Noetherian Ring

Theorem

Let \(R\) be a ring. The following are equivalent:

  1. Every ideal of \(R\) is finitely generated.

  2. For every sequence of ideals of \(R\) ordered by inclusion

    \(\)
    I0 \subseteq I1 \subseteq I2 \subseteq \dots \subseteq In \subseteq \dots
    \(\)
    there exists a \(k \in \mathbb{N}\) such that \(I_k = I_{k + l}\) for all \(l \in \mathbb{N}\). That is, the chain "terminates".

  3. 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 \((1) \implies (2) \implies (3) \implies (1)\).

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 \(I = \bigcup_{i = 1}^\infty I_i\) 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 \(\langle 0 \rangle \in S\). 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 \(\langle J, x \rangle\) 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.