Field tests for geometric regularity

The forward implication of clause 1 is immediate: taking $K=k$ in [F1] exhibits $A$ itself as regular, and every finite purely inseparable $k^{\mathrm{\prime }}\mathrm{/}k$ is a finitely generated field extension, so [F1] makes $A{\otimes }_k{k}^{\mathrm{\prime }}$ regular.

A base-change fact used below: if $B$ is a regular Noetherian $k$-algebra and $K\mathrm{/}k$ is a separably generated field extension, then $B{\otimes }_{k}K$ is regular. Indeed, choose a separating transcendence basis $t_1,\dots ,t_s$ and put $F:=k(t_1,\dots ,t_s)$, so that $K\mathrm{/}F$ is finite separable by [F5] and $K=F(\gamma )$ by [F6] with minimal polynomial $P\in F[T]$ satisfying $\gcd (P,P^{\mathrm{\prime }})=1$ by [F8]. The ring $B_1:=B{\otimes }_{k}F$ is a localisation of the polynomial extension $B[t_1,\dots ,t_s]$, hence regular by [F4]; let $q$ be a prime of $B_2:=B{\otimes }_{k}K=B_1{\otimes }_{F}K$ and $p:=q\cap B_1$, so that $(B_1{)}_p\to (B_2{)}_q$ is a flat local homomorphism of Noetherian local rings whose closed fibre is a localisation of $\kappa (p){\otimes }_{F}K\cong \kappa (p)[T]\mathrm{/}(\bar{P})$ by [F10]. Reducing the Bezout identity for $\gcd (P,P^{\mathrm{\prime }})=1$ modulo $p$ shows $\gcd (\bar{P},{\bar{P}}^{\mathrm{\prime }})=1$, so $\bar{P}$ is separable by [F8] and factors into pairwise distinct irreducibles by [F9], whence $\kappa (p)[T]\mathrm{/}(\bar{P})$ is a product of fields and its further localisation is a field, in particular regular; [F7] then makes $(B_2{)}_q$ regular. As $q$ was arbitrary, $B_2$ is regular by [F2].

The finite purely inseparable test implies the finitely generated case of clause 1: assume $A$ is regular and $A{\otimes }_k{k}^{\mathrm{\prime }}$ is regular for every finite purely inseparable $k^{\mathrm{\prime }}\mathrm{/}k$, and let $K\mathrm{/}k$ be finitely generated. By [F3] there are finite purely inseparable extensions $k^{\mathrm{\prime }}\mathrm{/}k$ and $K^{\mathrm{\prime }}\mathrm{/}K$ with $K^{\mathrm{\prime }}\mathrm{/}{k}^{\mathrm{\prime }}$ separably generated (in characteristic $0$ take both extensions trivial). The ring $A{\otimes }_k{k}^{\mathrm{\prime }}$ is regular by hypothesis, so step 1.2 applied to $B:=A{\otimes }_k{k}^{\mathrm{\prime }}$ and the separably generated extension $K^{\mathrm{\prime }}\mathrm{/}{k}^{\mathrm{\prime }}$ makes $A{\otimes }_k{K}^{\mathrm{\prime }}=(A{\otimes }_k{k}^{\mathrm{\prime }}){\otimes }_{k^{\mathrm{\prime }}}{K}^{\mathrm{\prime }}$ regular. Since $K^{\mathrm{\prime }}\mathrm{/}K$ is finite purely inseparable, the field extension $K\to K^{\mathrm{\prime }}$ is faithfully flat and local, and for every prime $\mathfrak{q}$ of $A{\otimes }_{k}K$ and every prime ${\mathfrak{q}}^{\mathrm{\prime }}$ of $A{\otimes }_k{K}^{\mathrm{\prime }}$ over it the induced map $(A{\otimes }_{k}K{)}_{\mathfrak{q}}\to (A{\otimes }_k{K}^{\mathrm{\prime }}{)}_{{\mathfrak{q}}^{\mathrm{\prime }}}$ is a flat local homomorphism of Noetherian local rings with regular target, so [F7] descends regularity; as $\mathfrak{q}$ was arbitrary, $A{\otimes }_{k}K$ is regular by [F2]. With step 1.1 this proves clause 1, since [F1] defines geometric regularity by the regular rings $A{\otimes }_{k}K$ over finitely generated $K\mathrm{/}k$.

Arbitrary field extensions. Assume $A$ is geometrically regular over $k$ and let $K\mathrm{/}k$ be any field extension, written as the filtered union of its finitely generated subextensions $K_i$. Then $A{\otimes }_{k}K={\bigcup }_{i}A{\otimes }_k{K}_i$ is a filtered union, and each $A{\otimes }_k{K}_i$ is regular by clause 1 as proved in step 2.1. Let $\mathfrak{q}$ be a prime of $A{\otimes }_{k}K$ and put $R:=(A{\otimes }_{k}K{)}_{\mathfrak{q}}$ and $R_i:=(A{\otimes }_k{K}_i{)}_{{\mathfrak{q}}_i}$ with ${\mathfrak{q}}_i:=\mathfrak{q}\cap (A{\otimes }_k{K}_i)$, so that $R={\bigcup }_i{R}_i$ is a filtered union of regular local rings. There is a uniform bound on dimensions: for $A\mathrm{\ne }0$, if $A$ is module-finite over $k[z_1,\dots ,z_d]$ by [F12], then base change along $k\to K_i$ presents $A{\otimes }_k{K}_i$ as a quotient of a finite free $K_i[z_1,\dots ,z_d]$-module by [F10], so it is integral over $K_i[z_1,\dots ,z_d]$ and has dimension $d$ by [F13]; hence $\dim R_i\le d$ by [F14]. For $A=0$ the ring is the zero ring, regular by [F2].

The filtered union is regular. In the notation of step 3.1 let $M$ be a finitely generated $R$-module; presenting $M$ by finitely many generators and relations, all structure constants lie in some stage, so $M\cong M_i{\otimes }_{R_i}R$ for a finitely generated $R_i$-module $M_i$ and some $i$. By [F11] the regular local ring $R_i$ of dimension at most $d$ has $\mathrm{gldim}{R}_i=\dim R_i\le d$, so $M_i$ has a projective resolution of length at most $d$ by [F15]; tensoring it with $R$, which is flat over $R_i$ because it is a localisation of the flat base change $R_i\to R_i{\otimes }_{K_i}K$ by [F16], gives an exact sequence of projective $R$-modules of length at most $d$ (a direct summand of a free module stays such after tensoring), hence ${\mathrm{pd}}_{R}M\le d$ by [F15]. As every finitely generated $R$-module has finite projective dimension, [F11] makes $R$ regular; therefore $A{\otimes }_{k}K$ is regular by [F2], which is clause 2.

Descent, clause 3. Let $K\mathrm{/}k$ be a field extension with $A{\otimes }_{k}K$ geometrically regular over $K$, and let $E\mathrm{/}k$ be a finitely generated field extension; we show $A{\otimes }_{k}E$ is regular. The $k$-algebra $E{\otimes }_{k}K$ is nonzero, so choose a prime of it and let $L$ be its residue field, a field receiving both $E$ and $K$ for which $L\mathrm{/}K$ is a finitely generated field extension. Then $A{\otimes }_{k}L=(A{\otimes }_{k}K){\otimes }_{K}L=(A{\otimes }_{k}E){\otimes }_{E}L$ is regular by clause 2, applied over $K$ to the geometrically regular $K$-algebra $A{\otimes }_{k}K$; and $A{\otimes }_{k}E\to A{\otimes }_{k}L$ is faithfully flat and local at corresponding primes because $E\to L$ is a field extension, so [F7] descends regularity to every local ring of $A{\otimes }_{k}E$, making it regular by [F2]. As $E$ was arbitrary, $A$ is geometrically regular over $k$ by [F1].

Clause 1 is step 2.1, clause 2 is step 4.1 and clause 3 is step 5.1; the zero algebra is covered by the vacuous regularity of [F2]. ∎