Suppose that X is a varity over an algebraically close field K. If I am not mistaken, a rational function over X is a function in K(X). And a function on X is regular at a point x∈X if there exists g,h∈K[X], and an open subset V⊆X containing x, such that for all y∈V,h(y)≠0 and f(y)=g(y)h(y). And f(x) is a regular function if it is regular at every point in X. I not understand why rational functions need not be regular, and I am wondering if there is any example of rational function being not regular.

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Suppose that X is a varity over an algebraically close field K. If I am not mistaken, a rational function over X is a function in K(X). And a function on X is regular at a point x∈X if there exists g,h∈K[X], and an open subset V⊆X containing x, such that for all y∈V,h(y)≠0 and f(y)=g(y)h(y). And f(x) is a regular function if it is regular at every point in X.    I not understand why rational functions need not be regular, and I am wondering if there is any example of rational function being not regular.

Kathleen Mcpherson · Accepted Answer

Here is an example: let X=∀K1, and let f(y)=1y. This is a rational function f:X→K that is regular at all y∈X,y≠0, but which is not defined at y=0. It cannot be extended to a regular function at y=0, for suppose    fg=1y    in some open neighborhood of y=0; this open neighborhood is in particular infinite, because the topology on ∀K1 is the cofinite topology and K (being algebraically closed) is infinite. Then f(y)y-g(y)=0 for infinitely many y∈K, but because fy-g can only have finitely many roots, we must actually have fy=g, hence fg=1y everywhere, and therefore it cannot be defined at y=0.    One of the things that I found a little counterintuitive about rational functions at first is that they are not functions, per se, because they might only be defined on an open subset of the variety in question.

Suppose that X is a varity over an algebraically close field K. If I am not mistaken, a rational fun

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