Let X be an integral and Noetherian scheme . We consider an open subset U of X and fϵΓ(U,OX). Hence we may regard (U,f) as a rational function in X. I would like to prove that if pϵX−U(and there is no rational function defined at p that is equivalent to (U,f)), there exists an open subset V of X containing p such that we may write the inverse rational function of (U,f) as (V,g) for certain gϵΓ(V,OX). The idea behind this is that if a rational function has a pole at a point, then its inverse rational function has a zero, and in particular it is defined at that point.

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Let X be an integral and Noetherian scheme . We consider an open subset U of X and fϵΓ(U,OX). Hence we may regard (U,f) as a rational function in X. I would like to prove that if pϵX−U(and there is no rational function defined at p that is equivalent to (U,f)), there exists an open subset V of X containing p such that we may write the inverse rational function of (U,f) as (V,g) for certain gϵΓ(V,OX).    The idea behind this is that if a rational function has a pole at a point, then its inverse rational function has a zero, and in particular it is defined at that point.

litoshypinaylh4 · Accepted Answer

This is not true. Morally, the point is that, in dimension at least 2, rational maps can fail to be defined at a point in ways that are more complicated than just &quot;having a pole&quot;.    Counterexample: let X=A2, let p=(0,0) , and let f=yx. Then the inverse rational function is g=xy but (every representative of) that function is not defined at p either.    A bit more conceptually, you can view a rational function on X as a rational map f:X→P1. If f has a pole at p, then f is a regular map in a neighbourhood of p, and it maps p to ∞ϵP1. But not every rational map to P1 is regular everywhere --- in fact, if X is a projective variety say, &quot;most&quot; rational maps to P1 will fail to be regular at some points.

Let X be an integral and Noetherian scheme . We consider an open subset U of X and f\epsilon\Gamm

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