Let $R$ be a commutative Noetherian ring with $1$. 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 $S\neq\varnothing$. Then any chain $I_{1}\subseteq I_{2}\subseteq\cdots$ 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 $I_{1}\subseteq I_{2}\subseteq\cdots$. 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 $S=\varnothing$ 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

 $\cdots\subseteq(b^{n})\subseteq\cdots\subseteq(b^{2})\subseteq(b),$

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

 $J_{1}\subseteq J_{2}\subseteq\cdots\subseteq J_{n}\subseteq\cdots$

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 $r=s+tb^{n}$, where $s\in I$ and $t\in R$, and $rb^{n}\in I$. So, $rb^{n}=sb^{n}+tb^{2n}$. Now, $t\in I\!:\!(b^{2n})$, so $t\in I\!:\!(b^{n})$. But this means that $r=s+tb^{n}\in I$, and this proves the other inclusion.

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

• 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, $a\in I\!:\!(b)=I$.

• If $I=K=(b^{n})+I$, then $b^{n}\in I$.

This completes the proof. ∎

Remarks.

• The above theorem can be generalized to any submodule 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 LaskerNoetherTheorem 2013-03-22 18:19:53 2013-03-22 18:19:53 CWoo (3771) CWoo (3771) 7 CWoo (3771) Theorem msc 13C99 Lasker ring