In Hartshorne's Algebraic Geometry the function field K(Y) of a variety Y is defined as the set of e

sunyerneq

sunyerneq

Answered question

2022-02-16

In Hartshorne's Algebraic Geometry the function field K(Y) of a variety Y is defined as the set of equivalence classes with f being a regular function on the open subset U. We have = if f and g agree on UV.
How does one evaluate a "rational function on Y'' f? Is it f(P) using the Evaluation map for polynomials? In this case, I don't see that this is well defined: Let PUV and =, then g(P) might not even be defined!

Answer & Explanation

twaza7fl

twaza7fl

Beginner2022-02-17Added 5 answers

The point is this: on a variety Y, given open subsets U and V and regular functions f:uk and g:Vk, if f| {UV}=g|UV, there is a unique regular function h:UVk such that f=hU and g=hV.  As a result, each maximal representative in each equivalence class of regular functions is distinct.
In particular, you can evaluate a rational function at a point P so long as there is some representative defined at P. They are called rational functions as a matter of tradition, not because they are actually functions. (It is tempting to think of rational functions on Y as being regular maps YPP1, but that doesn't work in general: for example, xy defines a regular function 2{(0,0)}k, hence a rational function 2k, but there is no way of extending it to a regular map 2PP1.)

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