# 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 $s = c$.