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

A minimal primary decomposition is an expressionin which every $Q_i$ is a primary ideal, the prime ideals $\sqrt{Q_i}$ are pairwise distinct, and the decomposition is irredundant: deleting any $Q_i$ changes the intersection.

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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.

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No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

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Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.

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