[proofplan] We compute $(I : x) = \bigcap_{i} (\mathfrak{q}_i : x)$ and analyse each factor: $(\mathfrak{q}_i : x) = R$ when $x \in \mathfrak{q}_i$, and $(\mathfrak{q}_i : x)$ is $\mathfrak{p}_i$-primary when $x \notin \mathfrak{q}_i$. For the inclusion $\supseteq$, we show that for each $\mathfrak{p}_j$ there exists $x$ with $(I : x) = \mathfrak{p}_j$: choosing $x \in \bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$ gives $(I : x) = (\mathfrak{q}_j : x) \subseteq \mathfrak{p}_j$, and maximising $(I:x)$ over such $x$ yields $(I:x) = \mathfrak{p}_j$. For the inclusion $\subseteq$, if $(I : x) = \mathfrak{p}$ is prime, then $\mathfrak{p} = \bigcap_{i \in S} \mathfrak{p}_i$ and primality forces $\mathfrak{p} = \mathfrak{p}_j$ for some $j$. [/proofplan] [step:Express $(I : x)$ in terms of the primary components and analyse each factor] For any $x \in R$: \begin{align*} (I : x) = \{r \in R : rx \in I\} = \bigcap_{i=1}^n \{r \in R : rx \in \mathfrak{q}_i\} = \bigcap_{i=1}^n (\mathfrak{q}_i : x). \end{align*} **Case 1: $x \in \mathfrak{q}_i$.** Then $rx \in \mathfrak{q}_i$ for all $r \in R$, so $(\mathfrak{q}_i : x) = R$. **Case 2: $x \notin \mathfrak{q}_i$.** We show $(\mathfrak{q}_i : x)$ is $\mathfrak{p}_i$-primary with radical $\mathfrak{p}_i$. For the radical: $(\mathfrak{q}_i : x) \supseteq \mathfrak{q}_i$ (since $r \in \mathfrak{q}_i$ implies $rx \in \mathfrak{q}_i$), so $\sqrt{(\mathfrak{q}_i : x)} \supseteq \mathfrak{p}_i$. Conversely, if $r^m \in (\mathfrak{q}_i : x)$ then $r^m x \in \mathfrak{q}_i$; since $x \notin \mathfrak{q}_i$ and $\mathfrak{q}_i$ is primary, $(r^m)^k \in \mathfrak{q}_i$ for some $k \geq 1$, giving $r \in \mathfrak{p}_i$. Hence $\sqrt{(\mathfrak{q}_i : x)} = \mathfrak{p}_i$. For the primary property: if $ab \in (\mathfrak{q}_i : x)$ with $b \notin (\mathfrak{q}_i : x)$, then $abx \in \mathfrak{q}_i$ with $bx \notin \mathfrak{q}_i$. Since $\mathfrak{q}_i$ is primary, $a^s \in \mathfrak{q}_i \subseteq (\mathfrak{q}_i : x)$ for some $s \geq 1$. Dropping the unit factors: \begin{align*} (I : x) = \bigcap_{\substack{i = 1 \\ x \notin \mathfrak{q}_i}}^n (\mathfrak{q}_i : x). \end{align*} [guided] The identity $(I : x) = \bigcap_i (\mathfrak{q}_i : x)$ follows from $I = \bigcap_i \mathfrak{q}_i$: $r$ annihilates $x$ modulo $I$ if and only if $rx \in \mathfrak{q}_i$ for every $i$. When $x \in \mathfrak{q}_i$, every $r$ satisfies $rx \in \mathfrak{q}_i$, so $(\mathfrak{q}_i : x) = R$ and that factor drops out. The interesting case is $x \notin \mathfrak{q}_i$: the factor $(\mathfrak{q}_i : x)$ inherits the primary structure of $\mathfrak{q}_i$. For the radical: $\mathfrak{q}_i \subseteq (\mathfrak{q}_i : x)$ gives $\mathfrak{p}_i \subseteq \sqrt{(\mathfrak{q}_i : x)}$. The reverse uses the primary condition: $r^m x \in \mathfrak{q}_i$ with $x \notin \mathfrak{q}_i$ forces $r^{mk} \in \mathfrak{q}_i$, so $r \in \mathfrak{p}_i$. The primary property transfers by applying the primary condition on $\mathfrak{q}_i$ to $a \cdot (bx)$ with $bx \notin \mathfrak{q}_i$. [/guided] [/step] [step:Prove the inclusion $\supseteq$: for each $\mathfrak{p}_j$, exhibit $x$ with $(I : x) = \mathfrak{p}_j$] Fix $j \in \{1, \ldots, n\}$. By minimality, $\bigcap_{i \neq j} \mathfrak{q}_i \not\subseteq \mathfrak{q}_j$ (otherwise $\mathfrak{q}_j$ is redundant). For any $x \in \bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$, we have $(I : x) = (\mathfrak{q}_j : x)$ and: **$(I : x) \subseteq \mathfrak{p}_j$:** If $r \in (I : x)$, then $rx \in I \subseteq \mathfrak{q}_j$. Since $x \notin \mathfrak{q}_j$ and $\mathfrak{q}_j$ is primary, $r^k \in \mathfrak{q}_j$ for some $k \geq 1$, so $r \in \mathfrak{p}_j$. Work in the quotient $\bar{R} = R/I$, which is a Noetherian ring (as a quotient of $R$). The set \begin{align*} \Sigma = \{\operatorname{Ann}_{\bar{R}}(\bar{y}) : y \in \textstyle\bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j\} \end{align*} is a non-empty collection of ideals of $\bar{R}$ (since $\bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j \neq \varnothing$), each contained in $\mathfrak{p}_j/I$ (by the inclusion above). Since $\bar{R}$ is Noetherian, $\Sigma$ has a maximal element $\operatorname{Ann}_{\bar{R}}(\bar{x})$ for some $x \in \bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$. We claim $(I : x) = \mathfrak{p}_j$. Suppose for contradiction that $(I : x) \subsetneq \mathfrak{p}_j$. Choose $r \in \mathfrak{p}_j \setminus (I : x)$, so $rx \notin I$. Since $x \in \mathfrak{q}_i$ for all $i \neq j$, we have $rx \in \mathfrak{q}_i$ for $i \neq j$. Since $r \notin (I : x) = (\mathfrak{q}_j : x)$, we have $rx \notin \mathfrak{q}_j$. Therefore $rx \in \bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$. Now $\operatorname{Ann}_{\bar{R}}(\overline{rx}) = ((I : rx))/I$. We show $\operatorname{Ann}_{\bar{R}}(\overline{rx}) \supsetneq \operatorname{Ann}_{\bar{R}}(\bar{x})$. If $\bar{s} \in \operatorname{Ann}_{\bar{R}}(\bar{x})$, then $sx \in I$, so $s(rx) = r(sx) \in rI \subseteq I$ (since $I$ is an ideal), giving $\bar{s} \in \operatorname{Ann}_{\bar{R}}(\overline{rx})$. Moreover, $\bar{r} \in \operatorname{Ann}_{\bar{R}}(\overline{rx})$ since $r(rx) = r^2 x$ and $r^2 \in (I : x)$ needs verification: since $r \in \mathfrak{p}_j = \sqrt{(\mathfrak{q}_j : x)}$ and $(I:x) = (\mathfrak{q}_j : x)$, there exists $m \geq 1$ with $r^m \in (I : x)$, so $r^m x \in I$, giving $r^{m-1} \cdot (rx) \in I$, hence $r^{m-1} \in (I : rx)$. In particular $r \in \sqrt{(I:rx)} = \mathfrak{p}_j$ (by the same analysis with $rx$ replacing $x$). We need $\bar{r} \in \operatorname{Ann}_{\bar{R}}(\overline{rx}) \setminus \operatorname{Ann}_{\bar{R}}(\bar{x})$, i.e., $r \in (I:rx) \setminus (I:x)$. We have $r \notin (I:x)$ by choice. We need $r \in (I:rx)$, i.e., $r^2 x \in I$. Since $r \in \mathfrak{p}_j = \sqrt{(I:x)}$, we have $r^m \in (I:x)$ for some $m \geq 1$, but $m$ could be larger than $2$. So $r$ need not lie in $(I : rx)$. To handle this: choose $m$ minimal with $r^m \in (I:x)$. Since $r \notin (I:x)$, we have $m \geq 2$. Set $x' = r^{m-1} x$. Then $x' \in \bigcap_{i \neq j} \mathfrak{q}_i$ (since $x \in \mathfrak{q}_i$ for $i \neq j$) and $x' \notin \mathfrak{q}_j$ (since $r^{m-1} \notin (\mathfrak{q}_j : x) = (I:x)$ by minimality of $m$, so $r^{m-1} x \notin \mathfrak{q}_j$). Also: - $\operatorname{Ann}_{\bar{R}}(\bar{x}) \subseteq \operatorname{Ann}_{\bar{R}}(\bar{x}')$: if $sx \in I$ then $s \cdot r^{m-1} x = r^{m-1} \cdot sx \in I$. - $r \in \operatorname{Ann}_{\bar{R}}(\bar{x}') \setminus \operatorname{Ann}_{\bar{R}}(\bar{x})$: $r \cdot x' = r^m x \in I$ (since $r^m \in (I:x)$), and $r \notin (I:x)$. So $\operatorname{Ann}_{\bar{R}}(\bar{x}') \supsetneq \operatorname{Ann}_{\bar{R}}(\bar{x})$, contradicting maximality of $\operatorname{Ann}_{\bar{R}}(\bar{x})$ in $\Sigma$. Therefore $(I : x) = \mathfrak{p}_j$, which is prime. This shows $\mathfrak{p}_j \in \{(I : y) \mid y \in R\} \cap \operatorname{Spec}(R)$. [guided] The forward inclusion needs to exhibit $x$ with $(I:x) = \mathfrak{p}_j$. The inclusion $(I:x) \subseteq \mathfrak{p}_j$ holds for any $x \in \bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$, but the reverse $\mathfrak{p}_j \subseteq (I:x)$ need not hold for an arbitrary such $x$. The strategy is to maximise $(I:x)$ over all valid choices of $x$. Since $R/I$ is Noetherian (as a quotient of the Noetherian ring $R$), the set of annihilators $\{\operatorname{Ann}_{\bar{R}}(\bar{x})\}$ admits a maximal element by ACC. The key argument: if the maximising $(I:x)$ is strictly smaller than $\mathfrak{p}_j$, we can find $r \in \mathfrak{p}_j \setminus (I:x)$. Let $m \geq 2$ be minimal with $r^m \in (I:x)$. Then $x' = r^{m-1} x$ lies in $\bigcap_{i \neq j} \mathfrak{q}_i \setminus \mathfrak{q}_j$ (minimality of $m$ ensures $r^{m-1} x \notin \mathfrak{q}_j$), and $\operatorname{Ann}_{\bar{R}}(\bar{x}') \supsetneq \operatorname{Ann}_{\bar{R}}(\bar{x})$ because $r$ annihilates $\bar{x}'$ but not $\bar{x}$. This contradicts maximality. Without the Noetherian hypothesis on $R$, this maximality argument would fail, and the proof would require a different approach. In the Noetherian setting, the ACC on ideals of $R/I$ provides the crucial tool. [/guided] [/step] [step:Prove the inclusion $\subseteq$: if $(I : x)$ is prime, it equals some $\mathfrak{p}_j$] Suppose $\mathfrak{p} = (I : x)$ is prime for some $x \in R$. Let $S = \{i : x \notin \mathfrak{q}_i\}$. Since $\mathfrak{p}$ is prime, $S \neq \varnothing$ (otherwise $x \in I$ and $(I : x) = R$, which is not prime). From the first step, $(I : x) = \bigcap_{i \in S} (\mathfrak{q}_i : x)$ where each $(\mathfrak{q}_i : x)$ is $\mathfrak{p}_i$-primary. Taking radicals and using $\sqrt{\bigcap_{i \in S} J_i} = \bigcap_{i \in S} \sqrt{J_i}$ (valid since each $J_i$ is primary): \begin{align*} \mathfrak{p} = \sqrt{\mathfrak{p}} = \sqrt{(I : x)} = \sqrt{\bigcap_{i \in S} (\mathfrak{q}_i : x)} = \bigcap_{i \in S} \mathfrak{p}_i. \end{align*} The first equality uses that $\mathfrak{p}$ is prime, hence radical. Since $\mathfrak{p}$ is prime and $\mathfrak{p} = \bigcap_{i \in S} \mathfrak{p}_i \supseteq \prod_{i \in S} \mathfrak{p}_i$ (a finite intersection of ideals contains their product), and a prime containing a finite product must contain one of the factors, there exists $j \in S$ with $\mathfrak{p} \supseteq \mathfrak{p}_j$. But $\mathfrak{p} = \bigcap_{i \in S} \mathfrak{p}_i \subseteq \mathfrak{p}_j$. Therefore $\mathfrak{p} = \mathfrak{p}_j$. [guided] If $(I:x) = \mathfrak{p}$ is prime, the component-wise formula gives $\mathfrak{p} = \bigcap_{i \in S} \mathfrak{p}_i$ where $S$ indexes the components not containing $x$. A prime that equals a finite intersection of primes must equal one of them. The argument: $\mathfrak{p} \subseteq \mathfrak{p}_i$ for all $i \in S$ (from the intersection). In the other direction, $\mathfrak{p} \supseteq \prod_{i \in S} \mathfrak{p}_i$ (since an intersection contains the product), and primality forces $\mathfrak{p} \supseteq \mathfrak{p}_j$ for some $j$. Combined: $\mathfrak{p} = \mathfrak{p}_j$. Together with the forward inclusion: \begin{align*} \{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\} = \{(I : x) \mid x \in R\} \cap \operatorname{Spec}(R). \end{align*} The right-hand side depends only on $I$, so the set of associated primes is an invariant of $I$, independent of the chosen minimal decomposition. [/guided] [/step]