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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 k-algbra, it is vital to know whether R is inte- gral, since the integral affine k-algebras 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 n-space 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 k-algebra R is geometrically integral iff the k-algebra 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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