Noetherian Ring

Theorem

Let $R$ be a ring. The following are equivalent:

Every ideal of $R$ is finitely generated.
For every sequence of ideals of $R$ ordered by inclusion

I0 \subseteq I1 \subseteq I2 \subseteq \dots \subseteq In \subseteq \dots

there exists a $k \in N$ such that $I_{k} = I_{k + l}$ for all $l \in N$ . That is, the chain "terminates".
Every non-empty set of ideals has a (not necessarily unique) maximal element when partially ordered by inclusion.

Definition

A ring is called Noetherian if it satisfies any of the equivalent conditions above.

Proof

We will prove that $(1) ⟹ (2) ⟹ (3) ⟹ (1)$ .

Firstly, assume $(1)$ , that is every ideal of $R$ is finitely generated. Now, let $I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots$ be a chain of ideals. By assumption $I = ⋃_{i = 1}^{\infty} I_{i}$ is finitely generated. Let $S$ be a finite generator of $I$ , and for each $a \in S$ , take $I_{i_{a}}$ to be the minimal ideal of the chain which contains $a$ . We can then take $I_{k}$ to be the maximal ideal of the form $I_{i_{a}}$ for each $a$ , and hence $I = I_{k}$ . This is possible in particular because $S$ is finite, so such a maximal element exists. This means that $I_{j + 1} = I_{j}$ for all $j \geq k$ .

Next, assume $(2)$ , that is, every sequence of ideals ordered by inclusion terminates. Now, let $S$ be a non-empty set of ideals of $R$ , and consider the partial order on $S$ given by set inclusion. If $S$ is finite, then it trivially has a (not necessarily unique) maximal element, so we consider only the case when $S$ is infinite. Now, let $I_{0} \in S$ . Define $S_{J} = {I \in S : I \supset J}$ for each ideal $J$ in $S$ , and, appealing to the axiom of choice, define a choice function $f : {S_{J} : J \in S} \to S$ such that

I_{i + 1} = {\begin{cases} f (S_{I_{i}}) & i \neq 0 \\ I_{0} & i = 0 \end{cases} .

This yields a strictly increasing, in terms of inclusion, sequence of ideals ${I_{i}}_{i = 0}^{\infty}$ . Appealing now to our assumption, this sequence of ideals must terminate at some $k$ such that $I_{j + 1} = I_{j}$ for all $j \geq k$ . This $I_{k}$ is a maximal element because if some ideal $J$ satisfies $I_{k} \subseteq J$ then $I_{k} = J$ .

Finally, assume $(3)$ , that is, every non-empty set of ideals partially ordered by inclusion has a maximal element. Let $I$ be an arbitrary ideal and let $S$ be the set of all finitely generated ideals of $R$ which are contained in $I$ . Clearly $S$ is non-empty because at least $⟨ 0 ⟩ \in S$ . So, $S$ has a maximal element $J$ . Suppose $I \neq J$ , in particular this means $J \subset I$ and there exists an $x \in I - J$ . This means $⟨ J, x ⟩$ is finitely generated and distinct from $J$ , which contradicts the fact that $J$ is a maximal element of $S$ . Hence we can conclude that $I = J$ , and $I$ is finitely generated.