Underlying set of affine schemes

Examples: (using division algorithm we can find points of the first three schemes)

• $\text{Spec} \mathbb{C}[x]$: closed points and generic pt;
• $\mathbb{A}^1_k$ for algebraically closed $k$: same;
• $\mathbb{A}^1_k$ for any $k$ has infinitely many point (imitate Euclid's proof of infinitude of primes). 
• $\text{Spec}\mathbb{Z}$;
• $\text{Spec} k$;
• $\text{Spec}k[\epsilon]/(\epsilon^2)$ where $k[\epsilon]/(\epsilon^2)$ is called the ring of dual numbers;
• $\mathbb{A}^1_{\mathbb{R}}= \text{Spec} \mathbb{R}[x] = \{(0), (x-a), (x^2 + ax + b) \mid \text{irreducible}\}$ (use the fact that $\mathbb{R}[x]$ is a UFD.)
  Note that the $(x-a)$ are maximal ideals $(x^2 + ax + b)$ as the corresponding quotients are always fields. In particulat, $\mathbb{R}[x]/(x^2 + ax + b) \cong \mathbb{C}$.
  So $\mathbb{A}^1_{\mathbb{R}} is the complex plane folded along real axis, where Galois-conjugate points are glued. 
• $\mathbb{A}^1_{\mathbb{Q}}$
• Closed points of $\mathbb{A}^n_{\mathbb{Q}}$ are just Galois-conjugate glued together
• $\mathbb{A}^1_{\mathbb{F}_p} = \{(0), (f(x))\mid \text{ irreducible }\}$ (use the fact that $\mathbb{F}_p[x]$ is a Euclidean domain)
  Think of $f \leftrightarrow$ roots of $f$ (i.e. set of Galois conjugates in $\overline{\mathbb{F}_p}.$
  $\mathbb{A}^1_{\mathbb{F}_p}$ is bigger than $\mathbb{F}_p$. A polynomial $f(x)$ is not determined by its value at $\mathbb{F}_p$ (e.g. $x^p - x$), but it is uniquely determined by its valued on $\mathbb{A}^1_{\mathbb{F}_p}$ (as it is a reduced scheme). 
• $\mathbb{A}^2_{\mathbb{C}} = \text{Spec} \mathbb{C}[x,y] = \{(0), (x-a, y-b), f(x,y) \mid f \text{ irreducible}\}.$
•  $\mathbb{A}^n_{\mathbb{C}} $

Interpreting Fibers as Fiber Product

Proposition. Let $f: Y \to Z$ be a continuous map of topological spaces.  Let $z$ be a point of $Z$. Then $f^{-1}(\{z\}) = \{z\} \times_Z Y$ as topological spaces.

More generally, let $g: X\to Z$ is a morphism and $x \in X$. Let $\pi: X \times_Z Y \to X$ be the pullback of $f$ along $g$. Then $\pi^{-1}(x) = f^{-1}(g(x))$ as topological spaces.

Explanation:  The scheme structure on $\{z\}$ is $\text{Spec}\kappa(z)$, spectrum of the residue field at $z$.
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Defn. Scheme-theoretic Fiber. Let $f: Y \to Z$ be a continuous map of topological spaces.  Let $z$ be a point of $Z$. Then we could assign $f^{-1}(\{z\}) $ the scheme structure of  $\{z\} \times_Z Y$ as topological spaces via the natural identification in the above proposition. We call $\{z\} \times_Z Y$  is scheme-theoretic preimage of $z$, or fiber of $f$ above $z$, and also denote it by $f^{-1}(z).$

If $Z$ is irreducible, the fiber above the generic point of $Z$ is called the generic fiber of $f$.

Note: Finite morphisms have finite fibers.

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Example. Projection of the parabola $y^2 = x$ to the $x$-axis over $\mathbb{Q}$. This corresponds to a map of Spec $f: \text{Spec} \mathbb{Q}[x,y]/(y^2 - x) \to \text{Spec} \mathbb{Q}[x]$ induces by the ring map $\mathbb{Q}[x] \to \mathbb{Q}[x,y]/(y^2 - x)$ given by $x \mapsto x$.

Preimage of $1$ is two points $\pm 1$.  $\kappa(x-1) = \mathbb{Q}[x]/(x-1)$ (since $(x-1)$ is maximal). So
$$f^{-1}(x-1) =  \text{Spec} \mathbb{Q}[x,y]/(y^2 - x) \times_{\text{Spec} \mathbb{Q}[x]$} \text{Spec}\mathbb{Q}[x]/(x-1)$$
i.e. $$ \text{Spec }\left( \mathbb{Q}[x,y]/(y^2 - x) \otimes_{ \mathbb{Q}[x]}\mathbb{Q}[x]/(x-1) \right).$$
The latter is equal to
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x- 1) = \text{Spec}\mathbb{Q}[y]/(y^2-1).$$
This is simply
$$\text{Spec} \mathbb{Q}[y]/(y-1) \sqcup \text{Spec} \mathbb{Q}[y]/(y+1).$$

Preimage of $0$ is a non-reduced point
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x) = \text{Spec}\mathbb{Q}[y]/(y^2).$$

Preimage of $-1$ is a non-reduced point  of "size 2 over the base field"
$$\text{Spec}\mathbb{Q}[x,y]/(y^2 -x, x+1 ) = \text{Spec}\mathbb{Q}[y]/(y^2+1) \cong \text{Spec}\mathbb{Q}[i] = \text{Spec}\mathbb{Q}(i).$$

Preimage of generic point is a non-reduced point of "size 2 over the residue field"
$$ \text{Spec }\left( \mathbb{Q}[x,y]/(y^2 - x) \otimes_{ \mathbb{Q}[x]}\mathbb{Q}(x)\right) \cong \text{Spec }\left( \mathbb{Q}[y] \otimes_{ \mathbb{Q}[y^2]}\mathbb{Q}(y^2)\right) .$$
The latter is simply $\text{Spec} \mathbb{Q}(y)$. Note that $[\mathbb{Q}(y): \mathbb{Q}(x)] = [\mathbb{Q}(y): \mathbb{Q}(y^2)] = 2$.


Notice: In all cases above, the fiber is an affine scheme whose vector space dimension over the residue field is $2$.

Normal Schemes

Normality = "not too far from smooth"

Recall that the punctured plane $U= \mathbb{A}^2- \{(0,0)\}$ viewed as an open subscheme of $X = \mathbb{A}^2$ is NOT affine. In proving this, we computed the global section of the punctured plane and found that $O_X(U) = O_X(X)$. This means in particular that every function on $U$ extends to the whole of $X$. This is an analogue of Hartogs's Lemma in complex geometry: we can extend a holomorphic function defined on the complement of a set of codimension at least two on a complex manifold over the missing set.

In algebraic geometry, there is an analogue. We can extend functions over points (in codim at least 2) not only if they are smooth, but also if they are "mildly singular"Hart, i.e. normal. Locally Noetherian normal schemes satisfy Hartogs's Lemma. Consequently rational functions without poles are defined everywhere.

Defn. A scheme $X$ is called normal if all of its stalks are normal, i.e. integral domains that are integrally closed in its field of fraction.

Claim. Normal schemes are reduced.
Proof. A scheme is reduced iff all of its stalks are reduced.
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Recall that if $A$ is integrally closed then so is its nontrivial localization $S^{-1}A$. From this we can deduce the following.

Claim. If $A$ is integrally closed then $\text{Spec} A$ is normal.

Claim. If $X$ is quasi-compact then $X$ is normal iff its normal at every closed point.
Proof. Here we simply use the fact that if a property $P$ is compatible with localization, then to check that $P$ holds for a quasi-compact scheme, it suffices to check it at closed points.  We replicate the proof from previous post here.

Suppose $X$ is normal at every closed point. Let $x \in X$. Then $\overline{\{x\}}$ contains a closed point $y$. Let $U= \text{Spec} A$ be an affine neighborhood of $y$. Then $U$ must contains $X$ as $y$ is in the closure of $x$. Thus we have $\mathbb{V}(\mathfrak{p}_x) \ni \mathfrak{p}_y$ so $\mathfrak{p}_y \supset \mathfrak{p}_x$ so we have a localization map
$$A_{\mathfrak{p}_y} \to A_{\mathfrak{p}_x}.$$
As $A_{\mathfrak{p}_y}$ is normal and normality is compatible with localization, we have $A_{\mathfrak{p}_x}$ is also normal.
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From commutative algebra, we know that for an integral domain $A$, TFAE:

• $A$ is integrally closed;
• $A_{\mathfrak{p}}$ is integrally closed for every $\mathfrak{p}$ prime in $A$;
• $A_{\mathfrak{m}}$ is integrally closed for every $\mathfrak{m}$ max in $A$.

Thus we have.

Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. For every affine open $U$ in $X$, $O_X(U)$ is normal; (in particular, $U$ itself is normal)
3. There is a cover of $X$ by open affine $U_{\alpha}$ such that $O_X(U_{\alpha})$ is normal

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Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. There exists an open covering $X = \bigcup X_i$ such that each open subscheme $X_i$ is normal;
3. Every open subscheme of $X$ is normal.

Proof.
(1) $\implies $ (2). Suppose $X$ is normal. Then for every open affine $U$ of $X$, $O_X(U)$ is normal. This implies $U$ itself is normal.
(2) $\implies$ (1). Cover $X$ by affine covers of $X_i$. Each of this affine piece is normal so $X$ is normal.
(1) $\implies$ (3). Every open subscheme of $X$ can be covered by affine opens of $X$. 

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Warning: Normal schemes are NOT necessarily integral.

Example. $X = \text{Spec} k \sqcup \text{Spec} k$ is a normal scheme. (Note: disjoint union of normal schemes is normal since normality is a stalk-local condition). However $X = \text{Spec} (k \times k) = \text{Spec} k[x]/(x(x-1))$ so its global section is not an integral domain.
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Claim: Global sections of an irreducible normal scheme is normal.

Proof.  Note a normal scheme is reduced so an irreducible normal scheme is integral. So we can embed all sections of $O_X$ to its function field $K(X)$ and all restriction maps are inclusion. In particular, we have $O_X(X) = \bigcap_{U \subset X, \text{open}}O_X(U) =\bigcap_{U \subset X, \text{affine}}O_X(U) $.

Suppose $s \in K(X)$ is integral over $O_X(X)$. Then it is integral over $O_X(U)$  for $U$ affine, so $s \in O_X(U)$, as it is normal by Proposition above. Thus $s \in \bigcap_{U \subset X, \text{affine}}O_X(U) = O_X(X).$

The Property of being Reduced

click here

A scheme $X$ is said to be reduced if $O_X(U)$ is reduced for all open $U$. 

Key point

Reduced-ness is stalk-local and affine local. However, it is not an open condition . 

Stalk-local.

Reduced-ness is a stalk-local property: $X$ is reduced iff $(O_X)_x$ is reduced for all $x$.

Paraphrasing this, a scheme is not reduced iff its talk at some point is not reduced. 

proof. (Trivial) Suppose $X$ is not reduced. Then there exists a section nonzero $s \in O_X(U)$ such that $s^n = 0$ for some $n$. As $s$ is not $0$ and $O_X$ is a sheaf, $s_x \neq 0$ for some $x \in U$, but $s_x^n = 0$ in $(O_X)_x$.

On the other hand, suppose $[(s, U)]^n = 0$ in $(O_X)_x$ and $[(s, U)] \neq 0$. Then $s|_W \neq 0$ on some $x \in W \subset U$.

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Affine Local

Affine Communication Lemma.

A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$.
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Reducedness is affine-local

Suppose $A$ is a ring generated by $f_1, \ldots, f_n$. Then $A$ is reduced iff each $A_i$ is.

$X$ is reduced iff X can be covered by affine open sets $\text{Spec} A$ where $A$ is reduced.

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Example: The scheme $X = \text{Spec} k[x,y]/(y^2, xy)$ is non-reduced. 

The ring $k[x,y]/(y^2, xy)$  contains the element $y \neq 0$ such that $y^2 = 0$. At the stalk of every point $p$, we will still have $y_p^2 = 0$. So $(O_X)_p$ is reduced iff $y_p = 0$. This is possible if we invert $x$, i.e. if $p \in D(x) = $ $x$-axis minus origin.

We show the origin is the only point at which $X$ is non-reduced.

Fuzzy Origin

Proof.  This is equivalent to saying that $\text{Spec} \left(k[x,y]/(y^2, xy)\right)_x$ is reduced, as 

the points of $X$ are prime ideals containing $y^2$ and $xy$, i.e. $(y)$- the $x$-axis.  We think of  $X = \text{Spec} k[x,y]/(y^2, xy)$ then as the $x$ axis with some derivatives at $(0,0)$. 

To say that the origin is the only non-reduced point is equivalent to saying that that on $x$-axis minus the origin, $X$ is reduced. In other words $O_X(D(x))$ is reduced. 

We have $A_x := \left(k[x,y]/(y^2, xy)\right)_x = k[x, y, x^{-1} ]/(y^2, xy)$. As $x$ is a unit in this ring and $xy = 0$, we must have $y = 0$ in this ring. So $A_x = k[x, y, x^{-1}]/(y) = k[x, x^{-1}]$ is reduced. 

At the origin, we have $$A_{(x, y)} = \left(k[x,y]/(y^2, xy)\right)_{(x, y)}  = \frac{k[x,y]_{(x,y)}}{(y^2, xy)}.$$

Thus $y$ is zero in $A_{(x,y)}$ iff the image of $y$ in $k[x,y]_{(x,y)}$ lies in $(y^2, xy)$. Thus for some $h$ outside of $(x, y) we have

$$h y = y^2 f + xy g.$$

Dividing both sides by $y$ gives $h \in (x, y)$ a contradiction. 

More Illuminating Explanation.   Remember that $\text{Spec} k[x,y]/(y^2, xy)$ is just the $x$-axis together with (the span of ) some extra derivative vectors at the origin $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ (cf. Visualizing Nilpotents). For $y$ to be $0$ at the stalk of $\mathfrak{p}$ it has to be zero in a neighborhood of $\mathfrak{p}$ (where neighborhood here include the derivative vectors). So for  $y$ to be zero at the stalk of the origin, we not only need $y|_{(0,0)} = 0$ but also that $\frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}$ and $\frac{\partial^2}{\partial_{x^2}}$ applied to $y$ is $0$. The later does not hold as $\frac{\partial}{\partial_y}(y) \neq 0.$  (It gives $y$, the $y$-term)

To say this rigorously, remember that $\frac{\partial}{\partial_y}(y)$ is just the image of $y$ in $k[x, y] \to k[x,y]/(y^2)$ and clearly the image of $y$ is not $0$.

Now how to formalize  "if $y$ vanishes at the stalk of the origin then $\frac{\partial}{\partial_y}(y) =0$"?  To view $\frac{\partial}{\partial_y}$ as an element of $\text{Spec} k[x,y]/(y^2, xy)$ is to have a morphism $k[x,y]/(y^2, xy) \to k[y]/(y^2)$ given by $x \mapsto 0$ and $y \mapsto y$, i.e every $F$ is mapped to its $y$-term.

 (Two polynomials $f, g \in k[x,y]$ are equal in the first ring iff $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ applied to them are equal, whereas in the second ring, functions are identified if they agree under $\frac{\partial}{\partial_y}$. As $x$ has no $y$ term, it becomes $0$ in the latter.)

The map $\text{Spec} k[y]/(y^2) \to X$ induces  a pull-back map on stalks, so we have
$$\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} \to \left(\frac{k[y]}{(y^2)}\right)_{(y)}$$  

which is a local homomorphism.
If $y$ is $0$ in $\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} $ then its image in $\left(\frac{k[y]}{(y^2)}\right)_{(y)}$ must also be $0$. In other words, as the latter is just $\frac{k[y]_{(y)}}{(y^2)}$, we must have the image of  $y$ in $k[y]_{(y)}$ lie in $y^2$ and thus $yh = y^2 f$ for some $h, f \not \in (y)$. This is impossible by unique factorization in $k[y]$. Another way to see this is to see that if $y$ is $0$ in the ring above then $\frac{k[y]_{(y)}}{(y^2)} =  k$, i.e. $y^2$ is kernel of $k \to k[y]_{(y)}$ which is not possible as the latter is injective. 

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Not an open condition:

Reducedness is in general, not an open condition (see example below). However if $X$ is a locally Noetherian scheme then reducedness is open.

Example.
Let $X$ be $\text{Spec} A$ where

$$A = \text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)}$$

Notice that prime ideals of $A$ are exactly prime ideals $\mathfrak{p}$ of $\mathbb{C}[x, y_1, y_2, \ldots]$ containing $ (y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)$. These are exactly prime ideals containing $(y_1, y_2, \ldots,)$.

So $\text{Spec} A$ as a set is homeomorphic to the $$\text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1, y_2, y_3, \ldots } = \text{Spec} \mathbb{C}[x].$$

Claim: the nonreduced points in $X$ correspond to the set $\{\text{x - m}\mid m \in \mathbb{Z}\}$ in $ \text{Spec} \mathbb{C}[x]$. In particular the set of  reduced points in $\text{Spec} A$ is therefore not open.

Proof:  Indeed similar to the above example, $A_{\mathfrak{p}}$ is not reduced iff one of $y_1, y_2 \ldots$ is not zero in the stalk at $\mathfrak{p}$.

Clearly, at points $\mathfrak{p}$ where $(x-m)$ is invertible in $\mathfrak{p}$, we have $y_m = 0$ in $\mathfrak{p}$, so $y_m$ is not nilpotent. Thus $A$ is reduced at $\mathfrak{p}$ if for all $m$, $x- m$ is invertible in $\mathfrak{p}$. This is true if $x-m$ all lie outside $\mathfrak{p}$, i.e. $\mathfrak{p} \neq x-m$ for any $m$. (As $\mathfrak{p}$ is maximal).

Let $\mathfrak{p}$ correspond to $(x-m)$ for some $m$. Then in $A_{\mathfrak{p}}$, we invert $(x - n)$ for every $n\neq m$, so for all such $n$, $y_n = 0$. Thus
$$A_{\mathfrak{p}} = \text{Spec}\frac{\mathbb{C}[x,y_m]}{(y_m^2, (x-m)y_m)},$$
which reduces the problem to the previous example.

Relationship with quasi-compact schemes: Recall that for every open condition $P$, to check if all points of a  quasi-compact scheme $X$ has $P$, it suffices to check that all closed points do. Though reducedness is not an open condition, it is still possible to check the reducedness of a quasi-compact schemes by checking it at closed points.

Closed Points of Quasi-Compact Schemes (Part 1)

Does every nonempty scheme contains a closed point?

Answer: No. There exist nonempty schemes with no closed points.
However, for quasi-compact schemes the statement is true.

Recall the following
Easy Lemma: A scheme $X$ is quasi-compact iff it can be written as a finite union of affine schemes.
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Quasi-Compact Schemes have Closed Points

Let $X$ be a quasi-compact scheme

Proof 1:  (Stack Exchange) As $X$ is quasi-compact, it is a finite union of affine pieces $X = \bigcup_{i=1}^n U_i$. Notice that a subset of $X$ is closed iff it is closed in every $U_i$.

Suppose $X$ does not contain any closed point.

Let $x_1$ be a closed point of $U_1$, then $\overline{\{ x_1\}} \cap U_1 = \{x_1\}$, i.e. $x_1$ is the unique point in $U_1$ of its closure.

As $x$ is not closed, $x$ must not be closed in some $U_i$. WLOG, assume $x_1$ is not closed in $U_2$.  Then $\overline{\{ x_1\}} \cap U_2$, being a closed subset of an affine scheme, must contain some closed point $x_2$ of $U_2$ other than $x_1$.

Repeating the argument, we can find $x_3 \in \overline{\{ x_2\}} \subset \overline{\{ x_1\}} $ such that $x_3$ is closed in $U_3$ but $x_3 \not \in U_2 \cup U_1$. Continuing the process we will get $x_{n+1} \not \in U_{n} \cup U_{n-1}\cup \cdots \cup U_1$, a contradiction.

Remark.  We constructed a sequence $x_1, \ldots, x_n, \ldots$ satisfying:

• $x_i$ is a closed point of $U_i$;
• $x_j$ is a specialization of $x_i$ for every $j > i$;
• $x_i$'s are all distinct
• $x_j \not \in U_i$ for all $j < i$. 

Proof 2: (Akhil) Using Quasi-compactness + Zorn's Lemma, we can show that $X$ has a minimal nonempty closed set $Z$. We can put a scheme structure on $Z$ such that the underlying topology is the subspace topology. As $Z$ is a scheme, it contains an affine piece $U$. As $Z - U$ is a closed subset of $Z$, by minimality of $Z$ we must have $Z- U = \emptyset$, i.e. $Z = U$ is affine so it must contains a closed point.

Proof 3: (Mine)

If $X$ has no closed points, then the closure of every point contains a point other than itself.
Thus we have a nested sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is some $x$ such that $\overline{\{x\}}$ does not contain any closed point other than $x$, contradiction.

This kind of argument is useful in proving some property (that might not be open) holds for every point of $X$ if it holds for closed point of $X$. For example, we use this argument to check that for a quasi-compact scheme $X$, it suffices to check reducedness at closed points. 

Main Usage

We say that property $P$ of points of a scheme  is open if whenever a point $x$ satisfies $P$, a neighborhood of it must satisfy $P$ also. 

Let $X$ be a quasi-compact scheme. To show that all points of $X$ satisfies $P$. It suffices to show that all \emph{closed} points of $X$ do. 

Explanation: As every closed subset of a quasi-compact scheme is quasi-compact, it must contains a closed point. In particular, for every $x \in X$, $\overline{\{x\}}$ contains a closed point $y$. By definition of closure, we have that every neighborhood of $y$ must contain $x$. So any covering of closed points must also be a cover of $X$. 

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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