TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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Polynomial functions on an affine algebraic set are its coordinate ring

Statement

Evaluation induces an isomorphism of $k$ -algebras from $k [X]$ to the algebra of polynomial functions $X \to k$ , including $X = \emptyset$ .

Facts & Assumptions

Given: An affine algebraic set $X \subseteq k^{n}$ over an algebraically closed field $k$ .

[F1]

The coordinate ring is $R / I (X)$ (The coordinate ring of a classical affine algebraic set).

[F2]

Evaluation is a unital homomorphism determined by coordinates (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[F3]

A ring modulo the kernel maps isomorphically onto the image (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

Proof

technique · direct

1.1F1F2givenalgebra

Give all functions $X \to k$ pointwise operations. Evaluating a polynomial at each point defines a unital $k$ -algebra homomorphism $e : R \to k^{X}$ , by iterated polynomial evaluation, since sums and products evaluate to sums and products. Its image is exactly the polynomial functions. Its kernel consists exactly of the polynomials vanishing on every point, namely $I (X)$ .

2.1F3step 1.1∎

F3 gives $R / I (X) ≅ im e$ by $[f] \mapsto (a \mapsto f (a))$ , and constants show it is an isomorphism over $k$ . When $X$ is empty there is one function, its function algebra is the zero ring, and $\ker e = R$ ; the same quotient identification applies.

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, §2i p. 48. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, Algebraic Geometry v6.10, §2i p. 48