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 \ \mathcal{O}_{X,x} then I say R has center x (this is not quite how Hartshorne says it).  In other words,

\ \mathcal{O}_{X,x} \subset R

\ m_R \cap \mathcal{O}_{X,x} = m_x.

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

  1.  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)
  2. 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:

\ a \in \Gamma(X, \mathcal{O}_X) \backslash k, b = a^{-1}

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, b is also not in k, so k[b] is a polynomial ring an \ k[b]_{(b)} is a local ring contained in K(X), so its dominated by some valuation ring R.  Let x \in X be its center.  Then by the previous paragraph a \in \mathcal{O}_{X,x} is nonzero.  If a \in m_x then a,b are both in the maximal ideal of R, which is a contradiction.  If a is not in m_x, then neither is b, but b is at least in the local ring \mathcal{O}_{X,x}, 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 s  considered as a regular function X \to k has constant value s(x) = c.  Look at s - c restricted to every affine patch, it must be zero, hence by sheaf axioms s = c.  


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