LaskerNoether theorem
Theorem 1 (LaskerNoether).
Let $R$ be a commutative^{} Noetherian ring^{} with $\mathrm{1}$. Every ideal in $R$ is decomposable^{} (http://planetmath.org/DecomposableIdeal).
The theorem can be proved in two steps:
Proposition 1.
Every ideal in $R$ can be written as a finite intersection^{} of irreducible ideals^{}
Proof.
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 $S\ne \mathrm{\varnothing}$. Then any chain ${I}_{1}\subseteq {I}_{2}\subseteq \mathrm{\cdots}$ in $S$ must terminate in a finite number of steps, as $R$ is Noetherian. Say $I={I}_{n}$ is the maxmimal element of this chain. Since $I\in S$, $I$ itself can not be irreducible^{}, so that $I=J\cap K$ where $J$ and $K$ are ideals strictly containing $I$. Now, if $J\in S$, then then $I$ would not be maximal in the chain ${I}_{1}\subseteq {I}_{2}\subseteq \mathrm{\cdots}$. Therefore, $J\notin S$. Similarly, $K\notin S$. By the definition of $S$, $J$ and $K$ are both finite intersections of irreducible ideals. But this would imply that $I\notin S$, a contradiction^{}. So $S=\mathrm{\varnothing}$ and we are done. ∎
Proposition 2.
Every irreducible ideal in $R$ is primary
Proof.
Suppose $I$ is irreducible and $ab\in I$. We want to show that either $a\in I$, or some power $n$ of $b$ is in $I$. Define ${J}_{i}=I:({b}^{i})$, the quotient of ideals $I$ and $({b}^{i})$. Since
$$\mathrm{\cdots}\subseteq ({b}^{n})\subseteq \mathrm{\cdots}\subseteq ({b}^{2})\subseteq (b),$$ 
we have, by one of the rules on quotients of ideals, an ascending chain of ideals
$${J}_{1}\subseteq {J}_{2}\subseteq \mathrm{\cdots}\subseteq {J}_{n}\subseteq \mathrm{\cdots}$$ 
Since $R$ is Noetherian, $J:={J}_{n}={J}_{m}$ for all $m>n$. Next, define $K=({b}^{n})+I$, the sum of ideals $({b}^{n})$ and $I$. We want to show that $I=J\cap K$.
First, it is clear that $I\subseteq J$ and $I\subseteq K$, which takes care of one of the inclusions. Now, suppose $r\in J\cap K$. Then $r=s+t{b}^{n}$, where $s\in I$ and $t\in R$, and $r{b}^{n}\in I$. So, $r{b}^{n}=s{b}^{n}+t{b}^{2n}$. Now, $t\in I:({b}^{2n})$, so $t\in I:({b}^{n})$. But this means that $r=s+t{b}^{n}\in I$, and this proves the other inclusion.
Since $I$ is irreducible, either $I=J$ or $I=K$. We analyze the two cases below:

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If $I=J=I:({b}^{n})$, then $I=I:(b)$ in particular, since $I\subseteq I:(b)\subseteq I:({b}^{n})$. As $ab\in I$ by assumption^{}, $a\in I:(b)=I$.

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If $I=K=({b}^{n})+I$, then ${b}^{n}\in I$.
This completes^{} the proof. ∎
Remarks.

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The above theorem can be generalized to any submodule^{} of a finitely generated module over a commutative Noetherian ring with 1.

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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  LaskerNoether theorem 

Canonical name  LaskerNoetherTheorem 
Date of creation  20130322 18:19:53 
Last modified on  20130322 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 