2.3 Varieties & Preschemes

Definition

Let $R$ be a ring, and let $X$ be a prescheme over $R$, denoted by $X/R$, if there exists a morphism from $X\to \mathrm{Spec}R$ (iff $R\to \Gamma (X,{\underline{o}}_X)$).

If $U\subseteq X$ is an open subset, then $U\to \mathrm{Spec}R$ induces a map $R\to \Gamma (U,{\underline{o}}_X)$. Since $R\to \Gamma (U,{\underline{o}}_X)$ is a ring homomorphism, $\Gamma (U,{\underline{o}}_X)$ is am $R$-algebra.

Definition

For two preschemes $X/R$ and $Y/R$, an $R$-morphism $f:X\to Y/R$ is a morphism, if $f:X\to Y$ makes the following diagram commutes.

Remark. This means for any $V\subseteq Y$ open set, the following diagram commutes.

Definition

We say $X/R$ is of finite type over $R$, if $X$ is quasi-compact and for any open affine subset $U\subseteq X$, $\Gamma (U,{\underline{o}}_X)$ is a finitely generated $R$-algebra, that is, there exists $x_1,\cdots ,x_n\in \Gamma (U,{\underline{o}}_X)$ such that $\Gamma (U,{\underline{o}}_X)=R[x_1,\cdots ,x_n]$.

Proposition

Let $X/R$ be a prescheme. If there exists $\{U_i{\}}_{1⩽i⩽\ell }$ such that $X={\cup }_{1⩽i⩽\ell }{U}_i$ with each $U_i$ an affine open subset of $X$ with each $U_i$ an affine open subset of $X$ and $\Gamma (U,{\underline{o}}_X)$ is a finitely generated $R$-algebra, then $X$ is of finite type over $R$.

\begin{proof} Each affine scheme $U_i$ is quasi-compact by ^fxk3gh, then $X$ is quasi-compact as it is a finite cover of $U_i$.

Let $U\subseteq X$ be an open affine subset. It suffices to show $\Gamma (U,{\underline{o}}_X)$ is a finitely generated $R$-algebra. Assume $R_i=\Gamma (U_i,{\underline{o}}_X)$, then for each $f_i\in R_i$, $\Gamma (U_{i,f_i})=(R_i{)}_{f_i}=R_i[1/f_i]$ is a finitely generated $R$-algebra. Remark that $(U_i{)}_{f_i}$ are a basis of the topology on $X$. As $U_i$ is quasi-compact, $U$ is a union of finitely many sets $U_i^{\prime }=\mathrm{Spec}{R}_i^{\prime }$, where $R_i^{\prime }$ is a finitely generated $R$-algebra. Hence, we can assume $U={\cup }_{\text{finite}}{U}_{i,f_i}={\cup }_{\text{finite}}{U}_i^{\prime }$.

We claim that $U_f\cap U_i^{\prime }=U_{i,\mathrm{res}(f)}^{\prime }$. Note that $[p]\in U_f\cap U_i^{\prime }$ $⟺$ $[p]\in U_i^{\prime }$ and $[{\mathrm{res}}^{-1}(p)]\in U_i^{\prime }$ $⟺$ $[p]\in U_i^{\prime }$, $f\notin {\mathrm{res}}^{-1}(p)$ $⟺$ $[p]\in U_i^{\prime }$ and $\mathrm{res}(f)\notin p$ iff $[p]\in U_{i,\mathrm{res}(f)}^{\prime }$. 这段我完全没懂 17:43?

As $U$ is affine, there is $U_i^{\prime }={\cup }_{f_j\in S,\text{ finite}}{U}_{f_j}$ and so

$$U={\cup }_{\text{finite}}{U}_i^{\prime }={\cup }_{\text{finite }{f}_j}{U}_{f_j}.$$

It deduces that $\Gamma (U,{\underline{o}}_X)=S=(f_j:\text{finite }j)$, and so we can assume that $1=\sum f_j{g}_j$ with $g_j\in S$. By claim we have $U_{f_j}=U_{f_j}\cap U_i^{\prime }=U_{i,\mathrm{res}(f_j)}^{\prime }$, which deduces that $S[1/f_j]=\Gamma (U_{f_j})=R_i^{\prime }[1/\mathrm{res}(f_j)]$, where and $R_i^{\prime }$ is a finitely generated $R$-algebra. Thus $S[1/f_j]$ is a finitely generated $R$-algebra, and we can assume $S[1/f_j]=R[(s_{\ell }/f_j^{n_{\ell }}{)}_{\ell =1}^m]$.

It remains to show $S$ is a finitely generated $R$-algebra. Define $\tilde{S}=R(s_{\ell },f_j)\subseteq S$. We claim that $\tilde{S}=S$. For any $\alpha \in S\subseteq S[1/f_j]$, $\alpha$ can be written as $\alpha =u/f_j^n$. Thus $f_j^m(\alpha \cdot f_j^n-u)=0\in S$ and $u\in \tilde{S}$. Then $\alpha \cdot f_j^{N_j}\in \tilde{S}$. Then

$$\alpha \cdot 1=\alpha \cdot (\sum f_j{g}_j{)}^N=\sum \alpha f_j^{N_j^{\prime }}(f_j^{\prime }\cdot g)\in \tilde{S}.$$

Hence $S=\tilde{S}$ is finitely generated. Now we finish the proof. \end{proof}

Corollary

If $X$ is an affine scheme and $U\subseteq X$ is an affine open subset, then $\Gamma (U,{\underline{o}}_X)$ is a finitely generated $\Gamma (X,{\underline{o}}_X)$-algebra.

Warning. If $X$ is of finite type over $R$, $\Gamma (X,{\underline{o}}_X)$ may not a finitely generated $R$-algebra. It is weird, as we can remark that $\Gamma (U,{\underline{o}}_X)$ is a finitely generated $R$-algebra and there is an embedding map $\Gamma (X,{\underline{o}}_X)\hookrightarrow \Gamma (U,{\underline{o}}_X)$. This is the fact that Hilbert 14th problem.

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下面是6.5的notes，我难得地很勤快，把它们整理完了。线上面的还没整理。

Definition

A prescheme $X$ is reduced if $\Gamma (U,{\underline{o}}_X)$ does not have nilpotent sections, i.e., there does not exists nonzero $\sigma \in \Gamma (U,{\underline{o}}_X)$ with open set $U$ such that ${\sigma }^n=0$ for some $n\in \mathbb{N}$.

Theorem

Assume $k$ is algebraic closed. There exists an equivalence of the following categories.

• Reduced, irreducible prescheme of finite type over $\mathrm{Spec}k$ & $k$-morphism.
• Prevarieties over $k$ and morphism between varieties.

\begin{proof} It is just a sketch of proof. Consider $X={\cup }_{\text{finite}}{S}_i$ with finitely generated $k$-algebra $S_i$ satisfying $\sqrt{0}=0$, then $S_i$ can be written as $S_i=k[x_1,\cdots ,x_n]/I$, then define $V_i=V(I)$ and $X^{\text{var}}=\cup V_i$. Conversely, for a prevariety $X=\cup U_i$ with affine $U_i$, $X^{\text{prescheme}}=\cup \mathrm{Spec}{U}_i$. \end{proof}

Definition

Assume $k$ is an algebraic closed field. A prevariety over $k$ is a reduced, irreducible prescheme of finite type over $k$.

Summary. For a prevariety, there is a $1$-$1$ correspondence between closed subvarieties and radical ideals $I$ with $\sqrt{I}=I$. For a prescheme, there is a $1$-$1$ correspondence between closed subpreschemes and arbitrary ideals $I$. Therefore, the concept of preschemes (or schemes) is more general than that of prevarieties (or varieties), as it allows structures with nilpotent elements.

Definition

Let $\mathcal{F},\mathcal{G}$ be sheaves if abelian groups on a topological space $X$, and let $\phi :\mathcal{F}\to \mathcal{G}$ be a morphism.

• Define $\ker \phi (U)=\ker ({\phi }_U)=\ker (\mathcal{F}(U)\to \mathcal{G}(U))$ for any $U\subseteq X$. We say $\phi$ is injective iff ${\phi }_U$ is injective for all $U\subseteq X$. Remark that $\phi$ is injective iff ${\phi }_{x_0}$ is injective for all $x_0\in X$, where ${\phi }_{x_0}:{\mathcal{F}}_{x_0}\to {\mathcal{G}}_{x_0}$.
• Define $\mathrm{coker}(\phi )=\tilde{C}$ as a sheafification, where $C(U)=\mathcal{G}(U)/\mathrm{im}(\phi (U))=\mathrm{coker}({\phi }_U)$ and $U$ is a sheaf. We say $\phi$ is surjective if $\mathrm{coker}(\phi )=0$. Note that it is NOT the same as ${\phi }_U$ surjective. Remark that $\phi$ is surjective iff ${\phi }_{x_0}:{\mathcal{F}}_{x_0}\to {\mathcal{G}}_{x_0}$ is surjective for any $x_0\in X$.

Example. Take $X=\mathbb{C}$. Assume $\mathcal{F}={\underline{o}}_{\mathbb{C}}^{\mathrm{an}}$ is the sheaf of holomorphic functions on $\mathbb{C}$, and $\mathcal{G}=({\underline{o}}_{\mathbb{C}}^{\mathrm{an}}{)}^{\ast }$ is the sheaf of nowhere vanishing holomorphic functions on $\mathbb{C}$. Define

$$\phi =\exp :{\underline{o}}_{\mathbb{C}}^{\mathrm{an}}\to ({\underline{o}}_{\mathbb{C}}^{\mathrm{an}}{)}^{\ast },\alpha \mapsto e^{\alpha },$$

and one can check that $\exp$ is a surjective sheaf morphism, because we can define $\log$ locally. In addition, remark that ${\exp }_{{\mathbb{C}}^{\ast }}$ is not surjective, as $g(z)=z$ is not in the image of it.

Definition

Assume that $f:(Y,{\underline{o}}_Y)\to (X,{\underline{o}}_X)$ is a morphism of preschemes. Then $f$ is a closed immersion (embedding) if

• $f$ is injective
• $f$ is a closed map
• $f_y^{\ast }:{\underline{o}}_{X,f(y)}\to {\underline{o}}_{Y,y}$ is surjective for all $y\in Y$.

Explanation. i)&ii) $⟺$ $f(Y)\subseteq X$ is closed and $Y\to f(Y)$ is homeomorphism. For iii), consider a submanifold $Y\subseteq X$ with $Y\subseteq {\mathbb{C}}^{n-k}$ and $X\subseteq {\mathbb{C}}^n$. Then $f_y^{\ast }$ is surjective means that locally functor of $Y$ come from $X$.

Example. Define $Y=\mathrm{Spec}(\mathbb{C}[t])$, $X=\mathrm{Spec}(\mathbb{C}[x,y]/(x^2-y^3))$, and $f:Y\to X,t\mapsto (t^3,t^2)$. Then $f$ corresponds to a ring homomorphism $\varphi :\mathbb{C}[x,y]/(x^2-y^3)\to \mathbb{C}[t],x\mapsto t^3,y\mapsto t^2$. One can check that $f$ satisfies i)&ii) as $X=f(Y)$ is homeomorphic to $Y$. However, $f_y^{\ast }$ is not surjective at $0$ and so $f$ is NOT a closed immersion. Notice that

$$f_0^{\ast }:(\mathbb{C}[x,y]/(x^2-y^3){)}_{(x,y)}\to (\mathbb{C}[t]{)}_{(t)},$$

and if $t\in \mathrm{im}{f}_0^{\ast }$, there exists $P(u,v),Q(u,v)\in \mathbb{C}[u,v]$ such that $Q(0,0)\ne 0$ and $t=P(t^3,t^2)/Q(t^3,t^2)$, which is impossible. Hence $f$ is not a closed immersion.

Proposition

Let $R$ be a ring, and let $I\subseteq R$ be an ideal. Then $\pi :R\to R/I$ is a quotient map, and $f:\mathrm{Spec}R/I\to \mathrm{Spec}R$ is a closed immersion.

\begin{proof} For $[Q]\in \mathrm{Spec}R/I$, we know $Q\subseteq R/I$ is a prime ideal and $f([Q])=[{\pi }^{-1}(Q)]$.

If $f([q_1])=f([q_2])$, then ${\pi }^{-1}(q_1)={\pi }^{-1}(q_2)$. It deduces that

$$q_1={\pi }^{-1}(q_1)/I=({\pi }^{-1}(q_2))/I=q_2$$

and so $f$ is injective. Assume that $C$ is a closed set in $\mathrm{Spec}(R/I)$, then $C=V(\mathcal{A})$ for some ideal $\mathcal{A}\subseteq R/I$. Then

$$f(C)=f(V(\mathcal{A}))=\{[{\pi }^{-1}(q):q\in V(\mathcal{A})]\}=\{[{\pi }^{-1}(q)]:q\supseteq \mathcal{A},q\text{ is a prime ideal of }R/I\}=V({\pi }^{-1}(\mathcal{A})).$$

Thus $f$ is a closed map. Also ${\underline{o}}_{\mathrm{Spec}R,[p]}\to {\underline{o}}_{\mathrm{Spec}R/I,[\overline{p}]}$ is surjective. Now we finish the proof. \end{proof}

Theorem

For $X=\mathrm{Spec}R$ and closed immersion $f:Y\to X$, let $I=\{r\in R:f_y^{\ast }(r)=0\in {\underline{o}}_{Y,y},\forall y\in Y\}$. Then $f$ is the same as $\mathrm{Spec}R/I\to \mathrm{Spec}R$.

\begin{proof} Firstly, it is easy to check $I$ is an ideal. The map $f:Y\to \mathrm{Spec}R$ induces $f^{\ast }:R\to \Gamma (Y,{\underline{o}}_Y)$.

We use the following fact without proof: If $f:Y\to X$ is a closed immersion and $X=\mathrm{Spec}R$, then $Y=\mathrm{Spec}A$ for some ring $A$. Thus $f$ is induces from $\varphi :R\to A$ where $\varphi$ is surjective, then we have $A\simeq R/\ker \varphi$. Define $J=\ker \varphi$, then $f$ is the same as $\mathrm{Spec}(R/J)\to \mathrm{Spec}R$. Thus, it suffices to show $I=J=\ker \varphi$.

Note that $r\in I$ iff for all $y\in Y$, $(\varphi (r){)}_q=0$ for some prime ideal $q$ in $A$ such that $[q]=y$. Since $\varphi (r{)}_q=0$ for all prime ideal $q$, one can check $\varphi (r)=0\in A$ by ^663ccb. Therefore, $r\in I$ iff $\varphi (r)=0\in A$ iff $I=\ker \varphi =J$. Now we finish the proof. \end{proof}

Remark. This proof relies on the established result that a closed subscheme of an affine scheme is itself affine. The argument then identifies the specific ideal defining this affine subscheme.