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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-06
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Closed projective embedding from a radical homogeneous ideal

Statement

Assume the Axiom of Choice. If J is homogeneous radical and V+(J)≠∅, then I+(V+(J))=J; its inclusion in Pkn is a closed projective embedding and its homogeneous coordinate ring is k[x0,…,xn]/J.

Facts & Assumptions

Given: The Axiom of Choice, the algebraically closed base field k fixed by projective space points, and a homogeneous radical ideal J⊆k[x0,…,xn] with nonempty X=V+(J).

[F1]

The projective space Pkn is defined over a fixed algebraically closed field k (projective space points); its projective algebraic sets are the homogeneous zero loci V+(J) (projective algebraic set).

[F2]

Over this algebraically closed field, under AC the projective radical-ideal correspondence gives I+(V+(J))=J for homogeneous radical J with nonempty zero locus (projective irreducibility homogeneous prime).

[F3]

The homogeneous coordinate ring of a projective algebraic set X is S(X)=k[x0,…,xn]/I+(X), with its induced grading (homogeneous coordinate ring).

[A1]

Assume the Axiom of Choice. Its use is the radical-ideal correspondence in [F2] (The Axiom of Choice).

Proof

technique · direct
1.1A1F1F2given

The base field is algebraically closed by the standing projective-space convention in [F1]. Since J is homogeneous and radical and X=V+(J) is nonempty, all hypotheses of [F2] hold; under [A1] it gives I+(X)=J.

2.1F1step 1.1

By [F1], X=V+(J) is projectively closed. Its inclusion identifies X with this closed algebraic subset and has the identity as inverse onto its image, so it is a closed projective embedding in the classical coordinate sense used here.

3.1F3step 1.1step 2.1algebra∎

Substituting I+(X)=J from step 1.1 into [F3] gives S(X)=k[x0,…,xn]/J. The quotient carries the induced grading because J is homogeneous. This proves the stated coordinate-ring formula together with the closed embedding in step 2.1.

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