Taken from: [1, Exercise 5.43].
Exercise 5.43
Let $I$ be a decomposable ideal of the commutative ring $R$. Let
\[I=Q_1 \cap \ldots \cap Q_n \quad \text { with } \sqrt{Q_i} =P_i \text { for } i=1, \ldots, n\]be a minimal primary decomposition of $I$. Let $\mathcal{P}$ be a non-empty subset of $\operatorname{ass}I$ with the property that whenever $P \in \mathcal{P}$ and $P^{\prime} \in \operatorname{ass}I$ are such that $P^{\prime} \subseteq P$, then $P^{\prime} \in \mathcal{P}$ too (such a subset of $\operatorname{ass}I$ is called an isolated subset of $\operatorname{ass}I$). Show that
\[\bigcap_{\substack{i=1 \\ P_i \in \mathcal{P}}}^n Q_i\]depends only on $I$ and not on the choice of minimal primary decomposition of $I$.
Proof
Let $I = Q_1’ \cap \dots \cap Q_n’$ be another minimal primary decomposition of $I$. Define $k = \lvert\mathcal{P}\rvert$. Without loss of generality, we can reorder the terms in each intersection so that $\sqrt{Q_i} = \sqrt{Q_i’} = P_i \in \mathcal{P}$ for all $i = 1, \dots, k$, and such that $\sqrt{Q_i} = \sqrt{Q_i’}$ for all $i = k+1, \dots, n$.
If $k = 1$, there is nothing to prove, since a unique element $P \in \mathcal{P}$ implies that $P$ is minimal, and the result follows directly from the 2nd Uniqueness Theorem of Primary Decomposition—[1, Th. 4.29]. For the remainder of the proof, we assume that $k \geq 2$.
Let $j \in {1, \dots, k}$. Observe that $P_i \not\subseteq P_j$ for every $i = k+1, \dots, n$; otherwise, $P_i$ would belong to $\mathcal{P}$. By [1, Lem. 3.55], we conclude that
\[\bigcap_{i=k+1}^{n} P_i \not\subseteq P_j.\]From this, applying the Prime Avoidance Theorem—[1, Lem. 3.61], we obtain
\[\bigcap_{i=k+1}^{n} P_i \not\subseteq \bigcup_{j=1}^{k} P_j.\]That is, there exists an element
\[a \in \left(\bigcap_{i=k+1}^{n} P_i\right) \mathbin{\big\backslash} \left(\bigcup_{j=1}^{k} P_j\right),\]which means that
\[a \in \bigcap_{i=k+1}^{n} P_i \setminus P_j\]for all $j = 1, \dots, k$.
Since each $P_i = \sqrt{Q_i} = \sqrt{Q_i’}$ for $i = k+1, \dots, n$, there exist integers $h_i, h_i’$ such that $a^{h_i} \in Q_i$ and $a^{h_i’} \in Q_i’$. Now, choose $t \in \mathbb{Z}^+$ such that
\[t \geq \max \{ h_{k+1}, h_{k+1}', \dots, h_n, h_n' \}.\]At this point, we note that $a^t \notin P_j$ for $j = 1, \dots, k$; otherwise, since $P_j$ is prime, this would imply $a \in P_j$, contradicting our earlier observation. Thus, by [1, Lem. 4.14], we deduce:
\[\begin{align*} (I : a^t) &= \left(\bigcap_{i=1}^{n} Q_i : a^t \right) \\ &= \left(\bigcap_{j=1}^{k} (Q_j : a^t) \right) \cap \left(\bigcap_{i=k+1}^{n} (Q_i : a^t) \right) \\ &= \left(\bigcap_{j=1}^{k} Q_j \right) \cap \left(\bigcap_{i=k+1}^{n} R \right) \\ &= \bigcap_{j=1}^{k} Q_j. \end{align*}\]A similar argument shows that $(I : a^t) = \bigcap_{j=1}^{k} Q_j’$, leading to the conclusion that
\[\bigcap_{j=1}^{k} Q_j = \bigcap_{j=1}^{k} Q_j',\]as claimed. $\blacksquare$
References
- Sharp, Rodney Y. (2001). Steps in Commutative Algebra (2nd ed.). Cambridge University Press. http://dx.doi.org/10.1017/CBO9780511623684