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

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Regular functions on a principal open are the principal localization

Statement

Assume the Axiom of Choice, inherited from the Nullstellensatz route. For every affine algebraic set $X$ , $A = k [X]$ and $f \in A$ , the map $A_{f} ⟶ O_{X} (D_{X} (f)), a / f^{r} ⟼ (x \mapsto a (x) / f (x)^{r})$ is an isomorphism of unital $k$ -algebras. If $D_{X} (f) = \emptyset$ , then $f = 0$ in the reduced ring $A$ , and both sides are zero rings.

Facts & Assumptions

Given: AC, an algebraically closed field $k$ , an affine algebraic set $X$ , $A = k [X]$ , and $f \in A$ .

[F1]

Relative Nullstellensatz gives $I_{X} (V_{X} (H)) = \sqrt{H}$ for every ideal of $A$ (Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals).

[F2]

Equality of polynomial functions on all of $X$ is equality in $A$ (Polynomial functions on an affine algebraic set are its coordinate ring).

[F3]

Principal opens form a basis (Principal opens form a basis and multiply under intersection).

[F4]

Regular sections have local quotient expressions (A regular function on an open subset of a classical affine variety).

[F5]

Regular sections form algebras (Classical regular functions satisfy locality and unique gluing).

[F6]

A map inverting the denominators extends uniquely to the localization (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[F7]

A fraction is zero when some allowed denominator annihilates its numerator (Equality, vanishing, and the kernel of the localisation map).

[F8]

An ideal membership has a finite sum expression (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

Proof

technique · direct

1.1

Restriction $A \to O_{X} (D (f))$ is a unital algebra map. The function $1 / f$ is regular on $D (f)$ , so F6 extends restriction to the displayed map. If $a / f^{r}$ maps to zero, $a$ vanishes on $D (f)$ ; hence $f a$ vanishes everywhere on $X$ , since $f = 0$ outside $D (f)$ . F2 gives $f a = 0$ in $A$ , and F7 makes the fraction zero. This proves injectivity without cancellation in $A$ .

F2F4F5F6F7givenalgebra

1.2

Fix $s \in O_{X} (D (f))$ . At each point take a local expression $s = g / h$ and refine its neighbourhood to $D (a) \subseteq D (f) \cap D (h)$ by F3. The inclusion $V_{X} (h) \subseteq V_{X} (a)$ gives $a \in \sqrt{(h)}$ by F1. Thus $a^{e} = h b$ for some $e \geq 1, b \in A$ . On $D (a)$ , set $t = a^{e}$ and $c = g b$ ; then $D (t) = D (a)$ and $s = c / t$ .

F1F3F4givenalgebra

2.1

Consider the set of all pairs $(t, c)$ obtained in step 1.2; their opens $D (t)$ cover $D (f)$ and are contained in it. Thus $f$ vanishes on the simultaneous zero locus of the $t^{2}$ . By F1, $f \in \sqrt{(t^{2} : (t, c) as above)}$ . F8 supplies finitely many of these pairs $(t_{i}, c_{i})$ and $u_{i} \in A$ with $f^{N} = \sum_{i = 1}^{m} u_{i} t_{i}^{2}$ , for some $N \geq 1$ . No compactness theorem or simultaneous choice of neighbourhoods is needed.

F1F8step 1.2

3.1

For $p \in D (f)$ and each selected pair, if $t_{i} (p) \neq 0$ then $c_{i} (p) = s (p) t_{i} (p)$ ; if $t_{i} (p) = 0$ then both $c_{i} (p) t_{i} (p)$ and $s (p) t_{i} (p)^{2}$ are zero. Hence $\sum_{i} u_{i} (p) c_{i} (p) t_{i} (p) = s (p) \sum_{i} u_{i} (p) t_{i} (p)^{2} = s (p) f (p)^{N}$ . Division by the nonzero scalar $f (p)^{N}$ shows that $(\sum_{i} u_{i} c_{i} t_{i}) / f^{N}$ maps to $s$ . This proves surjectivity.

step 1.2step 2.1algebra

4.1

If $D (f)$ is empty, $f$ is the zero function on $X$ , hence zero in $A$ by F2. Localizing at 0 is the zero ring by F7; the empty domain has one function and its algebra is zero. If $f = 1$ , the same construction gives global sections and $A_{1} = A$ . Together with injectivity and surjectivity this proves all cases.

F2F7step 1.1step 3.1∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, Lemma 3.10 and Proposition 3.11, pp. 61–62. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Used by

Dependency tree · two levels

31 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 v6.10, Lemma 3.10 and Proposition 3.11, pp. 61–62 (standard reference, not scraped)