[proofplan] We prove each part in turn. Part (1) reduces the primality of $\sqrt{I}$ to the primary condition on $I$. Part (2) uses the fact that in $R/I$, the radical $\mathfrak{m}/I$ is the unique maximal ideal, making every element either a unit or nilpotent, which forces every zero divisor to be nilpotent. Part (3) embeds $R/(\mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n)$ into $\prod R/\mathfrak{q}_i$ and uses this to show every zero divisor in the intersection quotient is nilpotent. Part (4) is a two-step reduction: merge components sharing a radical using (3), then discard redundant components. Part (5) relies on two facts about Noetherian rings: every ideal is a finite intersection of irreducible ideals, and every irreducible ideal is primary. [/proofplan] [step:Prove (1): if $I$ is primary then $\sqrt{I}$ is prime] Suppose $I$ is primary. Let $a, b \in R$ with $ab \in \sqrt{I}$. Then $(ab)^k \in I$ for some $k \geq 1$, i.e., $a^k b^k \in I$. Since $I$ is primary, either $a^k \in I$ or $(b^k)^m \in I$ for some $m \geq 1$. In the first case, $a^k \in I$ gives $a \in \sqrt{I}$. In the second case, $b^{km} \in I$ gives $b \in \sqrt{I}$. Therefore $\sqrt{I}$ is prime. [guided] We need to show $\sqrt{I}$ is prime, i.e., for any $a, b \in R$ with $ab \in \sqrt{I}$, either $a \in \sqrt{I}$ or $b \in \sqrt{I}$. The condition $ab \in \sqrt{I}$ means $(ab)^k = a^k b^k \in I$ for some $k \geq 1$. Now we apply the definition of primary: $I$ is primary means that whenever a product $xy \in I$ with $x \notin I$, then $y^n \in I$ for some $n \geq 1$. Taking $x = a^k$ and $y = b^k$, we have $x \cdot y = a^k b^k \in I$. So either $a^k \in I$ (which gives $a \in \sqrt{I}$ directly) or $a^k \notin I$, in which case the primary condition forces $(b^k)^m \in I$ for some $m \geq 1$, giving $b^{km} \in I$ and hence $b \in \sqrt{I}$. In either case, $a \in \sqrt{I}$ or $b \in \sqrt{I}$, confirming $\sqrt{I} \in \operatorname{Spec}(R)$. [/guided] [/step] [step:Prove (2): if $\sqrt{I}$ is maximal, then $I$ is primary] Let $\mathfrak{m} = \sqrt{I}$ be a maximal ideal. Consider the quotient ring $\bar{R} = R/I$. The radical $\sqrt{I}/I$ is the nilradical of $\bar{R}$, and $\sqrt{I}/I = \mathfrak{m}/I$. Since $\mathfrak{m}$ is maximal and $I \subseteq \mathfrak{m}$, the ideal $\mathfrak{m}/I$ is a maximal ideal of $\bar{R}$. We claim $\mathfrak{m}/I$ is the unique maximal ideal of $\bar{R}$. Let $\mathfrak{n}/I$ be any maximal ideal of $\bar{R}$ (where $\mathfrak{n} \supseteq I$ is a maximal ideal of $R$). Since $\mathfrak{n} \supseteq I$, we have $\mathfrak{n} \supseteq \sqrt{I} = \mathfrak{m}$ (because every prime ideal containing $I$ contains $\sqrt{I}$, and maximal ideals are prime). Since $\mathfrak{m}$ is maximal and $\mathfrak{n} \supseteq \mathfrak{m}$, we get $\mathfrak{n} = \mathfrak{m}$. Therefore $\bar{R}$ is a local ring with maximal ideal $\mathfrak{m}/I$. In a local ring, every element is either a unit or belongs to the maximal ideal. Since the maximal ideal $\mathfrak{m}/I$ equals the nilradical, every non-unit is nilpotent. Now let $\bar{a}, \bar{b} \in \bar{R}$ with $\bar{a}\bar{b} = \bar{0}$ (i.e., $ab \in I$) and $\bar{b} \neq \bar{0}$ (i.e., $b \notin I$). If $\bar{a}$ were a unit, then $\bar{b} = \bar{a}^{-1} \cdot \bar{0} = \bar{0}$, contradicting $\bar{b} \neq \bar{0}$. So $\bar{a}$ is not a unit, hence $\bar{a} \in \mathfrak{m}/I$, hence $\bar{a}$ is nilpotent: $\bar{a}^n = \bar{0}$ for some $n \geq 1$, i.e., $a^n \in I$. This shows $I$ is primary. [guided] The idea is to study the quotient $R/I$ and show that every zero divisor in $R/I$ is nilpotent, which is equivalent to $I$ being primary. We first identify the structure of $R/I$. The nilradical of $R/I$ is $\sqrt{I}/I = \mathfrak{m}/I$. Since $\mathfrak{m}$ is maximal, $\mathfrak{m}/I$ is a maximal ideal of $R/I$. Is it the only maximal ideal? Any maximal ideal of $R/I$ corresponds to a maximal ideal $\mathfrak{n}$ of $R$ containing $I$. Such $\mathfrak{n}$ is prime and contains $I$, so $\mathfrak{n} \supseteq \sqrt{I} = \mathfrak{m}$. By maximality of $\mathfrak{m}$, $\mathfrak{n} = \mathfrak{m}$. So $R/I$ is a local ring with unique maximal ideal $\mathfrak{m}/I$, which coincides with the nilradical. In such a ring, every element is either a unit (if it lies outside $\mathfrak{m}/I$) or nilpotent (if it lies inside $\mathfrak{m}/I$). Now take $\bar{a}\bar{b} = \bar{0}$ with $\bar{b} \neq \bar{0}$. Then $\bar{a}$ cannot be a unit (since $\bar{a}$ being a unit and $\bar{a}\bar{b} = \bar{0}$ would force $\bar{b} = \bar{0}$). So $\bar{a}$ lies in $\mathfrak{m}/I$, the nilradical, and is therefore nilpotent. This means $a^n \in I$ for some $n$, confirming that $I$ is primary. Note that the converse fails: the ideal $I = (x^2, xy)$ in $R = k[x, y]$ (for a field $k$) has $\sqrt{I} = (x)$, which is prime, but $I$ is not primary since $x \cdot y \in I$ with $x \notin I$ and $y^n \notin I$ for all $n$. [/guided] [/step] [step:Prove (3): a finite intersection of $\mathfrak{p}$-primary ideals is $\mathfrak{p}$-primary] Let $\mathfrak{q}_1, \ldots, \mathfrak{q}_n$ be $\mathfrak{p}$-primary ideals, and set $\mathfrak{q} = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n$. First, $\sqrt{\mathfrak{q}} = \sqrt{\mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n} = \sqrt{\mathfrak{q}_1} \cap \cdots \cap \sqrt{\mathfrak{q}_n} = \mathfrak{p} \cap \cdots \cap \mathfrak{p} = \mathfrak{p}$. It remains to show $\mathfrak{q}$ is primary. The canonical map \begin{align*} \varphi: R/\mathfrak{q} \to \prod_{i=1}^n R/\mathfrak{q}_i, \quad r + \mathfrak{q} \mapsto (r + \mathfrak{q}_1, \ldots, r + \mathfrak{q}_n), \end{align*} is injective: $\ker \varphi = \{r + \mathfrak{q} : r \in \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n\} = \{r + \mathfrak{q} : r \in \mathfrak{q}\} = 0$. Let $ab \in \mathfrak{q}$ with $b \notin \mathfrak{q}$. Then $b \notin \mathfrak{q}_j$ for some $j$. Since $ab \in \mathfrak{q} \subseteq \mathfrak{q}_i$ for all $i$, and each $\mathfrak{q}_i$ is $\mathfrak{p}$-primary: - For $i = j$: $ab \in \mathfrak{q}_j$ and $b \notin \mathfrak{q}_j$, so $a^{m_j} \in \mathfrak{q}_j$ for some $m_j \geq 1$. - For $i \neq j$: $ab \in \mathfrak{q}_i$, so either $a \in \mathfrak{q}_i$ (giving $a^1 \in \mathfrak{q}_i$) or $b^{m_i} \in \mathfrak{q}_i$ for some $m_i \geq 1$. In every case, $a \in \sqrt{\mathfrak{q}_i} = \mathfrak{p}$ or $b \in \sqrt{\mathfrak{q}_i} = \mathfrak{p}$. Since $b \notin \mathfrak{q}_j$, and $\mathfrak{q}_j$ is $\mathfrak{p}$-primary with $ab \in \mathfrak{q}_j$, we get $a \in \sqrt{\mathfrak{q}_j} = \mathfrak{p}$. Let $m = m_j$, so $a^m \in \mathfrak{q}_j$. Since $a \in \mathfrak{p} = \sqrt{\mathfrak{q}_i}$ for each $i$, there exists $m_i$ with $a^{m_i} \in \mathfrak{q}_i$. Set $M = \max(m_1, \ldots, m_n)$. Then $a^M \in \mathfrak{q}_i$ for all $i$, so $a^M \in \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n = \mathfrak{q}$. This shows $\mathfrak{q}$ is primary. [/step] [step:Prove (4): every primary decomposition can be reduced to a minimal one] Let $I = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n$ be a primary decomposition. We produce a minimal decomposition in two reduction steps. **Merge components with the same radical.** Group the $\mathfrak{q}_i$ according to their radical: for each distinct prime $\mathfrak{p}$ appearing among $\sqrt{\mathfrak{q}_1}, \ldots, \sqrt{\mathfrak{q}_n}$, replace the collection of all $\mathfrak{q}_i$ with $\sqrt{\mathfrak{q}_i} = \mathfrak{p}$ by their intersection. By part (3), this intersection is $\mathfrak{p}$-primary. Relabelling, we obtain a decomposition $I = \mathfrak{q}_1' \cap \cdots \cap \mathfrak{q}_m'$ where $\sqrt{\mathfrak{q}_1'}, \ldots, \sqrt{\mathfrak{q}_m'}$ are pairwise distinct. **Discard redundant components.** If $\mathfrak{q}_i' \supseteq \bigcap_{j \neq i} \mathfrak{q}_j'$ for some $i$, then $I = \bigcap_{j \neq i} \mathfrak{q}_j'$, and $\mathfrak{q}_i'$ is redundant. Remove all such redundant components. The resulting decomposition satisfies $I \subsetneq \bigcap_{j \neq i} \mathfrak{q}_j'$ for each remaining $i$ (since no component is redundant), and the radicals remain pairwise distinct (since we only removed components). This is a minimal primary decomposition. [/step] [step:Prove (5): in a Noetherian ring, every ideal has a primary decomposition] We proceed in two sub-steps. [claim:Every ideal of a Noetherian ring is a finite intersection of irreducible ideals] An ideal $I$ of $R$ is **irreducible** if $I = J_1 \cap J_2$ with $J_1, J_2 \supseteq I$ implies $I = J_1$ or $I = J_2$. [/claim] [proof] Suppose for contradiction that the set $\Sigma$ of ideals that are not finite intersections of irreducible ideals is non-empty. Since $R$ is Noetherian, $\Sigma$ has a maximal element $I$ (every non-empty set of ideals in a Noetherian ring has a maximal element). The ideal $I$ is not irreducible (otherwise $I$ would be a finite intersection of irreducible ideals, namely the intersection of the single ideal $I$, contradicting $I \in \Sigma$). So $I = J_1 \cap J_2$ with $J_1, J_2 \supsetneq I$. Since $J_1, J_2$ strictly contain $I$ and $I$ is maximal in $\Sigma$, neither $J_1$ nor $J_2$ belongs to $\Sigma$. Thus each $J_k$ is a finite intersection of irreducible ideals. But then $I = J_1 \cap J_2$ is also a finite intersection of irreducible ideals, contradicting $I \in \Sigma$. [/proof] [claim:Every irreducible ideal of a Noetherian ring is primary] [/claim] [proof] Let $I$ be an irreducible ideal of $R$. Suppose $ab \in I$ with $a \notin I$. We must show $b^n \in I$ for some $n \geq 1$. Consider the ascending chain of ideals \begin{align*} (I : b) \subseteq (I : b^2) \subseteq (I : b^3) \subseteq \cdots, \end{align*} where $(I : b^k) = \{r \in R : rb^k \in I\}$. Since $R$ is Noetherian, this chain stabilises: there exists $n \geq 1$ with $(I : b^n) = (I : b^{n+1}) = \cdots$. We claim $I = (I + (a)) \cap (I + (b^n))$. The inclusion $I \subseteq (I + (a)) \cap (I + (b^n))$ holds since $I \subseteq I + (a)$ and $I \subseteq I + (b^n)$. For the reverse, let $x \in (I + (a)) \cap (I + (b^n))$. Write $x = u + ra$ and $x = v + sb^n$ with $u, v \in I$ and $r, s \in R$. Then $xb \in I$ because $xb = ub + rab$ and $ub \in I$ (since $u \in I$) and $rab = r(ab) \in I$ (since $ab \in I$). Also $x = v + sb^n$, so $xb = vb + sb^{n+1}$. Since $xb \in I$, we get $sb^{n+1} = xb - vb \in I$, so $s \in (I : b^{n+1}) = (I : b^n)$, which gives $sb^n \in I$. Therefore $x = v + sb^n \in I$. Thus $I = (I + (a)) \cap (I + (b^n))$. Since $I$ is irreducible and $I \subsetneq I + (a)$ (because $a \notin I$), we must have $I = I + (b^n)$, which gives $b^n \in I$. [/proof] Combining the two claims: every ideal $I$ of the Noetherian ring $R$ can be written as $I = I_1 \cap \cdots \cap I_k$ where each $I_j$ is irreducible (by the first claim), hence primary (by the second claim). This is a primary decomposition of $I$. [guided] The proof of part (5) uses two structural facts about Noetherian rings that combine cleanly. The first fact is that every ideal is a finite intersection of irreducible ideals. This is proved by a maximal element argument: if some ideal $I$ fails to be such an intersection, pick one that is maximal among all such failures. Since $I$ is not irreducible, write $I = J_1 \cap J_2$ with $J_1, J_2 \supsetneq I$. By maximality of $I$, each $J_k$ is a finite intersection of irreducible ideals, so $I$ is too — a contradiction. The second fact is that irreducible ideals are primary. Given an irreducible ideal $I$ and $ab \in I$ with $a \notin I$, we need $b^n \in I$ for some $n$. The key construction is the chain $(I : b) \subseteq (I : b^2) \subseteq \cdots$, which stabilises by Noetherianness. Setting $n$ to be the stabilisation index, one shows $I = (I + (a)) \cap (I + (b^n))$ by a direct element chase: the non-trivial direction uses $xb \in I$ (from $ab \in I$) and the stabilisation $(I : b^n) = (I : b^{n+1})$ to conclude $x \in I$. Since $I$ is irreducible and $I \subsetneq I + (a)$, we must have $I = I + (b^n)$, giving $b^n \in I$. Combining these: write $I$ as a finite intersection of irreducible ideals, each of which is primary. The result is a primary decomposition. [/guided] [/step]