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Unformatted text preview: FUNCTION FIELDS AND GEOMETRIC INTEGRALITY PETE L. CLARK 1. Function fields If R is any integral domain, we have of course its field of fractions K = K ( R ). In particular, if R is finitely generated over a field k , then we write k ( R ) for K and call it the function field of R . As we have already seen, when k is not algebraically closed, one must keep in mind two different objects: the algebra R over k and the geometric algebra R = R k k . Especially, when given an integral kalgbra, it is vital to know whether R is inte gral, since the integral affine kalgebras are precisely the coordinate rings of affine algebraic varieties in the classical sense (i.e., the ring of all polynomial functions on an irreducible closed subset of affine nspace over k ). It is natural to ask how geometric integrality relates to the function field k ( R ). Let us first record a simple result to make sure we are all on the same page: Proposition 1. The integral kalgebra R is geometrically integral iff the kalgebra k ( R ) k k is a field, and in this case it is the fraction field of R : k ( R ) = k [ R ] k . Exercise: Prove it. Thus the question can be rephrased as one of pure field theory: given a field K (here K = k ( R )) which is finitely generated (in the sense of field extensions, not ring extensions) over a field k , give necessary and sufficient conditions for K k k to be a field. There is an easy necessary condition that we have seen before: if K contains a nontrivial algebraic extension l over k , then the nonintegral algebra l k k will be a subalgebra of K k k , which of course means that K k k itself cannot be integral....
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This note was uploaded on 10/26/2011 for the course MATH 8320 taught by Professor Clark during the Fall '10 term at University of Georgia Athens.
 Fall '10
 Clark
 Algebra, Fractions

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