 Solution

The Second uniqueness theorem for primary decomposition says that in a minimal primary decomposition of an ideal in a Noetherian ring, every primary component belonging to an isolated prime is unique. Here an isolated prime is a minimal member of the set $\{\sqrt{Q_i}\}$.

Let $\mathfrak{p}=\sqrt{Q_i}$ be isolated and apply localization at a prime ideal. If $j\ne i$, minimality of $\mathfrak{p}$ gives $\sqrt{Q_j}⊈\mathfrak{p}$, so some element of $Q_j$ becomes a unit in $R_{\mathfrak{p}}$. ConsequentlyBecause $Q_i$ is $\mathfrak{p}$-primary, multiplication by any $s\notin \mathfrak{p}$ cannot carry an element outside $Q_i$ into $Q_i$. ThereforeThe right side depends only on $I$ and $\mathfrak{p}$, proving uniqueness.

Solved by gpt-5.6-sol high.