Lasker–Noether theorem. Let RR be a commutative and let IRI\subsetneq R be a proper . Then there exist Q1,,QrQ_1,\dots,Q_r such that

I=Q1Qr.I = Q_1 \cap \cdots \cap Q_r .

One may choose the decomposition to be minimal: no component is redundant and the radicals Qi\sqrt{Q_i} are distinct prime ideals. For every minimal decomposition, the set

{Q1,,Qr}\{\sqrt{Q_1},\dots,\sqrt{Q_r}\}

depends only on II, although the primary components themselves need not be unique.

Primary ideals

An ideal QRQ\subsetneq R is primary if abQab\in Q and aQa\notin Q imply bnQb^n\in Q for some n1n\geq 1. Equivalently, every zero divisor in R/QR/Q is nilpotent. See for the general terminology.

Examples
  1. A reduced principal ideal in a polynomial ring. In k[x,y]k[x,y], the ideal (xy)(xy) decomposes as
    (xy)=(x)(y).(xy) = (x) \cap (y).
    Here (x)(x) and (y)(y) are prime ideals (hence primary).
  1. A decomposition with an embedded component. In k[x,y]k[x,y],
    (x2,xy)=(x)(x2,y).(x^2,xy) = (x) \cap (x^2,y).
    Indeed, if f(x)(x2,y)f \in (x) \cap (x^2,y), then f=xgf=xg and xg(x2,y)xg \in (x^2,y) forces g(x,y)g \in (x,y), so f(x2,xy)f \in (x^2,xy). The ideal (x)(x) is prime, and (x2,y)(x^2,y) is (x,y)(x,y)-primary since its radical is (x,y)(x,y).
  1. In the integers. In Z\mathbb Z,
    (12)=(4)(3).(12) = (4) \cap (3).
    The ideal (4)(4) is (2)(2)-primary and (3)(3) is prime (hence primary).