Lasker–Noether Theorem

Statement

Let $R$ be a Noetherian ring and $I\subseteq R$ an ideal . Then there exist primary ideals $Q_1,\dots,Q_r$ such that

Moreover, for each $i$, the radical $\sqrt{Q_i}$ is a prime ideal :

This is the existence of primary decomposition in Noetherian rings.

Cross-links

• Primary decomposition as a concept: primary decomposition
• Radicals and primes: radical of an ideal , prime ideal
• Noetherian hypothesis: Noetherian ring

Examples

1. Squarefree monomial ideal.
   In $R=k[x,y]$,

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

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

2. A non-prime primary component appears.
   In $R=k[x,y]$,

   $$(x^2,xy) = (x)\cap (x^2,y).$$

   Here $(x)$ is prime. The ideal $(x^2,y)$ is $(x,y)$-primary because $\sqrt{(x^2,y)}=(x,y)$ is maximal (hence prime), and powers of $(x,y)$ control nilpotence mod $(x^2,y)$.

3. Integers (PID case).
   In $R=\mathbb Z$,

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

   The ideal $(4)$ is $(2)$-primary and $(3)$ is $(3)$-primary (indeed prime), giving a primary decomposition of $(12)$.