# The closure of $\overline{\{x\}}$ being irreducible and relating the generic point to its associated irreducible scheme

If $x$ is a point in $X$ where $X$ is a scheme, we write $\overline{\{ x\}}$ for the closure of $x$ in $X$.

$\mathbf{Question \;1}$: I am a bit confused why $\overline{\{ x\}}$ is irreducible. According to some lecture notes, this scheme $\overline{\{ x\}}$ is irreducible since an open subset of $\overline{\{ x\}}$ that doesn't contain $x$ also doesn't contain any point of the closure of $x$ since the complement of an open set is closed. Therefore, every open subset of $\overline{\{ x\}}$ contains $x$, and is therefore dense in $\overline{\{x \}}$.

So how does every open subset of $\overline{\{x \}}$ being dense relate to $\overline{\{x \}}$ being irreducible?

$\mathbf{Question \;2}$: Suppose $X$ is a scheme and $U=\operatorname{Spec }A$ is a nonempty irreducible subset of $X$. Then $U$ has a unique generic point $\xi$ corresponding to the minimal prime of $A$. Why is it that $\{ \xi\}\not=U$, but instead $\overline{\{\xi \}}=U$?

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$A$ is irreducible if $A$ cannot be written as the union of two proper closed subsets. This is equivalent (by taking complements) to saying that $A$ does not contain two disjoint non-empty open sets, which in turn says that any non-empty open subset of $A$ is dense in $A$. – Brian M. Scott Jul 22 '12 at 22:24
@BrianM.Scott Thank you Brian. This helped. – math-visitor Jul 22 '12 at 22:29

## 3 Answers

$1)$ if $\overline{\{x\}}$ was reducible, you would have a non-trivial decomposition $\overline{\{x\}}=F\cup F'$ in a union of two closed subsets. The complementary of, says, $F$ give then a non dense open subset.

$2)$ $U$ can certainly by equal to $\{\xi \}$, but if $U$ is a subscheme of $X$, it refers usually to closed subschemes (just as subvarieties).

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Thank you Rolando. Can you explain a bit more when you say "it" refers usually to closed subschemes? – math-visitor Jul 22 '12 at 22:37
In (1), do you really want the word “disjoint”? – Dylan Moreland Jul 22 '12 at 22:57
Dylan is right: you are defining connected , not irreducible. – Georges Elencwajg Jul 22 '12 at 23:18

You already have two good answers; I just want to collect some general topological results with which one should become fluent when studying algebraic geometry.

A non-empty topological space $X$ is irreducible if the following equivalent conditions hold:

1. If $Y_1, Y_2$ are closed subsets of $X$ and $X = Y_1 \cup Y_2$, then $Y_1 = X$ or $Y_2 = X$.
2. If $U_1, U_2$ are non-empty open subsets of $X$ then $U_1 \cap U_2 \neq \emptyset$.
3. Any non-empty open subset of $X$ is dense in $X$.

Proving that these are equivalent will probably be straightforward. Whether or not $X$ is itself irreducible, if $Y \subset X$ is irreducible in the subspace topology then so is its closure $\overline Y$ in $X$. In your notation, $\{x\}$ is certainly irreducible.

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Excellent answer, Dylan. And +1 for mentioning that $X$ must be non-empty! – Georges Elencwajg Jul 22 '12 at 23:24

1) Since it implies that the intersection of two open subsets is not empty.

2) consider $X=U=\text{Spec} \; \mathbb{Z}$ with its generic point $\{ (0) \} \not = U$ while $\overline{\{ (0) \}}=V(0)=U.$

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Thank you ehsanmo! For 1), in Zariski topology, I thought two nonempty open sets in some scheme $X$ always have a nonempty intersection. Is this true only when $X$ is irreducible? – math-visitor Jul 22 '12 at 22:48
For your example in 2), which definitely helps by the way, what is an element in $U\setminus \{ (0)\}$? – math-visitor Jul 22 '12 at 22:50
@math-visitor To your first comment, that's equivalent to irreducibility. Not all schemes are irreducible, of course. The simplest example is probably $\operatorname{Spec} k[x]/(x(x - 1))$. – Dylan Moreland Jul 22 '12 at 22:53
Dear @math-visitor, to your second comment, take $(2)$ – Ehsan M. Kermani Jul 22 '12 at 22:56
@DylanMoreland and ehsanmo, thank you! – math-visitor Jul 22 '12 at 22:57