Regular and singular points of an analytic hypersurface

Fix $n\ge 1$ and a complex-analytic hypersurface germ $X$ at a point $p\in {\mathbb{C}}^n$, and fix a defining equation $f$ of $X$ together with a representative of $f$ on a connected open neighbourhood $U$ of $p$ (Complex-analytic hypersurface germ and its reduced equation). We keep the symbol $X$ for the corresponding representative zero set in $U$. Since the defining germ is nonzero, $f$ is not identically zero on $U$. The identity theorem implies that its germ at every $q\in U$ is nonzero (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically); at $q\in X$ it is also a nonunit. Thus $(X,q)$ is a hypersurface germ at every point under consideration.

For a point $q\in X$, write $I_q(X)$ for the vanishing ideal of $X$ at $q$: the ideal of germs $g\in {\mathcal{O}}_{{\mathbb{C}}^n,q}$ vanishing on $X$ near $q$. A local reduced equation of $X$ at $q$ is a germ $f_q\in {\mathcal{O}}_{{\mathbb{C}}^n,q}$ such that

$(f_q)=I_q(X)\text{and}{f}_q\text{ is reduced at }q\mathrm{.}$

Such equations exist and are well defined by the following two observations. First, taking the square-free reduction of the germ of any defining equation of $(X,q)$ produces a reduced germ generating $I_q(X)$, by the principal vanishing-ideal lemma applied with base point $q$ and by the square-free reduction lemma (The vanishing ideal of a reduced hypersurface germ is principal, Square-free reduction of a holomorphic equation). Second, if $f_q$ and $f_q^{\mathrm{\prime }}$ both generate the nonzero principal ideal $I_q(X)$ in the germ ring, then $f_q=u{f}_q^{\mathrm{\prime }}$ and $f_q^{\mathrm{\prime }}=v{f}_q$ for germs $u,v$, so $uv=1$ by cancellation in the integral domain ${\mathcal{O}}_{{\mathbb{C}}^n,q}$ (The ring of holomorphic germs is a UFD); thus $u$ and $v$ are units and any two local reduced equations differ by a unit (A germ is a unit exactly when its value at $0$ is nonzero, so ${\mathcal{O}}_{m,0}$ is local).

A point $q\in X$ is a regular point of $X$ when

$d{f}_q(q)\mathrm{\ne }0$

for one — equivalently, by the unit relation just noted, for every — local reduced equation $f_q$ of $X$ at $q$; here $d{f}_q(q)$ is the complex differential of the germ $f_q$ at its own base point $q$. A point that is not regular is a singular point of $X$. The regular locus $\mathrm{Reg}(X)$ and the singular locus $\mathrm{Sing}(X)$ are the subsets of $X$ consisting of its regular and of its singular points.

Independence of all choices. Let $f_q$ and $f_q^{\mathrm{\prime }}$ be local reduced equations of $X$ at $q$, with $f_q^{\mathrm{\prime }}=u{f}_q$ for a unit $u$ (A germ is a unit exactly when its value at $0$ is nonzero, so ${\mathcal{O}}_{m,0}$ is local). Since $f_q(q)=0$, the product rule gives

$d{f}_q^{\mathrm{\prime }}(q)=u(q)d{f}_q(q)+f_q(q)du(q)=u(q)d{f}_q(q),$

and $u(q)\mathrm{\ne }0$, so $d{f}_q^{\mathrm{\prime }}(q)$ vanishes exactly when $d{f}_q(q)$ does. Hence regularity at $q$ depends only on the set germ $X$: neither the global defining equation of $X$, nor the representative neighbourhood, nor the chosen local reduced equation enters the condition. In particular, if $X$ is described near $p$ by a reduced defining germ $f$ of Complex-analytic hypersurface germ and its reduced equation, then a local reduced equation at $q$ is the square-free reduction of the germ of $f$ at $q$, and the differential criterion can be tested with that germ.

A fixed equation near the base point. Let $f$ be reduced at $p$. Center at $p$ and choose an invertible complex-linear map $T$ so that the germ $F(z):=f(p+Tz)$ is regular in the last variable (After a linear coordinate change, every nonzero germ is regular in the last variable). Weierstrass preparation gives $F=uW$ for a unit $u$ and a Weierstrass polynomial $W$ (Weierstrass preparation theorem). The coordinate pullback is a ring isomorphism preserving irreducibles, so $F$ is reduced. If an irreducible square divided $W$, it would also divide $F=uW$; thus $W$ is reduced at the origin. Shrink so that $F=uW$ holds and $u$ is nowhere zero, hence the zero sets agree. After further shrinking to a product neighbourhood $V\times D$, the nearby-reduced lemma says that for every $q\in Z(W)\cap (V\times D)$ the germ of $W$ at $q$ is reduced (A reduced prepared hypersurface stays reduced nearby), and that germ generates $I_q(Z(W))$ by the principal vanishing-ideal lemma. So on that neighbourhood one fixed equation $W$ is a local reduced equation at every point of the hypersurface, and

$\mathrm{Sing}(X)=\{q\in Z(W)\cap (V\times D):dW(q)=0\}$

in the prepared coordinates. For $n\ge 2$ the condition $dW(q)=0$ is the system $W(q)=0$, ${\mathrm{\partial }}_{z_1}W(q)=\cdots ={\mathrm{\partial }}_{z_n}W(q)=0$ of holomorphic equations in $q$.

Biholomorphic invariance. Let $\mathrm{\Phi }:U\to U^{\mathrm{\prime }}$ be a biholomorphism of open sets and put $X^{\mathrm{\prime }}=\mathrm{\Phi }(X\cap U)$ near $q^{\mathrm{\prime }}=\mathrm{\Phi }(q)$. Since $\mathrm{\Phi }$ induces a ring isomorphism ${\mathcal{O}}_{{\mathbb{C}}^n,q^{\mathrm{\prime }}}\to {\mathcal{O}}_{{\mathbb{C}}^n,q}$, $g\mapsto g\circ \mathrm{\Phi }$, which preserves units and products, the local reduced equations of $X$ at $q$ correspond to those of $X^{\mathrm{\prime }}$ at $q^{\mathrm{\prime }}$: if $f_q$ is one for $X$, then $f_q\circ {\mathrm{\Phi }}^{-1}$ generates $I_{q^{\mathrm{\prime }}}(X^{\mathrm{\prime }})$ and is reduced. By the chain rule

$d(f_q\circ {\mathrm{\Phi }}^{-1})(q^{\mathrm{\prime }})=d{f}_q(q)\circ (D\mathrm{\Phi }(q){)}^{-1},$

and since $D\mathrm{\Phi }(q)$ is invertible the left side vanishes exactly when $d{f}_q(q)$ does. Hence regularity of points is a biholomorphic invariant; in particular, changing to the coordinates of the previous paragraph does not change the regular and singular loci. For $n=1$, near each $q\in X$ the zero germ is the singleton $\{q\}$; the square-free reduction is a unit multiple of $\zeta \mapsto \zeta -q$, whose differential is $1$. Thus every such point is regular.

Equivalence with a holomorphic graph. A subset $X$ of a domain in ${\mathbb{C}}^n$ is a holomorphic hypersurface graph near $q$ when, after relabelling the coordinates and shrinking to a product $A\times B\subseteq {\mathbb{C}}^{n-1}\times \mathbb{C}$ of polydiscs around $q$, there is a holomorphic function $\phi :A\to B$ with

$X\cap (A\times B)=\{(z^{\mathrm{\prime }},z_n):z_n=\phi (z^{\mathrm{\prime }})\}\mathrm{.}$

The point $q\in X$ is regular if and only if $X$ is a holomorphic hypersurface graph near $q$. For $n=1$ this follows from the singleton description above, viewed as a graph over ${\mathbb{C}}^0$. For $n\ge 2$, if $d{f}_q(q)\mathrm{\ne }0$, some partial derivative of $f_q$ at $q$ is nonzero; relabelling so that ${\mathrm{\partial }}_n{f}_q(q)\mathrm{\ne }0$, the holomorphic implicit function theorem applied to $f_q$ at $q$ gives polydiscs $A,B$ and a holomorphic $\phi :A\to B$ with $f_q(z^{\mathrm{\prime }},z_n)=0$ equivalent to $z_n=\phi (z^{\mathrm{\prime }})$ on $A\times B$ (The holomorphic implicit function theorem). Since the zero germ of $f_q$ is $X$ at $q$, this exhibits $X$ as a graph near $q$. Conversely, suppose that $X\cap (A\times B)$ is the graph of $\phi$, and put $G(z^{\mathrm{\prime }},z_n):=z_n-\phi (z^{\mathrm{\prime }})$. Then $G$ is holomorphic on $A\times B$, $Z(G)=X$ there, and $G(q)=0$. Its linear part at $q$ is $d{z}_n-{\sum }_{i<n}{\mathrm{\partial }}_i\phi (q^{\mathrm{\prime }})d{z}_i\mathrm{\ne }0$, so $G\mathrm{\notin }{\mathfrak{m}}_q^2$; a product of two nonunit germs lies in ${\mathfrak{m}}_q^2$, hence $G$ is not a product of two nonunits, that is, $G$ is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^n,q}$. An irreducible germ is reduced: if $r^2\mid G$ with $r$ irreducible, then $G=r\cdot (rh)$ with both factors nonunits, contradicting irreducibility. Therefore $G$ is a reduced germ whose zero germ is $X$ at $q$, so $I_q(X)=I_q(Z(G))=(G)$ by the principal vanishing-ideal lemma. Comparing with a local reduced equation $f_q$ of $X$, we get $f_q=uG$ for a unit $u$, and since $G(q)=0$,

$d{f}_q(q)=u(q)dG(q)\mathrm{\ne }0,$

because ${\mathrm{\partial }}_{n}G\equiv 1$. Hence $q$ is regular. This proves the claimed equivalence and shows that "regular point of a reduced hypersurface" is the coordinate-free notion of a smooth point of $X$.