Analytic Hypersurfaces and Local Parametrisation

By [F3] choose a factorisation $f=u{q}_1^{e_1}\cdots q_r^{e_r}$ with $u$ a unit, the $q_i$ pairwise nonassociate irreducible germs, $e_i\ge 1$ and $r\ge 1$, and set $f_{\mathrm{r}\mathrm{e}\mathrm{d}}:=q_1\cdots q_r$.

The germ $f_{\mathrm{r}\mathrm{e}\mathrm{d}}$ is a nonzero nonunit: it is a product of the nonunits $q_i$ in the domain of [F2], and a product of germs one of which is a nonunit cannot be a unit, while it is nonzero because a domain has no zero divisors and the $q_i\mathrm{\ne }0$.

The associate class of $f_{\mathrm{r}\mathrm{e}\mathrm{d}}$ depends only on $f$: if $f=v{r}_1^{g_1}\cdots r_s^{g_s}$ is another factorisation into pairwise nonassociate irreducibles, then by uniqueness in [F3] the multiset $\{(q_i,e_i)\}$ of associate classes with exponents equals $\{(r_j,g_j)\}$; hence the set of associate classes occurring, and therefore the product $f_{\mathrm{r}\mathrm{e}\mathrm{d}}=q_1\cdots q_r$ up to a unit, is the same for the two factorisations.

No irreducible germ divides $f_{\mathrm{r}\mathrm{e}\mathrm{d}}$ twice. Let $q$ be irreducible with $q^2\mid f_{\mathrm{r}\mathrm{e}\mathrm{d}}$. Then $q\mid q_1\cdots q_r$, and factoring the quotient $q_1\cdots q_r\mathrm{/}q$ into irreducibles exhibits $q\cdot (\text{quotient})$ and $q_1\cdots q_r$ as two irreducible factorisations of the same element; by uniqueness in [F3], $q$ is associate to one of the $q_i$, say $q_1$. But then $q_1^2\mid f_{\mathrm{r}\mathrm{e}\mathrm{d}}$, so writing $f_{\mathrm{r}\mathrm{e}\mathrm{d}}\mathrm{/}{q}_1^2$ as a product of irreducibles and comparing with $q_1\cdots q_r$ shows that the associate class of $q_1$ occurs at least twice among the classes of $q_1,\dots ,q_r$, contradicting their pairwise nonassociateness. Hence $f_{\mathrm{r}\mathrm{e}\mathrm{d}}$ is reduced by [F1].

For the zero sets, put $k:={\max }_i{e}_i$ and write the identities $f=f_{\mathrm{r}\mathrm{e}\mathrm{d}}\cdot u{\prod }_i{q}_i^{e_i-1}$ and $f_{\mathrm{r}\mathrm{e}\mathrm{d}}^k=\left(u^{-1}{\prod }_i{q}_i^{k-e_i}\right)f$ in the germ ring; after shrinking to a neighbourhood on which representatives of $f$ and $f_{\mathrm{r}\mathrm{e}\mathrm{d}}$ are both defined, the first identity gives $Z(f_{\mathrm{r}\mathrm{e}\mathrm{d}})\subseteq Z(f)$, and the second gives $Z(f)\subseteq Z(f_{\mathrm{r}\mathrm{e}\mathrm{d}})$, since a point with $f(x)=0$ has $f_{\mathrm{r}\mathrm{e}\mathrm{d}}(x{)}^k=0$ and $\mathbb{C}$ has no nilpotents. Therefore $Z(f_{\mathrm{r}\mathrm{e}\mathrm{d}})=Z(f)$ on that neighbourhood.

Steps 3.1, 2.2 and 4.1 establish all three asserted properties of the square-free reduction $f_{\mathrm{r}\mathrm{e}\mathrm{d}}=q_1\cdots q_r$.

By [F1] and [F2] write $f=u{\prod }_{i=1}^r{p}_i$ with $r\ge 1$, the $p_i$ irreducible and pairwise nonassociate, and no irreducible factor repeated.

For each $i$ factor $f=p_i{h}_i$ with $h_i:=u{\prod }_{j\mathrm{\ne }i}{p}_j$. By [F4] both $p_i$ and $h_i$ are regular in the last variable, the preparation $f=uW$ satisfies $W=W_i{V}_i$ where $p_i=u_i{W}_i$ and $h_i=v_i{V}_i$ are preparations, and $p_i$ is a nonunit, so its order $d_i=\mathrm{deg}{W}_i$ is at least $1$.

Applying the consequence in [F4] to the irreducible $p_i$ shows that $W_i$ is irreducible in ${\mathcal{O}}_{n-1,0}[T]$. Each $W_i$ is monic by [F3], hence primitive, so by Gauss's lemma [F5] $W_i$ is irreducible in $K[T]$.

The $W_i$ are pairwise distinct: if $W_i=W_j$ for $i\mathrm{\ne }j$, then $p_i=u_i{W}_i=u_i{u}_j^{-1}{p}_j$ makes $p_i$ and $p_j$ associates, contradicting step 1.1.

The product ${\prod }_{i=1}^r{W}_i$ is a Weierstrass polynomial: it is monic of degree ${\sum }_i{d}_i$ with coefficients in ${\mathcal{O}}_{n-1,0}$, and at $z^{\mathrm{\prime }}=0$ each factor equals $T^{d_i}$ by [F3], so the product equals $T^{{\sum }_i{d}_i}$. Since step 2.1 gives $f=\left(u{\prod }_i{u}_i\right){\prod }_i{W}_i$ and $f=uW$ is a preparation, uniqueness of the prepared polynomial [F3] yields $W={\prod }_{i=1}^r{W}_i$.

Hence $W$ is square-free in $K[T]$: an irreducible $P\in K[T]$ dividing $W$ twice would, by uniqueness of factorisation in the UFD $K[T]$ from [F6], be associate to two of the distinct monic irreducibles $W_i$; being monic it would equal both, contradicting step 4.1.

For the discriminant, let $E$ be a splitting field of $W$ over $K$ and write $W={\prod }_{k=1}^d(T-{\alpha }_k)$ as in [F7]. The roots of $W$ are the roots of the factors $W_i$. Two distinct factors $W_i,W_j$ are coprime in $K[T]$: their monic gcd divides the irreducible $W_i$, so it is $1$ or an associate of $W_i$, and in the second case it would also be an associate of $W_j$, forcing $W_i=W_j$; thus $1=A{W}_i+B{W}_j$ for some $A,B\in K[T]$ by [F9], and a common root would give $1=0$. A root of exactly one factor $W_i$ that were repeated for $W$ would be a repeated root of $W_i$, since the complementary product does not vanish there.

Each $W_i$ is separable by step 3.1 and [F8], so it has no repeated root in the extension $E$; combined with step 7.1, all roots ${\alpha }_1,\dots ,{\alpha }_d$ of $W$ in $E$ are pairwise distinct. The root formula in [F7] then gives ${\mathrm{Disc}}_T(W)={\prod }_{k<l}({\alpha }_k-{\alpha }_l{)}^2\mathrm{\ne }0$ in $K$.

Finally, ${\mathrm{Disc}}_T(W)$ is the coefficient expression $D_d(-a_1,\dots ,(-1{)}^d{a}_d)$ in the coefficients $a_j\in {\mathcal{O}}_{n-1,0}$ of $W$ by [F7], hence is a holomorphic base germ $D_W\in {\mathcal{O}}_{n-1,0}$; since it is nonzero as an element of $K=\mathrm{Frac}({\mathcal{O}}_{n-1,0})$ by step 8.1, it is a nonzero germ.

Center at $p$ by writing $\tilde{f}(z)=f(p+z)$. By [F2] choose an invertible complex-linear map $T$ with $\tilde{f}\circ T$ regular in the last variable of order $d$, and by [F3] prepare $\tilde{f}\circ T=uW$ with $W$ a Weierstrass polynomial of degree $d$. Since $f$ is a nonunit, $d>0$. Shrinking to a neighbourhood on which $u$ has no zeros, which [F4] permits, gives $Z(\tilde{f}\circ T)=Z(W)$ there; the map $z\mapsto p+Tz$ transports this local model to the original germ.

Apply [F5] to $W$ and shrink further: there are a base polydisc $V$ about $0\in {\mathbb{C}}^{n-1}$ and a radius $r>0$ such that for every $z^{\mathrm{\prime }}\in V$ the slice $T\mapsto W(z^{\mathrm{\prime }},T)$ has no zero on $\mathrm{\mid }T\mathrm{\mid }=r$ and exactly $d$ zeros in $D=\{T:\mathrm{\mid }T\mathrm{\mid }<r\}$, counted with multiplicity.

By [F6], the classes of $1,T,\dots ,T^{d-1}$ generate ${\mathcal{O}}_{n,0}\mathrm{/}(W)$ as an ${\mathcal{O}}_{n-1,0}$-module, so this quotient is finite.

The projection $\pi :Z(W)\cap (V\times D)\to V$ is surjective: for each $z^{\mathrm{\prime }}\in V$, the slice $W(z^{\mathrm{\prime }},\cdot )$ is monic of degree $d\ge 1$, hence has a root by [F10], and all its roots lie in $D$ by step 2.1. Each fibre is finite, with at most $d$ points by [F10].

Let $z_0^{\mathrm{\prime }}\in V$ with $D_W(z_0^{\mathrm{\prime }})\mathrm{\ne }0$. By [F9], $W(z_0^{\mathrm{\prime }},\cdot )$ has $d$ distinct roots ${\tau }_1,\dots ,{\tau }_d$, each simple. For each root [F7] gives a local holomorphic graph $T={\phi }_k(z^{\mathrm{\prime }})$ with ${\phi }_k(z_0^{\mathrm{\prime }})={\tau }_k$. Intersecting the finitely many base neighbourhoods and shrinking so the differences ${\phi }_k-{\phi }_l$ remain nonzero gives a common neighbourhood $U$ on which the graphs are defined and pairwise disjoint.

For compact $K\subseteq V$, let $E_K:=\{(z^{\mathrm{\prime }},T)\in K\times \bar{D}:W(z^{\mathrm{\prime }},T)=0\}$. Continuity of $W$ makes $E_K$ closed in the compact set $K\times \bar{D}$. By step 2.1 no slice has a zero on $\mathrm{\partial }D$, so $E_K={\pi }^{-1}(K)$ for $\pi :Z(W)\cap (V\times D)\to V$. Therefore ${\pi }^{-1}(K)$ is compact and $\pi$ is proper.

For $z^{\mathrm{\prime }}\in U$ the degree-$d$ polynomial $W(z^{\mathrm{\prime }},\cdot )$ vanishes at the $d$ distinct points ${\phi }_1(z^{\mathrm{\prime }}),\dots ,{\phi }_d(z^{\mathrm{\prime }})$ from step 3.2, so these are all its roots by [F10]. Thus ${\pi }^{-1}(U)={\bigcup }_{k=1}^d\{(z^{\mathrm{\prime }},{\phi }_k(z^{\mathrm{\prime }})):z^{\mathrm{\prime }}\in U\}$ is a disjoint union of graphs, each mapped biholomorphically onto $U$.

By [F8] the discriminant is not the zero germ, and by [F9] its complement is exactly the set of base points with distinct roots. For each such point step 4.1 gives a neighbourhood with $d$ disjoint sheets, so the restriction of $\pi$ over $V\setminus \{D_W=0\}$ is a $d$-sheeted holomorphic covering as defined in [F11].

If $n=1$, the base $V$ is a point. Steps 3.1 and 3.3 give surjectivity, finite fibres and properness; step 2.2 gives the finite quotient module. By [F8] and [F9] the reduced one-variable polynomial has nonzero discriminant and therefore $d$ distinct roots, so its finite zero set is a $d$-sheeted covering of the point.

By [F1] the germ $W$ is regular in the last variable of order $d$ and is its own preparation, so [F2] applies to it: $D_W$ is a nonzero base germ and [F3] identifies its vanishing with the existence of a repeated root in the slice $W(z^{\mathrm{\prime }},\cdot )$. Shrink the product representative so that $W$ and $D_W$ are defined on the connected base $V$ and the finite-projection conclusions hold.

Suppose for contradiction that some $q=(q^{\mathrm{\prime }},\tau )\in Z(W)\cap (V\times D)$ has a nonreduced germ: $W_q=g^{2}h$ for an irreducible germ $g$ at $q$. By [F5] the germ $g$ is a nonzero nonunit, so $g(q^{\mathrm{\prime }},\tau )=0$.

The vertical slice $\zeta \mapsto g(q^{\mathrm{\prime }},\zeta )$ is not identically zero near $\tau$: the identity $W=g^{2}h$ holds on a neighbourhood of $q$, so if that slice vanished identically then the slice $\zeta \mapsto W(q^{\mathrm{\prime }},\zeta )$ would vanish identically near $\tau$, contradicting that this slice is the monic polynomial of degree $d$ from step 1.1, which has only finitely many zeros. Hence $g$ is regular in the last variable of some order $m\ge 1$ at $q$, by [F1] and the vanishing of $g$ at $q$.

Choose a product neighbourhood $U_0\times \{\mathrm{\mid }\zeta -\tau \mathrm{\mid }<{\rho }_0\}$ of $q$ contained in the neighbourhood where $W=g^{2}h$ holds, and shrink ${\rho }_0$ so the slice $g(q^{\mathrm{\prime }},\cdot )$ has no zero on $\mathrm{\mid }\zeta -\tau \mathrm{\mid }={\rho }_0$. Prepare the regular germ $g$ at $q$: $g=u_{g}G$ with $G$ a Weierstrass polynomial of degree $m\ge 1$ in the translated coordinates. Apply the stability of the slice zero count [F6] on this chosen disc and shrink the base to a neighbourhood $U\subseteq U_0$ of $q^{\mathrm{\prime }}$; every slice $g(z^{\mathrm{\prime }},\cdot )$, $z^{\mathrm{\prime }}\in U$, then has exactly $m\ge 1$ zeros counted with multiplicity in $\mathrm{\mid }\zeta -\tau \mathrm{\mid }<{\rho }_0$. In particular the product used below remains inside the factorization neighbourhood.

Fix $z^{\mathrm{\prime }}\in U$ and let $\zeta$ be one of the $m\ge 1$ zeros of $g(z^{\mathrm{\prime }},\cdot )$ supplied by step 3.1. Because the identity $W=g^{2}h$ holds on a neighbourhood of $q$, the slices satisfy $W(z^{\mathrm{\prime }},\cdot )=g(z^{\mathrm{\prime }},\cdot {)}^{2}h(z^{\mathrm{\prime }},\cdot )$ near $\zeta$; by [F7] the slice of $g$ has a zero of some order $m_y\ge 1$ at $\zeta$, so the slice of $W$ vanishes there to order at least $2m_y\ge 2$. Thus $\zeta$ is a repeated root of $W(z^{\mathrm{\prime }},\cdot )$, and [F7] gives ${\mathrm{\partial }}_{T}W(z^{\mathrm{\prime }},\zeta )=0$ while [F3] gives $D_W(z^{\mathrm{\prime }})=0$.

Every $z^{\mathrm{\prime }}\in U$ therefore lies in the zero set of $D_W$. If $n=1$, the base $V$ is the single point of ${\mathbb{C}}^0$; step 4.1 gives $D_W=0$ there, contradicting the nonzero constant $D_W$ from step 1.1. If $n\ge 2$, $D_W$ vanishes on the nonempty open set $U$, contradicting [F4] on the connected base $V$ because $D_W$ is the nonzero base germ from step 1.1. Hence no point $q\in Z(W)\cap (V\times D)$ has a nonreduced local germ.

Shrinking to the product representative $V\times D$ fixed in step 1.1, every local equation germ of $W$ at a point of its zero set is reduced, which is the assertion.

Choose $T$ and $\mathrm{\Phi }(z)=p+Tz$ with $f\circ \mathrm{\Phi }=uW$ as in [F2]. The pullback ${\mathrm{\Phi }}^{\ast }:h\mapsto h\circ \mathrm{\Phi }$ is a ring isomorphism ${\mathcal{O}}_{{\mathbb{C}}^n,p}\to {\mathcal{O}}_{n,0}$, with inverse pullback by ${\mathrm{\Phi }}^{-1}$. It sends $(f)$ to $(W)$ by [F3] and sends $I_p(Z(f))$ to $I_0(Z(W))$, since $\mathrm{\Phi }$ carries the corresponding zero germs onto each other. Hence it suffices to prove $I_0(Z(W))=(W)$.

The reverse inclusion is immediate: if $h=cW$ then every representative of $h$ vanishes at every point where $W$ vanishes, so $(W)\subseteq I_0(Z(W))$.

Let $h\in I_0(Z(W))$. By [F4] write $h=qW+r$ with $r={\sum }_{j<d}{r}_j{T}^j$, $r_j\in {\mathcal{O}}_{n-1,0}$. Since $qW$ vanishes on $Z(W)$ and $h$ does too, the remainder $r=h-qW$ vanishes on $Z(W)$ near the origin.

Choose a common product $V\times \{\mathrm{\mid }T\mathrm{\mid }<\rho \}$ on which the division identity and vanishing of $r$ on $Z(W)$ hold. Write $W=T^d+{\sum }_{j<d}{a}_j(z^{\mathrm{\prime }})T^j$. Since $a_j(0)=0$, shrink the connected base polydisc $V$ until ${\sum }_{j<d}\mathrm{\mid }{a}_j(z^{\mathrm{\prime }})\mathrm{\mid }{\rho }^{j-d}<1$. For $\mathrm{\mid }T\mathrm{\mid }\ge \rho$ the lower terms have sum of absolute values strictly less than $\mathrm{\mid }T{\mathrm{\mid }}^d$, so no slice root lies there. For every $z^{\mathrm{\prime }}\in V$ with $D_W(z^{\mathrm{\prime }})\mathrm{\ne }0$, [F5] and [F7] therefore give $d$ distinct roots, all within the common product. The polynomial $r(z^{\mathrm{\prime }},\cdot )$ has degree less than $d$ and vanishes at all these roots, so [F7] makes it the zero polynomial. Thus $r_j(z^{\mathrm{\prime }})=0$ for every $j<d$.

If $n=1$, the base is the single point $V\subset {\mathbb{C}}^0$ and $D_W$ is a nonzero constant, so $V\setminus \{D_W=0\}=V$. Each $r_j$ is also a constant; step 3.1 says it vanishes at this sole point, hence $r_j=0$. If $n\ge 2$, then $V\setminus \{D_W=0\}$ is nonempty and open: $D_W$ is a nonzero holomorphic germ, so it cannot vanish on a nonempty open subset of $V$, and its zero set is closed. Since each $r_j\in {\mathcal{O}}_{n-1,0}$ vanishes on this nonempty open set, [F6] gives $r_j=0$ for every $j<d$. In either case $r=0$ and $h=qW\in (W)$; combined with step 1.2 this gives $I_0(Z(W))=(W)$.