From: wcw@math.psu.edu (William C Waterhouse) Subject: Re: affine plane minus a finite number of points Date: 24 Apr 2000 22:09:04 GMT Newsgroups: sci.math Summary: [missing] In article <8dn1k6$gq5@canteclaer.sci.kun.nl>, petervr@canteclaer.sci.kun.nl (Peter van Rossum) writes: > I wonder if someone could help me with the following, seemingly easy, > algebraic gemeometry question. > Let k be an algebraically closed field. > Let A be a k-algebra, a UFD, dimension 1, finitely generated over k. > Then: Spec(A) is (isomorphic to) the affine line minus a finite number > of points. > Now I can translate this as follows: > Then: Q(A) is a pure transcedental extension of k. > Does anyone have a clue? Here is one proof (I think), but it uses something complicated at the end. The hypotheses guarantee (first) that A is integrally closed of dimension 1 and finitely generated, so inside the field Q(A) it is precisely all rational functions with no poles at (closed) points of Spec(A). There is a finite set S containing all points of the complete nonsingular curve not in Spec(A). Now since A is actually a UFD and a Dedekind domain, it is a principal ideal domain. Thus for each point P of Spec(A) there is a function f_P with zero of order 1 at P and no other zeros or poles in Spec(A) -- that is, the poles and other zeros of f_P are all in S. Using this fact inductively, we see that all divisors of the complete curve are linearly equivalent to divisors involving only points in S. Thus the free abelian group generated by the points in S maps onto the divisor class group. But if Q(A) is not rational, then the genus is at least 1 and the divisor classes correspond to points on an abelian variety. That is in particular divisible, so it cannot be finitely generated. William C. Waterhouse Penn State