Lasker-Noether theorem

Theorem 1 (Lasker-Noether).

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:

• 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.