Free differentials imply regularity in characteristic zero

Setup. By [F1] and [F2], $R$ is a Noetherian local ring with residue field $\kappa$, and by [F3] the extension $\kappa \mathrm{/}k$ is finitely generated. Since $k$ has characteristic $0$, it is perfect by [F4], so by [F5] $\kappa \mathrm{/}k$ is separably generated; [F6] therefore gives the exact sequence $0\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\stackrel{\delta }{\to }{\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa \to {\mathrm{\Omega }}_{\kappa \mathrm{/}k}\to 0$ with $\delta ([x])=\mathrm{d}x\otimes 1$. Moreover $\dim R<\mathrm{\infty }$ by [F8]. The identification ${\mathrm{\Omega }}_{A\mathrm{/}k,\mathfrak{q}}\cong {\mathrm{\Omega }}_{R\mathrm{/}k}$ of [F7] makes the hypothesis say that $\mathrm{\Omega }:={\mathrm{\Omega }}_{R\mathrm{/}k}$ is a free $R$-module; It is finitely generated: differentials of a finite list of $k$-algebra generators of $A$ generate ${\mathrm{\Omega }}_{A\mathrm{/}k}$ by the polynomial product rule, and localization preserves finite generation by [F7]. Thus its free rank is finite; fix a basis $e_1,\dots ,e_r$ of $\mathrm{\Omega }$.

Induction claim. We prove by induction on the natural number $d=\dim R$ the assertion: for every finite-type $k$-algebra $A^{\mathrm{\prime }}$ and prime ${\mathfrak{q}}^{\mathrm{\prime }}$ with $A_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ Noetherian local of dimension $d$ and ${\mathrm{\Omega }}_{A^{\mathrm{\prime }}\mathrm{/}k,{\mathfrak{q}}^{\mathrm{\prime }}}$ free over $A_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$, the ring $A_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ is regular. The induction is over the two mutually exclusive cases for $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ analysed below; in the nonvanishing case a nonzerodivisor $f\in \mathfrak{m}$ will be produced whose existence forces $d\ge 1$ and permits descent to dimension $d-1$, and in the vanishing case regularity is obtained directly, so the case $d=0$ is covered as well.

The case $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2=0$. Then $\mathfrak{m}={\mathfrak{m}}^2$, and since $\mathfrak{m}\subseteq J(R)$ and $R$ is Noetherian with $M=R$ finite, the Krull intersection theorem [F12] gives $\mathfrak{m}\subseteq {\bigcap }_n{\mathfrak{m}}^n=0$. Hence $\mathfrak{m}=0$, so $R=\kappa$ is a field of dimension $0$ and embedding dimension $0$; by [F8] it is regular.

The case $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\mathrm{\ne }0$. Choose $f\in \mathfrak{m}$ with nonzero class in $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$; then $\delta ([f])=\mathrm{d}f\otimes 1\mathrm{\ne }0$ in $\mathrm{\Omega }\mathrm{/}\mathfrak{m}\mathrm{\Omega }$ by the injectivity of $\delta$ in step 1.1. Writing $\mathrm{d}f={\sum }_i{a}_i{e}_i$ in the basis of step 1.1, some coefficient $a_i$ lies outside $\mathfrak{m}$ and is therefore a unit of $R$; replacing $e_i$ by $\mathrm{d}f$ and keeping the other $e_j$ gives a second basis $\mathrm{d}f,e_j(j\mathrm{\ne }i)$ of $\mathrm{\Omega }$, since $e_i=a_i^{-1}(\mathrm{d}f-{\sum }_{j\mathrm{\ne }i}{a}_j{e}_j)$ and the $r$ elements are linearly independent by the unit coefficient. Define the $R$-linear map $\theta :\mathrm{\Omega }\to R$ by $\theta (\mathrm{d}f)=1$ and $\theta (e_j)=0$ for $j\mathrm{\ne }i$; the ring $R$ is a $\mathbb{Q}$-algebra because $k$ has characteristic $0$, so [F9] applies and shows that $f$ is a nonzerodivisor in $R$. Since $f\in \mathfrak{m}$, [F11] then gives $\dim (R\mathrm{/}fR)=\dim R-1$, so this case forces $\dim R\ge 1$.

The quotient $R\mathrm{/}fR$. Since $f$ is a nonzerodivisor in $\mathfrak{m}$, [F11] gives $\dim (R\mathrm{/}fR)=\dim R-1$, and $R\mathrm{/}fR\mathrm{\ne }0$. By [F10] applied to the ring map $k\to R$, the ideal $I=(f)$ and $B=R\mathrm{/}fR$, the sequence $(f)\mathrm{/}(f^2)\to (R\mathrm{/}fR){\otimes }_R\mathrm{\Omega }\to {\mathrm{\Omega }}_{(R\mathrm{/}fR)\mathrm{/}k}\to 0$ is exact, the first map sending the class of $f$ to $1\otimes \mathrm{d}f$; the first term is generated by the class of $f$ and its image is the cyclic submodule $(R\mathrm{/}fR)(\mathrm{d}f\otimes 1)$. Under the basis $\mathrm{d}f,e_j$ of step 2.2, the free module $(R\mathrm{/}fR){\otimes }_R\mathrm{\Omega }$ splits as $(R\mathrm{/}fR)(\mathrm{d}f\otimes 1)\oplus {⨁}_{j\mathrm{\ne }i}(R\mathrm{/}fR)(e_j\otimes 1)$, so the cokernel is free of rank $r-1$; that is, ${\mathrm{\Omega }}_{(R\mathrm{/}fR)\mathrm{/}k}\cong (R\mathrm{/}fR{)}^{r-1}$ and $(R\mathrm{/}fR,\mathfrak{m}\mathrm{/}fR)$ is a Noetherian local ring of dimension $d-1$ whose module of differentials at its maximal ideal is free.

Regularity is lifted and the induction closes. Write $f=a\mathrm{/}s$ with $a\in \mathfrak{q}$ and $s\in A\setminus \mathfrak{q}$, so $fR=aR$. Put $A^{\mathrm{\prime }}=A\mathrm{/}aA$ and let ${\mathfrak{q}}^{\mathrm{\prime }}\in \mathrm{Spec}{A}^{\mathrm{\prime }}$ be the prime with $A_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}\cong R\mathrm{/}fR$; by [F7] the localization of ${\mathrm{\Omega }}_{A^{\mathrm{\prime }}\mathrm{/}k}$ at ${\mathfrak{q}}^{\mathrm{\prime }}$ is ${\mathrm{\Omega }}_{(R\mathrm{/}fR)\mathrm{/}k}$, which step 3.1 shows is free of rank $r-1$. By step 1.2, whose induction hypothesis applies in dimension $d-1$, the ring $R\mathrm{/}fR$ is regular; then [F11] applied to the nonzerodivisor $f\in \mathfrak{m}$ makes $R$ regular. Steps 2.1 and 2.2 exhaust the two possibilities for $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, and the induction runs down from the finite value $\dim R$ of step 1.1, so every case is covered by steps 2.1, 3.1 and 4.1.