Singular locus of a reduced analytic hypersurface

Since $f$ is reduced we may apply [F3]: choose the invertible complex-linear change $T$, the unit $u$ and the Weierstrass polynomial $W$ of degree $d\ge 1$ with $\tilde{f}\circ T=uW$, and the product neighbourhood $V\times D$ on which the zero sets agree and every slice has all $d$ roots inside $D$ and none on its boundary; the affine pullback $h\mapsto (z\mapsto h(p+Tz))$ is a ring isomorphism from ${\mathcal{O}}_{{\mathbb{C}}^n,p}$ to ${\mathcal{O}}_{{\mathbb{C}}^n,0}$ and carries irreducibles to irreducibles, so $\tilde{f}\circ T$ is again reduced and is the reduced equation in the prepared coordinates. Replace the representative of $X$ by $Z(W)\cap (V\times D)$ and write $\pi (q)=z^{\mathrm{\prime }}$ for the base coordinate of a point $q=(z^{\mathrm{\prime }},T)$.

For every $q\in Z(W)\cap (V\times D)$ the germ $W_q$ of $W$ at $q$ is reduced by [F5], and $Z(W_q)$ is the germ of $Z(W)$, namely $X$, at $q$; applying [F6] with base point $q$ and $g=W_q$ gives $I_q(X)=(W_q)$. Hence for every $q\in X$ the fixed germ $W_q$ is a local reduced equation of $X$ at $q$, and by [F2] the point $q$ is regular exactly when $dW(q)\mathrm{\ne }0$ and singular exactly when $dW(q)=0$.

Let $z_0^{\mathrm{\prime }}\in V$ and let $\tau \in \mathbb{C}$ with $q=(z_0^{\mathrm{\prime }},\tau )\in Z(W)$. By [F8] the partial derivative ${\mathrm{\partial }}_{T}W(q)$ is the derivative at $\tau$ of the one-variable slice $W(z_0^{\mathrm{\prime }},\cdot )$ and equals the value of the formal derivative of that monic polynomial, so ${\mathrm{\partial }}_{T}W(q)=0$ exactly when this particular root $\tau$ is repeated. Thus ${\mathrm{\partial }}_{T}W(q)=0$ implies $D_W(z_0^{\mathrm{\prime }})=0$ by [F4]; conversely, $D_W(z_0^{\mathrm{\prime }})=0$ means that some root of the slice is repeated. Therefore $D_W(z_0^{\mathrm{\prime }})\mathrm{\ne }0$ exactly when all $d$ roots of the slice are distinct and simple.

Let $q=(z_0^{\mathrm{\prime }},\tau )\in Z(W)\cap (V\times D)$. The slice $W(z_0^{\mathrm{\prime }},\cdot )$ is a monic polynomial of degree $d$ vanishing at $\tau$, so the germ $W_q$ is regular in the last variable of some order $k$ with $1\le k\le d$; translate $q$ to the origin and let $A:={\mathcal{O}}_{{\mathbb{C}}^{n-1},z_0^{\mathrm{\prime }}}$, $K:=\mathrm{Frac}(A)$. By [F9] the Weierstrass preparation $W_q=vP$ has $P$ square-free over $K$ and $D_P\mathrm{\ne }0$ in $A$. Since $D_P\mathrm{\ne }0$, the roots of $P$ in any field in which $P$ splits are pairwise distinct by [F10]; a repeated root of $P$ in an extension field would therefore give a contradiction, so $P$ is separable over $K$, and the separability criterion in [F10] gives $\gcd (P,P^{\mathrm{\prime }})=1$ in $K[T]$. By [F11] choose $A_0,B_0\in K[T]$ with $A_{0}P+B_0{P}^{\mathrm{\prime }}=1$; clearing the denominators of the coefficients of $A_0$ and $B_0$ produces a nonzero $c\in A$ and polynomials $A_1,B_1\in A[T]$ with $c=A_{1}P+B_1{P}^{\mathrm{\prime }}$, so that $c\in (P,P^{\mathrm{\prime }})$ and $c\mathrm{\ne }0$.

Let $q_0=(z_0^{\mathrm{\prime }},{\tau }_0)\in Z(W)\cap (V\times D)$ and let $k_0\ge 1$ be the order of the zero of the slice $W(z_0^{\mathrm{\prime }},\cdot )$ at ${\tau }_0$. The germ $W_{q_0}$ is regular in the last variable of order $k_0$, so [F7] provides a base neighbourhood $U_0$ of $z_0^{\mathrm{\prime }}$ and a radius $\rho >0$ such that for every $z^{\mathrm{\prime }}\in U_0$ the slice $W(z^{\mathrm{\prime }},\cdot )$ has exactly $k_0$ zeros, counted with multiplicity, in the disc $\mathrm{\mid }\zeta -{\tau }_0\mathrm{\mid }<\rho$ and no zero on its boundary circle.

By steps 1.1 and 1.2, a point $q\in V\times D$ lies in $X$ exactly when $W(q)=0$, and then $q$ is singular exactly when $dW(q)=0$; by [F2] the condition $dW(q)=0$ is the vanishing of all partial derivatives. Hence $\mathrm{Sing}(X)=\{q\in V\times D:W(q)=0,{\mathrm{\partial }}_{1}W(q)=\cdots ={\mathrm{\partial }}_{n}W(q)=0\}$, the common zero set of the finitely many holomorphic functions $W,{\mathrm{\partial }}_{1}W,\dots ,{\mathrm{\partial }}_{n}W$, which is closed in $V\times D$ and hence in $X$ by [F18]. This is part 1, and it identifies the germ of $\mathrm{Sing}(X)$ at $q$ with the zero germ of $J_q$ for every $q\in \mathrm{Sing}(X)$.

Fix $q\in Z(W)\cap (V\times D)$ with the preparation $W_q=vP$, base ring $A$ and element $c\in A$ of step 1.4. By [F12] every germ $h\in {\mathcal{O}}_{{\mathbb{C}}^n,q}$ has a unique remainder $r_0+r_1\tilde{T}+\cdots +r_{k-1}{\tilde{T}}^{k-1}$ of degree less than $k$ modulo $P$, so the classes of $1,\tilde{T},\dots ,{\tilde{T}}^{k-1}$ form a basis of ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P)$ as an $A$-module; in particular $A\to {\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P)$ is injective, the identity $P\in (P,c)$ is trivial, and quotienting by $c$ shows that ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)$ is a free module of rank $k$ over $A\mathrm{/}(c)$, with the same basis.

With $J_q=(W_q,{\mathrm{\partial }}_1{W}_q,\dots ,{\mathrm{\partial }}_n{W}_q)$ as in part 2, one has $(P,c)\subseteq (P,P^{\mathrm{\prime }})\subseteq J_q$. Indeed $c\in (P,P^{\mathrm{\prime }})$ by step 1.4; $P\in J_q$ because $W_q=vP$ with $v$ a unit; and $P^{\mathrm{\prime }}\in J_q$, since the product rule [F17] gives ${\mathrm{\partial }}_{\tilde{T}}{W}_q=({\mathrm{\partial }}_{\tilde{T}}v)P+v{P}^{\mathrm{\prime }}$, so that $v{P}^{\mathrm{\prime }}\in J_q$ and, $v$ being a unit, $P^{\mathrm{\prime }}\in J_q$; here the last partial derivative of $W_q$ is its derivative in the last variable and $P^{\mathrm{\prime }}$ is the formal derivative of the polynomial $P$.

Every singular point of $X$ lies over the branch set: if $q=(z_0^{\mathrm{\prime }},\tau )\in \mathrm{Sing}(X)$, then $dW(q)=0$ by step 1.2, so in particular ${\mathrm{\partial }}_{T}W(q)=0$ and step 1.3 gives $D_W(z_0^{\mathrm{\prime }})=0$.

Conversely, every point of the zero set lying over the complement of the branch set is regular: if $z_0^{\mathrm{\prime }}\in V$ satisfies $D_W(z_0^{\mathrm{\prime }})\mathrm{\ne }0$ and $q=(z_0^{\mathrm{\prime }},\tau )\in Z(W)$, then the slice has $d$ distinct simple roots by step 1.3, so in particular ${\mathrm{\partial }}_{T}W(q)\mathrm{\ne }0$ and therefore $dW(q)\mathrm{\ne }0$; by step 1.2 the point $q$ is regular.

Let $q\in \mathrm{Sing}(X)$ and suppose $J_q$ is a proper ideal. By step 2.3 we have $(P,c)\subseteq J_q$, so $c$ is not a unit, and $A\mathrm{/}(c)$ and ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)$ are nonzero; by step 2.2 the ring ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)$ is a free module of rank $k\ge 1$ over $A\mathrm{/}(c)$, hence a module-finite, injective extension of it, which is integral by [F13]. Under the Axiom of Choice, [F13] therefore gives $\dim {\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)=\dim A\mathrm{/}(c)$. Every strict chain of primes of $A$ containing $(c)$ can be prepended with the zero ideal, which is prime because $A$ is a domain by [F14] and is strictly smaller than the first member because $c\mathrm{\ne }0$ lies in it; such a chain of length $j$ therefore yields a strict chain of length $j+1$ in $A$, and the chain description of dimensions in [F13] together with $\dim A=n-1$ gives $\dim A\mathrm{/}(c)\le n-2$. Hence $\dim {\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)\le n-2$.

If $n=1$, then the base ring of the preparation at a point $q$ is $A={\mathcal{O}}_{{\mathbb{C}}^0,\cdot }=\mathbb{C}$ by [F13], a field; the element $c\mathrm{\ne }0$ of step 1.4 is then a unit, so $(P,c)$ is the unit ideal and step 2.3 makes $J_q$ the unit ideal for every point $q$ of the prepared zero set. But by part 1 as proved in step 2.1 the germ of $\mathrm{Sing}(X)$ at $q$ is the zero germ of $J_q$, which is empty when $J_q={\mathcal{O}}_{{\mathbb{C}}^n,q}$; since every point of the representative lies in the prepared neighbourhood, the singular locus is empty.

The regular locus is dense in $X$. If $n=1$, step 3.2 gives $\mathrm{Sing}(X)=\mathrm{\varnothing }$, so $\mathrm{Reg}(X)=X$ is dense. Assume now $n\ge 2$. Let $U\subseteq X$ be a nonempty open subset of the representative and choose $q_0=(z_0^{\mathrm{\prime }},{\tau }_0)\in U$; since $U$ is open in the subspace topology there are a polydisc $V_0\subseteq V$ around $z_0^{\mathrm{\prime }}$ and a radius $\epsilon >0$ with $X\cap (V_0\times D_{\epsilon }({\tau }_0))\subseteq U$. Apply step 1.5 to $q_0$ and shrink $V_0$ and $\rho$ so that $\rho \le \epsilon$ and every slice over $V_0$ has exactly $k_0\ge 1$ zeros in $\mathrm{\mid }\zeta -{\tau }_0\mathrm{\mid }<\rho$. Since $D_W$ is a nonzero germ on the connected polydisc $V$, its zero set has empty interior: if $D_W$ vanished on a nonempty open subset of $V$, then [F15] would force $D_W$ to vanish identically on $V$, contradicting that $D_W\mathrm{\ne }0$ as a germ. Hence there is $z^{\mathrm{\prime }}\in V_0$ with $D_W(z^{\mathrm{\prime }})\mathrm{\ne }0$; for this $z^{\mathrm{\prime }}$ the slice has $k_0\ge 1$ zeros in the disc and all its roots are simple by step 1.3, so choosing one of them, say $\zeta$, gives a point $q=(z^{\mathrm{\prime }},\zeta )\in X\cap (V_0\times D_{\epsilon }({\tau }_0))\subseteq U$ that is regular by step 2.5. Thus every nonempty open subset of $X$ contains a regular point, that is, $\mathrm{Reg}(X)$ is dense in $X$.

For every $q\in \mathrm{Sing}(X)$ with $J_q$ proper one has $\dim {\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}{J}_q\le n-2$: by step 2.3 the quotient ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}{J}_q$ is a quotient of ${\mathcal{O}}_{{\mathbb{C}}^n,q}\mathrm{/}(P,c)$, and by the chain description of dimensions in [F13] passing to a quotient cannot increase the dimension, so step 3.1 gives the bound.

All parts are now established: part 1 and the closedness and local-ideal claims of part 2 are step 2.1, where the germ of $\mathrm{Sing}(X)$ at $q$ is the zero germ of $J_q$, which is nonempty when $J_q$ is proper because a proper ideal of the local ring is contained in its maximal ideal and all its elements then vanish at $q$, and empty when $J_q$ is the unit ideal; part 3 is step 4.2; part 5 is step 3.2; and part 4 follows because $\mathrm{Sing}(X)$ is closed in $X$ by step 2.1 while $\mathrm{Reg}(X)$ is dense by step 4.1, so the closure of $\mathrm{Sing}(X)$, namely $\mathrm{Sing}(X)$ itself, has empty interior in $X$. Finally, $\mathrm{Sing}(X)$ lies over the branch set by step 2.4, and since $X$ has pure local dimension $n-1$ by [F16] while every nonempty singular germ has dimension at most $n-2$ by step 4.2, such a germ has codimension at least two in the ambient ${\mathbb{C}}^n$.