TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-12

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Polynomial functions on an affine algebraic set are exactly its coordinate ring

Statement

Let $k$ be an algebraically closed field and let $X \subseteq A_{k}^{n}$ be an affine algebraic set. Evaluation induces a bijection $k [X] ⟶ {φ : X \to k : φ is the restriction of a polynomial on A_{k}^{n}} .$ Equivalently, two polynomials define the same polynomial function on $X$ if and only if their difference lies in $I (X)$ .

Facts & Assumptions

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

[L1]

The coordinate ring of $X$ is the quotient $k [x_{1}, \dots, x_{n}] / I (X)$ (The coordinate ring of an affine algebraic set).

[L2]

A polynomial lies in $I (X)$ exactly when it vanishes at every point of $X$ (The vanishing ideal of a subset of affine space).

Proof

technique · direct

1.1

Let $\overline{f} \in k [X]$ , represented by a polynomial $f$ . Define $Φ (\overline{f}) : X \to k$ by $Φ (\overline{f}) (x) = f (x)$ . If $\overline{f} = \overline{g}$ in $k [X]$ , then $f - g \in I (X)$ by [L1], so [L2] says $f (x) = g (x)$ for every $x \in X$ . Thus $Φ$ is well defined.

L1L2given

1.2

If $Φ (\overline{f}) = Φ (\overline{g})$ , then $f (x) = g (x)$ for every $x \in X$ , so $f - g \in I (X)$ by [L2]. Hence $\overline{f} = \overline{g}$ by [L1]. Therefore $Φ$ is injective.

L1L2

1.3

Every polynomial function on $X$ is, by definition, the restriction of some polynomial $f \in k [x_{1}, \dots, x_{n}]$ , and that function is exactly $Φ (\overline{f})$ . Hence $Φ$ is surjective.

given

2.1

Steps 1.1, 1.2, and 1.3 give the stated bijection, and step 1.2 is precisely the criterion that two polynomials define the same function on $X$ if and only if their difference lies in $I (X)$ .

step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry, Chapter 2i (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, §1.5 (standard reference, not scraped)