Noetherian Rings

Definition: A ring $R$ is said to be a Noetherian Ring if it satisfies the following condition (known as the Ascending Chain Condition): Every infinite ascending chain of ideals $I_1 \subseteq I_2 \subseteq ... \subseteq I_n \subseteq ...$ stabilizes, that is, there exists an $N \in \mathbb{N}$ such that for all $m \geq N$ then $I_m = I_N$.

There are a few different definitions of a ring to be Noetherian. The following theorem gives us another way to define Noetherian rings.

Theorem 1: Let $R$ be a ring. Then $R$ is a Noetherian ring if and only if every ideal in $R$ is finitely generated, that is, for every ideal $I$ there exists $x_1, x_2, ..., x_n \in I$ such that $I = (x_1, x_2, ..., x_n)$.

Here, $(x_1, x_2, ..., x_n) = \{ \sum_{i=1}^{n} a_ix_i : a_i \in R \}$.

• Proof: $\Rightarrow$ Let $R$ be a Noetherian ring and let $I$ be an ideal. Suppose that $I$ is not finitely generated. Choose any element $x_1 \in I$ and let $I_1 = (x_1)$. Since $I$ is not finitely generated, $I_1 \subset I$ and $I_1 \neq I$. So there exists an element $x_2 \in I \setminus I_1$. Let $I_2 = (x_1, x_2)$. Once again, since $I$ is not finitely generated, $I_2 \subseteq I$ and $I_2 \neq I$. So there exists an element $x_3 \in I \setminus I_2$. Let $I_3 = (x_1, x_2, x_3)$. We continue indefinitely to obtain a chain of ideals:

(1)

\begin{align} \quad (x_1) \subset (x_1, x_2) \subset (x_1, x_2, x_3) \subset ... \subset (x_1, x_2, x_3, ..., x_n) \subset ... \end{align}

• But by construction, this chain of ideals does not stabilize. Hence $R$ is not a Noetherian ring. This is a contradiction. So the assumption that $I$ is not finitely generated is false. So every ideal $I$ is finitely generated.

• $\Leftarrow$ Suppose that every ideal $I$ is finitely generated and consider an arbitrary infinite chain of ideals:

(2)

\begin{align} \quad I_1 \subseteq I_2 \subseteq ... \subseteq I_n \subseteq ... \end{align}

• Let $\displaystyle{I = \bigcup_{n=1}^{\infty} I_n}$. Then $I$ is itself an ideal. So by assumption, $I$ is finitely generated. So there exists elements $x_1, x_2, ..., x_n \in I$ such that $I = (x_1, x_2, ..., x_n)$. Each element $x_i$ is contained in some $I_j$. Since there are only finitely many such $x_i$ this means that there exists an $N \in \mathbb{N}$ such that $x_1, x_2, ..., x_n \in I_N$. But then $I_N = I$. This shows that $I_m = I_N$ for all $m \geq N$. So $R$ is a Noetherian ring. $\blacksquare$

There are many examples of Noetherian rings. Recall that a ring $R$ is said to be a principal ideal domain if every ideal in $R$ is a principal ideal, and an ideal is said to be a principal ideal if $I = (a)$ for some element $a \in R$. So by Theorem 1, we see that every principal ideal domain is a Noetherian ring.

Hence, an example of a Noetherian ring is the ring of integers $\mathbb{Z}$.

Not all rings are Noetherian though. For example, consider the ring of polynomials $\mathbb{R}[x_1, x_2, ...]$ of infinitely variables. This ring is not Noetherian. To show this, consider the following ascending chain of ideals:

(3)

\begin{align} \quad (x_1) \subset (x_1, x_2) \subset ... \subset (x_1, x_2, ..., x_n) \subset ... \end{align}

This infinite chain of ideals does not stabilize. Hence $\mathbb{R}[x_1, x_2, ...]$ is not Noetherian.