Primary Decomposition of Ideals

Definition.

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 $\mathfrak b$.

Proof

(⇒) 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

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

Proof

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 $\mathfrak q_i = \mathfrak q_i'$.

Proof

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.

A 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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