Locally standard smooth iff flat with geometrically regular fibres

Set-up and conventions. Write $F:=S{\otimes }_R\kappa (\mathfrak{p})$ for the fibre over $\mathfrak{p}$; by [F5] the fibre is defined because $R\to S$ is finitely presented, and "geometrically regular at $\mathfrak{q}$" means that for every field extension $K\mathrm{/}\kappa (\mathfrak{p})$ and every prime $Q$ of $F_K=F{\otimes }_{\kappa (\mathfrak{p})}K$ lying over the image of $\mathfrak{q}$ in $F$, the local ring $(F_K{)}_Q$ is regular. Since a standard smooth chart at $\mathfrak{q}$ is by definition a standard smooth presentation of some $S_h$, $h\mathrm{\notin }\mathfrak{q}$ [F1], and since $S_h$ is finitely presented over $R$ when $S$ is [F2, F12], both sides of clause 1 only concern finitely presented $R$-algebras.

Flatness from a principal cover. Suppose $g_1,\dots ,g_k\in S$ generate the unit ideal of $S$ and each $S_{g_i}$ is flat over $R$. Then $S$ is flat over $R$. Indeed, let $I\subseteq R$ be an ideal and let $K:=\ker (I{\otimes }_{R}S\to S)$; localising the map at $g_i$ gives the map $I{\otimes }_R{S}_{g_i}\to S_{g_i}$ by [F10], which is injective because $S_{g_i}$ is flat over $R$, so $K_{g_i}=0$ for every $i$ by [F10]. If $K\mathrm{\ne }0$, then $K_{\mathfrak{m}}\mathrm{\ne }0$ for some maximal ideal $\mathfrak{m}\subseteq S$ by [F8], and since the $g_i$ generate the unit ideal some $g_i\mathrm{\notin }\mathfrak{m}$; then $K_{\mathfrak{m}}=(K_{g_i}{)}_{\mathfrak{m}}=0$ by [F10], a contradiction. Hence $K=0$, so every such multiplication map is injective and $S$ is flat over $R$ by [F8].

Clause 1, only-if: flatness at $\mathfrak{q}$. Assume $R\to S$ is standard smooth at $\mathfrak{q}$, and choose $g\mathrm{\notin }\mathfrak{q}$ such that $A:=S_g$ has a standard smooth presentation over $R$ [F1]. Then $A$ is flat over $R$ by [F2], so for every ideal $I\subseteq R$ the map $I{\otimes }_{R}A\to A$ is injective by [F8]; the localisation $S_{\mathfrak{q}}=T^{-1}A$ at $T=S\setminus \mathfrak{q}$ is a localisation of the $R$-module $A$, so $I{\otimes }_R{S}_{\mathfrak{q}}\to S_{\mathfrak{q}}$ is the localisation of that injective map and is injective by [F10]; hence $S_{\mathfrak{q}}$ is flat over $R$ by [F8]. Because $R\setminus \mathfrak{p}$ maps into $T$, the ring $S_{\mathfrak{q}}$ is an $R_{\mathfrak{p}}$-algebra, so [F9] upgrades flatness over $R$ to flatness over $R_{\mathfrak{p}}$: the local homomorphism $R_{\mathfrak{p}}\to S_{\mathfrak{q}}$ is flat.

Clause 3, if direction. Conversely let $k\to A$ be locally standard smooth, and let $\mathfrak{q}\in \mathrm{Spec}A$; choose $g\mathrm{\notin }\mathfrak{q}$ with $A_g$ standard smooth over $k$ [F1]. For every field extension $K\mathrm{/}k$, [F3] applied to the standard smooth $k$-algebra $A_g$ with $\mathfrak{p}=(0)$ and fibre $(A_g{\otimes }_{k}k){\otimes }_{k}K=A_g{\otimes }_{k}K$ shows that every local ring of $A_g{\otimes }_{k}K$ is regular; a prime $Q\subseteq A{\otimes }_{k}K$ lying over $\mathfrak{q}$ does not contain $\bar{g}$, and [F10] identifies $(A{\otimes }_{k}K{)}_Q$ with the local ring of $A_g{\otimes }_{k}K$ at the corresponding prime, so it is regular. As $K$ was arbitrary, $A$ is geometrically regular at $\mathfrak{q}$ in the sense of [F5]; in particular, taking $K$ finitely generated over $k$, the finite-type $K$-algebra $A{\otimes }_{k}K$ has all its prime localisations regular, so it is a regular Noetherian ring by [F11] and $A$ is geometrically regular over $k$.

Clause 1, only-if: geometric regularity of the fibre at $\mathfrak{q}$. Keep the chart $A=S_g$ of step 1.3. Since $R\to S$ is finitely presented, so is the coefficient extension $R\to A$, and $A{\otimes }_R\kappa (\mathfrak{p})\cong (S{\otimes }_R\kappa (\mathfrak{p}){)}_g=F_g$ by [F10]; write $\bar{g}$ for the image of $g$ in $F$, so $\bar{g}\mathrm{\notin }{\mathfrak{q}}_F$, where ${\mathfrak{q}}_F\subseteq F$ is the image of $\mathfrak{q}$. Let $K\mathrm{/}\kappa (\mathfrak{p})$ be a field extension and let $Q\subseteq F_K$ be a prime lying over ${\mathfrak{q}}_F$; then $\bar{g}\mathrm{\notin }Q$, and [F10] identifies $(F_K{)}_Q$ with the local ring of $(F_K{)}_{\bar{g}}=(A{\otimes }_R\kappa (\mathfrak{p})){\otimes }_{\kappa (\mathfrak{p})}K$ at the corresponding prime. That local ring is regular by [F3] applied to the standard smooth $R$-algebra $A$ and the extension $K\mathrm{/}\kappa (\mathfrak{p})$. Since $K$ and $Q$ were arbitrary, $F$ is geometrically regular at $\mathfrak{q}$ by [F5].

Clause 1, if direction. If $R_{\mathfrak{p}}\to S_{\mathfrak{q}}$ is flat and the fibre $F$ is geometrically regular at $\mathfrak{q}$, then [F4] produces $g\mathrm{\notin }\mathfrak{q}$ with $S_g$ standard smooth over $R$, that is, $R\to S$ is standard smooth at $\mathfrak{q}$. Steps 1.3 and 2.1 give the converse, so clause 1 holds.

Clause 2, only-if. Assume $R\to S$ is locally standard smooth. For each $\mathfrak{q}\in \mathrm{Spec}S$ choose $g_{\mathfrak{q}}\mathrm{\notin }\mathfrak{q}$ with $S_{g_{\mathfrak{q}}}$ standard smooth over $R$ [F1]; the open sets $D(g_{\mathfrak{q}})$ cover the affine, hence quasi-compact, scheme $\mathrm{Spec}S$ [F14], so finitely many of them, say for ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_k$, already cover, and their elements $g_1,\dots ,g_k$ generate the unit ideal of $S$. Each $S_{g_i}$ is flat over $R$ by [F2], so $S$ is flat over $R$ by step 1.2. For the fibres, let $\mathfrak{p}\in \mathrm{Spec}R$ and let $\mathfrak{Q}\subseteq \mathrm{Spec}F$ be a point of the fibre, with image $\mathfrak{q}\in \mathrm{Spec}S$; choosing the chart at that $\mathfrak{q}$ and applying step 2.1 shows that the local ring of the fibre at the prime corresponding to $\mathfrak{Q}$ — after any field extension of $\kappa (\mathfrak{p})$ — is regular, so $F$ is geometrically regular at $\mathfrak{q}$ and hence the whole fibre over $\mathfrak{p}$ is geometrically regular by [F5]. As $\mathfrak{p}$ was arbitrary, $R\to S$ is flat with geometrically regular fibres.

Clause 2, if direction. Assume $R\to S$ is flat and every fibre $S{\otimes }_R\kappa (\mathfrak{p})$ is geometrically regular. Fix $\mathfrak{q}\in \mathrm{Spec}S$, $\mathfrak{p}=\mathfrak{q}\cap R$. Flatness of $S$ over $R$ localises: $S_{\mathfrak{q}}$ is flat over $R$ by [F9, F10] applied to the localisation of the flat $R$-module $S$, hence flat over $R_{\mathfrak{p}}$ by [F9] since $S_{\mathfrak{q}}$ is an $R_{\mathfrak{p}}$-module. The fibre condition is exactly hypothesis 2 of [F4] at $\mathfrak{q}$, because a fibre that is geometrically regular at each of its points is geometrically regular at $\mathfrak{q}$ [F5]. So [F4] gives $g\mathrm{\notin }\mathfrak{q}$ with $S_g$ standard smooth over $R$. As $\mathfrak{q}$ was arbitrary, $R\to S$ is locally standard smooth.

Clause 3, only-if. Let $k$ be a field and $A$ a finite-type, hence finitely presented [F16], $k$-algebra that is geometrically regular over $k$; then the local homomorphism $k\to A_{\mathfrak{q}}$ is flat for every prime $\mathfrak{q}\subseteq A$ because every $k$-module is flat [F7, F9], and the only prime of $k$ is $(0)$ with residue field $k$, so the fibre is $A{\otimes }_{k}k\cong A$. By [F6] the geometric regularity of $A$ over $k$ makes $A{\otimes }_{k}K$ a regular ring for every field extension $K\mathrm{/}k$, not only the finitely generated ones; by [F11] this says precisely that every local ring of $(A{\otimes }_{k}k){\otimes }_{k}K$ at a prime lying over a given prime of $A$ is regular, so $A$ is geometrically regular at every prime in the sense of [F5]. Clause 1 (step 3.1) then gives a standard smooth chart of $A$ over $k$ at every prime, that is, $k\to A$ is locally standard smooth.

The relative-dimension clause of clause 3. Let $\mathfrak{q}\in \mathrm{Spec}A$ be a $k$-rational point of the finite-type $k$-algebra $A$, that is $A\mathrm{/}\mathfrak{q}=k$ as $k$-algebras, and let $A_h\cong (k[x_1,\dots ,x_n]\mathrm{/}(f_1,\dots ,f_c){)}_g$, $h\mathrm{\notin }\mathfrak{q}$, be a standard smooth chart of $A$ over $k$ at $\mathfrak{q}$ with leading $c\times c$ minor a unit of the localisation [F1]. Write $Q^{\mathrm{\prime }}\subseteq k[x_1,\dots ,x_n]$ for the prime corresponding to $\mathfrak{q}$. The composite $k[x_1,\dots ,x_n]\to A_h\to A_{\mathfrak{q}}\to A_{\mathfrak{q}}\mathrm{/}\mathfrak{q}{A}_{\mathfrak{q}}=k$ is a $k$-algebra map whose kernel is $Q^{\mathrm{\prime }}$, so $k[x]\mathrm{/}{Q}^{\mathrm{\prime }}$ is a $k$-subalgebra of the field $k$ containing the image of $k$, hence equal to $k$; thus $Q^{\mathrm{\prime }}$ is maximal and $\mathrm{ht}(Q^{\mathrm{\prime }})=n$ by [F15]. Applying [F3] to the standard smooth $k$-algebra $A_h$ with $\mathfrak{p}=(0)$, $K=k$ and the local ring $(A_h{)}_{\mathfrak{q}}=A_{\mathfrak{q}}$ gives $\dim A_{\mathfrak{q}}=\mathrm{ht}(Q^{\mathrm{\prime }})-c=n-c$, which is the relative dimension of the chart, in the situation of step 1.4.