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Suppose you have a Noetherian topological space $X$, and its finitely many irreducible components are $X_1,\dots,X_n$. If $U\subseteq X$ is open, why are the irreducible components of $U$ precisely the intersections $U\cap X_i$ which aren't empty?

The thing that's tripping me up is I don't see how to go from components in $U$ to those in $X$. For suppose $U\cap X_i=(U\cap Z_1)\cup(U\cap Z_2)$, where $Z_1,Z_2$ are closed in $X$. To show $U\cap X_i$ is irreducible, it has to be either $U\cap Z_1$ or $U\cap Z_2$. If not, there exist points $x,y\in U$ where $x\in X_i\setminus Z_1$ and $y\in X_i\setminus Z_2$. My hunch says there should be a way to go up and find a proper decomposition of $X_i$ in $X$ as a union of closed sets to get a contradiction.

I run into the same issue trying to show $U\cap X_i$ is maximal among irreducible sets in $U$.

What's the right way to see this property?

Koji Hamada
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3 Answers3

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Let $X$ be any topological space. The irreducible components of $X$ meeting $U$ correspond to the irreducible components of any open subset $U$ of $X$, via intersecting with $U$.

In this paragraph, we will give a bijection between irreducible closed subsets of $X$ meeting $U$ and the irreducible closed subsets of $U$.

  • If $Z$ is an irreducible closed subset of $X$ meeting $U$, then $Z\cap U$ is irreducible. This is obvious because a nonempty space is irreducible if and only if any two nonempty open subspaces have nonempty intersection, by definition.
  • If $T = Z\cap U$ is irreducible, where $Z\subset X$ closed, then the closure $\overline{T}$ of $T$ in $X$ is irreducible. This is obvious because the closure of an irreducible set is irreducible as well, by Exercise 3.6.B(b).
  • If $Z$ is an irreducible closed subset of $X$ meeting $U$, then the closure of $Z\cap U$ in $X$ is just $Z$. This is true because in an irreducible topological space, any nonempty open subspace is dense.
  • If $T = Z\cap U$ irreducible, where $Z\subset X$ closed, then $\overline{T}\cap U = T$, where $\overline{T}$ is the closure of $T$ in $X$. On one hand, $\overline{T} = \overline{Z\cap U}\subset Z$ and thus $\overline{T}\cap U\subset Z\cap U = T$. On the other hand, $T\subset\overline{T}$ and $T\subset U$ and thus $T\subset\overline{T}\cap U$.

In this paragraph, we will show that the bijection above maps components to components in both directions.

  • If $Z$ is an irreducible component of $X$ meeting $U$, then $Z\cap U$ is an irreducible component of $U$. Let $S$ be an irreducible closed subset of $U$ containing $Z\cap U$. Then, $\overline{S}$ is an irreducible closed subset of $X$ containing $\overline{Z\cap U} = Z$. Since $Z$ is an irreducible component, $\overline{S} = Z$. Thus, $S = \overline{S}\cap U = Z\cap U$.
  • If $T$ is an irreducible component of $U$, then $\overline{T}$ is an irreducible component of $X$. Let $Y$ be an irreducible closed subset of $X$ containing $\overline{T}$. Then, $Y\cap U\supset\overline{T}\cap U = T$. Since $T$ is an irreducible component of $U$, $Y\cap U = T$. Thus, $Y = \overline{Y\cap U} = \overline{T}$.
Zhulin Li
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    Note that I didn't use the Noetherian property. – Zhulin Li Oct 15 '16 at 15:52
  • By Exercise 3.6.B(b), I refer to the Vakil's Foundations of Algebraic Geometry (Dec. 25, 2016 version). – Zhulin Li Oct 15 '16 at 15:54
  • I think Noetherian is just needed to write $X$ as a finite union of irreducibles. – hwong557 Nov 25 '16 at 22:47
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    (By convention, an irreducible subset is required to be non-empty.) Here's a clean argument to see that the one-to-one correspondence \begin{align} \newcommand{\testleftlong}{\longleftarrow!\shortmid} \begin{Bmatrix} \text{Irreducible closed}\ \text{subsets of } X \text{ meeting } U \end{Bmatrix} &\longleftrightarrow \begin{Bmatrix} \text{Irreducible closed}\ \text{subsets of } U \end{Bmatrix}\ Z&\longmapsto Z\cap U\ \overline{F}&\testleftlong F \end{align} preserves irreducible components: By definition, an irreducible component in a topological – Elías Guisado Villalgordo Jun 16 '23 at 09:42
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    space is a maximal irreducible subset. Irreducible components are always closed (see this). Since the previous one-to-one correspondence preserves inclusions, it's a poset isomorphism and, hence, it preserves maximal elements, i.e., the irreducible components. – Elías Guisado Villalgordo Jun 16 '23 at 09:42
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Replacing $X$ by $X_i$ and $U$ by $U \cap X_i$ you only have to prove that for any irreducible space $X$ and any nonempty open subset $U \subseteq X$, $U$ is also irreducible. If $U = (U \cap Z_1) \cup (U \cap Z_2)$ for some closed subsets $Z_1$, $Z_2$ of $X$, then the closure $\overline{U}$ of $U$ in $X$ is $\overline{U \cap Z_1} \cup \overline{U \cap Z_2}$. As $U$ is a nonempty open subset of an irreducible space, in particular dense in $X$, this implies $\overline{U \cap Z_1} \cup \overline{U \cap Z_2} = X$ and irreducibility gives us (say) $X = \overline{U \cap Z_1}$. Intersecting with $U$ now gives $U = U \cap Z_1$.

Matthias Klupsch
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Consider irreducible Topology space $X$ and $U$ is its open subset.

Use the fact that if $X$ is irreducible iff every two nonempty open subsets of $X$ have nonempty intersection.

Then we can prove $U$ is still irreducible.

But I also don't know how to show $U\cap X_{i}$ is maximal among irreducible sets in $U$.

@Koji Hamada Do you find the answer for this?

user44312
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