Regular Maps and Regular Functions

Definitions.

Definition (Regular Function). Let $X\subseteq {\mathbb{P}}^m$ be a quasiprojective variety over a field $K$. A function $f:X\to K$ is said to be regular on $X$ if for every point $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and two homogeneous polynomials $g,h\in K[x_0,\dots ,x_m]$ of the same degree such that $h(a)\ne 0$ for all $a\in U$, and for all $a\in U$, the function is given by $f(a)=g(a)/h(a)$.

Definition (Regular Map). Let $X\subseteq {\mathbb{P}}^m$ and $Y\subseteq {\mathbb{P}}^n$ be quasiprojective varieties over a field $K$. A map $F:X\to Y$ is a regular map if for every point $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and $n+1$ homogeneous polynomials $f_0,\dots ,f_n\in K[x_0,\dots ,x_m]$ of the same degree such that for every point $a\in U$, at least one $f_i(a)$ is non-zero, and $F(a)=[f_0(a):\cdots :f_n(a)]$.

Correspondence with the Affine Line.

Theorem (Regular Maps to the Affine Line). Let $X\subseteq {\mathbb{P}}^m$ be a quasiprojective variety over a field $K$, and let ${\mathbb{A}}^1$ be the affine line over $K$. There is a one-to-one correspondence between the set of regular maps $F:X\to {\mathbb{A}}^1$ and the set of regular functions $f:X\to K$.

Proof. We establish the correspondence by constructing a unique regular function from a regular map, and conversely.

First, let $F:X\to {\mathbb{A}}^1$ be a regular map. We view ${\mathbb{A}}^1$ as the open subset of ${\mathbb{P}}^1$ via the embedding $c\mapsto [1:c]$. Thus, $F$ is a regular map to ${\mathbb{P}}^1$ whose image lies in the open set where the first coordinate is non-zero. By definition, for any $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and homogeneous polynomials $f_0,f_1\in K[x_0,\dots ,x_m]$ of the same degree such that $F(a)=[f_0(a):f_1(a)]$ for all $a\in U$. Since the image of $F$ is contained in ${\mathbb{A}}^1$, we must have $f_0(a)\ne 0$ for all $a\in X$. We define a function $f:X\to K$ by setting $f(a)=f_1(a)/f_0(a)$. This local representation as a ratio of homogeneous polynomials of the same degree where the denominator is non-vanishing is the definition of a regular function. If $F(a)=[g_0(a):g_1(a)]$ is another local representation on some open set, then $[f_0(a):f_1(a)]=[g_0(a):g_1(a)]$ implies $f_0(a)g_1(a)=f_1(a)g_0(a)$. As $f_0(a)$ and $g_0(a)$ are non-zero, it follows that $f_1(a)/f_0(a)=g_1(a)/g_0(a)$, so the function $f$ is uniquely determined by $F$.

Conversely, let $f:X\to K$ be a regular function. By definition, for every point $p\in X$, there exists a neighborhood $U\subseteq X$ and two homogeneous polynomials $g,h\in K[x_0,\dots ,x_m]$ of the same degree such that $h(p)\ne 0$ and $f(a)=g(a)/h(a)$ for all $a\in U$. We define a map $F:X\to {\mathbb{P}}^1$ by $F(a)=[h(a):g(a)]$. Since $h(p)\ne 0$ for any $p$, at least one of the components of $[h(p):g(p)]$ is non-zero, so this defines a regular map. Furthermore, since $h(a)$ is locally non-zero, the image of $F$ lies in the open subset of ${\mathbb{P}}^1$ where the first coordinate is non-zero, which is isomorphic to ${\mathbb{A}}^1$. Thus, $F$ is a regular map from $X$ to ${\mathbb{A}}^1$. If $f(a)=g^{\prime }(a)/h^{\prime }(a)$ is another local representation, then $g(a)/h(a)=g^{\prime }(a)/h^{\prime }(a)$, which implies $g(a)h^{\prime }(a)=h(a)g^{\prime }(a)$. This is the condition for $[h(a):g(a)]$ and $[h^{\prime }(a):g^{\prime }(a)]$ to define the same point in ${\mathbb{P}}^1$, so the map $F$ is uniquely determined by $f$.

The two constructions are inverse to each other, establishing the one-to-one correspondence. $\square$

Example of a Regular Map on Projective Space.

Theorem (Swapping Map on ${\mathbb{P}}^1$). The map $F:{\mathbb{P}}^1\to {\mathbb{P}}^1$ defined by $F([x_0:x_1])=[x_1:x_0]$ is a regular map that is not induced by a regular function.

Proof. First, we show $F$ is a regular map. The map is defined by homogeneous polynomials $f_0(x_0,x_1)=x_1$ and $f_1(x_0,x_1)=x_0$. Both are homogeneous of degree 1. Their only common zero in ${\mathbb{A}}^2$ is $(0,0)$, which does not correspond to any point in ${\mathbb{P}}^1$. Thus, for any $p\in {\mathbb{P}}^1$, at least one of $f_0(p)$ or $f_1(p)$ is non-zero, so $F$ is a regular map.

Next, we prove by contradiction that $F$ cannot be induced by a single regular function. Assume $F$ is induced by a regular function $f:{\mathbb{P}}^1\to K$. According to the correspondence, this implies that the image of $F$ is contained in an affine chart of ${\mathbb{P}}^1$ and can be written as $F(p)=[1:f(p)]$ for all $p\in {\mathbb{P}}^1$. However, any regular function on a projective variety, such as ${\mathbb{P}}^1$, must be a constant function. Thus, there exists some $c\in K$ such that $f(p)=c$ for all $p\in {\mathbb{P}}^1$. This would mean $F$ is the constant map $F(p)=[1:c]$. But the given map $F([x_0:x_1])=[x_1:x_0]$ is not constant. For example, $F([1:0])=[0:1]$ and $F([1:1])=[1:1]$. Since $[0:1]$ and $[1:1]$ are different points in ${\mathbb{P}}^1$, the map is not constant. This is a contradiction. Therefore, the map $F$ is not induced by a regular function. $\square$