Commutative Algebra 60

Primary Decomposition of Ideals


Let \mathfrak a \subsetneq A be a proper ideal. A primary decomposition of \mathfrak a is its primary decomposition as an A-submodule of A:

\mathfrak a = \mathfrak q_1 \cap \mathfrak q_2 \cap \ldots \cap \mathfrak q_n,

where each \mathfrak q_i is \mathfrak p_i-primary for some prime \mathfrak p_i \subset A, i.e. \mathrm{Ass}_A (A/\mathfrak q_i) = \{\mathfrak p_i\}.

Here is a quick way to determine if an ideal is primary.

Proposition 1.

A proper ideal \mathfrak b \subsetneq A is \mathfrak p-primary for some \mathfrak p if and only if:

x, y \in A, xy \in \mathfrak b \implies x \in \mathfrak b \text{ or } y\in r(\mathfrak b),

in which case \mathfrak p = r(\mathfrak b).

Recall that r(\mathfrak b) is the radical of the ideal \mathfrak b.


(⇒) Suppose \mathrm{Ass}_A A/\mathfrak b = \{\mathfrak p\}. By proposition 3 here, the set of zero-divisors (in A) of A/\mathfrak b is \mathfrak p. Now if xy\in \mathfrak b and x\not\in \mathfrak b, then y\in A is a zero-divisor of A/\mathfrak b so y\in \mathfrak p.

To prove that r(\mathfrak b) = \mathfrak p, suppose x^n \in \mathfrak b; then x^n \in A is a zero-divisor for A/\mathfrak b so x^n \in \mathfrak p and we have x\in \mathfrak p. Conversely, if x\in \mathfrak p, then since every minimal prime in V(\mathfrak b) is an associated prime of A/\mathfrak b (by proposition 2 here), x is contained in every minimal prime of V(\mathfrak b). By proposition 5 here, x\in r(\mathfrak b).

(⇐) Suppose \mathfrak b satisfies the given condition; we first show that r(\mathfrak b) is prime. Suppose xy \in r(\mathfrak b) and x\not\in r(\mathfrak b). For some n > 0, x^n y^n \in \mathfrak b. But x^n \not \in r(\mathfrak b) so by the given condition y^n \in \mathfrak b and y\in r(\mathfrak b).

Let \mathfrak p = r(\mathfrak b); it remains to show \mathrm{Ass}_A (A/\mathfrak b) = \{\mathfrak p\}. First, \mathfrak p is the unique minimal element of V(\mathfrak b) so \mathfrak p \in \mathrm{Ass}_A (A/\mathfrak b).

Conversely, it remains to show any zero-divisor of A/\mathfrak b as an A-module lies in \mathfrak p. But if x\in A and y\in A - \mathfrak b are such that xy \in \mathfrak b, then by the given condition x \in r(\mathfrak b) = \mathfrak p. ♦

We thus say a proper ideal \mathfrak q \subsetneq A is primary if:

x, y\in A, xy \in \mathfrak q\implies x \in \mathfrak q \text{ or } y \in r(\mathfrak q).

Note that prime ideals are primary, and by proposition 1, the radical of a primary ideal is prime.

Exercise A

1. Prove that if f:A\to B is a ring homomorphism and \mathfrak q\subset B is a primary ideal, then f^{-1}(\mathfrak q) is a primary ideal of A. Also r(f^{-1}(\mathfrak q)) = f^{-1}(r(\mathfrak q)).

2. Prove that for an ideal \mathfrak a \subseteq A, there is a bijection between primary ideals of A containing \mathfrak a and primary ideals of A/\mathfrak a.


Primary Decomposition and Localization

Throughout this section S\subseteq A is a fixed multiplicative subset.

Proposition 2.

There is a bijection between:

  • primary ideals \mathfrak q \subset A such that \mathfrak q \cap S = \emptyset;
  • primary ideals \mathfrak q' \subset S^{-1}A

Furthermore, if \mathfrak q is in the first set and \mathfrak p = r(\mathfrak q), then

\mathfrak p\cdot S^{-1}A = r(\mathfrak q \cdot S^{-1}A).


By exercise A.2 above and proposition 3 here, it suffices to show: if \mathfrak q\subset A is primary and \mathfrak q \cap S = \emptyset then

  • \mathfrak q' := \mathfrak q (S^{-1}A) is primary, and
  • \mathfrak q' \cap A = \mathfrak q.

For the first claim, note that \mathfrak q' is a proper ideal since \mathfrak q \cap S = \emptyset. Suppose \frac x s, \frac y t \in S^{-1}A satisfy \frac{xy}{st} \in \mathfrak q'; then for some s'\in S we have s'xy \in \mathfrak q. Since \mathfrak q \cap S = \emptyset no power of s’ is contained in \mathfrak q so

xy\in \mathfrak q \implies x\in\mathfrak q \text{ or } y\in r(\mathfrak q) \implies \frac x s \in \mathfrak q' \text{ or } \frac y t \in r( \mathfrak q').

For the second claim, clearly \mathfrak q' \cap A \supseteq \mathfrak q. Conversely let a\in A satisfy \frac a 1 \in \mathfrak q'. For some s\in S we have sa \in \mathfrak q. As above s\not\in r(\mathfrak q) so a \in \mathfrak q. ♦

Finally, we will see later that unlike factorization of ideals in a Dedekind domain, primary decompositions are not unique. However we can still salvage the following.

Proposition 3.

Suppose we have minimal primary decompositions

\mathfrak a = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n = \mathfrak q'_1 \cap \ldots \cap \mathfrak q'_n

with \mathfrak p_i = r(\mathfrak q_i) = r(\mathfrak q_i') for each i. If \mathfrak p_i is minimal in V(\mathfrak a) then \mathfrak q_i = \mathfrak q_i'.


We localize \mathfrak a = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n at \mathfrak p := \mathfrak p_i. But since \mathfrak p is minimal in V(\mathfrak a), we have

j\ne i \implies \mathfrak p_j \not\subseteq \mathfrak p \implies \mathfrak q_j \not\subseteq \mathfrak p \implies \mathfrak q_j A_{\mathfrak p} = A_{\mathfrak p}.

Hence \mathfrak a A_{\mathfrak p} = \mathfrak q_i A_{\mathfrak p}. By proposition 2 this gives \mathfrak q_i = \mathfrak q_i A_{\mathfrak p} \cap A = \mathfrak a A_{\mathfrak p} \cap A which is uniquely determined by \mathfrak a. ♦

Exercise B

Prove the following.

1. If \mathfrak a \subseteq A is a reduced ideal, then it (i.e. A/\mathfrak a) has no embedded primes.

2. If \mathfrak m\subset A is maximal, then \mathfrak q is \mathfrak m-primary if and only if \mathfrak m^n \subseteq \mathfrak q \subseteq \mathfrak m for some n > 0.

warningA power of a prime ideal is not primary in general. E.g. let k be a field and A =k[X, Y, Z]/(Z^2 - XY) with \mathfrak p = (X, Z) \subset A, which is prime since

A/\mathfrak p \cong k[X, Y, Z]/(Z^2 - XY, X, Z) \cong k[Y].

Then \mathfrak p^2 is not primary because XY = Z^2 \in \mathfrak p^2 but X\not\in \mathfrak p^2 and Y\not\in r(\mathfrak p^2) = \mathfrak p.


Worked Examples

Throughout this section, k denotes a field.

Example 1

In an earlier example, we saw that for A = k[X, Y] and \mathfrak a = (X^2, XY), the A-module A/\mathfrak a has two associated primes: (X) and (XY). The following are primary decompositions, both of which are clearly minimal:

(X^2, XY) = (X) \cap (X^2, XY, Y^2) = (X) \cap (X^2, Y).

To check that these ideals are primary:

  • (X) is prime and hence primary;
  • (X^2, XY, Y^2) = (X, Y)^2 is a power of the maximal ideal (X, Y); by exercise B.2 it is primary;
  • (X, Y)^2 \subseteq (X^2, Y) \subseteq (X, Y) so (X^2, Y) is primary by exercise B.2.

Note that (X^2, XY, Y^2) \subsetneq (X^2, Y). Geometrically, the k-scheme with coordinate ring A/\mathfrak a looks like the following.


Example 2

Let A = k[X, Y, Z] with \mathfrak a = (X - YZ, XY). Then

k[X, Y, Z]/(X - YZ) \cong k[Y, Z], \ X \mapsto YZ \implies k[X, Y, Z]/\mathfrak a \cong k[Y, Z]/(Y^2 Z).

We have (Y^2 Z) = (Y^2) \cap (Z), an intersection of primary ideals with r((Y^2)) = (Y) and r((Z)) = (Z). This translates to

\mathfrak a = (X - YZ, Y^2) \cap (X - YZ, Z) = (X - YZ, Y^2) \cap (X, Z)

with r((X - YZ, Y^2)) = (X, Y) and (X, Z) is already prime.

Example 3

Let A = k[X, Y] with \mathfrak a = (X) \cap (X, Y)^2 \cap (X, Y-1)^2. Then each of (X), (X,Y)^2 and (X, Y-1)^2 is primary (the first ideal is prime; the remaining two ideals are powers of a maximal ideal). Hence this gives a primary decomposition of \mathfrak a so its associated primes are (X), (X, Y) and (X, Y-1), with the latter two embedded.

Geometrically, the k-scheme with coordinate ring A/\mathfrak a looks like:


Computing an explicit set of generators for \mathfrak a is not trivial, but it can be done with Buchberger’s algorithm.

Exercise C (from Atiyah & MacDonald, Exercise 4.5)

Let A = k[X, Y, Z] and \mathfrak p_1 = (X, Y), \mathfrak p_2 = (X, Z), \mathfrak m = (X, Y, Z) be ideals of A. Set \mathfrak a := \mathfrak p_1 \mathfrak p_2. Prove that

\mathfrak a = \mathfrak p_1 \cap \mathfrak p_2 \cap \mathfrak m^2

is a minimal primary decomposition of \mathfrak a.


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