Primary decomposition

Let $R$ be a commutative ring and let $I\subseteq R$ be an ideal.

Definition (primary ideal)

An ideal $Q\subsetneq R$ is primary if for all $a,b\in R$,

Equivalently, $Q$ is primary if $R/Q$ has the property that every zero-divisor is nilpotent.

If $Q$ is primary, then its radical $\sqrt{Q}$ is a prime ideal; one often says that $Q$ is $\mathfrak p$-primary when $\sqrt{Q}=\mathfrak p$.

Definition (primary decomposition)

A primary decomposition of $I$ is an expression

where each $Q_i$ is a primary ideal of $R$.

A primary decomposition is called minimal if (i) the radicals $\sqrt{Q_i}$ are pairwise distinct and (ii) no $Q_i$ can be omitted without changing the intersection. The primes $\sqrt{Q_i}$ that occur in a minimal decomposition are intrinsic invariants (the “associated primes” of $I$), even though the $Q_i$ themselves need not be unique.

Existence in Noetherian rings

Primary decompositions do not exist in arbitrary rings. The fundamental existence theorem is the Lasker–Noether theorem , often packaged as primary decomposition in Noetherian rings : if $R$ is a Noetherian ring , then every ideal $I\subseteq R$ admits a primary decomposition.

Examples

1. In $\mathbb{Z}$.
   In the PID $\mathbb{Z}$, primary ideals are exactly $(p^n)$ for primes $p$. Using $(a)\cap(b)=(\mathrm{lcm}(a,b))$, one gets

   $$(12) = (4)\cap(3),$$

   where $(4)$ is $(2)$-primary and $(3)$ is $(3)$-primary.

2. A squarefree monomial ideal.
   Over a field $k$, in $k[x,y]$ one has

   $$(xy) = (x)\cap(y).$$

   Here $(x)$ and $(y)$ are prime (hence primary), so this is a primary decomposition.

3. A “one-piece” primary decomposition.
   In $k[x,y]$, the ideal $Q=(x,y)^2=(x^2,xy,y^2)$ is $(x,y)$-primary. Thus it is already a primary decomposition of itself:

   $$(x,y)^2 = Q.$$