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Roland Waters 2022-06-25 Answered
Let X be some affine algebraic variety over k (i.e. some closed subset in A k n ). First suppose X to be irreducible. Then the algebra k [ X ] is a domain and we can consider the field of rational functions Q u o t k [ X ] = k ( X ). Could you explain me how to build an analogue of this field in the case when X is not necessarily irreducible? Then k [ X ] must not be a domain and we are to build some kind of localization?
Also, what is the destination of rational functions? Why we cannot be satisfied with only regular maps and regular functions?
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Answers (1)

Braylon Perez
Answered 2022-06-26 Author has 34 answers
The analogue of the quotient field for a ring with zero divisors is the total ring of fractions: basically, just invert everything that is not a zero divisor. Geometrically, an element of this ring can be viewed as a collection of rational functions, one on each irreducible component of X, such that they coincide on intersections.
Rational functions are important for a wide variety of reasons. Asking this question is like asking why Q is important. Why weren't we happy with Z ? Well, it wasn't big enough for what we wanted to do, so we enlarged it.

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How does this sound? Aside from the problem of discussing –     in the context of an arbitrary field, I am worried about not using the fact that 1 + 1 0, at least not explicitly. Where exactly is this assumed used, if at all?
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Suppose that q ( x ) = 1, and let f ( x ) = a n x n + . . . a 1 x + a 0 . Then 1 x 2 = f ( x ) 2 . If x = 0, 1 = a 0 2 and therefore a 0 must be a unit. Letting x = 1 we get 0 = ( a n + . . . + a 1 + a 0 ) 2 ; and letting x = 1 we get 0 = ( a n . . . a 1 + a 0 ) 2 . Since we are working in a field, there can be no nonzero nilpotent which means that a n + . . . + a 1 + a 0 and a n . . . a 1 + a 0 are both zero. Adding the two equations together yields 2 a 0 = 0, and since 1 + 1 0, a 0 = 0 which contradicts the fact that a 0 is a unit.
How does this sound?
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