Lasker-Noether theorem

Theorem 1 (Lasker-Noether).

Let R be a commutativePlanetmathPlanetmathPlanetmath Noetherian ringMathworldPlanetmath with 1. Every ideal in R is decomposableMathworldPlanetmathPlanetmath (

The theorem can be proved in two steps:

Proposition 1.

Every ideal in R can be written as a finite intersectionMathworldPlanetmathPlanetmath of irreducible idealsMathworldPlanetmath


Let S be the set of all ideals of a Noetherian ring R which can not be written as a finite intersection of irreducible ideals. Suppose S. Then any chain I1I2 in S must terminate in a finite number of steps, as R is Noetherian. Say I=In is the maxmimal element of this chain. Since IS, I itself can not be irreduciblePlanetmathPlanetmath, so that I=JK where J and K are ideals strictly containing I. Now, if JS, then then I would not be maximal in the chain I1I2. Therefore, JS. Similarly, KS. By the definition of S, J and K are both finite intersections of irreducible ideals. But this would imply that IS, a contradictionMathworldPlanetmathPlanetmath. So S= and we are done. ∎

Proposition 2.

Every irreducible ideal in R is primary


Suppose I is irreducible and abI. We want to show that either aI, or some power n of b is in I. Define Ji=I:(bi), the quotient of ideals I and (bi). Since


we have, by one of the rules on quotients of ideals, an ascending chain of ideals


Since R is Noetherian, J:=Jn=Jm for all m>n. Next, define K=(bn)+I, the sum of ideals (bn) and I. We want to show that I=JK.

First, it is clear that IJ and IK, which takes care of one of the inclusions. Now, suppose rJK. Then r=s+tbn, where sI and tR, and rbnI. So, rbn=sbn+tb2n. Now, tI:(b2n), so tI:(bn). But this means that r=s+tbnI, and this proves the other inclusion.

Since I is irreducible, either I=J or I=K. We analyze the two cases below:

  • If I=J=I:(bn), then I=I:(b) in particular, since II:(b)I:(bn). As abI by assumptionPlanetmathPlanetmath, aI:(b)=I.

  • If I=K=(bn)+I, then bnI.

This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof. ∎


  • The above theorem can be generalized to any submoduleMathworldPlanetmath of a finitely generated module over a commutative Noetherian ring with 1.

  • A ring is said to be Lasker if every ideal is decomposable. The theorem above says that every commutative Noetherian ring with 1 is Lasker. There are Lasker rings that are not Noetherian.

Title Lasker-Noether theorem
Canonical name LaskerNoetherTheorem
Date of creation 2013-03-22 18:19:53
Last modified on 2013-03-22 18:19:53
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type Theorem
Classification msc 13C99
Defines Lasker ring