# Noetherian topological space

(Redirected from Noetherian space)

In mathematics, a Noetherian topological space, named for Emmy Noether, is a topological space in which closed subsets satisfy the descending chain condition. Equivalently, we could say that the open subsets satisfy the ascending chain condition, since they are the complements of the closed subsets. It can also be shown to be equivalent that every open subset of such a space is compact, and in fact the seemingly stronger statement that every subset is compact.

## Definition

A topological space ${\displaystyle X}$ is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence

${\displaystyle Y_{1}\supseteq Y_{2}\supseteq \cdots }$

of closed subsets ${\displaystyle Y_{i}}$ of ${\displaystyle X}$, there is an integer ${\displaystyle m}$ such that ${\displaystyle Y_{m}=Y_{m+1}=\cdots .}$

## Relation to compactness

The Noetherian condition can be seen as a strong compactness condition:

• Every Noetherian topological space is compact.
• A topological space ${\displaystyle X}$ is Noetherian if and only if every subspace of ${\displaystyle X}$ is compact. (i.e. ${\displaystyle X}$ is hereditarily compact).

## Noetherian topological spaces from algebraic geometry

Many examples of Noetherian topological spaces come from algebraic geometry, where for the Zariski topology an irreducible set has the intuitive property that any closed proper subset has smaller dimension. Since dimension can only 'jump down' a finite number of times, and algebraic sets are made up of finite unions of irreducible sets, descending chains of Zariski closed sets must eventually be constant.

A more algebraic way to see this is that the associated ideals defining algebraic sets must satisfy the ascending chain condition. That follows because the rings of algebraic geometry, in the classical sense, are Noetherian rings. This class of examples therefore also explains the name.

If R is a commutative Noetherian ring, then Spec(R), the prime spectrum of R, is a Noetherian topological space. More generally, a Noetherian scheme is a Noetherian topological space. The converse does not hold, since Spec(R) of a one-dimensional valuation domain R consists of exactly two points and therefore is Noetherian, but there are examples of such rings which are not Noetherian.

## Example

The space ${\displaystyle \mathbb {A} _{k}^{n}}$ (affine ${\displaystyle n}$-space over a field ${\displaystyle k}$) under the Zariski topology is an example of a Noetherian topological space. By properties of the ideal of a subset of ${\displaystyle \mathbb {A} _{k}^{n}}$, we know that if

${\displaystyle Y_{1}\supseteq Y_{2}\supseteq Y_{3}\supseteq \cdots }$

is a descending chain of Zariski-closed subsets, then

${\displaystyle I(Y_{1})\subseteq I(Y_{2})\subseteq I(Y_{3})\subseteq \cdots }$

is an ascending chain of ideals of ${\displaystyle k[x_{1},\ldots ,x_{n}].}$ Since ${\displaystyle k[x_{1},\ldots ,x_{n}]}$ is a Noetherian ring, there exists an integer ${\displaystyle m}$ such that

${\displaystyle I(Y_{m})=I(Y_{m+1})=I(Y_{m+2})=\cdots .}$

Since ${\displaystyle V(I(Y))}$ is the closure of Y for all Y, ${\displaystyle V(I(Y_{i}))=Y_{i}}$ for all ${\displaystyle i.}$ Hence

${\displaystyle Y_{m}=Y_{m+1}=Y_{m+2}=\cdots }$ as required.