Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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global regular functions projective variety

Statement

Assume the Axiom of Choice. Every global regular function on a classical projective variety X is constant.

Proof

Given: The Axiom of Choice and a global regular function f on a nonempty irreducible projective algebraic set XPkn, where k is algebraically closed.

1.1

Put A=S(X). Irreducibility makes A a graded domain, so f is a [given, algebra] degree-zero element of Frac(A). If xi=0 in A, take Ni=1, and then xiNif=0ANi. Otherwise Ui=XD+(xi) is nonempty. Normalizing xi=1 identifies its defining ideal with the dehomogenizations of the homogeneous elements of I+(X); hence its affine coordinate ring is canonically the degree-zero localization A(xi). The affine global-functions theorem places fUi in A(xi), so it has the form a/xiNi with aANi. Therefore, for every i, there is Ni0 such that xiNifANi.

givenalgebra
2.1

Choose NNi for all i. If d>(n+1)(N1), every degree-d monomial is divisible by some xiN, so multiplication by f sends the finite-dimensional space Ad into itself. This space is nonzero: choose pX and a coordinate xi nonzero at p; then xid is nonzero in Ad.

step 1.1algebra
3.1

Cayley--Hamilton applied to the k-linear endomorphism afa of Ad gives a nonzero polynomial Pk[T] with P(f)a=0 for every aAd. Taking 0aAd and working in the field Frac(A) gives P(f)=0. Since k is algebraically closed, P splits into linear factors, and the domain property forces f=c for some ck. Thus every global regular function is constant.

step 2.1algebra

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