I'm self-learning commutative algebra from Introduction to Commutative Algebra of Atiyah and Macdonald and get frustrated about the second uniqueness primary decomposition theorem. I copy the theorem for you to reference (page 54):

Let $\mathfrak a$ be a decomposable ideal, let $\mathfrak a = \bigcap_{i=1}^nq_{i}$ be a minimal primary decomposition of $\mathfrak a$, and let $\{p_{i_1},...,p_{i_n}\}$ be an isolated set of prime ideals of $\mathfrak a$. Then $q_{i_1} \bigcap ...\bigcap q_{i_n}$ is independent of the decomposition.

Here is the proof:

We have $q_{i_1} \cap ...\cap q_{i_n} = S(\mathfrak a)$ where $S = A - p_{i_1} \cup ... \cup p_{i_n}$, hence depends only on $\mathfrak a$ (since the $p_i$ depend only on $\mathfrak a$).

What makes me confused is that: I understand that the set of ALL prime ideal of an ideal $\mathfrak a$ is independent of the decomposition, but should it still be true when we just get an isolated set from that set? If that isolated set is just a proper subset of the set of all prime ideal associated with ideal $\mathfrak a$, why "$p_i$ depend only on $\mathfrak a$"?

The second is that, I read from another the source which has another statement of this theorem. Here is the content:

Let $\mathfrak a$ be a decomposable ideal, let $\mathfrak a = \bigcap_{i=1}^nq_{i}$ be a minimal primary decomposition of $\mathfrak a$, and let $\{p_{1},...,p_{m}\}$ be the set of minimal prime ideals of $\mathfrak a$. Then $q_1, q_2, ..., q_m$ are independent of the decomposition.

In this statement, the selected set must be the set of ALL minimal prime ideal, not just an isolated set.

So which is the statement true. Please help me clarify this. Thanks so much. I really appreciate.

  • $\begingroup$ the first is clear. for the second see corollary 4.11 of Atiyah $\endgroup$
    – user147308
    Commented Jun 15, 2014 at 17:24
  • $\begingroup$ @72313: Can you please explain more about my first question: " If that isolated set is just a proper subset of the set of all prime ideal associated with ideal α, why pi depend only on α"? Thanks $\endgroup$ Commented Jun 15, 2014 at 17:59

2 Answers 2


The two statements you cite are related - the claim in Atiyah-Macdonald implies the second claim. Both address the uniqueness of certain primary components, rather than the primes themselves (which depend only on $\alpha$ in the sense that given $\alpha$, they can be fixed). Let me address them in order: $\DeclareMathOperator{\Ass}{Ass}$ $\DeclareMathOperator{\Min}{Min}$

1) Recall that in Atiyah-Macdonald, an isolated set of primes of $\mathfrak a$ is a downward-closed subset of $\Ass(\mathfrak a)$, i.e. $S \subseteq \Ass(\mathfrak a)$ is isolated if for $p, q \in \Ass(\mathfrak a)$, $p \subseteq q$, then $q \in S \implies p \in S$. The claim in Atiyah-Macdonald is that if $S$ is isolated, then the intersection of the primary components, corresponding to primes in $S$, depends only on $\mathfrak a$.

2) The claim in your other source is the following: each primary component, corresponding to a prime in $\Min(\mathfrak a)$, depends only on $\mathfrak a$. This follows from the claim in Atiyah-Macdonald, by taking $S = \{p\}$ for $p \in \Min(\mathfrak a)$, which is an isolated set.

The point is that generally speaking, the primary components are not unique: for a given ideal $\alpha$ and $p \in \Ass(\mathfrak a)$, there may be (infinitely) many $p$-primary ideals $q$ that can appear in (valid) minimal primary decompositions of $\mathfrak a$. In other words, there is no way in general to uniquely fix a $p$-primary component of $\mathfrak a$, even after giving $\mathfrak a$ and $p$. However, the second claim says that this behavior can only happen for embedded (i.e. non-minimal) primes.

  • $\begingroup$ Thanks, I understand, but what about the proof of the first statement, why $S = A - p_{i_1} \cup ... \cup p_{i_n}$ depends only on $\mathfrak a$. Please explain for me. Thanks $\endgroup$ Commented Jun 16, 2014 at 1:13
  • $\begingroup$ @leducquang: The associated primes of $\mathfrak{a}$ is uniquely determined by $\mathfrak{a}$, and thus so is any subset of them. This is only stated in contrast to the fact that a primary component of $\mathfrak{a}$ is not uniquely determined by $\mathfrak{a}$ and an associated prime $p$ of $\mathfrak{a}$. It is irrelevant that the $p_{i_1}, \ldots, p_{i_n}$ do not make up the entire set of associated primes $\endgroup$
    – zcn
    Commented Jun 16, 2014 at 20:06

The proof of the first statement: Let $q_i$ an isolated prime ideal. Then {$p_i$} is a isolated set because $p_i$ is minimal in the set of the associated primes. If you take S=A-$p_i$ then $q_i=S(a)$, where $S(a)=a^{ec}$. So, $q_i$ is uniquely determined by $a$.


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