Standard smooth algebras are finitely presented and flat

The encoded presentation. Let $Q:=R[x_1,\dots ,x_n,z]\mathrm{/}(f_1,\dots ,f_c,zg-1)$. The composite $P\to S$ sends $g$ to a unit of $S$, so by [F6] and [F7], and then [F8], there is a unique $R$-algebra homomorphism $Q\to S$ with $x_i\mapsto \overline{x_i}$ and $z\mapsto {\bar{g}}^{-1}$; here $z$ is the new variable and the relations $f_j\mapsto 0$, $zg-1\mapsto {\bar{g}}^{-1}\bar{g}-1=0$ hold. Conversely, the substitution $x_i\mapsto x_i$, $z\mapsto z$ gives a ring homomorphism $P\to Q$ whose kernel contains $I$ because each $f_j$ maps to $0$, and which sends $g$ to a unit of $Q$ with inverse $z$; hence [F7] and [F8] produce a unique $R$-algebra homomorphism $S=(P\mathrm{/}I{)}_g\to Q$ with $\bar{F}\mathrm{/}{g}^N\mapsto F{z}^N$. The two composites fix the generators $x_1,\dots ,x_n,z$ of $Q$ over $R$ and the generators $\overline{x_1},\dots ,\overline{x_n},{\bar{g}}^{-1}$ of $S$, so by the uniqueness clauses of [F6], [F7] and [F8] they are the respective identities; thus $S\cong Q$.

Reduction of flatness to the Noetherian local case. Assume first that $R$ is Noetherian. By [F10] it suffices to prove that $S{\otimes }_R{R}_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ for every prime $\mathfrak{p}\subseteq R$; by [F2] the algebra $S{\otimes }_R{R}_{\mathfrak{p}}$ is standard smooth over $R_{\mathfrak{p}}$ with the same $n,c$ and a minor that is still a unit. Hence it suffices to prove: if $(R,\mathfrak{m})$ is a Noetherian local ring and $S$ is standard smooth over $R$, then $S$ is flat over $R$.

Dependent choice is available. Given the Axiom of Choice [F33], let $X$ be a nonempty set with an entire relation $R$ and let $a\in X$; choosing an element of each nonempty subset of $X$ and setting $f(x)$ to be the chosen element of $\{y:xRy\}$ gives a function $X\to X$, so recursion produces $x\text{\hspace{0.17em}⁣}:\mathbb{N}\to X$ with $x_0=a$ and $x_{n}R{x}_{n+1}$. Thus the Dependent Choice supplier [F26] is available throughout this proof.

The local criterion, prepared. Let now $R$ be Noetherian local with maximal ideal $\mathfrak{m}$ and residue field $\kappa =R\mathrm{/}\mathfrak{m}$, and let $S=(P\mathrm{/}I{)}_g$ be standard smooth over $R$. Then $S$ is Noetherian: $P$ is Noetherian by [F21], $P\mathrm{/}I$ is Noetherian by [F20] and so is its localisation $S$ by [F20]. Fix a finitely generated ideal $J\subseteq R$ and put $K_J:=\ker (J{\otimes }_{R}S\to S)$; this is a finitely generated $S$-module by [F19], since $J{\otimes }_{R}S$ is a finitely generated $S$-module. For every maximal ideal $\mathfrak{n}\subseteq S$ the localisation $(K_J{)}_{\mathfrak{n}}$ equals the kernel of $J{\otimes }_R{S}_{\mathfrak{n}}\to S_{\mathfrak{n}}$, by [F27] and [F14] applied to the localisation $S\to S_{\mathfrak{n}}$. Hence if $S_{\mathfrak{n}}$ is flat over $R$, then $(K_J{)}_{\mathfrak{n}}=0$; and if that holds for every maximal $\mathfrak{n}$, then $K_J=0$ by [F17] and [F18]. Therefore, by [F11], it is enough to prove that $S_{\mathfrak{n}}$ is flat over $R$ for every maximal ideal $\mathfrak{n}\subseteq S$.

Set-up at the contracted prime. Fix a maximal ideal $\mathfrak{n}\subseteq S$ and put $\mathfrak{r}=\mathfrak{n}\cap R$. Before the local computation, replace the base and presentation by $R_{\mathfrak{r}}$ and $S{\otimes }_R{R}_{\mathfrak{r}}$, using [F2]. The ideal $\mathfrak{n}$ induces a maximal ideal there with the same local ring $S_{\mathfrak{n}}$. In steps 1.6, 2.2, 3.1 and 4.1 only, write $R,\mathfrak{m},\kappa ,P$ for this localized base, its maximal ideal and residue field, and its polynomial ring. Let ${\mathfrak{q}}^{\mathrm{\prime }}\subseteq P$ be the prime over $\mathfrak{n}$, so ${\mathfrak{q}}^{\mathrm{\prime }}\cap R=\mathfrak{m}$, $I\subseteq {\mathfrak{q}}^{\mathrm{\prime }}$ and $g\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$. Put $T=P_{{\mathfrak{q}}^{\mathrm{\prime }}}$ and $N_i=T\mathrm{/}(f_1,\dots ,f_i)T$; then $N_c=S_{\mathfrak{n}}$ by [F30]. The module $T$ is flat over this base: tensoring an injection with the free $R$-module $P$ preserves injectivity by [F13], and localizing the result preserves it by [F27], with the tensor identifications of [F14, F16]. Each $N_i$ is Noetherian local, and $\mathfrak{m}T\subseteq {\mathfrak{q}}^{\mathrm{\prime }}T$ makes $R\to N_i$ local. Once the computation proves $N_c$ flat over $R_{\mathfrak{r}}$, it is flat over the original base by clause 2 of Every localization is flat, and localizing a flat module preserves flatness.

The fibre at $\kappa$. The quotient map $P\to P\mathrm{/}(f_1,\dots ,f_i)$ tensored with $\kappa$ has cokernel $\kappa [x_1,\dots ,x_n]\mathrm{/}({\bar{f}}_1,\dots ,{\bar{f}}_i)$ by [F15]; localising this at the prime ${\bar{\mathfrak{q}}}^{\mathrm{\prime }}$ induced by ${\mathfrak{q}}^{\mathrm{\prime }}$ and applying [F14] twice together with [F16] gives $N_i{\otimes }_R\kappa \cong \kappa [x_1,\dots ,x_n{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}\mathrm{/}({\bar{f}}_1,\dots ,{\bar{f}}_i)$, where ${\bar{f}}_j$ denotes the image of $f_j$ in $\kappa [x_1,\dots ,x_n]$.

$\mathbb{Z}$ is Noetherian. Every ideal of $\mathbb{Z}$ is an additive subgroup, hence generated by one element by [F23] and therefore finitely generated; so $\mathbb{Z}$ is Noetherian by [F22].

The unit witness for the minor. Since $h$ maps to a unit of $S=(P\mathrm{/}I{)}_g$, there are $u\in P$ and $N\ge 0$ with $\bar{h}\cdot \bar{u}\mathrm{/}{\bar{g}}^N=1$ in $S$, that is, $\overline{hu-g^N}=0$ in $(P\mathrm{/}I{)}_g$; by [F29] applied to the localisation $(P\mathrm{/}I)\to S$ there is $M\ge 0$ with $g^M(hu-g^N)\in I$. Putting $w:=g^{M}u$ and $N^{\mathrm{\prime }}:=M+N$ gives $hw-g^{N^{\mathrm{\prime }}}\in I$, so there are $m_1,\dots ,m_c\in P$ with $hw-g^{N^{\mathrm{\prime }}}={\sum }_{j=1}^c{m}_j{f}_j$ in $P$.

Finite presentation. By step 1.1 the $R$-algebra $S$ is isomorphic to the quotient of the polynomial ring $R[x_1,\dots ,x_n,z]$ by the ideal generated by the finitely many elements $f_1,\dots ,f_c,zg-1$; by [F5] this exhibits $S$ as a finitely presented $R$-algebra.

The minor survives in the fibre. The element $h$ maps to a unit of $S$, hence its image in the localisation $S_{\mathfrak{n}}=N_c$ is a unit, hence its image in $N_c{\otimes }_R\kappa$ is a unit. By step 1.6 with $i=c$ this ring is $\kappa [x_1,\dots ,x_n{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}\mathrm{/}({\bar{f}}_1,\dots ,{\bar{f}}_c)$, and the image of $h$ there is the image of $\bar{h}$; since a unit of a local ring does not lie in the maximal ideal, $\bar{h}\mathrm{\notin }{\bar{\mathfrak{q}}}^{\mathrm{\prime }}$. So [F3] applies over the field $\kappa$: the images ${\bar{f}}_1,\dots ,{\bar{f}}_c$ form a regular sequence in the regular local ring $\kappa [x_1,\dots ,x_n{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}$, and consequently ${\bar{f}}_{i+1}$ is a nonzerodivisor on $\kappa [x_1,\dots ,x_n{]}_{{\bar{\mathfrak{q}}}^{\mathrm{\prime }}}\mathrm{/}({\bar{f}}_1,\dots ,{\bar{f}}_i)$ for every $i<c$. By step 1.6 this says that $f_{i+1}$ is a nonzerodivisor on $N_i{\otimes }_R\kappa$ for every $0\le i<c$.

Descent to a finitely generated subring. Let $R_0\subseteq R$ be the $\mathbb{Z}$-subalgebra generated by the finitely many coefficients occurring in the polynomials $f_1,\dots ,f_c$, $g$, $h$, $w$, $m_1,\dots ,m_c$. Then $R_0$ is a finitely generated $\mathbb{Z}$-algebra, hence Noetherian by steps 1.7 and [F24]. The polynomial identity of step 1.8 has all its coefficients in $R_0$, and $R_0[x_1,\dots ,x_n]\to R[x_1,\dots ,x_n]$ is injective by [F34], so $hw-g^{N^{\mathrm{\prime }}}={\sum }_j{m}_j{f}_j$ holds already in $R_0[x_1,\dots ,x_n]$; therefore $S_0:=(R_0[x_1,\dots ,x_n]\mathrm{/}(f_1,\dots ,f_c){)}_g$ is a standard smooth $R_0$-algebra with the same $n,c,f_j,g$ and the minor $h$ a unit.

Induction: a fibre nonzerodivisor lifts across a flat map. Suppose $N_i$ is flat over $R$ for some $i<c$, put $J:=\mathfrak{m}{N}_i$, and let $f:=f_{i+1}$. By step 2.2, multiplication by the image of $f$ on $N_i\mathrm{/}J$ is injective. For every $r\ge 0$, flatness of $N_i$ over $R$ applied to $0\to {\mathfrak{m}}^{r+1}\to {\mathfrak{m}}^r\to {\mathfrak{m}}^r\mathrm{/}{\mathfrak{m}}^{r+1}\to 0$ gives the natural isomorphism $J^r\mathrm{/}{J}^{r+1}\cong ({\mathfrak{m}}^r\mathrm{/}{\mathfrak{m}}^{r+1}){\otimes }_{\kappa }(N_i\mathrm{/}J),\kappa =R\mathrm{/}\mathfrak{m}\mathrm{.}$ After choosing a $\kappa$-basis of ${\mathfrak{m}}^r\mathrm{/}{\mathfrak{m}}^{r+1}$, this is a direct sum of copies of $N_i\mathrm{/}J$, so multiplication by $f$ is injective on each graded piece $J^r\mathrm{/}{J}^{r+1}$. If $fx=0$ in $N_i$, induction on $r$ now gives $x\in J^r$ for every $r$: injectivity modulo $J$ starts the induction, and injectivity on $J^r\mathrm{/}{J}^{r+1}$ advances it. The ring $N_i$ is Noetherian local and $J$ lies in its maximal ideal by step 1.5, so [F35] gives ${\bigcap }_{r\ge 0}{J}^r=0$ and therefore $x=0$. Thus $f_{i+1}$ is a nonzerodivisor on $N_i$, and $0\to N_i\stackrel{f_{i+1}}{\to }{N}_i\to N_{i+1}\to 0$ is a short exact sequence.

Induction: the Tor vanishing and flatness pass to the next quotient. In the situation of step 3.1 the long exact Tor sequence of [F26] for $0\to N_i\stackrel{f_{i+1}}{\to }{N}_i\to N_{i+1}\to 0$, together with ${\mathrm{Tor}}_1^R(\kappa ,N_i)=0$ from the flatness of $N_i$, exhibits ${\mathrm{Tor}}_1^R(\kappa ,N_{i+1})$ as the kernel of $N_i{\otimes }_R\kappa \stackrel{f_{i+1}}{\to }{N}_i{\otimes }_R\kappa$, which is zero by step 2.2. The ring $N_{i+1}$ is a Noetherian local ring, $R\to N_{i+1}$ is a local homomorphism by step 1.5 and $N_{i+1}$ is a finite $N_{i+1}$-module, so [F4] gives that $N_{i+1}$ is flat over $R$.

Conclusion of the induction and of the local case. Steps 3.1 and 4.1, starting with $N_0$ of step 1.5, prove that every $N_i$ is flat over the localized base $R_{\mathfrak{r}}$. In particular the original local ring $S_{\mathfrak{n}}$ is flat over $R_{\mathfrak{r}}$, hence over the original $R$ by step 1.5. As $\mathfrak{n}$ was arbitrary, step 1.4 gives that $S$ is flat over the original base. This proves flatness over every Noetherian local base, and step 1.2 proves it over every Noetherian base.

Flatness of $S_0$ and base change back to $R$. By step 2.3 the algebra $S_0$ is standard smooth over the Noetherian ring $R_0$, so $S_0$ is flat over $R_0$ by step 5.1; and by [F2] applied to the ring map $R_0\to R$ there is an $R$-algebra isomorphism $R{\otimes }_{R_0}{S}_0\cong (R[x_1,\dots ,x_n]\mathrm{/}(f_1,\dots ,f_c){)}_g=S$. Hence $S\cong R{\otimes }_{R_0}{S}_0$ is flat over $R$ by [F28]. This proves the second assertion for arbitrary $R$, and step 2.1 proves the first.