a}(d)$. Since $\mathfrak{b}(d) = \mathfrak{a}(d)$, there exists $g \in \mathfrak{b}$ of degree $d$ with leading coefficient $c$. (If $d \leq m$, such a $g$ is an $R$-linear combination of the $f_{d,j}$; if $d > m$, such a $g$ is an $R$-linear combination of $T^{d-i} f_{i,j}$ for appropriate $i \leq m$.) The polynomial $f - g$ lies in $\mathfrak{a}$ (since both $f \in \mathfrak{a}$ and $g \in \mathfrak{b} \subseteq \mathfrak{a}$) and has $\deg(f - g) < d$ because $f$ and $g$ have the same degree $d$ and the same leading coefficient $c$, so the degree-$d$ terms cancel. By minimality of $d$, since $f - g \in \mathfrak{a}$ has degree less than $d$, we must have $f - g \in \mathfrak{b}$. Since $g \in \mathfrak{b}$ as well, we obtain $f = (f - g) + g \in \mathfrak{b}$, contradicting $f \notin \mathfrak{b}$. Hence $\mathfrak{a} = \mathfrak{b}$, so $\mathfrak{a}$ is finitely generated. Since $\mathfrak{a}$ was an arbitrary ideal of $R[T]$, every ideal of $R[T]$ is finitely generated, which means $R[T]$ is noetherian. [guided] This is a standard minimal-degree argument, also called a "greedy" or "descent" argument. The idea is: if some polynomial $f \in \mathfrak{a}$ fails to lie in $\mathfrak{b}$, pick one of smallest possible degree $d$. We then use the equality $\mathfrak{b}(d) = \mathfrak{a}(d)$ to find a polynomial $g \in \mathfrak{b}$ that matches $f$ in degree and leading coefficient. Subtracting, $f - g$ has strictly smaller degree and still lies in $\mathfrak{a}$. By the minimality of $d$, $f - g$ must lie in $\mathfrak{b}$, and then $f = (f - g) + g \in \mathfrak{b}$ as well — a contradiction. What if $f - g = 0$, i.e., $f = g$? Then $f \in \mathfrak{b}$ directly, which is also a contradiction. What if $\deg(f - g) < 0$, i.e., $f - g = 0$? Same conclusion. In all cases we reach a contradiction, so the assumption $\mathfrak{a} \not\subseteq \mathfrak{b}$ is untenable. Since every ideal $\mathfrak{a} \trianglelefteq R[T]$ equals the finitely generated ideal $\mathfrak{b}$, every ideal of $R[T]$ is finitely generated, and $R[T]$ is noetherian. [/guided] [/step]