[proofplan] For part (1), we identify isolated primes as the minimal elements of $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$ and show these correspond to the minimal primes of $R/I$. For part (2), we compute the extension-contraction $I^{ec}$ along $R \to R_{\mathfrak{p}_i}$. Extending $I = \bigcap_j \mathfrak{q}_j$ to $R_{\mathfrak{p}_i}$ and contracting back, the components $\mathfrak{q}_j$ with $\mathfrak{p}_j \not\subseteq \mathfrak{p}_i$ extend to the unit ideal and vanish, while those with $\mathfrak{p}_j \subseteq \mathfrak{p}_i$ survive. When $\mathfrak{p}_i$ is isolated (minimal), only $\mathfrak{q}_i$ survives, giving $I^{ec} = \mathfrak{q}_i$. [/proofplan] [step:Prove (1): identify the isolated primes as the minimal primes over $I$] The associated primes of $I$ are $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$ by the [First Uniqueness Theorem](/theorems/2886). A prime $\mathfrak{p}_i$ is **isolated** if it is minimal in $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$ (i.e., $\mathfrak{p}_j \subseteq \mathfrak{p}_i$ implies $\mathfrak{p}_j = \mathfrak{p}_i$); otherwise $\mathfrak{p}_i$ is **embedded**. We show the isolated primes coincide with the minimal primes over $I$, i.e., the primes minimal among all $\mathfrak{p} \in \operatorname{Spec}(R)$ with $\mathfrak{p} \supseteq I$. **Every isolated prime is a minimal prime over $I$.** Let $\mathfrak{p}_i$ be isolated, and suppose $\mathfrak{p} \in \operatorname{Spec}(R)$ satisfies $I \subseteq \mathfrak{p} \subseteq \mathfrak{p}_i$. Since $I = \bigcap_j \mathfrak{q}_j \subseteq \mathfrak{p}$ and $\mathfrak{p}$ is prime, $\mathfrak{p} \supseteq \bigcap_j \mathfrak{q}_j$, so $\mathfrak{p} \supseteq \mathfrak{q}_k$ for some $k$ (a prime containing a finite intersection contains one of the intersectands). Taking radicals: $\mathfrak{p} \supseteq \sqrt{\mathfrak{q}_k} = \mathfrak{p}_k$. But $\mathfrak{p} \subseteq \mathfrak{p}_i$, so $\mathfrak{p}_k \subseteq \mathfrak{p}_i$. Since $\mathfrak{p}_i$ is isolated, $\mathfrak{p}_k = \mathfrak{p}_i$. Then $\mathfrak{p}_i = \mathfrak{p}_k \subseteq \mathfrak{p} \subseteq \mathfrak{p}_i$, so $\mathfrak{p} = \mathfrak{p}_i$. This shows $\mathfrak{p}_i$ is a minimal prime over $I$. **Every minimal prime over $I$ is an isolated prime.** Let $\mathfrak{p}$ be a minimal prime over $I$. Since $I \subseteq \mathfrak{p}$, we have $\bigcap_j \mathfrak{q}_j \subseteq \mathfrak{p}$. Since $\mathfrak{p}$ is prime, $\mathfrak{q}_k \subseteq \mathfrak{p}$ for some $k$, giving $\mathfrak{p}_k = \sqrt{\mathfrak{q}_k} \subseteq \mathfrak{p}$. Since $\mathfrak{p}_k \supseteq \mathfrak{q}_k \supseteq I$ and $\mathfrak{p}$ is minimal over $I$, we get $\mathfrak{p} = \mathfrak{p}_k$. It remains to check $\mathfrak{p}_k$ is isolated: if $\mathfrak{p}_j \subseteq \mathfrak{p}_k$ for some $j$, then $\mathfrak{p}_j \supseteq I$ (since $\mathfrak{p}_j \supseteq \mathfrak{q}_j \supseteq I$) and $\mathfrak{p}_j \subseteq \mathfrak{p}_k = \mathfrak{p}$, so by minimality of $\mathfrak{p}$, $\mathfrak{p}_j = \mathfrak{p}_k$. Hence $\mathfrak{p}_k$ is isolated. [guided] The isolated primes are the minimal elements of the set $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$ of associated primes. The claim is that these coincide with the minimal primes over $I$. For the first direction, take an isolated $\mathfrak{p}_i$ and suppose $\mathfrak{p}$ is a prime with $I \subseteq \mathfrak{p} \subseteq \mathfrak{p}_i$. The key step is that $\mathfrak{p} \supseteq I = \bigcap_j \mathfrak{q}_j$ and $\mathfrak{p}$ is prime, so $\mathfrak{p} \supseteq \mathfrak{q}_k$ for some $k$ (this is the standard fact that a prime containing a finite intersection contains one of the terms). Taking radicals, $\mathfrak{p} \supseteq \mathfrak{p}_k$. Combined with $\mathfrak{p} \subseteq \mathfrak{p}_i$, this gives $\mathfrak{p}_k \subseteq \mathfrak{p}_i$, which by isolatedness forces $\mathfrak{p}_k = \mathfrak{p}_i$, and then $\mathfrak{p} = \mathfrak{p}_i$. For the converse, take a minimal prime $\mathfrak{p}$ over $I$. The same argument ($\mathfrak{p} \supseteq \bigcap_j \mathfrak{q}_j$, $\mathfrak{p}$ prime) gives $\mathfrak{p} \supseteq \mathfrak{p}_k$ for some $k$. Since $\mathfrak{p}_k \supseteq I$ and $\mathfrak{p}$ is minimal, $\mathfrak{p} = \mathfrak{p}_k$. To see $\mathfrak{p}_k$ is isolated: any $\mathfrak{p}_j \subseteq \mathfrak{p}_k$ satisfies $\mathfrak{p}_j \supseteq I$ and $\mathfrak{p}_j \subseteq \mathfrak{p}$, so minimality of $\mathfrak{p}$ gives $\mathfrak{p}_j = \mathfrak{p}_k$. The upshot: isolated primes are determined by $I$ alone (as the minimal primes in $\operatorname{Spec}(R)$ containing $I$), confirming the intrinsic nature of this distinguished subset of associated primes. [/guided] [/step] [step:Compute the extension of $I$ to $R_{\mathfrak{p}_i}$] Let $\mathfrak{p}_i$ be an isolated prime. Consider the localisation map $\iota: R \to R_{\mathfrak{p}_i}$, where $R_{\mathfrak{p}_i} = S_i^{-1} R$ with $S_i = R \setminus \mathfrak{p}_i$. The extension of $I$ is \begin{align*} I^e = I \cdot R_{\mathfrak{p}_i} = \left(\bigcap_{j=1}^n \mathfrak{q}_j\right) R_{\mathfrak{p}_i}. \end{align*} Since [localisation is exact](/theorems/2847) (i.e., $R_{\mathfrak{p}_i}$ is a flat $R$-module), the extension distributes over finite intersections: \begin{align*} I^e = \bigcap_{j=1}^n \mathfrak{q}_j R_{\mathfrak{p}_i}. \end{align*} For each $j$, we analyse $\mathfrak{q}_j R_{\mathfrak{p}_i}$: **Case $\mathfrak{p}_j \not\subseteq \mathfrak{p}_i$:** There exists $s \in \mathfrak{p}_j \setminus \mathfrak{p}_i$, so $s \in S_i$. Since $s \in \mathfrak{p}_j = \sqrt{\mathfrak{q}_j}$, there exists $m \geq 1$ with $s^m \in \mathfrak{q}_j$. Then $s^m \in \mathfrak{q}_j \cap S_i$, so $\frac{1}{1} = \frac{s^m}{s^m} \in \mathfrak{q}_j R_{\mathfrak{p}_i}$, giving $\mathfrak{q}_j R_{\mathfrak{p}_i} = R_{\mathfrak{p}_i}$. **Case $\mathfrak{p}_j \subseteq \mathfrak{p}_i$:** Here $S_i = R \setminus \mathfrak{p}_i \subseteq R \setminus \mathfrak{p}_j$, so $\mathfrak{q}_j \cap S_i = \varnothing$ (since $\mathfrak{q}_j \subseteq \mathfrak{p}_j \subseteq \mathfrak{p}_i$ gives $\mathfrak{q}_j \cap S_i \subseteq \mathfrak{p}_i \cap S_i = \varnothing$). In this case $\mathfrak{q}_j R_{\mathfrak{p}_i} \subsetneq R_{\mathfrak{p}_i}$ is a proper ideal. Therefore: \begin{align*} I^e = \bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} \mathfrak{q}_j R_{\mathfrak{p}_i}. \end{align*} [/step] [step:Contract back to $R$ and use isolation to extract $\mathfrak{q}_i$] Contracting $I^e$ from $R_{\mathfrak{p}_i}$ back to $R$: \begin{align*} I^{ec} = \left(\bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} \mathfrak{q}_j R_{\mathfrak{p}_i}\right) \cap R = \bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} (\mathfrak{q}_j R_{\mathfrak{p}_i} \cap R), \end{align*} where we used that contraction distributes over intersections (since $\iota^{-1}(\bigcap A_k) = \bigcap \iota^{-1}(A_k)$ for any collection of ideals $A_k$ in $R_{\mathfrak{p}_i}$). For each $j$ with $\mathfrak{p}_j \subseteq \mathfrak{p}_i$, we compute $\mathfrak{q}_j R_{\mathfrak{p}_i} \cap R$. Since $\mathfrak{q}_j$ is $\mathfrak{p}_j$-primary and $\mathfrak{p}_j \subseteq \mathfrak{p}_i$ (so $S_i \cap \mathfrak{q}_j = \varnothing$), the contraction of the extension of $\mathfrak{q}_j$ is: \begin{align*} \mathfrak{q}_j R_{\mathfrak{p}_i} \cap R = \{r \in R : rs \in \mathfrak{q}_j \text{ for some } s \in S_i\} = \mathfrak{q}_j, \end{align*} where the last equality holds because if $rs \in \mathfrak{q}_j$ with $s \in S_i \subseteq R \setminus \mathfrak{p}_j$, then $s \notin \mathfrak{p}_j = \sqrt{\mathfrak{q}_j}$, so $s^k \notin \mathfrak{q}_j$ for any $k$; since $\mathfrak{q}_j$ is primary and $rs \in \mathfrak{q}_j$ with $s \notin \sqrt{\mathfrak{q}_j}$, we must have $r \in \mathfrak{q}_j$. Therefore: \begin{align*} I^{ec} = \bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} \mathfrak{q}_j. \end{align*} Since $\mathfrak{p}_i$ is isolated (minimal among $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$), the condition $\mathfrak{p}_j \subseteq \mathfrak{p}_i$ forces $\mathfrak{p}_j = \mathfrak{p}_i$ (by minimality). But the decomposition is minimal, so the radicals $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$ are pairwise distinct, meaning $j = i$ is the only index with $\mathfrak{p}_j = \mathfrak{p}_i$. Hence: \begin{align*} I^{ec} = \mathfrak{q}_i. \end{align*} Since the left-hand side $I^{ec}$ depends only on $I$ and $\mathfrak{p}_i$ (which is an isolated prime, determined by $I$ alone by part (1)), the primary component $\mathfrak{q}_i$ depends only on $I$. [guided] The contraction step is where isolation plays its decisive role. After extending and contracting, we obtain $I^{ec} = \bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} \mathfrak{q}_j$. The contraction $\mathfrak{q}_j R_{\mathfrak{p}_i} \cap R = \mathfrak{q}_j$ uses a key property of primary ideals under localisation: if $\mathfrak{q}$ is $\mathfrak{p}$-primary and $S \cap \mathfrak{q} = \varnothing$ (equivalently $S \cap \mathfrak{p} = \varnothing$), then $\mathfrak{q} = (\mathfrak{q} S^{-1}R) \cap R$. The proof: if $rs \in \mathfrak{q}$ with $s \notin \mathfrak{p} = \sqrt{\mathfrak{q}}$, then $s$ is not nilpotent mod $\mathfrak{q}$, so the primary condition forces $r \in \mathfrak{q}$. The critical observation: when $\mathfrak{p}_i$ is isolated, no other $\mathfrak{p}_j$ can be contained in $\mathfrak{p}_i$ (that would contradict minimality). So the intersection $\bigcap_{\mathfrak{p}_j \subseteq \mathfrak{p}_i} \mathfrak{q}_j$ has only the $j = i$ term, giving $I^{ec} = \mathfrak{q}_i$. By contrast, if $\mathfrak{p}_i$ were embedded (non-minimal), there would exist some $\mathfrak{p}_k \subsetneq \mathfrak{p}_i$, and $I^{ec}$ would equal $\mathfrak{q}_k \cap \mathfrak{q}_i$ (or a larger intersection), which need not equal $\mathfrak{q}_i$. This is why the theorem guarantees uniqueness only for isolated components: the embedded components can genuinely vary between different minimal decompositions. Since $I^{ec}$ with respect to $R \to R_{\mathfrak{p}_i}$ is an intrinsic invariant of $I$ (it depends only on $I$ and $\mathfrak{p}_i$, both of which are determined by $I$), the isolated primary component $\mathfrak{q}_i$ is uniquely determined by $I$. [/guided] [/step]