[Math] field of rational functions of a projective variety equal to that of an affine variety

algebraic-geometry

Let $Y \subset \mathbb{P}^n$ be a projective variety and let $U_i$ be the open set $x_i \neq 0$. Let $\phi_i : U_i \rightarrow \mathbb{A}^n$ be the isomorphism of varieties, defined e.g. in Hartshorne p. 10, that takes a point $P=(a_0,\cdots,a_n) \in \mathbb{P}^n$ to $(a_0/a_i,\cdots,a_n/a_i) \in \mathbb{A}^n$. Define $Y_i$ to be the image of the closed set $U_i \cap Y$ under $\phi_i$. Then $Y_i$ is an affine variety of $\mathbb{A}^n$.

Question: Hartshorne in the proof of Theorem 3.4(c), p. 18, says that $K(Y)=K(Y_i)$, where $K(\cdot)$ means field of rational functions (function field). Why is this true?

Best Answer

A rational function on $Y$ is by definition a regular function defined on an open (dense) subvariety of $Y$ (up to the equivalence relation of being equal to another locally defined regular function on the intersection of the open sets where they are defined). Since the intersection of two open dense subsets is open and dense, a rational function on $Y$ defines a rational function on $Y_i$ and vice-versa.