[proofplan] For part (1), we establish equivalence between $\mathfrak{q}$ being $\mathfrak{m}$-primary and the containment $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$. The forward direction uses the fact that in a Noetherian ring, every ideal contains a power of its radical (so $\sqrt{\mathfrak{q}} = \mathfrak{m}$ forces $\mathfrak{m}^t \subset \mathfrak{q}$), together with the containment $\mathfrak{q} \subset \sqrt{\mathfrak{q}} = \mathfrak{m}$. The reverse direction computes $\sqrt{\mathfrak{q}}$ using monotonicity of the radical. For part (2), we show that $A/\mathfrak{q}$ is a Noetherian ring whose unique prime ideal is $\mathfrak{m}/\mathfrak{q}$, hence has Krull dimension zero, which for Noetherian rings implies Artinian. [/proofplan] [step:Prove the forward direction of (1): $\mathfrak{m}$-primary implies $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$] Assume $\mathfrak{q}$ is $\mathfrak{m}$-primary, meaning $\sqrt{\mathfrak{q}} = \mathfrak{m}$. For the upper containment: since $\mathfrak{q} \subset \sqrt{\mathfrak{q}} = \mathfrak{m}$ (every ideal is contained in its radical), we have $\mathfrak{q} \subset \mathfrak{m}$. For the lower containment: since $A$ is Noetherian, the maximal ideal $\mathfrak{m}$ is finitely generated, say $\mathfrak{m} = (a_1, \ldots, a_s)$. Since $\sqrt{\mathfrak{q}} = \mathfrak{m}$, each generator $a_i$ satisfies $a_i^{n_i} \in \mathfrak{q}$ for some $n_i \geq 1$. Set $t = 1 + \sum_{i=1}^{s}(n_i - 1)$. Every monomial of degree $t$ in $a_1, \ldots, a_s$ has the form $a_1^{e_1} \cdots a_s^{e_s}$ with $e_1 + \cdots + e_s = t$. Since $t > \sum_{i=1}^{s}(n_i - 1)$, by the pigeonhole principle there exists some index $i$ with $e_i \geq n_i$, so $a_i^{e_i} \in \mathfrak{q}$ and hence $a_1^{e_1} \cdots a_s^{e_s} \in \mathfrak{q}$. Since $\mathfrak{m}^t$ is generated by all such monomials, we conclude $\mathfrak{m}^t \subset \mathfrak{q}$. [guided] We assume $\mathfrak{q}$ is $\mathfrak{m}$-primary, i.e., $\sqrt{\mathfrak{q}} = \mathfrak{m}$. We must show $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$ for some $t \geq 1$. The upper containment $\mathfrak{q} \subset \mathfrak{m}$ is immediate: for any ideal $\mathfrak{q}$ we have $\mathfrak{q} \subset \sqrt{\mathfrak{q}}$, and $\sqrt{\mathfrak{q}} = \mathfrak{m}$ by assumption. The lower containment requires more work and is where the Noetherian hypothesis enters. Since $A$ is Noetherian, $\mathfrak{m}$ is finitely generated: write $\mathfrak{m} = (a_1, \ldots, a_s)$. The condition $\sqrt{\mathfrak{q}} = \mathfrak{m}$ means every element of $\mathfrak{m}$ has some power lying in $\mathfrak{q}$. In particular, for each generator $a_i$, there exists $n_i \geq 1$ with $a_i^{n_i} \in \mathfrak{q}$. How large must $t$ be to guarantee $\mathfrak{m}^t \subset \mathfrak{q}$? The ideal $\mathfrak{m}^t$ is generated by all monomials $a_1^{e_1} \cdots a_s^{e_s}$ with $e_1 + \cdots + e_s = t$. We need each such monomial to lie in $\mathfrak{q}$. If $e_i \geq n_i$ for some $i$, then $a_i^{e_i} \in \mathfrak{q}$ (since $\mathfrak{q}$ is an ideal and $a_i^{n_i} \in \mathfrak{q}$), so the entire monomial lies in $\mathfrak{q}$. We want every tuple $(e_1, \ldots, e_s)$ with $\sum e_i = t$ to have at least one $e_i \geq n_i$. The worst case for avoiding this is $e_i = n_i - 1$ for all $i$, which gives $\sum e_i = \sum(n_i - 1)$. So choosing $t = 1 + \sum_{i=1}^{s}(n_i - 1)$ ensures that in every monomial of degree $t$, the pigeonhole principle forces some $e_i \geq n_i$. Hence $\mathfrak{m}^t \subset \mathfrak{q}$. [/guided] [/step] [step:Prove the reverse direction of (1): $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$ implies $\mathfrak{q}$ is $\mathfrak{m}$-primary] Assume $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$ for some $t \geq 1$. Taking radicals and using monotonicity of the radical operation: \begin{align*} \mathfrak{m} = \sqrt{\mathfrak{m}^t} \subset \sqrt{\mathfrak{q}} \subset \sqrt{\mathfrak{m}} = \mathfrak{m}. \end{align*} The first equality holds because $\sqrt{\mathfrak{m}^t} = \sqrt{\mathfrak{m}} = \mathfrak{m}$ (the radical of any power of a prime ideal equals the prime ideal itself, and $\mathfrak{m}$ is maximal hence prime). The last equality holds because $\mathfrak{m}$ is a prime ideal. Therefore $\sqrt{\mathfrak{q}} = \mathfrak{m}$. Since $\mathfrak{m}$ is a maximal ideal, $\sqrt{\mathfrak{q}} = \mathfrak{m}$ implies $\mathfrak{q}$ is $\mathfrak{m}$-primary: in a Noetherian ring, an ideal whose radical is maximal is primary. [guided] Assume $\mathfrak{m}^t \subset \mathfrak{q} \subset \mathfrak{m}$. We apply the radical operation to the chain of containments. Recall that the radical is monotone: if $I \subset J$ then $\sqrt{I} \subset \sqrt{J}$. Thus \begin{align*} \sqrt{\mathfrak{m}^t} \subset \sqrt{\mathfrak{q}} \subset \sqrt{\mathfrak{m}}. \end{align*} Since $\mathfrak{m}$ is a maximal ideal (hence prime), $\sqrt{\mathfrak{m}} = \mathfrak{m}$. For the left side, $\sqrt{\mathfrak{m}^t} = \mathfrak{m}$ because $x \in \sqrt{\mathfrak{m}^t}$ iff $x^k \in \mathfrak{m}^t \subset \mathfrak{m}$ for some $k$, which (since $\mathfrak{m}$ is prime) iff $x \in \mathfrak{m}$; and conversely $x \in \mathfrak{m}$ implies $x^t \in \mathfrak{m}^t$. So the chain becomes \begin{align*} \mathfrak{m} \subset \sqrt{\mathfrak{q}} \subset \mathfrak{m}, \end{align*} forcing $\sqrt{\mathfrak{q}} = \mathfrak{m}$. Why does $\sqrt{\mathfrak{q}} = \mathfrak{m}$ (a maximal ideal) imply $\mathfrak{q}$ is primary? This is a standard fact: if $\sqrt{\mathfrak{q}}$ is maximal, then $A/\mathfrak{q}$ is a local ring whose unique prime ideal is $\mathfrak{m}/\mathfrak{q}$, which is also the nilradical. If $ab \in \mathfrak{q}$ with $a \notin \mathfrak{q}$, then $\bar{a} \neq 0$ in $A/\mathfrak{q}$ and $\bar{a}\bar{b} = 0$. Since $\bar{b}$ is in the nilradical $\mathfrak{m}/\mathfrak{q}$ (the only prime of $A/\mathfrak{q}$ must contain all zero divisors), some power $\bar{b}^n = 0$, i.e., $b^n \in \mathfrak{q}$, so $b \in \sqrt{\mathfrak{q}} = \mathfrak{m}$. More precisely, every zero divisor of $A/\mathfrak{q}$ is nilpotent (since the nilradical is the unique prime), so $\mathfrak{q}$ is primary. [/guided] [/step] [step:Prove part (2): $A/\mathfrak{q}$ is Artinian for any $\mathfrak{m}$-primary ideal $\mathfrak{q}$] Let $\mathfrak{q}$ be an $\mathfrak{m}$-primary ideal. We show that $A/\mathfrak{q}$ is Artinian by establishing two properties: it is Noetherian and has Krull dimension zero. First, $A/\mathfrak{q}$ is Noetherian because $A$ is Noetherian and any quotient of a Noetherian ring is Noetherian (ideals of $A/\mathfrak{q}$ correspond bijectively to ideals of $A$ containing $\mathfrak{q}$, and any ascending chain of the latter stabilises). Second, $A/\mathfrak{q}$ has Krull dimension zero. Let $\mathfrak{p}/\mathfrak{q}$ be any prime ideal of $A/\mathfrak{q}$, where $\mathfrak{p}$ is a prime ideal of $A$ containing $\mathfrak{q}$. Since $\mathfrak{q} \subset \mathfrak{p}$, taking radicals gives $\mathfrak{m} = \sqrt{\mathfrak{q}} \subset \sqrt{\mathfrak{p}} = \mathfrak{p}$. Since $\mathfrak{m}$ is the unique maximal ideal and $\mathfrak{p} \subset \mathfrak{m}$ (as $\mathfrak{p}$ is a proper ideal in a local ring), we conclude $\mathfrak{p} = \mathfrak{m}$. Therefore $\mathfrak{m}/\mathfrak{q}$ is the unique prime ideal of $A/\mathfrak{q}$, giving $\dim(A/\mathfrak{q}) = 0$. A Noetherian ring of Krull dimension zero is Artinian. This classical equivalence holds because $\dim(A/\mathfrak{q}) = 0$ means every prime is maximal, so by the structure theory for Noetherian rings with finitely many primes (all maximal), the ring is a finite product of Artinian local rings, hence Artinian. Therefore $A/\mathfrak{q}$ is Artinian. [guided] We need to show that $A/\mathfrak{q}$ is Artinian. The strategy is to establish two properties that together force the Artinian condition: the ring is Noetherian and has Krull dimension zero. **Noetherian property.** Since $A$ is Noetherian by hypothesis, and $A/\mathfrak{q}$ is a quotient ring, $A/\mathfrak{q}$ is Noetherian. This follows because ideals of $A/\mathfrak{q}$ are in order-preserving bijection with ideals of $A$ containing $\mathfrak{q}$ (via the correspondence $I/\mathfrak{q} \leftrightarrow I$), so any ascending chain in $A/\mathfrak{q}$ corresponds to an ascending chain in $A$ and therefore stabilises. **Krull dimension zero.** We claim that $\mathfrak{m}/\mathfrak{q}$ is the only prime ideal of $A/\mathfrak{q}$. Let $\mathfrak{p}$ be any prime ideal of $A$ with $\mathfrak{q} \subset \mathfrak{p}$. Since $\mathfrak{q} \subset \mathfrak{p}$ and $\mathfrak{p}$ is prime, $\sqrt{\mathfrak{q}} \subset \sqrt{\mathfrak{p}} = \mathfrak{p}$ (the radical of a prime is itself). So $\mathfrak{m} = \sqrt{\mathfrak{q}} \subset \mathfrak{p}$. But $A$ is a local ring with unique maximal ideal $\mathfrak{m}$, and every proper ideal is contained in $\mathfrak{m}$, so $\mathfrak{p} \subset \mathfrak{m}$. Combined with $\mathfrak{m} \subset \mathfrak{p}$, we get $\mathfrak{p} = \mathfrak{m}$. Therefore the only prime of $A/\mathfrak{q}$ is $\mathfrak{m}/\mathfrak{q}$, so $\dim(A/\mathfrak{q}) = 0$. **Combining.** A Noetherian ring of Krull dimension zero is Artinian. The proof of this equivalence: if every prime is maximal, then the set of minimal primes is finite (Noetherian rings have finitely many minimal primes), and these minimal primes are also maximal. The nilradical $\operatorname{nil}(A/\mathfrak{q}) = \mathfrak{m}/\mathfrak{q}$ contains a power of $\mathfrak{m}/\mathfrak{q}$ that is zero (since $\mathfrak{m}^t \subset \mathfrak{q}$ by part (1)). The descending chain condition then follows from the fact that $A/\mathfrak{q}$ has a finite composition series: the chain $A/\mathfrak{q} \supset \mathfrak{m}/\mathfrak{q} \supset \mathfrak{m}^2/\mathfrak{q} \supset \cdots \supset \mathfrak{m}^t/\mathfrak{q} = 0$ has each successive quotient $\mathfrak{m}^i/\mathfrak{m}^{i+1}$ a finitely generated module over the field $A/\mathfrak{m}$, hence finite-dimensional, hence Artinian. So $A/\mathfrak{q}$ is Artinian. [/guided] [/step]