TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27

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Regular algebras over a perfect field are geometrically regular

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $k$ be a perfect field (Perfect fields: every irreducible polynomial is separable) and let $A$ be a finite-type $k$ -algebra that is regular (regular noetherian ring). Then $A \otimes_{k} K$ is a regular ring for every field extension $K / k$ , not only for the finitely generated ones. In particular a regular finite-type algebra over a perfect field is geometrically regular (Field tests for geometric regularity).

No assumption is made on the extension field $K$ : it need not be perfect. In characteristic $0$ the hypothesis that $k$ is perfect is automatic, and the statement says that regular finite-type algebras over such fields stay regular after arbitrary scalar extension.

Facts & Assumptions

Given: A perfect field $k$ , a regular finite-type $k$ -algebra $A$ , and the Axiom of Choice.

[F1]

A field is perfect exactly when it has characteristic zero or its Frobenius map is surjective: $F$ is perfect if and only if either $char F = 0$ , or $char F = p > 0$ and the Frobenius map $a \mapsto a^{p}$ is surjective; iterating, every element of a perfect field of characteristic $p$ is a $p^{e}$ -th power for every $e \geq 0$ .

[F2]

Perfect fields: every irreducible polynomial is separable: a field is perfect when every algebraic extension of it is separable, equivalently (in characteristic $p$ ) when its Frobenius endomorphism is surjective.

[F3]

Pure inseparability and its conjugate, embedding, and separable-degree criteria: for algebraic $K / F$ of characteristic $p$ , $K / F$ is purely inseparable if and only if every $α \in K$ has $α^{p^{e}} \in F$ for some $e \geq 0$ ; in characteristic $0$ a purely inseparable extension is trivial.

[F4]

The binomial theorem over an arbitrary commutative ring and A prime $p$ divides $(\binom{p}{k})$ for $0 < k < p$ : in a commutative ring of characteristic $p$ one has $(u + v)^{p} = \sum_{k} (\binom{p}{k}) u^{k} v^{p - k} = u^{p} + v^{p}$ and hence $(u - v)^{p^{e}} = u^{p^{e}} - v^{p^{e}}$ for every $e \geq 0$ ; in a field $t^{p^{e}} = 0$ forces $t = 0$ .

[F5]

Field tests for geometric regularity: under the Axiom of Choice, a finite-type $k$ -algebra $A$ is geometrically regular over $k$ if and only if $A$ is regular and $A \otimes_{k} k^{'}$ is regular for every finite purely inseparable $k^{'} / k$ ; and then $A \otimes_{k} K$ is regular for every field extension $K / k$ .

[F6]

regular noetherian ring: a commutative Noetherian ring is regular when all its prime localisations are regular local rings.

Proof

1.1

Every finite purely inseparable extension of $k$ is trivial. Let $k^{'} / k$ be finite purely inseparable. If the characteristic is $0$ , then $k^{'} = k$ by [F3]. If the characteristic is $p > 0$ , fix $a \in k^{'}$ ; by [F3] there is $e \geq 0$ with $a^{p^{e}} \in k$ , and since the Frobenius map of $k$ is surjective by [F1] we may write $a^{p^{e}} = b^{p^{e}}$ with $b \in k$ . Then $(a - b)^{p^{e}} = a^{p^{e}} - b^{p^{e}} = 0$ by [F4], and $k^{'}$ is a field, so $a = b \in k$ . Hence $k^{'} = k$ .

F1F3F4

2.1

Applying the field tests. The algebra $A$ is regular by hypothesis, and by step 1.1 the only finite purely inseparable extension of $k$ is $k$ itself, for which $A \otimes_{k} k = A$ is regular. Hence [F5] makes $A$ geometrically regular over $k$ , and its second clause makes $A \otimes_{k} K$ regular for every field extension $K / k$ .

F5step 1.1F6

3.1

Thus a regular finite-type algebra over a perfect field is geometrically regular, and every scalar extension $A \otimes_{k} K$ is regular, whether or not $K$ is perfect or finitely generated; in characteristic $0$ the perfectness hypothesis is automatic by

F1F2∎

Depends on

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Sources

Stacks Algebra 10.45.3–4 (standard reference, not scraped)