1.4 Sheaves & Affine Varieties

Presheaf, Sheaf and Sheafification

Definition

Let $X$ be a topological space. A presheaf $F$ on $X$ consists of

• for any open set $U\subseteq X$, $F(U)$ is a set;
• for any $U_1\subseteq U_2$, there exists ${\mathrm{res}}_{U_2,U_1}:F(U_2)\to F(U_1),s\mapsto s{|}_{U_1}$;

such that

• ${\mathrm{res}}_{U,U}={\mathrm{id}}_U$;
• for $U_1\subset U_2\subset U_3$, we have ${\mathrm{res}}_{U_3,U_1}={\mathrm{res}}_{U_2,U_1}\circ {\mathrm{res}}_{U_3,U_2}$.

Remark. We may write $\Gamma (U,F)$ for $F(U)$.

Definition

Let $F_1$ and $F_2$ be presheaves on $X$, a map $\phi :F_1\to F_2$ is a collection of maps $\phi (U):F_1(U)\to F_2(U)$ for each open $U$ such that if $U\subseteq V$, then

$$\text{require}AMScd\begin{array}{ccc}{F}_1(U)& \stackrel{\phi (U)}{\to }& F_2(U)\\ {\mathrm{res}}_{U,V}\downarrow & \#& \downarrow {\mathrm{res}}_{U,V}\\ F_1(V)& \underset{\phi (V)}{\to }& F_2(V)\end{array}$$

commutes.

Remark. Let $X,Y$ be topological spaces, and let $\theta :X\to Y$ be a continuous map. Then $\theta$ induces a map from ${\mathcal{F}}_{c,X}({\theta }^{-1}(U))$ to ${\mathcal{F}}_{c,Y}(U)$, where ${\mathcal{F}}_{c,Y}(U)=\{f:f\text{mbox}iscontinuous\}$. It is not a sheaf map, as they are defined on different topological spaces. However, later we can define "${\mathcal{F}}_{c,Y}$" on $X$.

Definition

A presheaf $\mathcal{F}$ is a sheaf if for every covering $U={\cup }_{i\in I}{U}_i$, the diagram

is exact. That is,

• $s\mapsto {\prod }_{i\in I}(s{|}_{U_i})$ is injective (if $s{|}_{U_i}=t{|}_{U_i}$ for all $i$, then $s=t$);
• for any $(s_i)\in {\prod }_{i\in I}F(U_i)$ such that $s_i{|}_{U_i\cap U_j}=s_j{|}_{U_i\cap U_j}$ for any $i,j$, then there exists $s\in F(U)$ such that $(s_i{)}_i=(s{|}_{U_i}{)}_i$.

Examples and Counterexamples.

• Let $X$ be a topological space, and let $F(U)=\{f:U\to \mathbb{R}\text{mbox}iscontinuous\}$. Then $F$ is a sheaf with ${\mathrm{res}}_{U,V}:F(U)\to F(V),f\mapsto f{|}_V$.
• $G(U):=\{f\in F(U):f(U)\text{mbox}isbounded\}$ is a presheaf but not a sheaf. If $X=\mathbb{R}$, $U_i=(-i,i)$ and $f_i={\mathrm{id}}_{(-i,i)}$, then there does not exist $f\in G(U)$ such that $f{|}_{U_i}=f_i$ as $f={\mathrm{id}}_{\mathbb{R}}$ is unbounded.
• $\mathcal{F}(U):=\{f:U\to \mathbb{R}\text{mbox}islocallyconstant\}/\{f:U\to \mathbb{R}\text{mbox}isconstant\}$ is a presheaf but not a sheaf. Take $[f_1]\ne 0$ and $[f_2]=0$, then we can take special $U_i$ such that $[f_1{|}_{U_i}]=[f_2{|}_{U_i}]$ for all $i$. So $s\mapsto {\prod }_{i\in I}{s}_i$ is not injective.

Sheafification of a presheaf

For any presheaf $\mathcal{F}$, there exists unique sheaf $\tilde{\mathcal{F}}$ such that

• $\mathcal{F}\stackrel{g}{\to }\tilde{\mathcal{F}}$ is a presheaf map;
• for any $\mathcal{F}\stackrel{h}{\to }\mathcal{G}$ where $\mathcal{G}$ is a sheaf, there exists unique $\theta :\tilde{\mathcal{F}}\to \mathcal{G}$ such that $h=\theta \circ g$.

Examples.

• $F(U)=\{f:U\to \mathbb{R}\text{mbox}iscontinuous\}$ is the sheafification of $G(U):=\{f\in F(U):f(U)\text{mbox}isbounded\}$.
  • need verification;
  • $G(U)\to \tilde{G}(U)$ is injective but not surjective.
• For $\mathcal{F}(U):=\{f:U\to \mathbb{R}\text{mbox}islocallyconstant\}/\{f:U\to \mathbb{R}\text{mbox}isconstant\}$, $\tilde{\mathcal{F}}(U)=\{e\}$.
  • $\mathcal{F}(U)\to \tilde{\mathcal{F}}(U)$ is not injective.

Stalks

Let $F$ be a presheaf. For any $x\in X$, define the stalk of $F$ at $x$

$$F_x=\underset{x\in U}{\underrightarrow{\lim }}F(U)=\{(s,U):s\in \mathcal{F}(U),x\in U\}/\sim$$

where $(s_1,U_1)\sim (s_2,U_2)$ if there exists $W\subseteq U_1\cap U_2$ such that $x\in W$ and $s_1{|}_W=s_2{|}_W$.

Example. Define ${\mathcal{O}}_{\mathbb{C}}(U)=\{f:f\text{mbox}isholomorphonU\}$, then the stalk at $0$ is

$${\mathcal{O}}_{\mathbb{C},0}=\{(s,U):s\text{mbox}isholomorphonU\}/\sim ,$$

where $(s_1,U_1)\sim (s_2,U_2)$ if there exists $W\subseteq U_1\cap U_2$ such that $s_1{|}_W=s_2{|}_W$ and $0\in W$. Thus ${\mathcal{O}}_{\mathbb{C},0}=\{\sum c_n{z}^n\text{mbox}convergesnear0\}$.

Example. Let $F(U)=$ continuous functions $U\to \mathbb{R}$. Then $F_x$ is the set of germs of continuous functions at $x$. It is ${\cup }_{x\in U}F(U)$ modulo an equivalence relation: $f_1\sim f_2$ if $f_1$ and $f_2$ agree in a neighborhood of $x$.

Structure Sheaf

For any $U\subseteq X$, if $\mathcal{F}(U)$ is a group/ring and $\mathcal{F}(U)\to \mathcal{F}(V)$ is a group/ring homomorphism, then ${\mathcal{F}}_x$ is also a group/ring. Note that ${\mathcal{F}}_x=\{[U,s]:s\in \mathcal{F}(U)\}$, and define $[(U,s)]+[(V,t)]=[U\cap V,s{|}_{U\cap V}+t{|}_{(U\cap V)}]$.

In classical algebraic geometry, one often wants to study functions that behave well under localization, which leads to the concept of a structure sheaf.

structure sheaf

Let $X\subseteq k^n$ be an irreducible algebraic set, and let $R=k[x_1,\cdots ,x_n]/I(X)$.

• Let $K(R)$ be its field of fractions.
  • Since $X$ is irreducible, $R$ is an integral domain.
• For $x\in X$, define $m_x=\{f\in R\mid f(x)=0\}$.
  • This is a maximal ideal, as it is the kernel of the homomorphism $R\to k$ given by $f\mapsto f(x)$.
• Let ${\underline{o}}_x=R_{m_x}=\{\alpha /\beta :\alpha \in R,\beta \notin m_x\}$.
  • It is a localization of a ring.
  • Recall that $R_f=\{r/f^n:n\in \mathbb{Z}\}$.

We have then ${\underline{o}}_x=\{f/g\mid f,g\in R,g(x)\ne 0\}\subset K$. Now for a Zariski open set $U\subseteq X$, let

$${\underline{o}}_X(U)=\bigcap _{x\in U}{\underline{o}}_x.$$

All the ${\underline{o}}_X(U)$ are subrings of $K(R)$. If $V\subset U$, then ${\underline{o}}_X(U)\subset {\underline{o}}_X(V)$; if we take the inclusion as the restriction map, this defines a sheaf ${\underline{o}}_X$.

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 $F(x_0)=\alpha (x_0)/\beta (x_0)\in k$.

Proposition

Let $X$ be an irreducible algebraic set. Define $R=\Gamma (X)$. For $f\in R$, let $X_f=\{x\in X:f(x)\ne 0\}$. Then ${\underline{o}}_X(X_f)=R_f$.

\begin{proof} Take $\alpha /f^n\in R_f$. For any $x_0\in X_f$, $f^n(x_0)\ne 0$ and so $f^n\notin m_{x_0}$. Then $\alpha /f_n\in R_{m_{x_0}}$ and $\alpha /f^n\in {\cap }_{x_0\in X_f}{R}_{m_{x_0}}$.

For any $F\in {\underline{o}}_X(X_f)$, define $B_F:=\{h\in R:hF\in R\}$ and then $B_F$ is an ideal of $R$. If there exists $n$ such that $f^n\in B_F$, then $f^{n}F=r\in R$ and $F=r/f^n\in R_f$. For some $F\in {\cap }_{x_0\in X_f}{R}_{m_{x_0}}$ and for any $x_0\in X_f$, we have $F\in R_{m_{x_0}}$ and $F=\alpha /\beta$ with $\beta \notin m_{x_0}$. Since $\beta F=\alpha \in R$, we have $\beta \in B_F$ and $x_0\notin V(B_F)$. It deduces that $V(B_F)\cap X_f=\varnothing$ and $V(B_F)\subseteq V((f))$. Then $\sqrt{B_F}\supseteq \sqrt{(f)}$ containing $f$ and so there exists $n⩾0$ such that $f^n\in B_F$. \end{proof}

Example. In particular, ${\underline{o}}_X(X)=R$. Take $f\equiv 1$, and then $R_f=R$ and $X_1=\{x\in X:1\ne 0\}=X$. It deduces that

$${\underline{o}}_X(X_1)={\underline{o}}_X(X)={\cap }_{x\in X}{\underline{o}}_x=R=\Gamma (X,{\underline{o}}_X).$$

Remarks.

• If $f\in {\underline{o}}_X(U)$ and $f(x)\ne 0$ for all $x\in U$, then $1/f\in O_X(U)$, as $f\ne 0$ yields $1/f\ne 0$ and $1/f\notin m_X$.

• $h\in {\underline{o}}_X(U)$, $h$ may not be $\alpha /\beta$ with $\alpha ,\beta \in R$. However, it is true when $U=X_f$ by ^9d743b.

• The stalk of ${\underline{o}}_X$ at $x$ is ${\underline{o}}_x$. Recall that ${\underline{o}}_{X,x_0}=\underset{x\in U}{\underrightarrow{\lim }}{\underline{o}}_X(U)$. Since $\{X_f:f\in R\}$ is a topological basis, by ^9d743b we have

  $${\underline{o}}_{X,x_0}=\underset{x_0\in U}{\underrightarrow{\lim }}{\underline{o}}_X(U)=\underset{x_0\in X_f}{\underrightarrow{\lim }}{\underline{o}}_X(X_f)=\underset{x_0\in X_f}{\underrightarrow{\lim }}{R}_f.$$

  As all restriction maps in our sheaf are injective, it is just ${\cup }_{f(x_0)\ne 0}{R}_f={\underline{o}}_{x_0}$.

Proposition

Let $X\subset k^n,Y\subset k^m$ be irreducible algebraic sets, and let $f:X\to Y$ be a continuous map. The following conditions are equivalent:

• i) $f$ is a morphism
• ii) for all $g\in \Gamma \left(Y,{\underline{o}}_Y\right),g\cdot f\in \Gamma \left(X,{\underline{o}}_X\right)$
• iii) for all open $U\subset Y$, and $g\in \Gamma \left(U,{\underline{o}}_Y\right)⟹g\cdot f\in \Gamma \left(f^{-1}U,{\underline{o}}_X\right)$
• iv) for all $x\in X$, and $g\in {\underline{o}}_{f(x)}⟹g\cdot f\in {\underline{o}}_x$.

\begin{proof} i) → ii) is easy. Take $f_i=y_i\circ f$ for ii)→i).

iii)→ii) is trivial by taking $U=Y$, and iv) $⟹$ iii) by the definition of ${\underline{o}}_X$.

ii)→iv) Let $g\in {\underline{o}}_{f(x)}$. We write $g=a/b$ with $a,b\in \Gamma \left(Y,{\underline{o}}_Y\right)$ and $b(f(x))\ne 0$. By ii), $a\cdot f,b\cdot f\in$ $\Gamma \left(X,{\underline{o}}_X\right)$; hence $g\cdot f=a\cdot f/b\cdot f\in {\underline{o}}_x$, since we have $b\cdot f(x)\ne 0$. \end{proof}

Affine Variety

Definition

An affine variety is a topological space $X$ plus a sheaf of $k$-valued function ${\underline{o}}_X$ on $X$, which is isomorphic to an irreducible algebraic subset $\Sigma$ of some $k^n$ plus the sheaf ${\underline{o}}_{\Sigma }$, where $O_{\Sigma }(U)={\cap }_{x\in U}{R}_{m_x}$.

Affine Variety Definitions

• Classical Definition: An Affine Variety is an irreducible closed subset of affine space $k^n$ defined by the vanishing of a set of polynomials.
• Modern Definition: An Affine Variety is a topological space $X$ together with a sheaf of $k$-valued functions ${\underline{o}}_X$, which is isomorphic to a classical Affine Variety $\Sigma$ together with its structure sheaf ${\underline{o}}_{\Sigma }$.

The modern definition adds the structure sheaf to the classical definition. This allows for a more sophisticated study of functions on the variety.

Definition

$(k^n,{\underline{o}}_{k^n})$ is called an affine $n$-space, denoted by ${\mathbb{A}}^n$.

Remark. If there is $\theta :X\to \Sigma$ such that $(X,{\underline{o}}_X)$ is an affine variety, then

• $\theta$ is a bijective map between topological spaces $X$ and $\Sigma$;
• both $\theta$ and ${\theta }^{-1}$ induce the isomorphism ${\underline{o}}_X(U)\simeq {\underline{o}}_Y(\theta (U))$ for all open sets $U\subseteq X$.

By ^dbd10f, $\theta$ is a morphism in the category of affine varieties. However, a bijective morphism in this category may not be an isomorphism.

• It is correct, for example, in the category of compact topological spaces, of Banach spaces, and of complex analytic manifolds.
  • In the category of complex manifolds, a bijective holomorphic map is a morphism, and its inverse is also holomorphic by the inverse function theorem. Thus, in this category, bijective morphisms are isomorphisms.
• It is false for the following cases.
  • For the category of differentiable manifolds, consider the map $f:\mathbb{R}\to \mathbb{R},x\mapsto x^3$, it is not a diffeomorphism because $f^{-1}(y)=y^{1/3}$ is not smooth at $y=0$.
  • For a field $k$ with $\mathrm{char}k=p\ne 0$, define $F:{\mathbb{A}}_1^1\to {\mathbb{A}}_2^1,t\mapsto t^p$. It is bijective and corresponds to the inclusion map in the pair of rings $k[X]\hookleftarrow k\left[X^p\right]$. Thus, $F$ is not an isomorphism since these rings are not equal.
  • Let $k$ be any algebraically closed field. Define the morphism $f:{\mathbb{A}}^1\to {\mathbb{A}}^2,t\mapsto (t^2,t^3)$. The image of this morphism is the irreducible closed curve $C:X^3=Y^2$. The morphism $f$ from ${\mathbb{A}}^1$ to $C$ is a bijection which corresponds to the inclusion map in the pair of rings $k[T]\hookleftarrow k[T^2,T^3]$. These rings are not equal, so $\Gamma ({\mathbb{A}}^1,{\underline{o}}_{{\mathbb{A}}^1})\not\simeq \Gamma (C,{\underline{o}}_C)$ and $f$ is not an isomorphism.

Induced Variety Structure

Definition

Let $(X,{\underline{o}}_X)$ be an affine variety, and let $Y\subseteq X$ be an irreducible closed subset. Define

$${\underline{o}}_Y(V)=\{k\text{mbox}-valuedfunction\alpha :V\to k\mid \forall y_0\in Y,\exists U\subseteq X\text{mbox}open,F\in {\underline{o}}_X(U),\alpha {|}_{Y\cap U}=F{|}_{Y\cap U}\}.$$

Proposition

$(Y,{\underline{o}}_Y)$ is an affine variety.

\begin{proof} Suppose $X=V(I)$ and $I\subseteq k[x_1,\cdots ,x_n]$. For closed irreducible $Y\subseteq X$, assume that $Y=V(J)$ and then $I\subseteq J\subseteq k[x_1,\cdots ,x_n]$.

It suffices to show ${\underline{o}}_Y={\underline{o}}_{V(J)}$. Define $R=k[x_1,\cdots ,x_n]/I=\Gamma (X)$ and $\overline{R}=R/J=\Gamma (Y)$. For any open set $V\subseteq V(J)$ and any $\alpha \in {\underline{o}}_{V(J)}(V)={\cap }_{y\in V}\overline{R_{m_y}}$, $\alpha$ can be written as $\alpha =\overline{r}/\overline{g}$ where $\overline{r}\in \overline{R}$ and $\overline{g}\in \overline{R\setminus m_y}$. Let $U=X_g$, then $r/g\in \Gamma (X_g,{\underline{o}}_X)$ and $r/g{|}_{X_g\cap V}=\overline{r}/\overline{g}=\alpha$ as $\overline{r}=r{|}_y$. Now we have proved that ${\underline{o}}_{V(J)}(V)\subseteq {\underline{o}}_Y(V)$.

Conversely, for any $\alpha \in {\underline{o}}_Y(V)$ and any $y\in Y$, there is $\alpha {|}_{Y\cap U}=r/g{|}_{Y\cap U}$ for some $r/g\in {\underline{o}}_X(U)$. Then $\overline{r}/\overline{g}\in \overline{R_{m_y}}$ and so ${\underline{o}}_{V(J)}(V)\supseteq {\underline{o}}_Y(V)$. \end{proof}

Proposition

Let $(X,{\underline{o}}_X)$ be an affine variety, and let $f\in \Gamma (X,{\underline{o}}_X)$. Then $(X_f,{\underline{o}}_X{|}_{X_f})$ is an affine variety, where $X_f=\{x\in X:f(x)\ne 0\}$.

\begin{proof} We identify $X$ with $V(I)$ and define $R=k[x_1,\cdots ,x_n]/I$. Since ${\underline{o}}_X(X_f)=R_f$, we define

$$k[x_1,\cdots ,x_n,x_{n+1}]/(I,1-f{x}_{n+1})\to R_f,x_{n+1}\mapsto 1/f.$$

Thus $R_f$ is an affine algebraic set and $X_f$ can be identified with $V((I,1-f{x}_{n+1}))\subseteq k^{n+1}$. Now we aim to show the identification works.

Define $\pi :V((I,1-x_{n+1}f))\to X_f,(x_1,\cdots ,x_{n+1})\mapsto (x_1,\cdots ,x_n)$, then $\pi$ is a morphism. Since $(x_1,\cdots ,x_n)\in V(I)$ and $1-x_{n+1}f(x_1,\cdots ,x_n)=0$, we have $f(x_1,\cdots ,x_n)\ne 0$ and $(x_1,\cdots ,x_n)\in X_f$. It is easy to check $\pi$ is bijective and is a homeomorphism.

Note that $\Gamma (V,{\underline{o}}_V)=k[x_1,\cdots ,x_{n+1}]/(I,1-x_{n+1}f)\simeq R_f$ and by ^9d743b there is $\Gamma (X_f,{\underline{o}}_X{|}_{X_f})={\cap }_{f(x)\ne 0}{R}_{m_x}=R_f$. Then $\Gamma (V,{\underline{o}}_V)\simeq \Gamma (X_f,{\underline{o}}_X{|}_{X_f})$ and so $\pi$ is an isomorphic. Now we finish the proof. \end{proof}

Check

Although $X_f$​ is a distinguished open set, it can be viewed as an affine variety in a higher-dimensional affine space. In this space, $X_f$ is isomorphic to a closed irreducible algebraic set.

Example. Define ${\mathbb{A}}^2\setminus \{(0,0)\}\to {\mathbb{A}}^2,(x,y)\mapsto (x,y)$, then $\Gamma ({\mathbb{A}}^2)\stackrel{\not\simeq }{\to }\Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})$.

\begin{proof} Let $\mathcal{P}:=\{p:p\text{mbox}isaprimeideal,\dim V(p)=1\}$. For any $\alpha \in \Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})$, we claim that $\alpha \in R_p$ for any $p\in \mathcal{P}$, where $R=k[x_1,x_2]=\Gamma ({\mathbb{A}}^2)$.

If the claim holds, then we can prove that $\Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})={\cap }_{p\in \mathcal{P}}{R}_p$. By claim we have $\Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})\subseteq {\cap }_{p\in \mathcal{P}}{R}_p$. For any $\beta \in {\cap }_{p\in \mathcal{P}}{R}_p$, to show $\beta$ is regular on ${\mathbb{A}}^2\setminus \{(0,0)\}$, it suffices to show $\beta \in {\underline{o}}_{(a,b)}=R_{m_{(a,b)}}$ for any $(a,b)\ne (0,0)$, where $m_{(a,b)}=(x-a,y-b)$ is dimension $0$. If $\beta \notin R_{m_{(a,b)}}$, define $I=\{s\in R:\beta s\in R\}$ and then $I\subseteq m_{(a,b)}$. For any $V(p)$ contains $(a,b)$, we have $\beta \in R_p$ and so $\beta =r_p/s_p$ with $s_p\notin p$, i.e., $s_p(a,b)\ne 0$. It contradicts with $I\subseteq m_{(a,b)}$. Therefore, $\beta \in R_{m_{(}a,b)}$. Now we prove that $\Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})={\cap }_{p\in \mathcal{P}}{R}_p$.

Now it is enough to prove the claim. Take $\alpha \in \Gamma ({\mathbb{A}}^2\setminus \{(0,0)\}$, and assume that $\alpha \notin R_p$. For any $(r,s)$ such that $\alpha =r/s$, there is $s\in p$ and so $V(s)\supseteq V(p)$ with $\dim V(p)=1$. However, $V(s)=\{(0,0)\}$ by $\alpha$ regular on ${\mathbb{A}}^2\setminus \{(0,0)\}$ being dimension $0$, which is a contradiction. Now we proved the claim. \end{proof}

Check

It took me a long time to fill in the proof for this example because I had difficulty proving $\Gamma ({\mathbb{A}}^2\backslash$ $\{(0,0)\})={\cap }_{p\in \mathcal{P}}{R}_p$.

The most crucial point is that to prove that elements in ${\cap }_{p\in \mathcal{P}}{R}_p$ are regular on ${\mathbb{A}}^2\backslash \{(0,0)\}$, it suffices to show that they belong to $R_{m_{(a,b)}}$ for any $(a,b)\in {\mathbb{A}}^2\setminus \{(0,0)\}$ by ^436480.

In summary, for an affine variety $X$ and its coordinate ring $A(X)$, the following three statements regarding a function $f$ defined on $X$ are equivalent:

1. $f\in A(X)$.
   • (That is, $f$ is an element of the coordinate ring, representable by a global polynomial defined on the ambient affine space and restricted to $X$).
2. $f$ is a global regular function on $X$.
   • (That is, $f\in {\mathcal{O}}_X(X)$, also denoted $f\in \Gamma (X,{\mathcal{O}}_X)$).
3. For every open subset $U\subseteq X$, the restriction $f{|}_U$ is a regular function on $U$.

   • (That is, $f{|}_U\in {\mathcal{O}}_X(U)$ holds for all open sets $U$).

Remark on Non-Affine Varieties does not hold for non-affine varieties, such as most projective varieties.

This equivalence generally

Example: For the projective line ${\mathbb{P}}^1$ (assuming it is connected and defined over an algebraically closed field), the only global regular functions are the constant functions, i.e., ${\mathcal{O}}_{{\mathbb{P}}^1}({\mathbb{P}}^1)\cong k$. However, there are many non-constant functions on ${\mathbb{P}}^1$ that are regular on their domains of definition (a proper open subset). For instance, the function defined by $x_1/x_0$ is regular on the affine open set $U_0=\{[x_0:x_1]\mid x_0\ne 0\}$, but it is not constant and thus not globally regular on ${\mathbb{P}}^1$.