Theorem 9.3.4 (Local parametrization)833B11F874F9

Proof. We prove more generally the following statement for every prime ideal $J\subset \mathcal O_ n$: the germ $V(J)$ has coordinates and polydiscs as in the theorem, and $I_{V(J),0}=J$. The theorem then follows by taking $J=I_{A,0}$, which is prime by proposition 9.3.1 and satisfies $V(J)=(A,0)$. If $J=0$, then $V(J)$ is a neighborhood of $0$; we take $d=n$, $q=1$, $u=0$ and $W(z',T)=T$, so that $\delta =\operatorname {Disc}(W)=1$, and $I_{V(0),0}=0$ by the identity theorem. Assume that $J\neq 0$. Steps 1–3 construct the covering and prove that $I_{V(J),0}=J$; Steps 4 and 5 prove the density and the connectedness of the covering locus.

Step 1: triangular equations. Since $J$ is prime, it is a proper ideal, so its elements vanish at $0$. Choose a nonzero element of $J$, of order $s_ n\geq 1$, and linear coordinates in which it is distinguished in $z_ n$. By the preparation theorem it is a unit times a Weierstrass polynomial $P_ n\in J$ of degree $s_ n$ in $z_ n$. Since a unit has order $0$, the polynomial $P_ n$ also has order $s_ n$, so its coefficient of $z_ n^{s_ n-j}$ has order at least $j$ in $(z_1,\ldots ,z_{n-1})$. The ideal $J\cap \mathcal O_{n-1}$ of germs in the first $n-1$ coordinates is again prime. If it is nonzero, we repeat the construction with a linear change of $z_1,\ldots ,z_{n-1}$ only, which preserves the properties of $P_ n$; we continue until $J\cap \mathcal O_ d=0$ for some $d$. This gives coordinates and Weierstrass polynomials

with $a_{jk}=O(|(z_1,\ldots ,z_{k-1})|^ j)$. At each step only the zero set of a nonzero homogeneous polynomial is excluded in choosing the direction, so the coordinates can be chosen arbitrarily close to given ones. Fix generators of $J$, holomorphic on a neighborhood $U$ of $0$, and let $V(J)$ also denote their common zero set in $U$. Applying (9.2.1) successively to $P_{d+1},\ldots ,P_ n$, we find a constant $C$ with

Consequently, if $\Delta ''$ is a small polydisc and $\Delta '$ a polydisc whose radius is small compared with that of $\Delta ''$, the points of $V(J)\cap (\Delta '\times \Delta '')$ over each compact subset of $\Delta '$ form a compact set. Hence $\pi $ is proper, its fiber over $0$ is $\{ 0\} $, and its fibers are finite, since they lie in the sets of successive roots of $P_{d+1},\ldots ,P_ n$.

Successive division by $P_ n,\ldots ,P_{d+1}$ (proposition 9.2.2) shows that every germ is congruent modulo $J$ to a polynomial in $z_{d+1},\ldots ,z_ n$ of degree less than $s_ k$ in $z_ k$, with coefficients in $\mathcal O_ d$. Since $J\cap \mathcal O_ d=0$, the ring $\mathcal O_ d$ injects into

and $B$ is a finitely generated $\mathcal O_ d$-module, generated by the classes of the monomials $\prod _{k>d}z_ k^{\alpha _ k}$ with $0\le \alpha _ k<s_ k$.

Step 2: a primitive element and division by the discriminant. The ring $B$ is a domain, because $J$ is prime, and by Step 1 it is generated as an $\mathcal O_ d$-algebra by the classes of $z_{d+1},\ldots ,z_ n$. We apply lemma 9.3.3 to $B$ and these generators, and let $q$ be the dimension of the field of fractions of $B$ over $K=\operatorname {Frac}(\mathcal O_ d)$. Choose $c$ as in lemma 9.3.3(2), and let $u=\sum _{k>d}c_ kz_ k$, which will be the primitive linear form of the statement; let $W(z',T)\in \mathcal O_ d[T]$ be its minimal polynomial, of degree $q$, and let $\delta =\operatorname {Disc}(W)\in \mathcal O_ d\setminus \{ 0\} $ be its discriminant. By lemma 9.3.3(3) there are polynomials $B_ k(z',T)\in \mathcal O_ d[T]$ of degree less than $q$ with

The class of each $z_ k$, $k>d$, lies in $B$ and is therefore integral over $\mathcal O_ d$, so by lemma 9.3.3(1) its minimal polynomial $W_ k(z',T)$ over $K$ is monic of degree at most $q$ with coefficients in $\mathcal O_ d$. These polynomials, and $W$, are Weierstrass polynomials in $T$. Indeed, the germ $W_ k(z',z_ k)$ belongs to $J$, so it vanishes at the origin, and $0$ is a root of $W_ k(0,\cdot )$ of some multiplicity $\mu _ k\geq 1$. By lemma 9.2.5 in the variables $(z',T)$ at the origin, $W_ k=V_ kQ_ k$, where $V_ k$ is a Weierstrass polynomial in $T$ of degree $\mu _ k$ and $Q_ k$ is a polynomial in $T$ with $Q_ k(0,0)\neq 0$. Substituting $T=z_ k$, which vanishes at the origin, $Q_ k(z',z_ k)$ becomes a unit of $\mathcal O_ n$, so $V_ k(z',z_ k)\in J$. By minimality $\deg V_ k\geq \deg W_ k$, hence $W_ k=V_ k$. The same argument applies to $W$ and $u$.

Let $G\subset J$ be the ideal generated by $W(z',u)$ and the germs $\delta z_ k-B_ k(z',u)$. We shall use the following divisibility property several times: if $H\in \mathcal O_ d[T]$ and $H(z',u)\in J$, then $H$ is divisible by $W$ in $\mathcal O_ d[T]$; in particular, $H(z',u)\in G$, and $H=0$ if $\deg H<q$. Indeed, $H$ vanishes at the class of $u$ in $\operatorname {Frac}(B)$, and so does the remainder of its Euclidean division by the monic polynomial $W$; this remainder has degree less than $q$, so it is zero, because $1,u,\ldots ,u^{q-1}$ are linearly independent over $K$.

We claim that, with $m=\max \{ q,(n-d)(q-1)\} $, every germ $f\in \mathcal O_ n$ satisfies

for a unique polynomial $R_ f(z',T)\in \mathcal O_ d[T]$ of degree less than $q$. To prove this, divide $f$ successively by the Weierstrass polynomials $W_ n(z',z_ n),\ldots ,W_{d+1}(z',z_{d+1})$, whose coefficients depend only on $z'$. This writes $f$ as a polynomial $\sum _\alpha e_\alpha (z')z''^\alpha $ with $\alpha _ k\leq q-1$ for all $k$, plus multiples of the $W_ k(z',z_ k)$. Multiplying by $\delta ^ m$ and substituting $\delta z_ k\equiv B_ k(z',u)$ modulo $G$, which is possible because $|\alpha |\leq (n-d)(q-1)\leq m$, we obtain a polynomial in $u$ with coefficients in $\mathcal O_ d$; dividing it by $W$ gives $R_ f$. The discarded multiples of $W_ k(z',z_ k)$ disappear modulo $G$, because $m\geq q$ and $\delta ^ qW_ k(z',z_ k)\in G$. Indeed, the polynomial

has coefficients in $\mathcal O_ d$, since $\deg W_ k\leq q$. The same substitution gives $\delta ^ qW_ k(z',z_ k)\equiv H_ k(z',u)$ modulo $G$; hence $H_ k(z',u)\in J$, and by the divisibility property $H_ k$ is divisible by $W$ and $H_ k(z',u)\in G$. Two admissible polynomials $R_ f$ differ by a polynomial $H$ of degree less than $q$ with $H(z',u)\in G\subset J$, so they agree, by the divisibility property. If $f\in J$, then $R_ f(z',u)\in J$, and $R_ f=0$ for the same reason.

Step 3: the covering off the discriminant. Choose polydiscs as in Step 1 so small that all the germs above and the generators of $J$ are holomorphic near the closure of $\Delta '\times \Delta ''$, and such that the following identities between germs hold there: (9.3.5) for the generators of $J$, the expressions of the generators of $G$ as combinations of the generators of $J$, and the divisions of $H_ k$ and of $\sum _ kc_ kB_ k(z',T)-\delta T$ by $W$ used below. Let $z'\in \Delta '$ with $\delta (z')\neq 0$. Then $W(z',\cdot )$ has $q$ distinct roots $\tau _1,\ldots ,\tau _ q$, by (9.3.2), and the points of $V(G)$ over $z'$ are exactly the $q$ points

Indeed, a point of $V(G)$ satisfies these equations with $T=u(z'')$. Conversely, the polynomial $\sum _ kc_ kB_ k(z',T)-\delta T$ takes at $T=u$ a value congruent to $\delta u-\delta u=0$ modulo $J$, so it is divisible by $W$, by the divisibility property of Step 2, and vanishes at each $\tau _ i$; hence the point defined by $\tau _ i$ satisfies $u(z'')=\tau _ i$ and lies in $V(G)$. These $q$ points are distinct, because their values of $u$ are distinct.

The simple roots $\tau _ i$, and hence the points, depend holomorphically on $z'$ locally: if $\tau $ is a simple root of $W(z'_1,\cdot )$, the map $(z',T)\mapsto (z',W(z',T))$ has invertible derivative at $(z'_1,\tau )$, and its holomorphic local inverse (proposition 5.2.4(1)) yields a holomorphic function $\tau (z')$ near $z'_1$ with $W(z',\tau (z'))=0$ and $\tau (z'_1)=\tau $. Since $H_ k$ is divisible by $W$, we have $\delta (z')^ qW_ k(z',B_ k(z',\tau _ i)/\delta (z'))=H_ k(z',\tau _ i)=0$, so the coordinate $z_ k=B_ k(z',\tau _ i)/\delta (z')$ is a root of the Weierstrass polynomial $W_ k(z',\cdot )$; by (9.2.1) it is small when $z'$ is small. Shrinking $\Delta '$, we may therefore assume that these points lie in $\Delta '\times \Delta ''$. They belong to $V(J)$, since $\delta ^ mf\in G$ for every generator $f$ of $J$ by (9.3.5), and $\delta (z')\neq 0$. As $G\subset J$, we conclude that over $\Delta '\setminus \{ \delta =0\} $ the set $V(J)\cap (\Delta '\times \Delta '')$ is the union of the $q$ disjoint local holomorphic graphs just described, on which $u$ takes the values $\tau _1,\ldots ,\tau _ q$; this gives the assertion about $u$ and $W$, which persists when the polydiscs are shrunk later. Thus $A_\delta $ is a complex submanifold of dimension $d$, open in $V(J)\cap (\Delta '\times \Delta '')$, and a $q$-sheeted covering of $\Delta '\setminus \{ \delta =0\} $; this base is connected by lemma 9.2.9.

Now let $f\in \mathcal O_ n$ vanish on $V(J)$. By (9.3.5), the polynomial $R_ f(z',\cdot )$ vanishes at the $q$ distinct roots $\tau _ i$ for every $z'$ with $\delta (z')\neq 0$ near $0$, because $\delta ^ mf$ and the elements of $G$ vanish at the corresponding points of $V(J)$. Its degree is less than $q$, so $R_ f=0$. Hence $\delta ^ mf\in G\subset J$, and since $\delta \notin J$ and $J$ is prime, $f\in J$. This proves $I_{V(J),0}=J$.

Step 4: density of the covering locus and the bound on the fibers. For a complex linear form $\lambda $ on $\mathbb C^{n-d}$ and $z'\in \Delta '\setminus \{ \delta =0\} $, put

Its coefficients are elementary symmetric functions of the values of $\lambda $ on the sheets, hence holomorphic and bounded on $\Delta '\setminus \{ \delta =0\} $; by lemma 9.2.9 they extend holomorphically to $\Delta '$. The function $\delta (z')P_\lambda (z',\lambda (z''))$ vanishes on $V(J)\cap (\Delta '\times \Delta '')$: on $A_\delta $ the second factor vanishes, and elsewhere the first. By Step 3 its germ belongs to $J$, and since $\delta \notin J$, the germ of $P_\lambda (z',\lambda (z''))$ belongs to $J$; so this function vanishes on $V(J)$ near $0$.

Let $z'_0\in \Delta '$ and let $F$ be the fiber over $z'_0$ of the closure $\overline{A_\delta }$ of $A_\delta $ in $\Delta '\times \Delta ''$. We claim that the roots of $P_\lambda (z'_0,\cdot )$ are exactly the values $\lambda (y'')$, $y\in F$. Indeed, every $y\in F$ is a limit of points $y_ j\in A_\delta $ with $\pi (y_ j)\to z'_0$, and $\lambda (y_ j'')$ is a root of $P_\lambda (\pi (y_ j),\cdot )$; since the coefficients of $P_\lambda $ are continuous, $\lambda (y'')$ is a root of $P_\lambda (z'_0,\cdot )$. Conversely, choose $z'_ j\to z'_0$ with $\delta (z'_ j)\neq 0$. The $q$ points of $A_\delta $ over $z'_ j$ lie in a compact subset of $\Delta '\times \Delta ''$ by Step 1, so after passing to a subsequence they converge to points of $F$; their values of $\lambda $, which are the $q$ roots of $P_\lambda (z'_ j,\cdot )$, converge to the $q$ roots of $P_\lambda (z'_0,\cdot )$, counted with multiplicity. It follows that $F$ has at most $q$ points: a linear form taking distinct values at $q+1$ points of $F$ would give $q+1$ distinct roots of a polynomial of degree $q$.

It remains to show that $V(J)$ is contained in $\overline{A_\delta }$ near $0$; after shrinking the polydiscs once more, this gives the density of $A_\delta $ and the bound $q$ for all fibers. Suppose not. Then there are points $y_ j\in V(J)\setminus \overline{A_\delta }$ with $y_ j\to 0$. For each $j$, the forms $\lambda $ with $\lambda (y_ j'')=\lambda (w'')$ for some point $w$ of the finite fiber $F_ j$ of $\overline{A_\delta }$ over $\pi (y_ j)$ form finitely many proper linear subspaces. We choose one form $\lambda $ outside all these countably many subspaces: for instance, $\lambda $ with coefficients $(1,t,\ldots ,t^{n-d-1})$, where $t$ avoids the countably many roots of the nonzero polynomials $\sum _ kv_ kt^{k-1}$ attached to the vectors $v=y_ j''-w''$. Then $\lambda (y_ j'')$ is not a root of $P_\lambda (\pi (y_ j),\cdot )$ for any $j$, by the previous paragraph. On the other hand, $P_\lambda (z',\lambda (z''))$ vanishes on $V(J)$ near $0$, hence at $y_ j$ for all large $j$, a contradiction.

Step 5: connectedness of the covering locus. The covering $A_\delta $ of the connected manifold $\Delta '\setminus \{ \delta =0\} $ has finitely many connected components, each of which is itself a covering with a positive number of sheets. Indeed, over a small ball $N\subset \Delta '\setminus \{ \delta =0\} $ the set $A_\delta $ is the disjoint union of $q$ connected holomorphic graphs, each of which lies in a single connected component of $A_\delta $. Hence the number of points of a given component over $z'$ is locally constant, hence constant on the connected set $\Delta '\setminus \{ \delta =0\} $, and it is positive because the component is not empty.

Suppose that there are at least two components. For each component $A^{(i)}$ and each linear form $\lambda $, the product $P^{(i)}_\lambda $ of the factors $T-\lambda (z'')$ over the points of $A^{(i)}$ again has bounded holomorphic coefficients on $\Delta '\setminus \{ \delta =0\} $, which extend to $\Delta '$, and $P_\lambda =\prod _ iP^{(i)}_\lambda $. Choose points $z'_ j\to 0$ with $\delta (z'_ j)\neq 0$, and, by the countable choice of Step 4, a form $\lambda $ that takes distinct values at the $q$ points of $A_\delta $ over each $z'_ j$. Since the germ of $P_\lambda (z',\lambda (z''))$ belongs to the prime ideal $J$, so does the germ of some factor $P^{(i)}_\lambda (z',\lambda (z''))$, which then vanishes on $V(J)$ near $0$. Choose $k\neq i$. For large $j$ the points of $A^{(k)}$ over $z'_ j$ are close to $0$, by Step 1, so $P^{(i)}_\lambda (z'_ j,\cdot )$ vanishes at their values of $\lambda $; but these values differ from the roots of $P^{(i)}_\lambda (z'_ j,\cdot )$, which are the values of $\lambda $ at the points of $A^{(i)}$ over $z'_ j$. This contradiction shows that $A_\delta $ is connected, and completes the proof. $\square$