2.1 Spec(R)

For an affine variety, define $R=\Gamma (X)$, and then we can associate a “geometric” objective to $R$, which is called $\mathrm{Spec}(R)$. If $R$ is a finitely generated integral domain over an algebraically closed field, $\mathrm{Spec}(R)$ is nearly the same as $X$. In this section, we assume that $R$ is a commutative ring with identity. One can define a point set

$$\begin{array}{rl}\text{ Spec }(R)=& \text{ the set of prime ideals }P⊈R\\ & [R\text{ itself is not counted as a prime ideal }\\ & \text{ but }(0),\text{ if prime, is counted }].\end{array}$$

and define Zariski topology

$$\begin{array}{rl}\text{ Closed sets }=& \text{ Sets of the form }\{[P]\mid P\supseteq A\}\text{ for some ideal }\\ & A\subseteq R.\text{ This set will be denoted }V(A).\end{array}$$

It is easy to verify the collection of closed subsets $\{V(A):A\subseteq R\text{ is an ideal}\}$ does define a topology, which is proved in ^vf0aa5. Furthermore, define $\mathrm{Spec}(R{)}_f=\mathrm{Spec}(R)-V((f))$. Later we can check $\mathrm{Spec}(R{)}_f\simeq \mathrm{Spec}(R_f)$. In addition, we have $\mathrm{Spec}(R)={\cup }_{f\in A}\mathrm{Spec}(R{)}_f$ and $\{\mathrm{Spec}(R{)}_f:f\in R\}$ is a set of topology basis of $\mathrm{Spec}R$, called distinguished open sets.

For any $[p]\in \mathrm{Spec}(R)$, there is

$$V(I)=\overline{\{[p]\}}=V(p).$$

• $[p]$ is called a closed point if $\overline{\{[p]\}}=V(p)=\{[p]\}$. Note that $[p]$ is a closed point iff $p$ is a maximal ideal.
• If $\overline{[p]}=\mathrm{Spec}R$, then $[p]$ is called a generic point.

Definition

Let $Z\subseteq \mathrm{Spec}R$ be an irreducible closed subset. Then $z_0\in Z$ is a generic point of $Z$ if $\overline{\{z_0\}}=Z$, i.e. any open subset of $Z$ contains $z_0$.

Proposition

If $x\in \mathrm{Spec}(X)$, then $\overline{\{x\}}$ is irreducible.

\begin{proof} Assume that $x\in \overline{\{x\}}=W_1\bigsqcup W_2$, WLOG $x\in W_1$, then $\overline{\{x\}}={\overline{W}}_1=W_1$ and so $W_2=\varnothing$. \end{proof}

Proposition

Every irreducible subset $Z\subseteq \mathrm{Spec}(R)$ equals $V(p)$ as set for some prime ideal $p$, and $[p]$ is its unique generic point.

\begin{proof} Suppose $Z=V(A)=V(\sqrt{A})$ as set. Assume $A=\sqrt{A}$. We will show if $Z$ is irreducible, then $A$ is prime.

Otherwise, there exists $f,g\notin A$ such that $fg\in A$. We claim $B\cap C=A$ with $B=A+(f)$ and $C=A+(g)$. Take $t\in B\cap C$, then $t$ can be written as $t=a_1+f\cdot \alpha =a_2+h\cdot \beta$. Note that

$$(a_1+f\alpha )(a_2+h\beta )\in A$$

then we have $a_1+f\alpha \in A$ as $A$ is radical. Now we proved the claim.

It deduces that $V(B\cap C)=V(B)\cup V(C)=V(A)$. Since $A={\cap }_{p\supseteq A}p$, there exists $p\supseteq A$ such that $f\notin p$ and $q\supseteq A$ such that $g\notin q$. Thus $p\notin V(B)$ and $q\notin V(C)$. In the other word, $V(B),V(C)⊊V(A)$, which contradicts with $V(A)$ irreducible. Therefore, $A$ is a prime ideal.

If there exist two distinct generic point $p,q$, then $V(p)=\overline{\{[p]\}}=Z=\overline{\{[q]\}}=V(q)$ and so $p=q$. \end{proof}

Example. Let $R=k[x]/(x^2)=\{a[x]+b:a,b\in k,[x{]}^2=0\}$. Then $\mathrm{Spec}(R)=\{([x])\}$ and $R/([x])\simeq k$. Note that $\{([x])\}=V([x])=V((0))$, and $\sqrt{(0)}=([x])$ by ^bgejpl.

Proposition

Let $\{f_{\alpha }:\alpha \in S\}\subseteq R$. Then $\mathrm{Spec}R={\cup }_{\alpha \in S}\mathrm{Spec}(R{)}_{f_{\alpha }}$ iff $1\in (f_{\alpha }{)}_{\alpha \in S}$.

\begin{proof} Assume that $\mathrm{Spec}R={\cup }_{\alpha \in S}\mathrm{Sp}ec(R{)}_{f_{\alpha }}$. Then for any prime ideal $p\subseteq R$, there exists $f_{\alpha }$ such that $f_{\alpha }\notin p$ and so $p⊉(f_{\alpha }:\alpha \in S)$. If $I:=(f_{\alpha }{)}_{\alpha \in S}\ne R$, then $I$ is contained in some maximal ideal $M$. However, $M$ is a prime ideal and $M\supseteq (f_{\alpha }{)}_{\alpha \in S}$, which is impossible. Therefore, $(f_{\alpha }{)}_{\alpha \in S}=R$ and so $1\in (f_{\alpha }{)}_{\alpha \in S}$.

Conversely, assume that $1\in (f_{\alpha }:\alpha \in S)$. If there exists a prime ideal $p$ such that $f_{\alpha }\in p$ for all $p$, then $I\subseteq p$, which is impossible. Thus $p⊉(f_{\alpha }:\alpha \in S)$ and so $p\in \mathrm{Spec}(R{)}_{f_{\alpha }}$ for some $\alpha$. \end{proof}

Remark. Since $1\in (f_{\alpha }:\alpha \in S)=\{{\sum }_{\text{finite}}{h}_{\alpha }{f}_{\alpha }:\alpha \in S\}$, there exist ${\alpha }_1,\cdots ,{\alpha }_k$ such that $1={\sum }_{i=1}^k{h}_i{f}_{{\alpha }_i}$ (partition of unity). Hence $\mathrm{Spec}R={\cup }_{i=1}^k\mathrm{Spec}(R{)}_{f_{{\alpha }_i}}$.

Corollary

$\mathrm{Spec}R$ is quasi-compact. (same as compactness without Hausdorff in topology)

\begin{proof} Suppose that $\mathrm{Spec}R={\cup }_i{U}_i$, and we aim to find its finite subcover. For each open set $U_i$, $U_i$ can be written as $U_i=\mathrm{Spec}R\setminus V(A_i)$ for some $A_i\subseteq R$. For any point $[p]\in U_i$, we have $[p]\notin V(A_i)$ and so $A_i⊈p$. Thus, there exists $f_{ij}\in A_i\setminus p$, that is, $[p]\in \mathrm{Spec}(R{)}_{f_{ij}}$. Also note that

$$\mathrm{Spec}(R{)}_{f_{ij}}=\mathrm{Spec}R\setminus V((f_{ij}))\subseteq \mathrm{Spec}R\setminus V(A_i)=U_i,$$

so for each $[p]\in U_i$, there exists $f_{ij}$ such that $[p]\in \mathrm{Spec}(R{)}_{f_{ij}}\subseteq U_i$. It deduces that

$$\mathrm{Spec}R={\cup }_i{U}_i={\cup }_i({\cup }_j\mathrm{Spec}(R{)}_{f_{ij}}).$$

By ^05e2eb, $\mathrm{Spec}R={\cup }_{i,j\text{ finite}}\mathrm{Spec}(R{)}_{f_{ij}}\subseteq {\cup }_{i\text{ finite}}{U}_i$. \end{proof}

Remark. Since $\mathrm{Spec}(R_f)\simeq \mathrm{Spec}(R{)}_f$ as topology spaces, one can check $\mathrm{Spec}(R{)}_f$ is also quasi-compact. Remark that $\{\mathrm{Spec}(R{)}_f:f\in R\}$ is a set of topology basis of $\mathrm{Spec}R$. However, not all $U\subseteq \mathrm{Spec}R$ are quasi-compact, and here is an counterexample:

• For a ascending chain $P_1⊊P_2⊊\cdots ⊊q\ne R$, define $U_i=\mathrm{Spec}R\setminus V(P_i)$ and $U={\cup }_{i\in \mathbb{N}}{U}_i$. There does not exist finite $i$ such that $U={\cup }_{\text{finite}}{U}_i$.

---

15:30: Let $X=\mathrm{Spec}R$ and $X_f=\mathrm{Spec}(R{)}_f$. The followings hold:

• $X_f\cap X_g=X_{fg}$
• $X_f\supseteq X_g$ iff $g\in \sqrt{(f)}$

For $X_g\subseteq X_f$, we aim to define a map $R_f\to R_g$. Since $X_f\supseteq X_g$ iff $g\in \sqrt{(f)}$, there exists $n$ such that $g^n=f\cdot h$. So the map can be defined as

$$R_f\to R_g,r/f^m=(r{h}^m)/(f^m{h}^m)\mapsto r{h}^m/g^{mn}.$$

It is a functor(这个词是这么用吗),

If $[p]\in X_f$, then $f\notin p$, $\{f,f^2,\cdots \}\subseteq R\setminus p$. $R_f\to R[R-p]=R_p$. 好像是和stalk的对应，但是这里符号好怪。。

Lemma

Let $X_f={\cup }_{\alpha \in S}{X}_{f_{\alpha }}$. If there is $g\in R_f$ such that $g{|}_{X_{f_{\alpha }}}=0$ for all $\alpha$, then $g=0\in R_f$.

\begin{proof} Let $g=b/f^n$. Let $A=\{c\in R\mid cb=0\}$. The following are clearly equivalent:

• i) $g=0$ in $R_f$

• ii) $\exists m$ such that $f^m\cdot b=0$ in $R$

• iii) $f\in \sqrt{A}$

• iv) $P\supset A\Rightarrow f\in P$.

• iv)→i) ?

Therefore if $g\ne 0$, we can choose a prime ideal $P\supset A$ with $f\notin P$, i.e., $[P]\in X_f$. Then $[P]\in X_{f_{\alpha }}$ for some $\alpha$. Using the commutative diagram it follows that $g$ goes to 0 in $R_p$. Since $b=g\cdot f^n$, so does $b$. Therefore there is some $c\in R-P$ with $c\cdot b=0$, i.e., $c\in A$. This contradicts the fact that $P\supset A$ and the lemma is proven. \end{proof}

---

Lemma

$X_f={\cup }_{\alpha \in S}{X}_{f_{\alpha }}$, if $g_{\alpha }\in R_{f_{\alpha }}$ satisfies $g_{\alpha }{|}_{X_{f_{\alpha }{f}_{\beta }}}=g_{\beta }{|}_{X_{f_{\alpha },f_{\beta }}}\in R_{f_{\alpha }{f}_{\beta }}$, then there exists $g\in R_f$ such that $g{|}_{X_{f_{\alpha }}}=g_{\alpha }$ for all $\alpha \in S$.

\begin{proof} Take $f=1$. Since $X$ is quasi-compact, $X={\cup }_{\text{finite}}{X}_{f_{\alpha }}$ and it suffices to show for finite covering. If the finite case hold, then by the argument above we finish the proof: Since $X_{f_{\beta }}={\cup }_{\text{finite}}{X}_{f_{\beta }{f}_{\alpha }}$, by finite case, there exists $g\in R$ such that $g{|}_{X_{f_{\alpha }}}=g_{\alpha }$, and so $g{|}_{X_{f_{\beta }{f}_{\alpha }}}=g_{\alpha }{|}_{X_{f_{\alpha }{f}_{\beta }}}=g_{\beta }{|}_{X_{f_{\beta }{f}_{\alpha }}}$. By ^376c07, $g{|}_{X_{\beta }}=g_{\beta }$.

• Now suppose $X={\cup }_{i=1}^{\ell }{X}_i$, where $X_i=X_{f_i}$. For $g_i=b_i/f_i^n\in R_{f_i}$ with $b_i\in R$ and $n$ is independent of $i$.

坏消息，一点没听；好消息，和课本一模一样 \end{proof}

(this part prove that $X_f$ is a sheaf.)

Structure Sheaf

Let $X=\mathrm{Spec}R$, and let $U\subseteq X$ be an open set. Define

$${\underline{o}}_X(U)=\left\{(s_{[p]})\in \prod _{[p]\in U}{R}_p:\forall [p]\in U,\exists [p]\in X_f\subseteq U,\text{ and }r\in R_f\text{ such that }{s}_{[q]}\text{ is the image of }r\text{ under }{R}_f\to R_q,\forall [q]\in X_f\right\}$$

and remark that

$$\underset{[p]\in X_f}{\underrightarrow{\lim }}{R}_f=\underset{f\notin p}{\underrightarrow{\lim }}{R}_f=\underset{f\in R\setminus p}{\underrightarrow{\lim }}{R}_f=R_p=R[R-p].$$

In fact, this definition is equivalent to Mumford, that is,

Proposition

• ${\underline{o}}_X$ is a sheaf of rings;
• $\Gamma (X_f,{\underline{o}}_X)=R_f$;
• ${\underline{o}}_{X,[p]}=R_p$, where ${\underline{o}}_{X,[p]}=\underset{[p]\in U}{\underrightarrow{\lim }}{\underline{o}}_X(U)$.

\begin{proof} i) If $s\in {\underline{o}}_X(U)$ with $s{|}_{U_i}=0$, then $s_{[p]}=0\in R_p$ and so $s=0\in {\underline{o}}_X(U)$. Furthermore, if $\{s_i\}$ satisfies $s_i{|}_{U_i\cap U_j}=s_j{|}_{U_i\cap U_j}$, then there exists $s\in {\underline{o}}_X(U)$. Recall the definition of the sheaf, and we finish the proof.

• ii) Define $R_f\to \Gamma (X_f,{\underline{o}}_X),r\mapsto s=(r_p)$. It is easy to check this map is injective: if $r_p=0$ for any $[p]\in X_f$, then $r=0\in X_f$. 17:27 后面还有两行，没看懂

Now we check surjective. 17:33-17:35

iii) Notice that

$${\underline{o}}_{X,[p]}=\underset{[p]\in U}{\underrightarrow{\lim }}{\underline{o}}_X(U)=\underset{[p]\in X_f}{\underrightarrow{\lim }}{\underline{o}}_X(X_f)=\underset{f\notin p}{\underrightarrow{\lim }}{R}_f=R_p,$$

then we finish the proof. \end{proof}

Examples.

• $\mathrm{Spec}(k[x]):={\mathbb{A}}_k^1$, and all points of ${\mathbb{A}}_k^1$ are $\{(0)\}\cup \{(f):f\text{ is irreducible}\}$.
  • when $k=\mathbb{Q}$, then $(x^2+1)\in \mathrm{Spec}(\mathbb{Q}[x])$;
  • if $k=\overline{k}$, then $\mathrm{deg}f=1$ and $f=bx-a$ with $a/b\in k$.
• $X=\mathrm{Spec}\mathbb{Z}$, then $\mathfrak{p}=(p)$ with prime number $p$ or $\mathfrak{p}=(0)$, where $(0)$ is a generic point. Note that
  • ${\underline{o}}_{X,(p)}={\mathbb{Z}}_p$, which has the maximal ideal $p{\mathbb{Z}}_p$ and ${\mathbb{Z}}_p/p{\mathbb{Z}}_p=\mathbb{Z}/(p\mathbb{Z})$
  • ${\underline{o}}_{X,(0)}={\mathbb{Z}}_{(0)}=\mathbb{Q}$
• ${\mathbb{A}}_k^2=\mathrm{Spec}(k[x,y])$ with $\overline{k}=k$, then all prime ideals are
  • maximal ideals $(x-a,y-b)$
  • principal ideal $(f)$ with irreducible polynomial $f$
  • $(0)$ generic point
  • 
• $\mathrm{Spec}(\mathbb{R}[x,y]/(x^2+y^2+1))$ has prime ideal $p=(x^2+1,y)\supseteq (x^2+y^2+1)$, where $\mathbb{R}[x,y]/(x^2+1,y)\simeq \mathbb{R}[x]/(x^2+1)$ is an integral domain. Also $(x^2+a,y+1-a)$ is also a prime ideal.
• $\mathrm{Spec}(\mathbb{Z}[x])$ arithmetic surface, see ^6bfhl3.
  • $V((3))=\mathrm{Spec}\mathbb{Z}[x]/(3)=\mathrm{Spec}\mathbb{Z}/3\mathbb{Z}[x]={\mathbb{A}}_{\mathbb{Z}/3\mathbb{Z}}^1$.
  • blabla
• $\mathrm{Spec}(k[x]/(x^2))$ has prime ideal $([x])$. For any $a[x]+b\in p$, note that $(a[x]+b)(a[x]-b)=-b^2\in p$, then $b=0$ and one can set $\mathrm{Spec}(k[x]/(x^2))=\mathrm{Spec}k$. In fact,
  • ${\underline{o}}_X(X)=\{\text{linear functions}\}$
  • $\mathrm{Hom}(\mathrm{Spec}(k[x]/(x^2))\to Y,p\mapsto y)\simeq T_{Y,y}$.

---

Values of Sections

For $s\in \Gamma (U,{\underline{o}}_X)$, define $s([p])=s_p\in R_p/p{R}_p:=k([p])$. For any $x\in U$, stalk ${\underline{o}}_{X,x}$ is a local ring with the maximal ideal $m_x\subseteq {\underline{o}}_{X,x}$. Then $k(x):={\underline{o}}_{X,x}/m_x$ is a residue field. Note that when $x=[p]\in U$, we have ${\underline{o}}_{X,x}=R_p$, $m_{[p]}=p{R}_p$ and

$${\underline{o}}_{X,[p]}/m_{[p]}=R_p/p{R}_p\simeq (R/p{)}_p=\mathrm{Frac}R/p.$$

This definition is coincides with

Remark. For $F\in {\underline{o}}_X(U)$ and $x_0\in U$, there is $F=\alpha /\beta$ with $\alpha \in R$ and $\beta \notin m_{x_0}$. Then $F(x_0)=\alpha (x_0)/\beta (x_0)\in k$.

Link to original

However, for affine varieties, all values are in the same field $k$, which not always hold here. For example, consider $X=\mathrm{Spec}\mathbb{Z}$, then $r\in \Gamma (X,{\underline{o}}_X)\simeq \mathbb{Z}$ and image of $r$ at $[p]$ is

$$r([p])\in {\mathbb{Z}}_p/p{\mathbb{Z}}_p=\mathrm{Frac}(\mathbb{Z}/p\mathbb{Z}).$$

Proposition

For $X=\mathrm{Spec}R$ and $f\in R$, we have isomorphism of ringed space

(\mathrm{Spec}R_f,\underline{o}_X|_{\mathrm{Spec}R_f})\simeq(\mathrm{Spec}(R_f),\underline{o}_{\mathrm{Spec}R_f}). $$ ^pzdwqy

\begin{proof} For any $[p]\in \mathrm{Spec}{R}_f$, we know $f\notin [p]\subseteq R$. Thus there is a $1$-$1$ corresponding

$$\begin{array}{rl}[p]\in \mathrm{Spec}{R}_f& ⟷[p\cdot R_f]\in \mathrm{Spec}(R_f)\\ V(I)& ⟷V(I\cdot R_f)\end{array}$$

Also it is easy to check for any $U$, ${\underline{o}}_X(U)={\underline{o}}_{\mathrm{Spec}(R_f)}$, because $(R_f{)}_{p\cdot R_f}=R_p$ when $f\notin p$. \end{proof}

prescheme

A presheme is a topological space $X$, plus a sheaf of rings ${\underline{o}}_X$ provided that there exists open covering $\{U_{\alpha }{\}}_{\alpha }$ of $X$ such that $(U_{\alpha },{\underline{o}}_X{|}_{})U_{\alpha }\simeq (\mathrm{Spec}{R}_{\alpha },{\underline{o}}_{\mathrm{Spec}(R_{\alpha })})$ for some commutative ring $R_{\alpha }$.

Remark. For a prescheme $(X,{\underline{o}}_X)$ and open subset $U\subseteq X$, one can check $(U,{\underline{o}}_X{|}_U)$ is also a prescheme by ^pzdwqy.

Definition

For preschemes $(X,{\underline{o}}_X)$ and $(Y,{\underline{o}}_Y)$, a morphism consists of:

• a continuous map $f:X\to Y$

• a morphism of rings between sheaves ${\underline{o}}_Y\stackrel{f^{\ast }}{\to }{f}_{\ast }{\underline{o}}_X$: for any $U\subseteq Y$, the following diagram commutes

  

  where $V\subseteq U\subseteq Y$ and "$\downarrow$" are restriction maps.

• for any $U\subseteq Y$, $x\in f^{-1}(U)$ and $r\in \Gamma (U,{\underline{o}}_Y)$, if $r(f(x))=0\in k(f(x))$, then $(f_U^{\ast }(r))(x)=0\in k(x)$ (morphism of locally ringed space).

• Example. Here is an example to show there exists map satisfies i) and ii) but not iii), see Pasted image 20250522153005.png.

• 然后在解释第三条有什么用: 好像是极大理想射到极大理想/residue field是单射

Theorem

For a prescheme $X$ and a ring $R$, there is a $1$-$1$ map

$$\{\mathrm{Hom}(X,\mathrm{Spec}R)\}\stackrel{1\text{-}1}{⟷}\mathrm{Hom}(R,\Gamma (X,{\underline{o}}_X)),$$

where $R=\Gamma (\mathrm{Spec}R)$.

\begin{proof} "→" Define $f:X\to \mathrm{Spec}R\in \text{LHS}$ and define $A_f\in \text{RHS}$, where $A_f$ is defined as follows.

For $[p]=f(x_0)$ and $r\in p$, there is $r(f(x_0))=0\in R_p/p{R}_p$. Note that $r\in p$ iff $r=0\in R_p/r\cdot R_p$. Define $A_f(x_0)=[p]$, where $p=\{r\in R:A_f(r)(x_0)=0,x_0\in {\underline{o}}_{X,x_0}/m{x}_0\}$ is a prime ideal.

To check $f\mapsto A_f$ is a morphism:

• $f$ is continuous: if $V(I)\subseteq \mathrm{Spec}R$ closed, $f^{-1}(V(I))$ is closed.
  • $X={\cup }_{\alpha }{X}_{\alpha }$ with affine $X_{\alpha }$, then it suffices to check $f^{-1}(V(I))\cap X_{\alpha }$ closed. Define $A_{f_{\alpha }}:R\stackrel{A_f}{\to }\Gamma (X,{\underline{o}}_X)\to \Gamma (X_{\alpha },{\underline{o}}_{X_{\alpha }})$ and it induces $f_{\alpha }:X_{\alpha }\to \mathrm{Spec}R$. One can check that $f_{\alpha }=f{|}_{\alpha }$, by comparing $f_{\alpha }(x_0)$. Then $f^{-1}(V(I))\cap X_{\alpha }=f_{\alpha }^{-1}(V(I))$.
  • So we can assume that $X$ is affine.
  • in affine case, $f:R\to S$ and $\mathrm{Spec}S\to \mathrm{Spec}R$ 的关系:14:34 下黑板 ^pnaept
    • $S\cap q{S}_q=q$.
    • $p=\{r\in R:A_f(r)\in q\}$, $p=A_f^{-1}(q)$
    • $f([q])=[A_f^{-1}(q)]$.
  • 14:39 下半黑板。验证continuous
• Consider $f^{\ast }$ on distinguish open sets. For $Y_g\subseteq Y=\mathrm{Spec}R$, we aim to find $f^{\ast }:\Gamma (Y_g,{\underline{o}}_Y)\to \Gamma (f^{-1}(Y_g),{\underline{o}}_X)$.
  • Recall that $A_f:R=\Gamma (Y,{\underline{o}}_Y)\to \Gamma (X,{\underline{o}}_X)$, we can define $r/g^n\mapsto A_f(r)/A_f(g^n)$.
  • It remains to show $A_f(r)/A_f(g^n)\in \Gamma (f^{-1}(Y_g),{\underline{o}}_X)$, and it is enough to show $A_f(g)(x)\ne 0$ for any $x\in f^{-1}(Y_g)$.
  • For any $x\in f^{-1}(Y_g)$, then $f(x)\in Y_g$ and so $g\notin \{r\in R:A_f(r)(x)=0\}$. Thus $A_f(g)(x)\ne 0$ and $A_f(g)\notin m_x$.
• Verify if $r(f(x))=0$, then $(f^{\ast }(r))(x)=0$.

Finally, the map is bijective (the proof is omitted).

This argument is important, as we get a corollary from it.

Corollary

The following categories are isomorphic

$$\{\text{affine schemes}\}\leftrightarrow \{\text{commutative rings with unit}\}.$$

Corollary

Let $X$ be a prescheme, then there exists unique $X\to \mathrm{Spec}\mathbb{Z}$, i.e. $\mathrm{Spec}\mathbb{Z}$ is the final object in the category of schemes.

\begin{proof} 15:32 上黑板 \end{proof}

Proposition

Let $X$ be a prescheme, and let $Z\subseteq X$ be an irreducible closed subset. Then there exists unique $z_0\in Z$ such that $\overline{\{z_0\}}=Z$.

\begin{proof} There exists affine open set $U$ of $X$ such that $Z\cap U\ne \varnothing$. Then $Z\cap U$ is irreducible. Hence, there exists $z_0\in Z\cap U$ such that ${\overline{\{z_0\}}}_U=Z\cap U$. Since $Z\supseteq {\overline{\{z_0\}}}_X={\overline{(Z\cap U)}}_X=Z$, we have $\overline{\{z_0\}}=Z$.

For uniqueness, there exists another point $z_1$ such that $\overline{\{z_1\}}=Z$, then $z_1\in U$ as $Z\setminus U$ is closed. Then $z_1,z_0\in Z\cap U$. By affine case (generic point unique), $z_1=z_0$. \end{proof}

Definition

$z$ above is called the generic point of $Z$.

Example. ${\mathbb{P}}_{\mathbb{Z}}^n$. Gluing $(n+1)$-pieces of ${\mathbb{A}}_{\mathbb{Z}}^n=\mathrm{Spec}\mathbb{Z}[x_0/x_i,\cdots ,x_{i-1}/x_i,x_{i+1}/x_i,\cdots ,x_n/x_i]$.

Fiber Product

Theorem

If $A,B$ are $C$-algebras. Let the diagram of affine schemes

be associated to the natural ring. Then $\mathrm{Spec}(A{\otimes }_{C}B)\simeq \mathrm{Spec}A{\otimes }_{\mathrm{Spec}C}\mathrm{Spec}B$.

\begin{proof} 15:50 下黑板

6.3 update: by ^og85o0, the construction is trivial. 16:29 \end{proof}

---

In general, we glue these affine pieces. Remark that $S_i$ may not an affine variety, and we can write $S=\cup S_i$ and consider preimage of each pieces. 16:40

universal surjective

If $r:X\to S$ is surjective, then $p:Y{\times }_{S}S\to Y$ is surjective.

\begin{proof} Let $y_0\in Y$, then there exists $x_0\in X$ such that $r(x_0)=s(y_0)$. Since $y_0\in Y$, there is $\mathrm{Spec}R\subseteq Y$ such that $y_0\in \mathrm{Spec}R$. Note that there is an affine open set $V$ such that $s(y_0)\in V=\mathrm{Spec}Q\subseteq S$. Then $y_0\in U=\mathrm{Spec}R\subseteq s^{-1}(U)$. Define $k(y_0)=R_{p_{y_0}}/p_{y_0}{R}_{p_{y_0}}$ and $k(s_0)=Q_{p_{s_0}}/p_{s_0}{Q}_{p_{s_0}}$. Consider $s^{\ast }:\mathrm{Spec}R\to \mathrm{Spec}Y$, which induces a map from $k(s_0)$ to $k(y_0)$.

• Then by the universal property, there exists a unique map $\theta$ from $\mathrm{Spec}k$ to $Y{\times }_{S}X$. Then $p(\theta )$? is what we desired. 17:02

Now we finish the proof. \end{proof}