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I know how to prove this using a local equation for the non-regularity locus and the that the local ring at a smooth point is a UFD, and today I saw an example of this happening in practice:

Let $V : X^2 + Y^2 - Z^2 = 0$. Then the map $\phi: V \to \mathbb{P}^1$, $\phi([X:Y:Z]) = [X + Z : Y]$ is not defined at $x = [1:0:-1]$, but since $(X - Z)(X + Z) = Y^2$ on $V$: $[X + Z : Y] = [(X + Z)(X - Z) : Y(X - Z)] = [Y^2 : Y(X - Z)] = [Y : X - Z]$ so we can glue these rational functions functions together to get a regular function on the entire variety.

I have two questions:

  1. Can someone point out how smoothness is being used (implicitly) in this example? I don't get it.
  2. Is there a geometric (complex varieties) interpretation of extending rational functions to smooth points? Is some kind of limit being taken that wouldn't be possible otherwise?

Thanks. This is pretty weird to me.

Viktor Vaughn
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Elle Najt
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    The intuition is "multiply to clear denominators". Of course, for this to work at a point, one needs to say multiply by what? The answer is a coordinate at that point, or, a uniformizer of the local ring. This is where we use smoothness, to know that we have a uniformizer. The geometric intuition is that we want to be able to define $\phi(x)$ for a point $x$ for which $\phi$ is not defined as $\lim \phi(x_n)$ for a sequence of points where $\phi$ is defined and converge to $x$. Of course, for this to work we need that $\phi(x_n)$ converges (that's where we're using properness) and that it's – Alex Youcis Sep 13 '15 at 22:51
  • independent of the sequence $x_n$ (that's where we're using that it's a curve—there's only one dimension to approach $x$). – Alex Youcis Sep 13 '15 at 22:52
  • Funny, your example popped up in this thread. – Viktor Vaughn Mar 25 '20 at 18:37

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