# More Global sections of Projective Stuff

Last time I discussed why projective varieties have only constant global sections. This time I’ll do the same for the more general projective variety (abstract variety that is projective over a field k). This is basically the content of problem II.4.5 in Hartshorne, and seems to be a good application of the valuation criterion of properness.

The setup: X is integral, finite type over k (so if X is separated, then its an abstract variety). Terminology: let x be a point of X, and R a valuation ring of K(X)/k. If R dominates then I say R has center x (this is not quite how Hartshorne says it). In other words,

.

The valuation criterion of separatedness/properness give pretty easily that:

- If X is separated over k, then any valuation ring R has at most one center in X (i.e. if it exists its unique)
- If X is proper over k, then any valuation ring has a unique center in X.

2 follows from the proof of 1, and the proof of 1 is basically two different centers for R give two different maps from Spec R to X but the valuation criterion says there is at most one map (and it exists in the case X is proper).

2 can be used to how if X is proper then its global sections are constant. The idea is simple. Suppose there is a nonconstant section, then use it to show there is a valuation ring with no center in X. Let’s start:

Quick result to use: since X is integral, any nonzero global section remains nonzero in the stalk at every point (pf: 1. show its true for affines (sub pf: use A –> A_f is injective) 2. if a maps to 0 in some stalk, it means it maps to 0 is some nbd of that point, then intersect with some affine cover, and show element is zero in every affine, i.e. it was 0 to begin with)

Now, is also not in k, so is a polynomial ring an is a local ring contained in K(X), so its dominated by some valuation ring R. Let be its center. Then by the previous paragraph is nonzero. If then are both in the maximal ideal of R, which is a contradiction. If is not in , then neither is b, but b is at least in the local ring , so $b \in m_R \cap \mathcal{O}_{X,x} = m_x$, which again gives a contradiction. So we’re done!

There’s actually another (shorter) proof when X is a (abstract) variety, which relies on the fact that any map into the affine line from a complete variety (proper over k) is constant. But a map to the affine line corresponds to a global section, hence any global section considered as a regular function has constant value . Look at restricted to every affine patch, it must be zero, hence by sheaf axioms .

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- Published:
- May 16, 2009 / 11:29 pm

- Category:
- alg. geo.

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