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 $\sqrt{Q_j}\nsubseteq\mathfrak p$, so some element of $Q_j$ becomes a unit in $R_{\mathfrak p}$. Consequently
$$
IR_{\mathfrak p}=Q_iR_{\mathfrak p}.
$$
Because $Q_i$ is $\mathfrak p$-primary, multiplication by any $s\notin\mathfrak p$ cannot carry an element outside $Q_i$ into $Q_i$. Therefore
$$
Q_i=Q_iR_{\mathfrak p}\cap R=IR_{\mathfrak p}\cap R.
$$
The right side depends only on $I$ and $\mathfrak p$, proving uniqueness.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 2 / b / Solution

In a Noetherian ring, the primary component belonging to each isolated prime $\mathfrak p$ is uniquely determined by
$$
Q(\mathfrak p)=IR_{\mathfrak p}\cap R.
$$
Embedded primary components need not be unique.


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