Generic Points (Part 1)

Correspondence between points and irreducible closed subsets

Affine Schemes

We have a bijection $X=\text{Spec} A \leftrightarrow \{\text{Irreducible closed subsets of } X\}$ given by

$$x \mapsto \overline{\{x\}};$$

$$\text{generic point of } Z \leftarrow Z.$$

Why is $\overline{\{x\}}$ irreducible? That is because it is equal to $\mathbb{V}(\mathfrak{p}_x).$

Why does every irreducible closed subset $Z$ of $X$ has a  generic point? Let $Z = \mathbb{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p} \in A$. Then $[\mathfrak{p}]$ is the generic point of $Z$ (i.e. its closure in $X$ is $Z$). '

Why is this generic point unique? $\mathbb{V}(\mathfrak{p}) = \mathbb{V}(\mathfrak{q})$ iff $\mathfrak{p}= \mathfrak{q}$.

General Schemes

Let $X$ be any scheme. We claim that there is a bijection

$$X \leftrightarrow \{\text{Irreducible closed subsets of } X\},$$

given by the same map as above.

Why is $\overline{\{x\}}$ irreducible? This is true for every topological space. Suppose $\overline{\{x\}} = Y_1 \cup Y_2$ for some closed $Y_1, Y_2$ of $X$. Then one of the $Y_i$ must contain $x$ and therefore contain $\overline{\{x\}}$ by minimality of closure.

Why does every irreducible closed subset $Z$ of $X$ has a generic point?

Let $U \subset X$ be affine such that $U \cap Z \neq \emptyset$. Then $Z \cap U$ is irreducible in $U$ (by Lemma 1 below). So it has a relative generic point, i.e. a point $x \in U$ such that the closure of $x$ in $U$ is $Z \cap U$. (Note that the closure of $x$ in $U$ is $\overline{\{x\}} \cap U$.)

As $Z$ is closed, we have  the closure of $Z\cap U$ in $X$ is the same as its closure in $Z$, which is $Z$ as $Z$ is irreducible. As $\overline{{\{x\}}}$ is a closed subset of $X$ containing $Z \cap U$, it must contains $\overline{Z\cap U} = Z$.  

Thus $x$ is a generic point of $Z$. 

Lemma 1. Let $Z$ be an irreducible topological space and $U$ be a nonempty open subset of $Z$. Then $U$ is irreducible.

Proof.  Indeed, let $V$ be any nonempty open subset of $U$. Then $V = V' \cap U$ for some open subset  $V'$ of $Z$ so $V$ itself is open in $Z$. As $Z$ is irreducible, $V$ is dense in $Z$, i.e. $\overline{V} = Z$. Thus the closure of $V$ in $U$ is $\overline{V} \cap U = U$, so $V$ is dense in $U$.

Note here we use the following Lemma

Lemma 2. Let $A$ be an arbitrary subset of $X$. For all subset $B \subset A$, we have   $\overline{B} \cap A$ is the closure $B'$ of $B$ in $A$. 

Proof. $$B' = \bigcap_{F \text{ closed in } A, F \supset B } F = \bigcap_{F \text{ closed in }X, F \supset B} (F \cap A) = \left(\bigcap_{F \text{closed in X}, F \supset B} F\right) \cap A = \overline{B} \cap A.$$

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