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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?

Manos
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1 Answers1

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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.

Andrew
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  • Dear Andrew, i would like some clarification on your last sentence. I can see that a rational function of $Y$, say a pair $(V,f)$ gives a rational function of $Y_i$ by $(\phi_i(V),\alpha(f))$, where $a$ is the "dehomogenization" morphism. Where i have trouble is the converse. If $(W,g)$ is a rational function of $Y_i$, this is mapped to $(\phi_i^{-1}(W),\beta(g))$, where $\beta$ is the "homogenization" morphism. However, $\phi_i^{-1}(W) \subset U_i$, so it seems that i can not reach all rational functions of $Y$ from $K(Y_i)$. Thanks! – Manos Jul 31 '13 at 19:24
  • Dear @Manos, remember that the open sets in the Zariski topology are very big, and intersect nontrivially on (irreducible) varieties! So $\phi_i^{-1}(W)$ is open and thus dense in $Y$ -- it is clearly open in $Y\cap U_i\cong Y_i$ by isomorphism, thus there exists open $\widetilde W\subseteq Y$ such that $\widetilde W\cap U_i =\phi_i^{-1}(W)$ by definition of the subspace topology on $Y\cap U_i$. This implies that $\beta(g)$ is defined on an open set of $Y$, which must be dense. – Andrew Jul 31 '13 at 19:44
  • In other words, $g$ is defined on a nontrivial open subset $W$ of $Y_i$, and $Y_i$ is open in $Y$, thus $W$ is open and nontrivial in $Y$. – Andrew Jul 31 '13 at 19:49
  • Ah! This is the point of rational functions -- they are not meant to be defined everywhere! They are generally only regular on an open subset. For example, $K(\mathbb P^1) = K(\mathbb A^1)\cong K(t)$ but $K[\mathbb P^1] = K$. There are many more rational functions than regular functions, namely those that can be extended to all of $\mathbb P^1$ (the constants). – Andrew Jul 31 '13 at 20:03
  • One final question and i am covered: what is the significance of the fact that $\beta(g)$ is defined on a dense set? – Manos Jul 31 '13 at 20:04
  • So let me see if i understand: you are saying that if $(V,f)$ is a rational function of $Y$, then $(V,f) = (V\cap U_i, f)$ and the latter gives the rational function $(\phi(V), \alpha(f))$ of $Y_i$? – Manos Jul 31 '13 at 20:09
  • Yep, that's correct (though to be very precise the latter should be $(V\cap U_i, f|_{V\cap U_i})$). The way I interpret the significance is that regular functions do not always give the information we want (espcially on projective/complete objects), so we need to consider a "looser" class of functions. (Rational functions are the algebro-geometric version of meromorphic functions on Riemann surfaces.) – Andrew Jul 31 '13 at 20:17