Surface derivations and regular hypersurfaces

The derivation extends uniquely through localization. For a multiplicative set $S\subset T$, define $D(a\mathrm{/}s)=(sD(a)-aD(s))\mathrm{/}{s}^2$. The quotient rule is independent of the representative and satisfies the Leibniz rule, so it gives the unique extension to $S^{-1}T$.

Let $I\subseteq T$ be any ideal. The Leibniz rule gives $D(I^{n+1})\subseteq I^n$: differentiating a product of $n+1$ factors from $I$ leaves at least $n$ such factors in every term. For a compatible system , define $\widehat{D}(a{)}_n$ to be $D({\tilde{a}}_{n+1})<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>I^n$, where ${\tilde{a}}_{n+1}\in T$ lifts $a_{n+1}$. The containment just proved makes this independent of the lift and compatible in $n$. Applying the Leibniz rule modulo each $I^n$ shows that $\widehat{D}$ is a derivation; the defining formula also proves uniqueness.

Extend $D$ to $T[z]$ by setting $D(z)=0$. Let $h=z^r-f$ and let $\mathfrak{q}\subset T[z]$ be any prime containing $(h)$. The ambient local ring $B=T[z{]}_{\mathfrak{q}}$ is regular by [F4]. By step 1.1, the derivation extends from $T[z]$ to $B$, and in $B$ it satisfies $D(h)=-D(f)$, a unit. We keep this derivation on the ambient ring $B$; it is not asserted to descend to $B\mathrm{/}(h)$.

For the general criterion, localize $T$ at any prime containing $h$. If $h$ were in the square of that local maximal ideal, Leibniz would put $D(h)$ in the maximal ideal, contradicting its unit image modulo $h$. Thus $h$ is a parameter of the regular local ring, and [F2] makes its quotient regular. This holds at every prime of $T\mathrm{/}(h)$.

In the regular local ring $B$ of step 2.2, let $\mathfrak{n}$ be its maximal ideal. Since $h\in \mathfrak{n}$, suppose toward a contradiction that $h\in {\mathfrak{n}}^2$. Write $h$ as a finite sum of products of elements of $\mathfrak{n}$. The Leibniz rule then gives $D(h)\in \mathfrak{n}$, because in each differentiated product the undifferentiated factor lies in $\mathfrak{n}$. This contradicts the unit $D(h)=-D(f)$ from step 2.2. Thus $h\mathrm{\notin }{\mathfrak{n}}^2$; its class in $\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2$ is nonzero and it is part of a regular system of parameters of $B$.

The local ring of $S=T[z]\mathrm{/}(h)$ at the prime corresponding to $\mathfrak{q}$ is $B\mathrm{/}(h)$. By step 3.1, $h$ is a member of a regular system of parameters of the regular local ring $B$, so [F2] makes $B\mathrm{/}(h)$ regular. As this holds at every prime of $S$, the scheme $S$ is regular.

Hence $T[z]\mathrm{/}(z^r-f)$ is regular as a scheme; the sole choice hypothesis is AC, already included in the regular-quotient and parameter suppliers.