This may possibly be completely obvious. Suppose that $C$ is a smooth projective curve over a finite field $k$ and let $\infty$ be a $k$-rational point of $C$. I want to show that $C\setminus\{\infty\}$ is an affine open subset whose co-ordinate ring is integrally closed.
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What kind of background do you have? A rough argument can be as follows: removing the point of infinity gives us a smooth affine variety. To show the co-ordinate ring is integrally closed is equivalent to showing the local ring at each point is integrally closed. However, since it is smooth, each local ring is regular and so by the Auslander–Buchsbaum theorem it is a UFD but UFDs are integrally closed. – Leon Sot Dec 07 '16 at 10:58
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I have read the first chapter of Mumford's "Introduction to Algebraic Geometry". I just need to see how we can see that the complement of the point at infinity is affine. If we intersect the curve with the complement of a hypersurface in the ambient projective space, then the result will be affine. But how do we know that there is a hypersurface intersecting the curve in just one point? Or is there some other way to see that the complement of the point is affine? It may be really obvious, but I can't see how to show that the co-ordinate ring isn't too small for it to be an affine variety. – Rupert Dec 07 '16 at 11:46
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I found another answer on this website that addresses my question. Thanks for explaining how we can show the co-ordinate ring to be integrally closed. – Rupert Dec 07 '16 at 12:00
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@Rupert You should answer your own question then. – Alex Youcis Dec 08 '16 at 06:06
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The part of my question asking how we know that $C\setminus\{\infty\}$ is affine is addressed in a comment to this question:
Curve minus finite number of points affine
and the other part about knowing the co-ordinate ring is integrally closed has been addressed by the first comment here.
