# Unicity of regular map on quasiprojective variety

Let $X\subseteq \mathbb{P}^n$ and $Y\subseteq \mathbb{P}^m$ be quasiprojective varieties with $X$ irreducible and $f:X\rightarrow Y$ a regular map. If there is an open subset $U\subseteq X$ and homogeneous polynomials of the same degree $F_0,...,F_m$ such that $f|_U=(F_0,...,F_m)|_U$. Is it true that $f=(F_0,...,F_m)$ on $X?$?

In my book a regular map $f:X\rightarrow Y$ bwteen quasiprojective varieties is defined by the following condition: for every $x\in X$ and for some affine piece $\mathbb{A}^m_i$ containing $f(x)$ there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq\mathbb{A}^m_i$ and the map $f:U\rightarrow \mathbb{A}^m_i$ is regular, that is to say, if $f=(f_1,...,f_m)$ on $U$, then for every $x\in X$, there exists homogeneous polynomials $P_i$ and $Q_i$ of the same degree with $Q_i(x)\neq 0$ such that $f_i|_U=\frac{P_i}{Q_i}|_U$. The closed subsets of a quasiprojective variety are defined to be the zeros of homogeneous polynomials.

## 1 Answer

Let $y$ be another point in $X$. By definition of regular maps, $y$ has a neighborhood $V$ in $X$ such that $f(V)\subseteq \mathbb{A}^m_i$ and $f:V\rightarrow \mathbb{A}^m_i$ is regular. Thus there exists polynomials $G_0,...,G_m$ such that $f=(G_0,...,G_m)$ on $V$. We only need to show that $F_i G_j=F_j G_i$ on $X$ for $i,j\in\{0,...,m\}$. Since $X=(X\cap Z(F_i G_j-F_j G_i))\cup X\backslash (U \cap V)$ and $X$ is irreducible, $X\subseteq Z(F_i G_j-F_j G_i)$.