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Self studying commutative algebra. Ran into a small problem.

Theorem 4.10 of Atiyah & Macdonald reads

Let $\mathfrak a$ be a decomposable ideal, let $\mathfrak a=\bigcap_{i=1}^n\mathfrak q_i$ be a minimal primary decomposition of $\mathfrak a,$ and let $\{\mathfrak p_{i_1},\ldots,\mathfrak p_{i_m}\}$ be an isolated set of prime ideals belonging to $\mathfrak a.$ Then $\mathfrak q_{i_1}\cap\cdots\cap\mathfrak q_{i_m}$ is independent of the decomposition.

The corollary following this reads

The isolated primary components (i.e., the primary components $\mathfrak q_i$ corresponding to minimal prime ideals $\mathfrak p_i$) are uniquely determines by $\mathfrak a.$

My question is: why does the corollary follow from the theorem. The reason is probably simple, given that all of the proofs in the book are around six lines long and they didn't even bother writing this proof. I just don't see why uniqueness of the intersection gives uniqueness of the ideals themselves.

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"Isolated" is a property of a set of prime ideals, not a property of an individual prime ideal. (Notice that it's an "isolated set of prime ideals", not a "set of isolated prime ideals".) A set $S$ of prime ideals (among those associated to the given minimal decomposition of $\mathfrak{a}$) is isolated if it is downward closed. This means that if $\mathfrak{p}'$ and $\mathfrak{p}$ are both associated to the given decomposition of $\mathfrak{a}$, and $\mathfrak{p}' \subset \mathfrak{p}$, and $\mathfrak{p} \in S$, then $\mathfrak{p}' \in S$.

In particular, if $\mathfrak{p}$ is a minimal prime ideal among the prime ideals of the given minimal decomposition of $\mathfrak{a}$, then $\{\mathfrak{p}\}$ is an isolated set of prime ideals, and we may apply the theorem.

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  • $\begingroup$ Are you sure? That would make the proof easy, but the wording implies that there can be multiple isolated prime ideals. If it's a typo then I'll accept that, but I'm still doubtful. $\endgroup$ – D. Brogan Jul 9 '19 at 14:42
  • $\begingroup$ I edited my answer to clarify. Does that help? $\endgroup$ – Ted Jul 10 '19 at 2:57
  • $\begingroup$ Yes that last sentence answers my question. Thanks a lot. $\endgroup$ – D. Brogan Jul 10 '19 at 3:17

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