1.5 Definition of Prevarieties and Morphisms

Basic Concepts of Prevarieties

Definition

Let $X$ be a topological space, and let ${\underline{o}}_X$ be the sheaf of $k$-valued function on $X$. Then $(X,{\underline{o}}_X)$ is a prevariety if

• $X$ is connected;
• there exists finite open covering $\{U_i{\}}_{i=1}^{\ell }$ such that $(U_i,{\underline{o}}_X{|}_{U_i})$ is an irreducible affine variety.

Definition

$X$ is irreducible iff one of the followings hold:

• if $X=Z_1\cup Z_2$ with $Z_i$ closed, then there exists $Z_i=X$;
• for any open sets $U_1,U_2$, there is $U_1\cap U_2\ne \varnothing$;
• any open set is dense.

\begin{proof} i)→ii) Assume that there exists open sets $U_1,U_2$ such that $U_1\cap U_2=\varnothing$, then we have $U_1^c\cup U_2^c=X$ and $U_i^c$ are closed, which is impossible.

ii)→iii) If there exists open $U$ such that $\overline{U}\ne X$, then ${\overline{U}}^c$ is an open set and $U\cap {\overline{U}}^c=\varnothing$, leading to a contradiction.

iii)→i) If there exist closed set $Z_1$ and $Z_2$ such that $X=Z_1\cup Z_2$ and WLOG suppose $Z_1\ne X$, then $Z_1^c\subseteq Z_1\cup Z_2$ and so $Z_1^c\subseteq Z_2$. Since $Z_1^c$ is open, there is $\overline{Z_1^c}=X$. It deduces that $\overline{Z_2}=Z_2=X$. \end{proof}

Note that irreducible set is connected.

Proposition

Prevariety $X$ is irreducible.

\begin{proof} Define $\mathcal{U}=\{U\subseteq X:U\text{mbox}isanaffineopenset\}$. For nonempty open $V\subseteq X$, define $U_1={\cup }_{U\in \mathcal{U},U\cap V\ne \varnothing }U$ and $U_2={\cup }_{U\in \mathcal{U},U\cap V=\varnothing }U$. Then $X=U_1\cup U_2$. If there exists $x\in U_1\cap U_2$, then there is $W_1\subseteq U_1$ such that $W_1$ is irreducible and $x\in W_1$. Similarly, we can find affine $W_2\subseteq U_2$ such that $x\in W_2$. Thus $W_1\cap W_2\ne \varnothing$. Since $W_1\cap W_2$ is an open set of $W_1$ and $W_1$ is irreducible, we have $(W_1\cap W_2)\cap (W_1\cap V)\ne \varnothing$. Then $W_2\cap V\ne \varnothing$, contradiction. Therefore, $X=U_1\bigsqcup U_2$. By connectedness of $X$, $X=U_1$ and $U_2=\varnothing$. Hence, for any open set $U\in \mathcal{U}$, $U\cap V\ne \varnothing$ and so $X$ is irreducible. \end{proof}

Proposition

Let $X$ be a prevariety. The closed subsets are DCC.

\begin{proof} For a chain of closed set $Z_1\supseteq \cdots \supseteq Z_k\supseteq \cdots$, since $X={\cup }_{i=1}^{\ell }{U}_i$ with $U_i$ irreducible, there is $\{U_i\cap Z_j\}$ DCC and so for $\{Z_i\}$. \end{proof}

Proposition

Let $X$ be a Noetherian topological space, i.e., closed sets satisfy DCC. For any closed set $Z$, there exists unique decomposition $Z={\cup }_{i=1}^{\ell }{Z}_i$ such that $Z_i$ is irreducible, and $Z_i$ are called irreducible component.

\begin{proof} Existence: assume $Z$ is the minimal element such that the above property does not hold. Then $Z$ is not irreducible, and $Z=Z_1\cup Z_2$ where closed $Z_i⊊Z$. It deduces that $Z_i={\cup }_j{Y}_j$ and $Z_j={\cup }_j{W}_j$ can be decomposed as union of finitely many irreducible closed sets. So the above property also holds for $Z$, contradiction.

Uniqueness: let ${\cup }_{j\in J}{Z}_j={\cup }_{i\in I}{W}_i=Z$ be distinct decompositions, then $W_i={\cup }_{j\in J}(W_i\cap Z_j)$. Since $W_i$ is irreducible, there exists $j$ such that $W_i=W_i\cap Z_j$ and so $W_i\subseteq Z_j$. Conversely $Z_j\subseteq W_{i^{\prime }}$ for some $i^{\prime }$ and $i=i^{\prime }$. Therefore, the decomposition is unique. \end{proof}

Structure and Functions on Prevarieties

functional field

Let $X$ be a prevariety. The functional field $k(X)$ is the generic stalk of ${\underline{o}}_X$, i.e. $k(X)=\underset{\text{all open }U\subseteq X}{\underrightarrow{\lim }}{\underline{o}}_X(U)$.

The elements of $k(X)$ are called rational functions on $X$, although they are only functions on open dense subsets of $X$.

Fact. For affine set $V\subseteq X$, define $\Gamma (V,{\underline{o}}_X)=R$ and it is an integral domain and there is $R\stackrel{\varphi }{\hookrightarrow }k(X)$. We claim that $\varphi (\mathrm{Frac}(R))=k(X)$. For any $f\in \mathrm{Frac}(R)$, $f=g/h$ for some $g,h\in R$ with $h\ne 0$. Then $f\in {\underline{o}}_X(U)$ for an open dense $U\subseteq X$ and so $\mathrm{Frac}(R)\subseteq k(X)$. Let $\alpha \in k(X)$. Then $\alpha =[(U,f_U)]$ for some non-empty open set $U\subseteq X$ and $f_U\in {\underline{o}}_X(U)$. Consider the intersection $W=U\cap V$, which is open because $U$ and $V$ are open. Let $f_W=f{|}_{U\cap W}\in {\underline{o}}_V(W)$, then there exist $g,s\in R$ such that $s(x)\ne 0$ for all $x\in W$ and $f_W(x)=g(x)/s(x)$ for all $x\in W$. Notice that the fraction $q=g/s$ is an element of $\text{Frac}(R)$ and $\alpha =[(U,f_U)]$ is equivalent to $[(W,f_W)]$. Therefore, $\alpha =\varphi (g/s)\in \mathrm{Frac}(R)$ in $k(X)$. Now we finish the proof.

Local Ring of $X$ along $Y$

Let $Y\subseteq X$ be an irreducible closed subset of $X$, and let

$${\underline{o}}_{Y,X}=\underset{\begin{array}{c}\text{ open sets }U\\ \text{ such that }U\cap Y\ne \varnothing \end{array}}{\underrightarrow{\lim }}{\underline{o}}_X(U).$$

Note that ${\underline{o}}_{X,X}=k(X)$ and ${\underline{o}}_{\{x_0\},X}={\underline{o}}_{x_0}$ (see here), so ${\underline{o}}_{Y,X}$ is intermediate between the $\underset{⟶}{\lim }$ that leads to ${\underline{o}}_x$ and that which leads to $k(X)$.

For a given affine set $V\subseteq X$ with $V\cap Y\ne \varnothing$, define $\Gamma (V,{\underline{o}}_V)=R$. Then $V\cap Y$ is closed in $V$ and $V\cap Y=V(P)$ for some prime ideal $P\subseteq R$. As $Y$ is irreducible, $V\cap Y$ is also irreducible and $\Gamma (V\cap Y,{\underline{o}}_{V\cap Y})=R/P$. Note that

$${\underline{o}}_{Y,X}=\underset{\begin{array}{c}\text{ open sets }U\\ \text{ such that }U\cap Y\ne \varnothing \end{array}}{\underrightarrow{\lim }}{\underline{o}}_X(U)=\underset{\begin{array}{c}W\subseteq V,W\cap Y\ne \varnothing \end{array}}{\underrightarrow{\lim }}{\underline{o}}_X(W).$$

Since $W\cap Y\ne \varnothing$ and $W\subseteq V$, there exists $f$ such that $f\notin P$ and $V_f\subseteq W$ as $\{V_f\}$ is a set of topological basis. Then

$${\underline{o}}_{Y,X}=\underset{\begin{array}{c}\text{ open sets }U\\ \text{ such that }U\cap Y\ne \varnothing \end{array}}{\underrightarrow{\lim }}{\underline{o}}_X(U)=\underset{\begin{array}{c}W\subseteq V,W\cap Y\ne \varnothing \end{array}}{\underrightarrow{\lim }}{\underline{o}}_X(W)=\underset{f\notin P}{\underrightarrow{\lim }}{\underline{o}}_X(V_f)=\underset{f\notin P}{\underrightarrow{\lim }}{R}_f=R_P$$

by ^9d743b.

Geometric Meaning

Just as the stalk ${\underline{o}}_x$ consists of germs of functions defined in neighborhoods of a point $x,o_{Y,X}$ consists of germs of functions defined in open sets that approach or meet the entire subvariety $Y$. Two functions $(U,f)$ and $(V,g)$ represent the same element in $o_{Y,X}$ if they agree on some smaller open set $W\subseteq U\cap V$ which still intersects $Y$.

an Example

Example. For $U\simeq V\simeq {\mathbb{A}}^1$ with $u\in U,v\in V$, there is $U_u=V_v={\mathbb{A}}^1\setminus \{0\}$.

• The corresponding coordinate rings are
  • $\Gamma (U)=k[u]$ and $\Gamma (U_u)=k[u,1/u]$
  • $\Gamma (V)=k[v]$ and $\Gamma (U_v)=k[v,1/v]$
• Construct projective line $\mathbb{P}=U{\cup }_{\psi }V$.
  • Define $\psi :U_u\to V_v,u\mapsto v=1/u$ and $\Gamma (V_v)\to \Gamma (U_u),v\mapsto 1/u,1/v\mapsto u$
  • For any open subset $\Omega \subseteq \mathbb{P}$, define ${\underline{o}}_{\mathbb{P}}(\Omega )={\cap }_{x\in \Omega }{\underline{o}}_x$.
  • Then $(\mathbb{P},{\underline{o}}_{\mathbb{P}}):={\mathbb{P}}^1$ is a prevariety, as $\mathbb{P}=U\cup V$ with $(U,{\underline{o}}_{\mathbb{P}}{|}_U)\simeq (U,{\underline{o}}_U)$ and $(V,{\underline{o}}_{\mathbb{P}}{|}_V)\simeq (V,{\underline{o}}_V)$.
• On the other hand, construct the non-separated prevariety $W=U{\cup }_{\theta }V$. ^v1yo0z
  • Define the isomorphism $\theta :U_u\to V_v$ by $u\mapsto v=u$ and ${\theta }^{\ast }:\Gamma \left(V_v,{\underline{o}}_V\right)\to \Gamma \left(U_u,{\underline{o}}_U\right)$ maps $v\mapsto u$ and $1/v\mapsto 1/u$.
  • For any open subset $\Omega \subseteq W$, define ${\underline{o}}_W(\Omega )={\cap }_{x\in \Omega }{\underline{o}}_{W,x}$, where ${\underline{o}}_{W,x}$ is the stalk at $x\in W$.
  • Then $\left(W,{\mathcal{O}}_W\right)$ is a prevariety, as $W=U\cup V$ is a finite union of affine open sets such that the structure sheaf restricts correctly: $\left(W,{{\mathcal{O}}_W|}_U\right)\simeq \left(U,{\mathcal{O}}_U\right)$ and $\left(W,{{\mathcal{O}}_W|}_V\right)\simeq \left(V,{\mathcal{O}}_V\right)$.
  • This space $W$ is not a variety because it fails the separation axiom, i.e., it is not “Hausdorff”. It is often called the “affine line with the origin doubled”. See here.

Substructures of Prevarieties

Proposition

Let $(X,{\underline{o}}_X)$ be a prevariety. For any open set $U\subseteq X$, $(U,{\underline{o}}_X{|}_U)$ is also a prevariety.

\begin{proof} Note that $U$ is connected. Otherwise $U=U_1\bigsqcup U_2$ with open $U_i$. Then $X$ is irreducible yields that $U_1\cap U_2\ne \varnothing$, contradiction.

Assume that $\{U_i{\}}_{i=1}^{\ell }$ is a finite open cover of $X$. For any $x\in U$, there exists $i\in \{1,\cdots ,\ell \}$ such that $x\in U_i$. Then we can find $f$ such that $(U_i{)}_f\subseteq U$. Notice that $(U_i,{\underline{o}}_X{|}_{U_i})$ is affine, and then $((U_i{)}_f,{\underline{o}}_X{|}_{(U_i{)}_f})$ is affine. Hence, $U$ has an open cover $U={\cup }_{j\in J}{V}_j$ with $(V_i,{\underline{o}}_X{|}_{V_i})$ affine.

Then $U={\cup }_{\text{finite}}{V}_i$, as we cannot have strictly increasing sequence of open sets in $X$ by ^uia9mm. \end{proof}

Proposition

Now let $Y$ be a closed irreducible subset of a prevariety $X$. The sheaf ${\underline{o}}_X$ induces a sheaf ${\underline{o}}_Y$ on $Y$ as follows: If $V$ is open in $Y$, define

$${\underline{o}}_Y(V)=\left\{\begin{array}{ll}k\text{-valued functions }f\text{ on }V& \begin{array}{l}\forall x\in V,\exists \text{ a neighbourhood }U\text{ of }x\text{ in }X\\ \text{ and a function }F\in {\underline{o}}_X(U)\text{ such that }\\ f=\text{restriction to }U\cap V\text{ of }F\end{array}\end{array}\right\}.$$

Then the pair $\left(Y,{\underline{o}}_Y\right)$ is a prevariety.

\begin{proof}

• $Y$ connected
• $X={\cup }_{\text{finite}}{U}_i$ with $U_i$ affine. There is $Y={\cup }_{\text{finite}}{U}_i\cap Y$ where $(U_i\cap Y,{\underline{o}}_Y{|}_{U_i\cap Y})\simeq (U_i\cap Y,{\underline{o}}_{U_i\cap Y})$ \end{proof}

Combining ^695715 and ^ee0fb8 we can even give a prevariety structure to every locally closed subset of a prevariety $X$: For any $Z\subseteq X$ irreducible, $U\subseteq X$ open, then $Z\cap U\subseteq U$ (locally closed subset) admits a prevariety structure.

Introduction to Projective Varieties

Let $P\subseteq k[x_0,\cdots ,x_n]$ be a homogeneous prime ideal. Define $X=V(P)\subseteq {\mathbb{P}}_n(k)$ and $R=k[x_0,\cdots ,x_n]/P$. We aim to make $X$ into a prevariety.

Define graded integral domain $R={\oplus }_{i=0}^{\infty }{R}_i$ with

$$R_i=\{\overline{f}:\mathrm{deg}f=i,f\text{mbox}isahomogeneouspolynomial\}.$$

Define $k(X)=\{f/g:f,g\text{mbox}arehomogeneous,\mathrm{deg}f=\mathrm{deg}g\}$. Note that for $x\in X$ and $g\in R_n$, $g(x)\ne 0$ does not depend on the choice of representative of $x$. Hence we can define $X_g=\{x\in X:g(x)\ne 0\}$ for $g\in R_d$ for some $d$ and then define a ring

$${\underline{o}}_x=\{f/g\in k(x):g(x)\ne 0\}.$$

The set $m_{x_0}=\{f/g\in k(x):f(x_0)=0,g(x_0)\ne 0\}$ is the unique maximal ideal, as any element not in $m_{x_0}$ is invertible in ${\underline{o}}_{x_0}$. Hence, $({\underline{o}}_X,m_{x_0})$ is a local ring.

We now define a sheaf ${\underline{o}}_X$ on $X$ by

$${\underline{o}}_X(U)={\cap }_{x\in U}{\underline{o}}_x,\text{ for all open }U\subseteq X.$$

Remark that ${\mathbb{P}}_{x_j}=\{x:x_j\ne 0\}\simeq {\mathbb{A}}^n$. We claim that $(X\cap {\mathbb{P}}_n(k{)}_{x_j},{\underline{o}}_X{|}_{X\cap {\mathbb{P}}_{x_j}})$ is an affine variety. Here we only check the case of $j=0$.

For every homogeneous polynomial $F\in P$, $F^{\prime }:=F/x_0^{\mathrm{deg}F}$ is a polynomial in the variables $x_1/x_0,\cdots ,x_n/x_0$. Let $P^{\prime }\subset k\left[y_1,\cdots ,y_n\right]$ be the ideal generated by all these $F^{\prime }$. We can map $k\left[y_1,\dots ,y_n\right]$ into a subring of $k(X)$ by taking $y_i$ to the function given by $x_i/x_0$; the kernel of this map is exactly $P^{\prime }$, so $P^{\prime }$ is prime. Note that we get an isomorphism

$$\phi :X\cap {\mathbb{P}}_n(k{)}_{x_n}\to X^{\prime }=V\left(P^{\prime }\right)\subset k^n,x\mapsto (x_1/x_0,\cdots ,x_n/x_0)$$

and $\phi$ is actually a homeomorphism.

Now for $x\in X\cap {\mathbb{P}}_n(k{)}_{x_0}$, the local ring ${\underline{o}}_x$ is the set of all elements of $k(X)$ having the form $f/g$ for $f,g$ in some $R_m,g(x)\ne 0$. Define $R^{\prime }=k[y_1,\cdots ,y_n]/P^{\prime }$.

If $k(X^{\prime })=\mathrm{Frac}(R^{\prime })$, then ${\underline{o}}_{\phi (x)}$ is the set of all elements in $k\left(X^{\prime }\right)$ having the form $F/G$ for $F,G\in R^{\prime },G(\phi (x))\ne 0$. Then

$$\theta :R^{\prime }\to k(x),y_i\mapsto x_i/x_0$$

extends to an isomorphism $k(X^{\prime })\simeq k(X)$. Moreover, $\theta$ takes ${\underline{o}}_{\phi (x)}$ precisely onto ${\underline{o}}_x$.

• First of all it clearly maps ${\underline{o}}_{\phi (x)}$ into ${\underline{o}}_x$; and
• if $f,g\in R_m,g(x)\ne 0$, then $f/g=f/x_0^m/g/x_0^m$ in $k(X)$ and $f/x_0^m,g/x_0^m$ come from $F,G$ in $R^{\prime }$ with $G(\phi (x))\ne 0$.
  • Thus the local rings correspond;
  • since the sheaves were defined by intersecting local rings, they also correspond, and $X\cap {\mathbb{P}}_n(k{)}_{x_0}$ is indeed an affine variety.

Morphism between Varieties

Definition

Let $X$ and $Y$ be prevarieties. A map $f:X\to Y$ is a morphism if $f$ is continuous and, for all open sets $V$ in $Y$,

$$g\in \Gamma \left(V,{\underline{o}}_Y\right)⟹g\cdot f\in \Gamma \left(f^{-1}V,{\underline{o}}_X\right).$$

Remark. The definition is compatible with ^4524hh, see here.

Proposition

Let $f:X\to Y$ be any map. Let $\left(V_i\right)$ be a collection of open affine subsets covering $Y$. Suppose that $\left\{U_i\right\}$ is an open covering of $X$ such that

• $f\left(U_i\right)\subset V_i$
• $f^{\ast }$ maps $\Gamma \left(V_i,{\underline{o}}_Y\right)$ into $\Gamma \left(U_i,{\underline{o}}_X\right)$.

Then $f$ is a morphism.

\begin{proof} See Mumford, page 42. \end{proof}