Re: Is an algebraic curve minus a finite number of points always affine?

On 2006-12-24 03:38:32 -0500, Jannick Asmus <jannick.news@xxxxxx> said:

On 23.12.2006 16:08, Kira Yamato wrote:
On 2006-12-23 09:35:54 -0500, Jannick Asmus <jannick.news@xxxxxx> said:

On 23.12.2006 15:21, Kira Yamato wrote:
On 2006-12-23 08:58:35 -0500, Jannick Asmus <jannick.news@xxxxxx> said:

On 23.12.2006 13:31, Kira Yamato wrote:
Suppose C is a 1-dimensional algebraic set, and F is a proper
algebraic
subset of C. Then, is
C \ F
always affine?

I think I can prove this to be true if C is a rational smooth
curve. In
this case, the field of rational maps of C
K(C) = K(X)
where K(X) is the field of fractions of polynomials in one variable
X. So, suppose
F := { p_1, ..., p_n }.
Then, the coordinate ring of C \ F can be written as
k[X, 1/(X-p_1), ..., 1/(X-p_n)],
which is clearly a finitely-generated k-algebra also. This shows
that C
\ F is affine.

But can this be extended to the general setting to all algebraic
curves
where the curve may not be rational nor smooth?

For a reference I was thinking of, e.g., Hartshorne, Algebraic Geometry,
Exc. III.4.2.

Best wishes,
J.

Hmm... That's gonna take me awhile before I can attempt that exercise.

Chevalley's theorem says that if f: X -> Y is a surjective finite
morphism of separated noetherian schemes then X is affine iff Y is affine.

Let's assume that C is integral w.l.o.g. Then let Y be your original
curve C and C' = X be the normalization of Y in its quotient field. Then
C' is smooth (NB: for integral curves smooth and normal are the same).
By Chevalley, you are reduced to the case that C can be assumed to be
smooth. But you said that you had completed this case already.

J.

Well, I haven't really completed any case that's useful. I only proved it in the simplest case: C is smooth affine rational curve.

In this case, C is isomorphic to A^1. So,
A^1 \ { p_1, ..., p_n }
is isomorphic to the curve with coordinate ring
k[X, 1/((X-p_1)(X-p_2)...(X-p_n))]
which is a k-algebra generated by two elements X and 1((X-p_1)(X-p_2)...(X-p_n)).

This is just the standard way of showing an affine variety minus a hypersurface is always affine.

But thanks for that hint with the Chevalley's theorem. I will look deeper into this.

--

-kira

.

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