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 U∩V.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 P∈UV and =, then g(P) might not even be defined!

Question

In Hartshorne&#039;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 U∩V.How does one evaluate a &quot;rational function on Y&#039;&#039; f? Is it f(P) using the Evaluation map for polynomials? In this case, I don&#039;t see that this is well defined: Let P∈UV and =, then g(P) might not even be defined!

twaza7fl · Accepted Answer

The point is this: on a variety Y, given open subsets U and V and regular functions&amp;nbsp;f:u→k&amp;nbsp;and&amp;nbsp;g:V→k, if&amp;nbsp;f|&amp;nbsp;{U∩V}=g|U∩V, there is a unique regular function&amp;nbsp;h:U∪V→k&amp;nbsp;such that&amp;nbsp;f=h∣U&amp;nbsp;and&amp;nbsp;g=h∣V. &amp;nbsp;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&amp;nbsp;Y→PP1, but that doesn&#039;t work in general: for example,&amp;nbsp;xy&amp;nbsp;defines a regular function&amp;nbsp;∀2{(0,0)}→k, hence a rational function&amp;nbsp;∀2→k, but there is no way of extending it to a regular map&amp;nbsp;∀2→PP1.)

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

Answered question

Answer & Explanation

New Questions in Algebra II