The components of a Noetherian space are its maximal irreducible closed subsets.

general-topologynoetheriansolution-verification

This is Exercise 1.2.6(1) from page 3 of Springer's, "Linear Algebraic Groups (Second Edition)". I apologise for the length of this post. Also, (I think) I answered the question myself in writing this up, so it is now a question.

The Details:

Since definitions vary:

A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that

  • $\varnothing, X\in\tau$,
  • The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and
  • The union of finitely many closed sets is closed.

Note that $\tau$ is omitted sometimes when the context is clear.

On page 2 of Springer's book, paraphrased, we have this:

A topological space $X$ is Noetherian if any family of closed subsets of $X$ contains a minimal one.

Again on page 2, 1.2.1 reads,

A topological space $X\neq\varnothing $ is reducible if it is the union of two proper closed subsets. Otherwise $X$ is irreducible. A subset $A\subset X$ is irreducible if it is irreducible for the induced topology. Notice that $X$ is irreducible if and only if any two nonempty open subsets of $X$ have a nonempty intersection.

The next relevant thing is:

1.2.3. Lemma: Let $X$ be a topological space.

  1. $A\subset X$ is irreducible if and only if its closure $\overline{A}$ is irreducible;

  2. Let $f:X\to Y$ be a continuous map to a topological space $Y$. If $X$ is irreducible then so is the image $f(X)$.

Another theorem, page 3, is:

1.2.4 Proposition: Let $X$ be a Noetherian topological space. Then $X$ has finitely many maximal irreducible subsets. These are closed and cover $X$.

The maximal irreducible subsets of $X$ are called the (irreducible) components of $X$.

The Question:

Let $X$ be a Noetherian space. The components of $X$ are its maximal irreducible closed subsets.

Thoughts:

Let $A\subseteq X$ be a component of a Noetherian space $X$. Then, by definition, $A$ is a maximal irreducible subset of $X$. But now 1.2.4 Proposition gives that $A$ is closed. $\square$

I'm on shaky ground. Topology has never been a strong point for me. I think what I have is a proof. I'm not sure though. It seems too simple.

Please help 🙂

Best Answer

Yes, this is true - the key fact is that if $X$ is a topological space and a subset $S$ with the induced topology is irreducible, then $\overline{S}$ is also irreducible. As $S\subset\overline{S}$, we see that any maximal element of the poset of irreducible subsets must also be closed.

Let me also point out that none of this relies on noetherianity - that's useful for saying that a noetherian topological space has finitely many irreducible components, but it is completely immaterial to the proof above.

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