Isolated primary components are recovered by localization and contraction

Fix an isolated component $Q_i$. By [L1], if $j\mathrm{\ne }i$ then ${\mathfrak{p}}_j⊈{\mathfrak{p}}_i$, since otherwise minimality of ${\mathfrak{p}}_i$ would force ${\mathfrak{p}}_j={\mathfrak{p}}_i$, contradicting the distinct-radical part of [L1]. Hence for each $j\mathrm{\ne }i$ we may choose $a_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$. Localizing at $R\setminus {\mathfrak{p}}_i$, fact [L2] gives $(Q_j{)}_{{\mathfrak{p}}_i}=M_{{\mathfrak{p}}_i}$ for $j\mathrm{\ne }i$, while $(Q_i{)}_{{\mathfrak{p}}_i}$ remains a proper primary submodule. Using [L4], $N_{{\mathfrak{p}}_i}={\bigcap }_{j=1}^r(Q_j{)}_{{\mathfrak{p}}_i}=(Q_i{)}_{{\mathfrak{p}}_i}\mathrm{.}$

Fact [L3] recovers $Q_i$ by contracting $(Q_i{)}_{{\mathfrak{p}}_i}$ back to $M$. Since step 1.1 identifies $(Q_i{)}_{{\mathfrak{p}}_i}$ with $N_{{\mathfrak{p}}_i}$, this gives $Q_i=\{m\in M:m\mathrm{/}1\in N_{{\mathfrak{p}}_i}\}\mathrm{.}$ By [L5], the set of component primes depends only on $M\mathrm{/}N$, so ${\mathfrak{p}}_i$ occurs and is isolated in every minimal primary decomposition. Applying the same localization-and-contraction argument to any such decomposition recovers its ${\mathfrak{p}}_i$-component from the same right-hand side. The assumed Axiom of Choice supplies the choice hypotheses of the primary-characterization and associated-prime results used through [L2], [L3], and [L5]; the finitely many selections in step 1.1 require only finite choice. Hence the isolated component is unique.

Therefore every isolated component is recovered by localization and contraction.