Theorem 1 (Lasker-Noether)   Let $R$ be a commutative Noetherian ring with $1$ . Every ideal in $R$ is decomposable.

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 $S\ne \varnothing$ . 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 $J_i=\quo{I}{(b^i)}$ , 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, $t\in \quo{I}{(b^{2n})}$ , so $t\in \quo{I}{(b^n)}$ . 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 $I=J=\quo{I}{(b^n)}$ , then $I=\quo{I}{(b)}$ in particular, since $I\subseteq \quo{I}{(b)} \subseteq \quo{I}{(b^n)}$ . As $ab\in I$ by assumption, $a\in \quo{I}{(b)}= I$ .
• If $I=K=(b^n)+I$ , then $b^n\in I$ .

This completes the proof.