[proofplan] We prove $d(G_\mathfrak{m}(A)) \ge \dim(A)$ by induction on $d = d(G_\mathfrak{m}(A))$. In the base case $d = 0$, the Hilbert function of $G_\mathfrak{m}(A)$ eventually vanishes, forcing $\mathfrak{m}^n = \mathfrak{m}^{n+1}$ for large $n$. [Nakayama's Lemma](/theorems/2935) then gives $\mathfrak{m}^n = 0$, so $A$ is Artinian and $\dim(A) = 0$. For the inductive step $d > 0$, we take an arbitrary prime chain $\mathfrak{p}_0 \subsetneq \cdots \subsetneq \mathfrak{p}_r$ in $A$ with $r \ge 1$, pass to the quotient domain $A/\mathfrak{p}_0$, pick a non-zero-divisor $x \in \mathfrak{p}_1 \setminus \mathfrak{p}_0$, and apply the [Reduction by a Non-Zero-Divisor](/theorems/2124) to drop $d$ by at least one. The inductive hypothesis then bounds $r - 1$, yielding $r \le d$, and since the chain was arbitrary, $\dim(A) \le d$. [/proofplan] [step:Handle the base case $d = 0$: show $A$ is Artinian and $\dim(A) = 0$] Suppose $d = d(G_\mathfrak{m}(A)) = 0$. By definition, $d(G_\mathfrak{m}(A))$ is the degree of the Hilbert-Samuel polynomial $\ell(A/\mathfrak{m}^n)$ for large $n$. A polynomial of degree $0$ is eventually constant, so there exists $n_0 \ge 1$ such that $\ell(A/\mathfrak{m}^n) = \ell(A/\mathfrak{m}^{n+1})$ for all $n \ge n_0$. Equivalently, $\ell(\mathfrak{m}^n / \mathfrak{m}^{n+1}) = 0$ for all $n \ge n_0$, which means $\mathfrak{m}^n = \mathfrak{m}^{n+1}$ for all $n \ge n_0$. Fix such an $n \ge n_0$. The equality $\mathfrak{m}^{n+1} = \mathfrak{m}^n$ can be rewritten as $\mathfrak{m} \cdot \mathfrak{m}^n = \mathfrak{m}^n$. Since $A$ is Noetherian, $\mathfrak{m}^n$ is a finitely generated $A$-module, and $\mathfrak{m} \subseteq \operatorname{Jac}(A)$ (because $(A, \mathfrak{m})$ is local, so $\mathfrak{m}$ is the unique maximal ideal and $\operatorname{Jac}(A) = \mathfrak{m}$). By [Nakayama's Lemma](/theorems/2935), $\mathfrak{m}^n = 0$. Since $\mathfrak{m}^n = 0$, every prime ideal $\mathfrak{p}$ of $A$ satisfies $\mathfrak{m}^n \subset \mathfrak{p}$, so $\mathfrak{m} = \sqrt{\mathfrak{m}^n} \subset \sqrt{\mathfrak{p}} = \mathfrak{p}$, giving $\mathfrak{p} = \mathfrak{m}$. The only prime ideal of $A$ is $\mathfrak{m}$, so every prime chain has length $0$, and $\dim(A) = 0 = d$. [guided] The quantity $d(G_\mathfrak{m}(A))$ measures the polynomial growth rate of $\ell(A/\mathfrak{m}^n)$ as $n \to \infty$. When $d = 0$, this function is eventually constant: there exists $n_0 \ge 1$ such that $\ell(A/\mathfrak{m}^n)$ takes the same value for all $n \ge n_0$. In particular, for $n \ge n_0$: \begin{align*} \ell(\mathfrak{m}^n / \mathfrak{m}^{n+1}) = \ell(A/\mathfrak{m}^{n+1}) - \ell(A/\mathfrak{m}^n) = 0. \end{align*} Since the length of $\mathfrak{m}^n / \mathfrak{m}^{n+1}$ is zero, the module $\mathfrak{m}^n / \mathfrak{m}^{n+1}$ is the zero module, i.e., $\mathfrak{m}^{n+1} = \mathfrak{m}^n$. Rewriting: $\mathfrak{m} \cdot \mathfrak{m}^n = \mathfrak{m}^n$. We want to conclude $\mathfrak{m}^n = 0$ from the self-reproduction equation $\mathfrak{m} \cdot \mathfrak{m}^n = \mathfrak{m}^n$. This is exactly the setup for [Nakayama's Lemma](/theorems/2935): the module $M = \mathfrak{m}^n$ is finitely generated (since $A$ is Noetherian) and satisfies $\mathfrak{a}M = M$ where $\mathfrak{a} = \mathfrak{m}$. Since $(A, \mathfrak{m})$ is local, $\operatorname{Jac}(A) = \mathfrak{m}$, so $\mathfrak{a} \subseteq \operatorname{Jac}(A)$. Nakayama's Lemma gives $\mathfrak{m}^n = 0$. Now $\mathfrak{m}^n = 0$ means $A$ has a unique prime ideal. Indeed, if $\mathfrak{p}$ is any prime of $A$, then $0 = \mathfrak{m}^n \subset \mathfrak{p}$, so $\mathfrak{m} \subset \mathfrak{p}$ (since $\mathfrak{p}$ is prime and $\mathfrak{m}^n \subset \mathfrak{p}$ implies some factor $\mathfrak{m} \subset \mathfrak{p}$). But $\mathfrak{m}$ is maximal, so $\mathfrak{p} = \mathfrak{m}$. With only one prime ideal, no strictly ascending chain of primes of length $\ge 1$ exists, so $\dim(A) = 0 = d$. [/guided] [/step] [step:Set up the inductive step: reduce to a quotient domain by passing to $A/\mathfrak{p}_0$] Now suppose $d = d(G_\mathfrak{m}(A)) > 0$ and that the inequality holds for all Noetherian local rings with smaller $d$-value. Let $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_r$ be an arbitrary chain of prime ideals in $A$ with $r \ge 1$. We need to show $r \le d$. Pass to the quotient ring $\bar{A} = A / \mathfrak{p}_0$. Since $\mathfrak{p}_0$ is prime, $\bar{A}$ is an integral domain. Since $A$ is Noetherian and local with maximal ideal $\mathfrak{m}$, the quotient $\bar{A}$ is a Noetherian local ring with maximal ideal $\bar{\mathfrak{m}} = \mathfrak{m} / \mathfrak{p}_0$. The surjection $A \to \bar{A}$ induces a surjection $A / \mathfrak{m}^n \to \bar{A} / \bar{\mathfrak{m}}^n$ for each $n \ge 1$ (since $\bar{\mathfrak{m}}^n = (\mathfrak{m}^n + \mathfrak{p}_0)/\mathfrak{p}_0$). Since surjections do not increase length: \begin{align*} \ell(\bar{A} / \bar{\mathfrak{m}}^n) \le \ell(A / \mathfrak{m}^n). \end{align*} Taking polynomial degrees, $d(G_{\bar{\mathfrak{m}}}(\bar{A})) \le d(G_\mathfrak{m}(A)) = d$. [guided] We are given a chain of primes $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_r$ in $A$ with $r \ge 1$, and we need to bound $r$ from above by $d$. The strategy is to reduce to an integral domain and then apply the non-zero-divisor reduction. The quotient $\bar{A} = A/\mathfrak{p}_0$ is an integral domain (since $\mathfrak{p}_0$ is prime). It is Noetherian (as a quotient of a Noetherian ring) and local with maximal ideal $\bar{\mathfrak{m}} = \mathfrak{m}/\mathfrak{p}_0$. The images $\bar{\mathfrak{p}}_1 \subsetneq \cdots \subsetneq \bar{\mathfrak{p}}_r$ form a chain of length $r$ in $\bar{A}$ (with $\bar{\mathfrak{p}}_0 = 0$). Why does the $d$-invariant not increase upon passage to $\bar{A}$? The canonical surjection $\pi: A \to \bar{A}$ sends $\mathfrak{m}^n$ onto $\bar{\mathfrak{m}}^n$ and induces a surjection $A/\mathfrak{m}^n \to \bar{A}/\bar{\mathfrak{m}}^n$ for each $n$. For modules of finite length, surjections do not increase length, so $\ell(\bar{A}/\bar{\mathfrak{m}}^n) \le \ell(A/\mathfrak{m}^n)$. Since both sides are eventually polynomial in $n$, the degree of the left-hand side is at most the degree of the right-hand side: $d(G_{\bar{\mathfrak{m}}}(\bar{A})) \le d$. [/guided] [/step] [step:Pick a non-zero-divisor and apply the Reduction by a Non-Zero-Divisor to drop $d$ by one] Write $\bar{\mathfrak{p}}_i = \mathfrak{p}_i / \mathfrak{p}_0$ for the image of $\mathfrak{p}_i$ in $\bar{A}$. Since $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1$, we can choose $x \in \mathfrak{p}_1 \setminus \mathfrak{p}_0$, and let $\bar{x}$ denote the image of $x$ in $\bar{A}$. Since $\bar{A}$ is an integral domain and $\bar{x} \neq 0$ (because $x \notin \mathfrak{p}_0$), the element $\bar{x}$ is a non-zero-divisor in $\bar{A}$. Moreover, $\bar{x} \in \bar{\mathfrak{p}}_1 \subset \bar{\mathfrak{m}}$. The images of $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ in $\bar{A} / \bar{x}\bar{A}$ form a chain of length $r - 1$ (the image of $\bar{\mathfrak{p}}_1$ contains $\bar{x}$, so the chain $\bar{\mathfrak{p}}_1/\bar{x}\bar{A} \subsetneq \cdots \subsetneq \bar{\mathfrak{p}}_r/\bar{x}\bar{A}$ has $r - 1$ strict inclusions). Thus \begin{align*} \dim(\bar{A} / \bar{x}\bar{A}) \ge r - 1. \end{align*} By [Reduction by a Non-Zero-Divisor](/theorems/2124), since $\bar{x} \in \bar{\mathfrak{m}}$ is a non-zero-divisor in the Noetherian local ring $\bar{A}$: \begin{align*} d(G_{\bar{\mathfrak{m}}/\bar{x}\bar{A}}(\bar{A}/\bar{x}\bar{A})) \le d(G_{\bar{\mathfrak{m}}}(\bar{A})) - 1 \le d - 1. \end{align*} [guided] We now exploit the fact that $\bar{A}$ is a domain. Pick any $x \in \mathfrak{p}_1 \setminus \mathfrak{p}_0$ and let $\bar{x}$ be its image in $\bar{A}$. Why is $\bar{x}$ a non-zero-divisor? Because $\bar{A} = A/\mathfrak{p}_0$ is an integral domain and $\bar{x} \neq 0$ (since $x \notin \mathfrak{p}_0$). Every nonzero element of an integral domain is a non-zero-divisor. Also $\bar{x} \in \bar{\mathfrak{m}}$ because $x \in \mathfrak{p}_1 \subset \mathfrak{m}$. The chain $\bar{\mathfrak{p}}_1 \subsetneq \cdots \subsetneq \bar{\mathfrak{p}}_r$ descends to the quotient $\bar{A}/\bar{x}\bar{A}$: since $\bar{x} \in \bar{\mathfrak{p}}_1 \subset \bar{\mathfrak{p}}_i$ for all $i \ge 1$, each $\bar{\mathfrak{p}}_i$ survives in the quotient. The images $\bar{\mathfrak{p}}_1/\bar{x}\bar{A} \subsetneq \cdots \subsetneq \bar{\mathfrak{p}}_r/\bar{x}\bar{A}$ form a chain of $r - 1$ strict inclusions (they remain distinct because $\bar{\mathfrak{p}}_i \subsetneq \bar{\mathfrak{p}}_{i+1}$ and both contain $\bar{x}\bar{A}$). So $\dim(\bar{A}/\bar{x}\bar{A}) \ge r - 1$. Now apply the [Reduction by a Non-Zero-Divisor](/theorems/2124). This theorem states: if $(B, \mathfrak{n})$ is a Noetherian local ring and $y \in \mathfrak{n}$ is a non-zero-divisor, then $d(G_{\mathfrak{n}/yB}(B/yB)) \le d(G_\mathfrak{n}(B)) - 1$. We apply it with $B = \bar{A}$, $\mathfrak{n} = \bar{\mathfrak{m}}$, $y = \bar{x}$. The hypotheses are satisfied: $\bar{A}$ is Noetherian local, $\bar{x} \in \bar{\mathfrak{m}}$, and $\bar{x}$ is a non-zero-divisor. The conclusion gives $d(G_{\bar{\mathfrak{m}}/\bar{x}\bar{A}}(\bar{A}/\bar{x}\bar{A})) \le d(G_{\bar{\mathfrak{m}}}(\bar{A})) - 1 \le d - 1$. [/guided] [/step] [step:Apply the inductive hypothesis and conclude $\dim(A) \le d$] The ring $\bar{A}/\bar{x}\bar{A}$ is a Noetherian local ring with $d$-invariant at most $d - 1$. By the inductive hypothesis (applied to this ring): \begin{align*} \dim(\bar{A}/\bar{x}\bar{A}) \le d(G_{\bar{\mathfrak{m}}/\bar{x}\bar{A}}(\bar{A}/\bar{x}\bar{A})) \le d - 1. \end{align*} Combining with $\dim(\bar{A}/\bar{x}\bar{A}) \ge r - 1$ from the previous step: \begin{align*} r - 1 \le d - 1, \end{align*} so $r \le d$. Since the chain $\mathfrak{p}_0 \subsetneq \cdots \subsetneq \mathfrak{p}_r$ was an arbitrary prime chain of length $r \ge 1$ in $A$, and we have also handled the case $r = 0$ (which satisfies $r = 0 \le d$ since $d \ge 1$ in the inductive step), we conclude $\dim(A) \le d = d(G_\mathfrak{m}(A))$. [/step]