9.3 Analytic sets and their local branchesA49B7CFC7FBF

Proof. A decreasing sequence of analytic germs is eventually constant, because the corresponding ideals increase and are eventually constant by proposition 9.2.4, and $V(I_{A,x})=(A,x)$. Split a reducible germ into two analytic subgerms different from it, split each reducible piece again, and so on. If this process did not terminate, the resulting tree, in which every vertex has at most two children, would be infinite, and choosing at each level a child with infinitely many descendants would produce a strictly decreasing infinite sequence of analytic germs, which is impossible. Hence a finite decomposition into irreducible germs exists; removing members contained in others, we may assume that no member is contained in another. An irreducible germ contained in a finite union of analytic germs is contained in one of them, since it is the union of its intersections with them. Now let $\bigcup _ iA_ i=\bigcup _ jB_ j$ be two such decompositions. Each $A_ i$ is contained in some $B_ j$, which in turn is contained in some $A_ k$; then $A_ i\subset A_ k$ forces $k=i$ and $A_ i=B_ j$. By symmetry the two decompositions agree.

Let $(A,x)$ be irreducible. Its ideal is proper, since $x\in A$. If $fg$ vanishes on $(A,x)$, then $(A,x)$ is the union of its intersections with $\{ f=0\} $ and $\{ g=0\} $, so one of them is all of $(A,x)$, and $f$ or $g$ belongs to $I_{A,x}$; thus $I_{A,x}$ is prime. Conversely, suppose that $(A,x)$ is the union of two analytic subgerms $(B,x)$ and $(C,x)$ different from it. Since $V(I_{B,x})=(B,x)\neq (A,x)$, there is a germ $f\in I_{B,x}\setminus I_{A,x}$, and similarly $g\in I_{C,x}\setminus I_{A,x}$. Then $fg$ vanishes on $(A,x)$ while neither factor does, so $I_{A,x}$ is not prime. $\square$

Proof. We represent elements of $L$ by multiplication matrices over $K$ and evaluate these at points of $\mathbb C^ d$; integrality bounds their eigenvalues, and lemma 9.2.9 turns bounded elements of $K$ into elements of $\mathcal O_ d$.

We first recall the determinant criterion for integrality. Let $R$ be a subring of an integral domain $D$, let $b\in D$, and suppose that multiplication by $b$ maps an $R$-submodule of $D$ generated by elements $v_1,\ldots ,v_ N$, not all zero, into itself: $bv_ i=\sum _ ja_{ij}v_ j$ with $a_{ij}\in R$. Multiplying the matrix $bI-(a_{ij})$ by its adjugate shows that the monic polynomial $\det (bI-(a_{ij}))$ in $b$ annihilates every $v_ i$; since $D$ is a domain, this determinant is zero, and $b$ is integral over $R$. In particular, every element of $B$ is integral over $\mathcal O_ d$; take for the $v_ i$ generators of the $\mathcal O_ d$-module $B$. Conversely, if $x_1,\ldots ,x_ k\in D$ are integral over $R$, then $R[x_1,\ldots ,x_ k]$ is a finitely generated $R$-module, since powers of each $x_ i$ can be reduced with its monic equation; by the criterion, sums and products of integral elements are integral.

The ring $KB=\{ b/a:b\in B,\ a\in \mathcal O_ d\setminus \{ 0\} \} $ is a domain, and it is spanned over $K$ by finitely many module generators of $B$, so it is a finite-dimensional vector space over $K$. Multiplication by a nonzero element is an injective linear map of $KB$, hence surjective; so $KB$ is a field, and it equals $L$. Choose a basis $e_1=1,e_2,\ldots ,e_ q$ of $L$ over $K$ consisting of elements of $B$; it can be extracted from $1$ and the module generators of $B$. For $x\in L$ let $[x]\in K^{q\times q}$ be the matrix of the $K$-linear map $y\mapsto xy$ in this basis. The map $x\mapsto [x]$ is an injective homomorphism of $K$-algebras, because the first column of $[x]$ is the coordinate vector of $xe_1=x$. Let $\operatorname {Tr}(x)=\operatorname {tr}[x]$, which does not depend on the basis, and let $\chi _ x(T)=\det (TI-[x])\in K[T]$, a monic polynomial of degree $q$; by the Cayley–Hamilton theorem $[\chi _ x(x)]=\chi _ x([x])=0$, so $\chi _ x(x)=0$. The minimal polynomial $\mu _ x$ of $x$ exists, because $1,x,\ldots ,x^ q$ are linearly dependent over $K$, and $\deg \mu _ x\leq q$. By Euclidean division in $K[T]$, $\mu _ x$ divides every polynomial in $K[T]$ that vanishes at $x$.

Values of elements of $K$. Let $U$ be a connected neighborhood of $0$ in $\mathbb C^ d$. An element of $K$ is defined at $z'\in U$ if it can be written as $a/h$ with $a,h$ holomorphic on $U$ and $h(z')\neq 0$; its value at $z'$ is then $a(z')/h(z')$. This value does not depend on the representation, since two representations $a/h=a'/h'$ satisfy $ah'=a'h$ near $0$, hence on $U$ by the identity theorem. If finitely many elements of $K$ are defined at $z'$, so are their sums and products, and their quotients by those with nonzero value at $z'$, with the corresponding values. Finitely many elements of $K$ can be written with a common denominator $h\neq 0$ and numerators holomorphic on some connected neighborhood $U$ of $0$; they are then all defined at the points of $U\setminus \{ h=0\} $, a dense open subset of $U$. We shall use the following boundedness criterion: if the values of an element $a/h$ of $K$ at the points of $V\setminus \{ h=0\} $ are bounded, for some neighborhood $V\subset U$ of $0$, then $a/h\in \mathcal O_ d$. Indeed, $|a|\leq C|h|$ on $V\setminus \{ h=0\} $, hence on $V$, since both sides are continuous functions on $V$ and $V\setminus \{ h=0\} $ is dense in $V$; so $a/h\in \mathcal O_ d$ by lemma 9.2.9. (For $d=0$, the space $\mathbb C^0$ is a point and $K=\mathcal O_0=\mathbb C$.)

Proof of (1). Let $x\in L$ be integral over $\mathcal O_ d$, say $f(x)=0$ with $f(T)=T^ m+f_1T^{m-1}+\cdots +f_ m$ and $f_ j\in \mathcal O_ d$. Then $f([x])=[f(x)]=0$. Since $\chi _ x(x)=0$, we can write $\chi _ x=\mu _ x\rho $ with $\rho \in K[T]$. Write the entries of $[x]$ and the coefficients of $\mu _ x$ and $\rho $ with a common denominator $h$, so that they are all defined at the points of $U\setminus \{ h=0\} $ for a connected neighborhood $U$ of $0$; the coefficients of $\chi _ x$, polynomials in the entries of $[x]$, can then be written with the denominator $h^ q$.

At the points $z'\in U\setminus \{ h=0\} $ near $0$ the complex matrix $[x](z')$ satisfies $f(z',[x](z'))=0$. Applying this to an eigenvector, we see that every eigenvalue $\lambda $ of $[x](z')$ satisfies $f(z',\lambda )=0$, so that $|\lambda |\leq C_ f$ by (9.2.1), where $C_ f$ is an upper bound of $2\max _ j|f_ j|^{1/j}$ near $0$. The coefficients of $\chi _ x(z',T)$, the characteristic polynomial of $[x](z')$, are up to sign the elementary symmetric functions of these eigenvalues, so they are bounded at the points of $V\setminus \{ h=0\} $ for a neighborhood $V\subset U$ of $0$, and $\chi _ x\in \mathcal O_ d[T]$ by the boundedness criterion; in particular $\operatorname {Tr}(x)\in \mathcal O_ d$. The roots of $\mu _ x(z',\cdot )$ are roots of $\chi _ x(z',\cdot )=\mu _ x(z',\cdot )\rho (z',\cdot )$, so they are also bounded by $C_ f$; hence the coefficients of $\mu _ x(z',\cdot )$, which are up to sign the elementary symmetric functions of these roots, are bounded at the points of $V\setminus \{ h=0\} $. Therefore $\mu _ x\in \mathcal O_ d[T]$, again by the boundedness criterion, which proves (1).

Proof of (2). We evaluate the multiplication matrices at a point $z'_0$, where they span, in a suitable basis, the algebra of all diagonal matrices; then a generic combination of the matrices $[b_ k](z'_0)$ has distinct eigenvalues. We first show that the trace form is nondegenerate. Let $\Theta =\det (\operatorname {Tr}(e_ ie_ j))_{i,j}\in K$. If $\Theta =0$, there are $x_1,\ldots ,x_ q\in K$, not all zero, with $\sum _ ix_ i\operatorname {Tr}(e_ ie_ j)=0$ for all $j$; then $x=\sum _ ix_ ie_ i\neq 0$ satisfies $\operatorname {Tr}(xy)=0$ for all $y\in L$, by linearity, and $y=x^{-1}$ gives $\operatorname {Tr}(1)=q=0$, which is absurd. Write $e_ ie_ j=\sum _ l\gamma _{ijl}e_ l$ and $b_ k=\sum _ l\beta _{kl}e_ l$ with $\gamma _{ijl},\beta _{kl}\in K$, and, since $B$ is generated by the $b_ k$ as an $\mathcal O_ d$-algebra, $e_ l=G_ l(b_1,\ldots ,b_ r)$ for polynomials $G_ l$ with coefficients in $\mathcal O_ d$. Choose a connected neighborhood $U$ of $0$ and a point $z'_0\in U$ at which the finitely many elements $\Theta $, $\gamma _{ijl}$, $\beta _{kl}$, the entries of the matrices $[e_ l]$ and $[b_ k]$, and the coefficients of the $G_ l$ are defined, such that $\Theta (z'_0)\neq 0$.

Let $\mathcal N$ be the complex linear span of the matrices $N_ l=[e_ l](z'_0)$, $1\leq l\leq q$. Since $[e_ i][e_ j]=[e_ ie_ j]=\sum _ l\gamma _{ijl}[e_ l]$, the set $\mathcal N$ is a commutative algebra of complex $q\times q$ matrices, and it contains $N_1=I$. The matrix $(\operatorname {tr}(N_ iN_ j))_{i,j}$ is the value of $(\operatorname {Tr}(e_ ie_ j))_{i,j}$ at $z'_0$, which is invertible; hence the $N_ l$ are linearly independent, $\dim \mathcal N=q$, and the bilinear form $(N,N')\mapsto \operatorname {tr}(NN')$ on $\mathcal N$ is nondegenerate. If $N\in \mathcal N$ is nilpotent, then $NN'$ is nilpotent for every $N'\in \mathcal N$, because $N$ and $N'$ commute; so $\operatorname {tr}(NN')=0$ for all $N'$, and $N=0$. Consequently every $N\in \mathcal N$ is diagonalizable. Indeed, if $\lambda _1,\ldots ,\lambda _ p$ are the distinct eigenvalues of $N$, then the minimal polynomial of $N$ divides a power of $(T-\lambda _1)\cdots (T-\lambda _ p)$, so the element $(N-\lambda _1I)\cdots (N-\lambda _ pI)$ of $\mathcal N$ is nilpotent, hence zero, and the minimal polynomial of $N$ has simple roots. Commuting diagonalizable matrices are simultaneously diagonalizable; so in a suitable basis of $\mathbb C^ q$ all elements of $\mathcal N$ are diagonal, and since $\dim \mathcal N=q$, the algebra $\mathcal N$ consists of all diagonal matrices.

The matrices $[b_ k](z'_0)=\sum _ l\beta _{kl}(z'_0)N_ l$ belong to $\mathcal N$; let $\lambda ^{(i)}_ k$ be the $i$th diagonal entry of $[b_ k](z'_0)$ in this basis, and $\lambda ^{(i)}=(\lambda ^{(i)}_1,\ldots ,\lambda ^{(i)}_ r)$. For $i\neq i'$ we have $\lambda ^{(i)}\neq \lambda ^{(i')}$: otherwise every matrix $N_ l=G_ l(z'_0)([b_1](z'_0),\ldots ,[b_ r](z'_0))$ would have equal $i$th and $i'$th diagonal entries, and so would every element of $\mathcal N$. Therefore, for $c$ outside the finitely many proper linear subspaces $\{ c:\sum _ kc_ k(\lambda ^{(i)}_ k-\lambda ^{(i')}_ k)=0\} $, $i<i'$, the matrix $[u](z'_0)=\sum _ kc_ k[b_ k](z'_0)$ is diagonal with $q$ distinct diagonal entries $\tau _1,\ldots ,\tau _ q$.

Fix such a $c$, and let $\Lambda \in K^{q\times q}$ be the matrix whose columns are the coordinate vectors of $1,u,\ldots ,u^{q-1}$; its entries are polynomials in the entries of the $[b_ k]$, so they are defined at $z'_0$. The linear map $\mathcal N\to \mathbb C^ q$ that takes a matrix to its first column, its product with the coordinate vector $(1,0,\ldots ,0)^{\mathsf T}$ of $e_1=1$, maps $N_ l$ to the $l$th unit vector, the coordinate vector of $e_ le_1=e_ l$; so it is bijective. It maps $[u](z'_0)^ j=[u^ j](z'_0)$ to the $j$th column of $\Lambda (z'_0)$, counting from $j=0$. In the diagonal basis, $[u](z'_0)^ j=\operatorname {diag}(\tau _1^ j,\ldots ,\tau _ q^ j)$, and these matrices are linearly independent for $0\leq j<q$, by the Vandermonde determinant, since the $\tau _ i$ are distinct. Hence $\det \Lambda (z'_0)\neq 0$, so $\det \Lambda \neq 0$, and $1,u,\ldots ,u^{q-1}$ is a basis of $L$. Consequently $\mu _ u$ has degree $q$, and since $u\in B$ is integral over $\mathcal O_ d$, its coefficients lie in $\mathcal O_ d$ by (1). This proves (2).

Proof of (3). In the basis $1,u,\ldots ,u^{q-1}$, multiplication by $u$ has the companion matrix $C_ W$, because $u\cdot u^{q-1}=-\sum _ jw_ ju^{q-j}$. Hence $[u]=\Lambda C_ W\Lambda ^{-1}$, and $\operatorname {Tr}(u^ k)=\operatorname {tr}(C_ W^ k)=S_ k$ for $k\geq 0$, with $S_ k$ as in the definition of $\operatorname {Disc}(W)$. The $w_ j$, being the coordinates of $-u^ q$ in the basis $1,\ldots ,u^{q-1}$, are defined at $z'_0$, because $\det \Lambda (z'_0)\neq 0$; and $C_ W(z'_0)=\Lambda (z'_0)^{-1}[u](z'_0)\Lambda (z'_0)$ has the $q$ distinct eigenvalues $\tau _1,\ldots ,\tau _ q$. Since the characteristic polynomial of $C_ W(z'_0)$ is $W(z'_0,\cdot )$, these are its roots, and (9.3.2) shows that the value of $\delta $ at $z'_0$ is not zero; so $\delta \neq 0$. Now let $g\in L$ be integral over $\mathcal O_ d$, and write $g=\sum _{j<q}a_ ju^ j$ with $a_ j\in K$. Multiplying by $u^ l$ and taking traces, we obtain

The elements $gu^ l$ are integral over $\mathcal O_ d$, as products of integral elements, so $\operatorname {Tr}(gu^ l)\in \mathcal O_ d$ by the proof of (1). The matrix $(S_{j+l})$ has entries in $\mathcal O_ d$ and determinant $\delta $; multiplying the system by its adjugate gives $\delta a_ j\in \mathcal O_ d$, which is (3) with $d_ j=\delta a_ j$. $\square$

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. $\square$

Proof. The set $B\cap A_\delta $ is an analytic subset of the connected manifold $A_\delta $. It is a proper subset, for otherwise the closed set $B$ would contain the closure $A\cap (\Delta '\times \Delta '')$ of $A_\delta $, and the two germs would agree. By lemma 9.2.8, $A_\delta \setminus B$ is dense in $A_\delta $, and hence in $A\cap (\Delta '\times \Delta '')$, in which $A_\delta $ is dense. $\square$

Proof. Let $p\in Z$; near points outside $Z$ there is nothing to prove, since $\overline{Z\setminus Y}\subset Z$. Choose representatives $Z_1,\ldots ,Z_ k$ of the branches of $(Z,p)$ on an open neighborhood $U$ of $p$ with $Z\cap U=\bigcup _ iZ_ i$, so small that $Z_ i\subset Y$ whenever $(Z_ i,p)\subset (Y,p)$, and that, for each of the other branches, $Z_ i\setminus Y$ is dense in $Z_ i$; the latter is possible by corollary 9.3.6, applied to $(Z_ i,p)$ and its subgerm $(Z_ i\cap Y,p)$, which differs from $(Z_ i,p)$. Then $(Z\setminus Y)\cap U$ is the union of the sets $Z_ i\setminus Y$ over the branches not contained in $(Y,p)$, and its closure in $U$ is the union of the corresponding sets $Z_ i$, which is analytic in $U$. $\square$

Proof. Every element of $\sqrt I$ vanishes on $V(I)$. For the converse we first show that $\sqrt I$ is the intersection of the prime ideals containing $I$. A prime ideal containing $I$ contains $\sqrt I$. Let $f\notin \sqrt I$, so that the ideals containing $I$ and disjoint from $S=\{ 1,f,f^2,\ldots \} $ form a nonempty family; by proposition 9.2.4 it has a maximal element $\mathfrak p$, since otherwise there would be a strictly increasing infinite sequence of ideals. The ideal $\mathfrak p$ is prime: if $a,b\notin \mathfrak p$, then $\mathfrak p+(a)$ and $\mathfrak p+(b)$ meet $S$, say $f^ k=p_1+xa$ and $f^ l=p_2+yb$ with $p_1,p_2\in \mathfrak p$; multiplying, we see that $ab\in \mathfrak p$ would give $f^{k+l}\in \mathfrak p$, which is impossible. Thus $f$ lies outside some prime ideal containing $I$. Now let $g$ vanish on $V(I)$. For every prime ideal $J\supset I$ we have $V(J)\subset V(I)$, so $g$ vanishes on $V(J)$ and therefore $g\in J$, by the equality $I_{V(J),0}=J$ proved in theorem 9.3.4. Hence $g$ lies in the intersection of all these prime ideals, which is $\sqrt I$. $\square$

Proof. Let $A$ be the representative on a parametrization polydisc. Two local parametrizations of $(A,0)$ have covering loci which are open and dense in $A$ near $0$; their intersection is therefore nonempty, and at its points $A$ is a manifold whose dimension equals both base dimensions. Hence $d$ is determined by the germ. A regular point of $A$ has a neighborhood in $A$ which is a manifold; this neighborhood meets the dense covering locus, which has dimension $d$, so the regular point has dimension $d$ as well. Now let $y\in A$, let $B_1,\ldots ,B_ p$ be representatives of the branches of $(A,y)$, and fix $i$. Since $B_ i$ is not contained in $\bigcup _{k\neq i}B_ k$, corollary 9.3.6 gives points of the covering locus of a parametrization of $B_ i$ outside the other branches. Near such a point, $A$ coincides with $B_ i$ and is a manifold of dimension $\dim B_ i$; such points form a nonempty open subset of $A$, which meets the covering locus of $A$. Hence $\dim B_ i=d$.

Let $(B,0)\subset (A,0)$ be germs with both irreducible, and let $J_ A\subset J_ B$ be their prime ideals. For linear coordinates $z$ and $0\leq k\leq n$, let $\mathcal O_ k$ denote the ring of germs depending only on $z_1,\ldots ,z_ k$. For a nonzero prime ideal $J$, Step 1 of the proof of theorem 9.3.4 chooses the coordinates downward from $k=n$. As long as $J\cap \mathcal O_ k\neq 0$, it makes a linear change of $z_1,\ldots ,z_ k$ in which a chosen nonzero element of $J\cap \mathcal O_ k$ is distinguished in $z_ k$; it stops at the largest $d$ with $J\cap \mathcal O_ d=0$, which is the base dimension. The rest of that proof applies to every such choice. Since finitely many nonzero germs become distinguished in the last coordinate after one suitable linear change of coordinates, we may make these changes for $J_ B$ and, at the stages $k$ with $J_ A\cap \mathcal O_ k\neq 0$, simultaneously for $J_ A$. This gives parametrizations of both germs in common coordinates (if $J_ A=0$, the germ $(A,0)$ is the germ of $\mathbb C^ n$, with $d=n$). Since $J_ A\subset J_ B$, the ideal $J_ B\cap \mathcal O_ k$ is nonzero whenever $J_ A\cap \mathcal O_ k$ is, so $\dim B\leq \dim A$.

Suppose that the dimensions are equal, so that both germs are finite over the same base, and let $\delta _ A$, $\delta _ B$ be the two discriminants. By lemma 9.2.9, the set $A'$ of points of the connected covering locus of $A$ lying over $\{ \delta _ B\neq 0\} $ is a connected manifold of dimension $d$, dense in $A$. The points of $B$ over $\{ \delta _ A\delta _ B\neq 0\} $ form a nonempty submanifold of $A'$ of the same dimension, hence an open subset of $A'$, since the inclusion is an injective immersion between manifolds of the same dimension and therefore open by the inverse function theorem. Thus $B\cap A'$ is an analytic subset of the connected manifold $A'$ with nonempty interior; since proper analytic subsets of connected manifolds are nowhere dense (lemma 9.2.8), $B\supset A'$. Since $B$ is closed and $A'$ is dense in $A$, we get $(B,0)=(A,0)$. Hence a proper subgerm has strictly smaller dimension. For a general proper subgerm $B$ of $(A,0)$, each branch of $B$ is a proper subgerm of $(A,0)$, and the same conclusion follows. $\square$

Proof. The assertion is local. Near a point of $A$, the set $A$ is the union of representatives $A_1,\ldots ,A_ p$ of the branches of its germ, and $\mathcal I_ A=\bigcap _ i\mathcal I_{A_ i}$. The intersection of two coherent subsheaves $\mathcal F,\mathcal G$ of $\mathcal O_ M$ is the kernel of the morphism $\mathcal F\oplus \mathcal G\to \mathcal O_ M$, $(a,b)\mapsto a-b$, and is coherent by theorem 9.2.7, lemma 9.2.6. It therefore suffices to prove that $\mathcal I_ A$ is coherent near $0$ when $(A,0)\subset (\mathbb C^ n,0)$ is irreducible. We use the parametrization of theorem 9.3.4, with its primitive linear form $u$, its Weierstrass polynomial $W$ and $\delta =\operatorname {Disc}(W)$, and we put $m=\max \{ q,(n-d)(q-1)\} $. Since $A$ may have several branches at points near $0$, we describe all stalks near $0$ at once: we construct holomorphic functions $G_1,\ldots ,G_ N$ vanishing on $A$ and a holomorphic function $h\not\equiv 0$ of $z'$ such that, at every point $x$ near $0$, the stalk $\mathcal I_{A,x}$ consists of the germs $f$ with $hf\in (G_{1,x},\ldots ,G_{N,x})$; such a sheaf is coherent by Oka's theorem.

Step 1: dependence on the primitive form. For $c=(c_{d+1},\ldots ,c_ n)\in \mathbb C^{n-d}$ put $u_ c=\sum _ kc_ kz_ k$. For $z'\in \Delta '\setminus \{ \delta =0\} $ let $z^{(1)},\ldots ,z^{(q)}$ be the points of $A$ over $z'$, and define

where $D_ c$ is the Vandermonde determinant $\det (u_ c(z^{(i)})^ j)_{i,j}$ and $D_{kcj}$ is obtained from it by replacing the $j$th column with the coordinates $z_ k^{(i)}$. These expressions do not change under permutations of the points $z^{(i)}$, since $D_ c$ and $D_{kcj}$ both change sign under a transposition. Hence their coefficients in $T$ are polynomials in $c$ whose coefficients are bounded holomorphic functions on $\Delta '\setminus \{ \delta =0\} $, and these extend holomorphically to $\Delta '$ by lemma 9.2.9. On $A_\delta $ we have $W_ c(z',u_ c(z''))=0$ and, by Cramer's rule in the form $\sum _ jD_{kcj}u_ c(z^{(i)})^ j=D_ cz_ k^{(i)}$, also $\delta _ c(z')z_ k=B_{kc}(z',u_ c(z''))$, for every $c$. Let $G_1,\ldots ,G_ N$ be the coefficients, as polynomials in $c$, of the functions $W_ c(z',u_ c)$ and $\delta _ cz_ k-B_{kc}(z',u_ c)$, $d<k\leq n$. They are holomorphic on $\Delta '\times \Delta ''$ and vanish on $A_\delta $, hence on $A$, which is the closure of $A_\delta $. For the coefficient vector $c$ of the primitive linear form $u$, the values $u(z^{(i)})$ are the roots of $W(z',\cdot )$, by theorem 9.3.4, so $\delta _ c=\operatorname {Disc}(W)=\delta $ by (9.3.2), and $\delta _ c\neq 0$ off $\{ \delta =0\} $; so the polynomial $\delta _ c^ m=\sum _\alpha h_\alpha (z')c^\alpha $ in $c$ is not identically zero, and we choose a coefficient $h=h_\alpha \not\equiv 0$.

Step 2: a description of the stalks. We claim that on a small polydisc around $0$,

for every point $x$. If $hf\in (G_{1,x},\ldots ,G_{N,x})$, then $hf$ vanishes on $A$ near $x$, so $f$ vanishes on $A\setminus \{ h=0\} $ near $x$. This set is dense in $A$: the covering locus $A_\delta $ is dense in $A$, and since $\pi $ is a local biholomorphism on $A_\delta $, the set where $h\circ \pi $ vanishes is nowhere dense in $A_\delta $. Hence $f\in \mathcal I_{A,x}$, which proves the inclusion from right to left.

Step 3: the inclusion from left to right at points of $A$. Fix $x\in A$ and $f\in \mathcal I_{A,x}$. Call $c$ admissible if $\delta _ c\not\equiv 0$ and $u_ c$ takes distinct values at the points of the finite fiber of $A$ over $x'$; the admissible $c$ form a dense open subset of $\mathbb C^{n-d}$, the complement of an algebraic subset and finitely many proper linear subspaces. Fix an admissible $c$. Since $W_ c(z',u_ c(z''))$ vanishes on $A$ by Step 1, the number $u_ c(x'')$ is a root of $W_ c(x',\cdot )$. By lemma 9.2.5, $W_ c=W_ c^ xQ_ c^ x$ near $(x',u_ c(x''))$, where $W^ x_ c$ is a Weierstrass polynomial in $T-u_ c(x'')$ and $Q^ x_ c$ is a polynomial in $T$ with $Q^ x_ c(x',u_ c(x''))\neq 0$.

We call $z'$ generic if $\delta (z')\delta _ c(z')\neq 0$. For generic $z'$ near $x'$, the roots of $W^ x_ c(z',\cdot )$ are distinct and are exactly the values of $u_ c$ at the points of $A$ over $z'$ that are close to $x$; let $q_ x$ be their number, which is the degree of $W^ x_ c$. Indeed, the roots of $W_ c(z',\cdot )$ are the $q$ distinct values of $u_ c$ at the points of $A$ over $z'$, and by properness these points are close to the points of the fiber over $x'$. The values of $u_ c$ at the points close to $x$ are close to $u_ c(x'')$, where $Q^ x_ c(z',\cdot )$ does not vanish, so they are roots of $W^ x_ c(z',\cdot )$. The values at the points close to the other points of the fiber are close to values of $u_ c$ different from $u_ c(x'')$, since $c$ is admissible, whereas the roots of the Weierstrass polynomial $W^ x_ c(z',\cdot )$ in $T-u_ c(x'')$ tend to $u_ c(x'')$ as $z'\to x'$, by (9.2.1).

Let $\mathfrak g_ c\subset \mathcal O_{M,x}$ be the ideal generated by the germs at $x$ of $W_ c(z',u_ c)$ and $\delta _ cz_ k-B_{kc}(z',u_ c)$, $d<k\leq n$. These functions vanish on $A$, and they are complex linear combinations of $G_1,\ldots ,G_ N$, so $\mathfrak g_ c\subset (G_{1,x},\ldots ,G_{N,x})$. The germ of $W^ x_ c(z',u_ c)$ belongs to $\mathfrak g_ c$, because it equals $Q^ x_ c(z',u_ c)^{-1}W_ c(z',u_ c)$ and $Q^ x_ c(z',u_ c)$ is a unit at $x$. We use the following divisibility property: if $H(z',T)$ is a polynomial in $T$ whose coefficients are germs at $x'$, and $H(y',u_ c(y''))=0$ at all points $y\in A$ close to $x$ with $y'$ generic, then the germ of $H(z',u_ c)$ belongs to $\mathfrak g_ c$. Indeed, the remainder of the division of $H$ by the monic polynomial $W^ x_ c$ has degree less than $q_ x$, and for generic $z'$ near $x'$ it vanishes at the $q_ x$ distinct roots of $W^ x_ c(z',\cdot )$; so it is zero, and $H(z',u_ c)$ is a multiple of $W^ x_ c(z',u_ c)$.

For $d<k\leq n$ and the unit vector $e_ k$ we have $u_{e_ k}=z_ k$, so by Step 1 the coordinate $z_ k$ of each point of $A$ is a root of the monic polynomial $W_{e_ k}(z',\cdot )$ of degree $q$. Let $W^ x_ k$ be the Weierstrass polynomial in $T-x_ k$ given by lemma 9.2.5 for $W_{e_ k}$ at the root $x_ k$ of $W_{e_ k}(x',\cdot )$, so that $W_{e_ k}=W^ x_ kQ^ x_ k$ near $(x',x_ k)$ with $Q^ x_ k(x',x_ k)\neq 0$. Then $W^ x_ k(z',z_ k)$ vanishes at the points of $A$ close to $x$, and its degree $q_ k$ is at most $q$. Substituting $\delta _ cz_ k\equiv B_{kc}(z',u_ c)$ modulo $\mathfrak g_ c$, we get $\delta _ c^{q_ k}W^ x_ k(z',z_ k)\equiv H_ k(z',u_ c)$ modulo $\mathfrak g_ c$, where

is a polynomial in $T$. At a point $y\in A$ close to $x$ with $y'$ generic, Step 1 gives $B_{kc}(y',u_ c(y''))=\delta _ c(y')y_ k$, so $H_ k(y',u_ c(y''))=\delta _ c(y')^{q_ k}W^ x_ k(y',y_ k)=0$. By the divisibility property, $H_ k(z',u_ c)\in \mathfrak g_ c$; hence $\delta _ c^ qW^ x_ k(z',z_ k)\in \mathfrak g_ c$.

Now divide $f$ successively by the Weierstrass polynomials $W^ x_ k(z',z_ k)$ at $x$, for $k=n,\ldots ,d+1$ (proposition 9.2.2, in coordinates centered at $x$). This writes $f$ as a polynomial $\sum _\alpha e_\alpha (z')z''^\alpha $, with germs $e_\alpha $ at $x'$ and $\alpha _ k<q_ k\leq q$ for all $k$, plus multiples of the $W^ x_ k(z',z_ k)$. After multiplication by $\delta _ c^ m$, these multiples lie in $\mathfrak g_ c$, because $m\geq q$. Substituting $\delta _ cz_ k\equiv B_{kc}(z',u_ c)$, which is possible because $|\alpha |\leq (n-d)(q-1)\leq m$, we obtain $\delta _ c^ mf\equiv H(z',u_ c)$ modulo $\mathfrak g_ c$ for a polynomial $H(z',T)$ whose coefficients are germs at $x'$. Since $f$ and the elements of $\mathfrak g_ c$ vanish on $A$ near $x$, the divisibility property applies to $H$, and we conclude that

The polynomials in $c$ of degree at most $\deg _ c\delta _ c^ m$ form a finite-dimensional space, and the evaluations at points of the dense open set of admissible $c$ span its dual, since a polynomial vanishing on a nonempty open set is zero. Hence each coefficient $h_\alpha $ of $\delta _ c^ m$ is a complex linear combination of finitely many values $\delta _{c^{(l)}}^ m$ with admissible $c^{(l)}$, and $h_\alpha f\in (G_{1,x},\ldots ,G_{N,x})$. This proves (9.3.11) at points of $A$.

Step 4: points outside $A$, and conclusion. Let $x\notin A$. Choose $c$ such that $u_ c(x'')$ differs from the values of $u_ c$ at the finitely many points of $A$ over $x'$. The roots of $W_ c(x',\cdot )$ are values of $u_ c$ at points of $A$ over $x'$. Indeed, choose $z'_ j\to x'$ with $\delta (z'_ j)\neq 0$; by properness, after passing to a subsequence, the $q$ points $z^{(1)}_ j,\ldots ,z^{(q)}_ j$ of $A$ over $z'_ j$ converge to points $y^{(1)},\ldots ,y^{(q)}$ of $A$ over $x'$, so that $W_ c(x',T)=\lim _ j\prod _ i(T-u_ c(z^{(i)}_ j))=\prod _ i(T-u_ c(y^{(i)}))$. Hence $W_ c(x',u_ c(x''))\neq 0$, so one of the coefficients $G_ l$ does not vanish at $x$, and both sides of (9.3.11) are the unit ideal. The right side of (9.3.11) is the projection to the first coordinate of the kernel of the row matrix $(h,-G_1,\ldots ,-G_ N)$, which is coherent by theorem 9.2.7, lemma 9.2.6. Hence $\mathcal I_ A$ is coherent. $\square$

Proof. If $y\in U\setminus N$, then some $w_ l$ does not vanish at $y$, so it is a unit, and $(N,y)$ is empty because $N$ is closed in $U$; both ideals are the unit ideal. Let $y\in N$, and let $g$ be holomorphic on a polydisc centered at $y$ in the coordinates $(v,w)$, vanishing on $N$. Then $g(v,0)=0$, and

where the integrals are holomorphic by differentiation under the integral sign. Hence the germ of $g$ lies in the ideal generated by $w_1,\ldots ,w_ k$; conversely, the $w_ l$ vanish on $N$. Consequently, the map $g\mapsto g(v,0)$ identifies the quotient of $\mathcal O_{M,y}$ by this ideal with the ring of germs of holomorphic functions of $v$ at $v(y)$, which is an integral domain (section 9.2); so $I_{N,y}$ is prime. Since $y\in N$, the germ $(N,y)$ is nonempty, and it is irreducible by proposition 9.3.1. All points of $N$ near $y$ are regular points of $N$ of dimension $\dim _ yM-k$, so this is the dimension of $(N,y)$, by proposition 9.3.9. $\square$

Proof. (1) Let $A$ first be a small representative of an irreducible germ of dimension $e$ in $\mathbb C^ n$, pure-dimensional by proposition 9.3.9, and choose generators $g_1,\ldots ,g_ N$ of $\mathcal I_ A$ on a neighborhood, by theorem 9.3.10. At a regular point, these generate the ideal of the manifold $A$, which is generated by $n-e$ normal coordinates (lemma 9.3.12); differentiating, we see that the matrix of differentials of the $g_ j$ has rank $n-e$ there. Every minor of size $n-e+1$ therefore vanishes on the dense covering locus, hence on $A$, so the rank is at most $n-e$ everywhere on $A$. Conversely, if it equals $n-e$ at $y\in A$, choose $n-e$ of the $g_ j$ with independent differentials at $y$. Together with $e$ suitable coordinate functions they form holomorphic coordinates near $y$, by the holomorphic inverse function theorem (proposition 5.2.4(2)), so their common zero set is an $e$-dimensional submanifold $N$ near $y$, and it contains $A$. Every branch of $(A,y)$ has dimension $e$ and is contained in the germ $(N,y)$, which is irreducible of dimension $e$ by lemma 9.3.12, so it equals $(N,y)$ by proposition 9.3.9; hence $A$ is a manifold near $y$. Thus $A_{\mathrm{sing}}$ is the subset of $A$ where all minors of size $n-e$ of this matrix vanish, an analytic subset.

For a general analytic set $A$ near a point $x$, let $A_1,\ldots ,A_ p$ be small representatives of the branches of $(A,x)$. A point $y$ near $x$ lying on only one $A_ i$ is regular for $A$ if and only if it is regular for $A_ i$. A point $y\in A_ i\cap A_ j$, $i\neq j$, is singular. Indeed, otherwise $(A,y)$ would be irreducible, hence equal to some $(A_ k,y)$, and we may assume $k\neq i$; then $(A_ i,y)\subset (A_ i\cap A_ k,y)$. But $\dim (A_ i\cap A_ k,x)<\dim A_ i$ by proposition 9.3.9, so $\dim (A_ i\cap A_ k,y)<\dim A_ i=\dim (A_ i,y)$ by upper semicontinuity and purity, contradicting monotonicity. This proves the formula for $A_{\mathrm{sing}}$ near $x$, which is therefore analytic by the irreducible case. Finally, let $y\in A$ and let $B$ be a branch of $(A,y)$. By corollary 9.3.6, $B$ contains points of the covering locus of a parametrization of $B$ outside the other branches of $(A,y)$, arbitrarily close to $y$, and these are regular points of $A$. Hence the germ of $A_{\mathrm{sing}}\cap B$ at $y$ differs from $B$, and it has smaller dimension than $B$, by proposition 9.3.9.

(2) By (1), each $A^{(j)}$ is analytic, and $\dim (A^{(j+1)},x)<\dim (A^{(j)},x)$ at each point $x$ of $A^{(j+1)}$, because every branch of $(A^{(j+1)},x)$ lies in a branch of $(A^{(j)},x)$ and in the singular locus. Hence $A^{(j)}$ has dimension at most $d-j$ at each of its points, and $A^{(d+1)}=\varnothing $, a set of dimension $-1$ being empty. Each $A^{(j)}_{\mathrm{reg}}=A^{(j)}\setminus A^{(j+1)}$ is a complex submanifold of $M\setminus A^{(j+1)}$ of dimension at most $d-j$, and $A$ is the union of these sets.

(3) Each of the submanifolds in (2) is locally a countable union of coordinate charts, so $A$ is locally a countable union of real submanifolds of dimensions at most $2d$. Each such chart is a countable union of images of compact cubes in $\mathbb R^ r$, $r\leq 2d$, under Lipschitz maps, because the image of the chart in $\mathbb R^ r$ is a countable union of compact cubes, on each of which the smooth inverse of the chart is Lipschitz, by the mean value inequality. Such an image is $\mathcal H^ s$-null for $s>r$: covering the cube by $O(\delta ^{-r})$ cubes of side $\delta $, whose images have diameter $O(\delta )$, we cover the image by $O(\delta ^{-r})$ balls of radii $O(\delta )$, and the sum of the $s$th powers of these radii is $O(\delta ^{s-r})$, which tends to zero with $\delta $. A countable union of $\mathcal H^ s$-null sets is $\mathcal H^ s$-null, which completes the proof. $\square$

Proof. After permuting the coordinates of the source and of the target, we may assume that the matrix $(\partial F_ i/\partial x_ j(p))_{i,j\leq r}$ is invertible. The map $\phi (x)=(F_1(x),\ldots ,F_ r(x),x_{r+1},\ldots ,x_ m)$ has invertible derivative at $p$, so by the holomorphic inverse function theorem (proposition 5.2.4(1)) it maps a neighborhood of $p$ biholomorphically onto a polydisc $Q'\times Q''$ centered at $\phi (p)$, on which the rank of $G=F\circ \phi ^{-1}$ is $r$. This map has the form $G(y)=(y',G''(y))$ with $y'=(y_1,\ldots ,y_ r)$ and $G''=(G_{r+1},\ldots ,G_ K)$. For $i,j>r$, the minor of the Jacobian matrix of $G$ formed by the rows $1,\ldots ,r,i$ and the columns $1,\ldots ,r,j$ equals $\partial G_ i/\partial y_ j$, because the first $r$ rows are $(I,0)$; since the rank is $r$, these derivatives vanish. As $Q''$ is convex, $G''(y',y'')=G''(y',c'')$, where $c''$ is the center of $Q''$. The map $\psi (w',w'')=(w',w''+G''(w',c''))$ is a biholomorphism of $Q'\times \mathbb C^{K-r}$ onto itself, with inverse $(w',w'')\mapsto (w',w''-G''(w',c''))$, and $G(y',y'')=\psi (y',0)$. The fibers of $G$ are the slices $\{ y'\} \times Q''$, and its image is $\psi (Q'\times \{ 0\} )$, a closed submanifold of $Q'\times \mathbb C^{K-r}$. $\square$

Proof. We use the properties of $\mathcal H^ s$-null sets recalled in section 5.11: subsets and countable unions of such sets, and their images under Lipschitz maps, are again $\mathcal H^ s$-null, and an $\mathcal H^ s$-null subset of $\mathbb R^ b$ with $s\leq b$ is Lebesgue null. Moreover, the image of a compact cube in $\mathbb R^ r$ under a Lipschitz map is $\mathcal H^ s$-null for $s>r$, by corollary 9.3.13(3), and a $C^1$ map is Lipschitz on every compact convex subset of its domain, by the mean value inequality.

(1) We argue by induction on the dimension of $N$. Since $N$ has at most countably many connected components, each of them open with a countable basis, we may assume that $N$ is connected, of dimension $n$. If $n=0$, then $F(N)$ is a point. Let $n\geq 1$, and let $r\leq k$ be the maximal rank of $\mathrm dF$ on $N$. The set $N_1$ of points where the rank is less than $r$ is the common zero set of the minors of size $r$ of the Jacobian matrix in local coordinates, so it is an analytic subset of $N$ different from $N$. Hence it is nowhere dense, and its germs, being proper subgerms of the irreducible manifold germs of dimension $n$ (lemma 9.3.12), have dimension at most $n-1$ by proposition 9.3.9. On the open set $N\setminus N_1$ the rank is constantly $r$.

Let $p\in N\setminus N_1$, and choose $\phi $, $\psi $ and $Q'\times Q''$ as in lemma 9.3.14, in a chart around $p$. Let $C'\subset Q'$ be a compact cube, a product of closed squares, containing the first $r$ coordinates of $\phi (p)$ in its interior, and let $V=\phi ^{-1}(\operatorname{Int}C'\times Q'')$. Then $F(V)\subset \psi (C'\times \{ 0\} )$, the image of a compact cube in $\mathbb R^{2r}$ under the restriction of $\psi $, which is Lipschitz there because $C'\times \{ 0\} $ is a compact convex subset of the convex domain of $\psi $. Hence $F(V)$ is $\mathcal H^ s$-null for $s>2r$, in particular for $s>2k$. Countably many such sets $V$ cover $N\setminus N_1$, because $N$ has a countable basis. By corollary 9.3.13(2), $N_1$ is the union of finitely many complex submanifolds of open subsets of $N$, of dimensions at most $n-1$; they have countable bases, and the restriction of $F$ to each of them has rank at most $k$. By the induction hypothesis their images are $\mathcal H^ s$-null for $s>2k$, and therefore so is $F(N)$.

(2) Suppose that $\dim (Y,y)\geq k+1$ for some $y\in Y$, and let $Y_1$ be a branch of $(Y,y)$ of dimension $e\geq k+1$. By corollary 9.3.6, $Y_1$ contains points of the covering locus of a parametrization of $Y_1$ outside the other branches of $(Y,y)$; near such a point $p$, the set $Y$ is a complex submanifold of dimension $e$, so it contains a complex submanifold $P$ of dimension $k+1$ through $p$, for instance a coordinate slice. The orthogonal projection $\Pi $ of $\mathbb C^ K$ onto the tangent space $T_ pP$, a complex subspace of dimension $k+1$, is $1$-Lipschitz, and its restriction to $P$ has invertible derivative at $p$. By the inverse function theorem, $\Pi (P)$ contains a nonempty open subset of $T_ pP\cong \mathbb R^{2k+2}$. On the other hand, $\Pi (P)\subset \Pi (Y)$ is $\mathcal H^{2k+2}$-null and hence Lebesgue null. This contradiction proves (2).

(3) By corollary 9.3.13(2), $A$ is the union of finitely many complex submanifolds $A_ l$ of open subsets of $N$, of dimensions at most $d$. Let $(V,\chi )$ be a holomorphic chart of $M$ around a point of $F(A)$, with values in $\mathbb C^ K$. The restriction of $\chi \circ F$ to the open subset $A_ l\cap F^{-1}(V)$ of $A_ l$ has rank at most $\dim A_ l\leq d$, so by (1) the set $\chi (F(A)\cap V)=\bigcup _ l\chi \circ F(A_ l\cap F^{-1}(V))$ is $\mathcal H^{2d+2}$-null. Near each of its points it is an analytic subset of an open subset of $\mathbb C^ K$, so (2) shows that $\dim (F(A),w)\leq d$ for $w\in F(A)\cap V$. $\square$

Proof. Fix $x\in A$, let $A_1,\ldots ,A_ p$ be representatives of the branches of $(A,x)$, and for each $i$ choose a parametrization polydisc $P_ i$ of $(A_ i,x)$ so small that $A\cap P_ i=\bigcup _ j A_ j\cap P_ i$. By corollary 9.3.13(1), $A_{\mathrm{sing}}\cap A_ i$ is an analytic subset of $A_ i$ whose germ at $x$ differs from $(A_ i,x)$. Hence $A_{\mathrm{sing}}$ meets the connected covering locus of $A_ i$ in a proper analytic subset, and by lemma 9.2.9 and corollary 9.3.6, the covering locus minus $A_{\mathrm{sing}}$ is connected and dense in $A_ i\cap P_ i$. The set $R_ i=A_ i\cap P_ i\cap A_{\mathrm{reg}}$ lies between this connected set and its closure, so it is connected and dense in $A_ i\cap P_ i$. In particular $A_{\mathrm{reg}}$ is dense in $A$.

Let $\{ S_\alpha \} $ be the connected components of the manifold $A_{\mathrm{reg}}$, and $Z_\alpha =\overline{S_\alpha }$. Each $R_ i$ above is connected, so it lies in a single component $S_{\alpha (i)}$. Let $U=\bigcap _ iP_ i$. Then $S_\alpha \cap U$ is contained in the union of the $A_ i$ with $\alpha (i)=\alpha $, because points of $A_{\mathrm{reg}}\cap A_ i\cap U$ lie in $R_ i$; conversely $A_ i\cap U\subset \overline{R_ i}\subset Z_{\alpha (i)}$. Hence

Thus each $Z_\alpha $ is analytic, only finitely many of them meet $U$, and their union is $A$. Since $A_{\mathrm{reg}}$ is open in $A$ and its connected components are open in it, $S_\alpha $ is open in $A$. So every point of $S_\alpha $ has a neighborhood that meets no other $S_\beta $, and therefore does not lie in $Z_\beta =\overline{S_\beta }$ for $\beta \neq \alpha $. Thus $S_\alpha \cap Z_\beta =\varnothing $ for $\beta \neq \alpha $, and no $Z_\alpha $ is contained in the union of the others. The same remark shows that $S_\alpha \subset (Z_\alpha )_{\mathrm{reg}}$: near a point of $S_\alpha $, the set $Z_\alpha $ coincides with the manifold $A$. Hence $(Z_\alpha )_{\mathrm{reg}}$ lies between the connected set $S_\alpha $ and its closure, and is connected.

If $Z_\alpha $ were the union of two analytic subsets $Z'$, $Z''$ different from it, then $S_\alpha \cap Z'$ and $S_\alpha \cap Z''$ would be analytic subsets of the connected manifold $S_\alpha $ different from $S_\alpha $, since otherwise their closure would be $Z_\alpha $. As proper analytic subsets of a connected manifold, they would be closed and nowhere dense in $S_\alpha $ (lemma 9.2.8), and so would their union, which is $S_\alpha $. This contradiction shows that $Z_\alpha $ is irreducible.

If $M$ has a countable basis, so does $A_{\mathrm{reg}}$; each $S_\alpha $ is open in $A_{\mathrm{reg}}$ and contains a basic open set, and disjoint components contain different ones, so there are at most countably many components. Finally, a holomorphic function vanishing on a nonempty open subset of the connected manifold $(Z_\alpha )_{\mathrm{reg}}$ vanishes on all of it by the identity theorem, and hence on its closure $Z_\alpha $.

It remains to prove uniqueness. Let $A=\bigcup _\beta Y_\beta $ be a locally finite decomposition into irreducible analytic subsets, none contained in another. A connected manifold is not a locally finite union of closed nowhere dense subsets, since such a union is closed and nowhere dense. Since proper analytic subsets of the connected manifold $S_\alpha $ are nowhere dense (lemma 9.2.8), each $S_\alpha $ lies in some $Y_\beta $, and then $Z_\alpha \subset Y_\beta $. Conversely, fix $\beta $. By what we have proved, applied to the analytic set $Y_\beta $, the set $Y_\beta $ is the union of its irreducible components, and it has only one, for otherwise it would be the union of one of them and of the union of the others, two analytic subsets different from $Y_\beta $. Hence $(Y_\beta )_{\mathrm{reg}}$ is a connected manifold, dense in $Y_\beta $ and covered by the locally finite family of closed analytic subsets $(Y_\beta )_{\mathrm{reg}}\cap Z_\alpha $; as before, $Y_\beta \subset Z_\alpha $ for some $\alpha $. Choosing $\beta '$ with $Z_\alpha \subset Y_{\beta '}$, we get $Y_\beta \subset Y_{\beta '}$, so $\beta '=\beta $ and $Y_\beta =Z_\alpha $. Thus every $Y_\beta $ is an irreducible component of $A$; and every component $Z_\alpha $, being contained in some $Y_\beta =Z_{\alpha '}$, equals it, because $S_\alpha \cap Z_{\alpha '}=\varnothing $ for $\alpha '\neq \alpha $. $\square$

Proof. By the uniqueness in corollary 9.3.16, applied to the family consisting of $Z$ alone, $Z$ is its own only irreducible component; so $Z_{\mathrm{reg}}$ is connected and dense in $Z$, and its dimension is constant. (1) Let $B$ be a branch of $(Z,x)$. By corollary 9.3.6, $B$ contains points of the covering locus of a parametrization of $B$ outside the other branches of $(Z,x)$. Near such a point, $Z$ coincides with $B$ and is a manifold of dimension $\dim B$, so the point lies in $Z_{\mathrm{reg}}$, and $\dim B=\dim Z_{\mathrm{reg}}$.

(2) The analytic subset $Y\cap Z_{\mathrm{reg}}$ of the connected manifold $Z_{\mathrm{reg}}$ has empty interior in $Z_{\mathrm{reg}}$: otherwise it would be all of $Z_{\mathrm{reg}}$, by lemma 9.2.8, and the closed set $Y$ would contain the closure $Z$ of $Z_{\mathrm{reg}}$. Since $Z_{\mathrm{reg}}$ is open and dense in $Z$, it follows that the closed set $Y$ has empty interior in $Z$. Now let $x\in Y$. Every branch $Y_1$ of $(Y,x)$ is contained in a branch $B$ of $(Z,x)$, and it differs from $B$: otherwise $Y$ would contain the points of $B$ near $x$ outside the other branches of $(Z,x)$, which form a nonempty open subset of $Z$ by corollary 9.3.6. By proposition 9.3.9 and (1), $\dim Y_1<\dim B=\dim Z$.

(3) Let $e=\dim Z=\dim Z'$. By corollary 9.3.13(1) and by (1), $Z'_{\mathrm{sing}}$ has dimension less than $e$ at each of its points, so by (2), applied to $Z'_{\mathrm{sing}}\cap Z\subset Z$, there is a point $p\in Z_{\mathrm{reg}}\setminus Z'_{\mathrm{sing}}$. Near $p$, the set $Z$ is a submanifold of dimension $e$ of the manifold $Z'$ of dimension $e$; since the germ of a manifold is irreducible (lemma 9.3.12) and proper subgerms of irreducible germs have smaller dimension (proposition 9.3.9), $Z$ and $Z'$ coincide near $p$. Thus the analytic subset $Z$ of $Z'$ has interior points in $Z'$, and $Z=Z'$ by (2).

(4) Let $Z\subset \bigcup _\beta Y_\beta $ with $(Y_\beta )$ locally finite. The sets $Z_{\mathrm{reg}}\cap Y_\beta $ are closed analytic subsets of the connected manifold $Z_{\mathrm{reg}}$ with union $Z_{\mathrm{reg}}$. A locally finite union of closed nowhere dense sets is closed and nowhere dense, so one of them has interior points and is therefore all of $Z_{\mathrm{reg}}$. Then $Z\subset Y_\beta $, because $Y_\beta $ is closed and $Z_{\mathrm{reg}}$ is dense in $Z$. $\square$

Proof. The assertion is local, and extensions defined on overlapping charts agree, since their values on $A$ are given by the same formula. Near a point where $M$ has dimension at most one, the codimension condition forces $A=\varnothing $. Otherwise, let $\Omega $ be a connected coordinate chart of dimension $n\geq 2$. Then $\dim A\leq n-2$, and corollary 9.3.13(3) shows that $A\cap \Omega $ is locally $\mathcal H^{2n-3}$-null, since $2n-3>2(n-2)$. By theorem 5.11.3, $u$ extends uniquely to a plurisubharmonic function on $\Omega $ with the stated values on $A$; being upper semicontinuous, the extension is locally bounded above. $\square$

Proof. By corollary 9.3.17, $D_{\mathrm{reg}}$ is a connected manifold, dense in $D$, and all branches of $D$ at all points have the dimension $n-1$ of $D_{\mathrm{reg}}$. Let $x\in D$ and let $D_1,\ldots ,D_ k$ be representatives of the branches of $(D,x)$. We construct a generator $f_ i$ of the ideal of each $D_ i$ near $x$, and show that $f=f_1\cdots f_ k$ has the required properties.

Step 1: an equation of one branch. Fix $i$, and take a parametrization of $(D_ i,x)$ as in theorem 9.3.4, in coordinates centered at $x$. Its base dimension is $d=n-1$, by proposition 9.3.9, so $z''=z_ n$ is a single coordinate. For $z'\in \Delta '\setminus \{ \delta =0\} $, the $q$ points of $D_ i$ over $z'$ are distinct, and therefore so are their coordinates $z_ n$; put $P(z',T)=\prod _{z\in \pi ^{-1}(z')}(T-z_ n)$. The coefficients of $P$ are, up to sign, the elementary symmetric functions of these coordinates; they are holomorphic, because the sheets are graphs of holomorphic maps, and bounded, because $\Delta ''$ is bounded, so they extend holomorphically to $\Delta '$ by lemma 9.2.9. As $z'\to 0$, the points of $D_ i$ over $z'$ tend to $0$, because $\pi $ is proper and $\pi ^{-1}(0)=\{ 0\} $; hence the extended coefficients vanish at $0$, and $P$ is a Weierstrass polynomial of degree $q$ in $z_ n$. The function $P(z',z_ n)$ vanishes on the covering locus, hence on $D_ i\cap (\Delta '\times \Delta '')$, by density and continuity.

The germ of $P(z',z_ n)$ generates the prime ideal $I_{D_ i,x}$. Indeed, let $g\in I_{D_ i,x}$, and let $R$ be the remainder of the division of $g$ by $P$ (proposition 9.2.2), a polynomial in $z_ n$ of degree less than $q$ with coefficients holomorphic near $0$ in $\mathbb C^{n-1}$; it vanishes on $D_ i$ near $0$, since $g$ and $P(z',z_ n)$ do. For $z'$ near $0$ with $\delta (z')\neq 0$, the $q$ points of $D_ i$ over $z'$ are close to $0$, so $R(z',\cdot )$ vanishes at their $q$ distinct coordinates $z_ n$ and is zero. The coefficients of $R$ thus vanish on a dense subset of a neighborhood of $0$, and $R=0$.

Let $f_ i$ be a holomorphic function near $x$ with the germ of $P(z',z_ n)$. The sheaf $\mathcal I_{D_ i}$ has finitely many generators near $x$ by theorem 9.3.10, and each of them is a multiple of $f_ i$ at $x$, hence on a neighborhood; so $f_ i$ generates $\mathcal I_{D_ i,y}$ for $y$ near $x$. In particular, at every point near $x$ outside $D_ i$ the ideal of $D_ i$ is the unit ideal, so $f_ i$ does not vanish there, and $f_ i^{-1}(0)=D_ i$ near $x$.

Step 2: the product $f=f_1\cdots f_ k$. Let $y\in D_{\mathrm{reg}}$ be close to $x$. By corollary 9.3.13(1), $y$ lies on only one $D_ i$, and $D=D_ i=\{ \tau =0\} $ near $y$ for a local coordinate $\tau $; then $f_ i$ is a unit times $\tau $, since both generate $\mathcal I_{D_ i,y}$ (lemma 9.3.12), and $f_ j(y)\neq 0$ for $j\neq i$, by Step 1. Thus $f$ vanishes to order one along $D_{\mathrm{reg}}$ near $x$, and $f^{-1}(0)=D$ near $x$. Choose a neighborhood $V$ of $x$ on which all this holds and on which, in addition, $D_ i\setminus \bigcup _{j\neq i}D_ j$ is dense in $D_ i$ for every $i$; this is possible by corollary 9.3.6, since the branches are irredundant. Let $y\in V$ and $g\in \mathcal I_{D,y}$; we show by induction on $l$ that $g\in (f_1\cdots f_ l)_ y$. For $l=0$ there is nothing to prove. If $g=f_1\cdots f_ lh$ with $l<k$, then $h$ vanishes near $y$ on $D_{l+1}\setminus (D_1\cup \cdots \cup D_ l)$, where the $f_ i$, $i\leq l$, have no zeros; this set is dense in $D_{l+1}$, so $h$ vanishes on $D_{l+1}$ near $y$, and $h\in (f_{l+1})_ y$. Hence $f$ generates $\mathcal I_{D,y}$.

Finally, in the notation of section 5.5, the regular part of the zero set of $f$ is $D_{\mathrm{reg}}\cap V$, with multiplicity one. By theorem 5.5.6, $[D]=\mathrm{dd}^{\mathrm{c}}\log |f|^2$ on $V$; hence $[D]$ is closed and positive, with measure coefficients of locally finite mass. $\square$

Proof. Working on each connected component of $M$ separately, we may assume that $M$ is connected, of dimension $n$.

Step 1: supports of high codimension. Suppose first that, near a point, the support of $R$ lies in a real submanifold $\{ g_1=\cdots =g_ k=0\} $ with independent differentials $\mathrm dg_1,\ldots ,\mathrm dg_ k$, $k\geq 3$. Since $R$ has measure coefficients and $g_ j$ vanishes on its support, $g_ jR=0$; since $R$ is closed, $\mathrm dg_ j\wedge R=\mathrm d(g_ jR)=0$. Take real coordinates $x$ whose first three are $g_1,g_2,g_3$. For $l\leq 3$ and $l\notin \{ a,b\} $, the coefficient of $\mathrm dx_ l\wedge \mathrm dx_ a\wedge \mathrm dx_ b$ in $\mathrm dx_ l\wedge R$ is, up to sign, the coefficient of $\mathrm dx_ a\wedge \mathrm dx_ b$ in $R$, so this coefficient vanishes. Since no pair $\{ a,b\} $ contains all three indices $1,2,3$, every coefficient of $R$ vanishes, and $R=0$ near the point. Now let $A$ have codimension at least two. Near a point of $A_{\mathrm{reg}}$, the set $A$ is a complex submanifold of real codimension at least four, so $R$ vanishes there, and the support of $R$ lies in $A_{\mathrm{sing}}$. This is analytic of smaller dimension near each point by corollary 9.3.13(1), and again of codimension at least two; repeating the argument, we find after finitely many steps near each point that $R=0$. This proves the last assertion.

Step 2: regular points of codimension one. Let $y\in A_{\mathrm{reg}}$ be a point where $A$ is a complex hypersurface. Choose holomorphic coordinates $(\zeta ,w)$ near $y$ with $A=\{ \zeta =0\} $, write $\zeta =\xi +\mathrm i\eta $, and let $t\in \mathbb R^{2n-2}$ be the real coordinates of $w$. The identities $\mathrm d\xi \wedge R=\mathrm d\eta \wedge R=0$ of Step 1 force $R=\mu \, \mathrm d\xi \wedge \mathrm d\eta $ for a real measure $\mu $ supported on $\{ \xi =\eta =0\} $, so $\mu =\delta _0(\xi ,\eta )\otimes \sigma (t)$ for a real measure $\sigma $. Since $R$ is closed, $\mathrm dR=\sum _ l(\partial \mu /\partial t_ l)\, \mathrm dt_ l\wedge \mathrm d\xi \wedge \mathrm d\eta $ vanishes, the derivatives in $\xi $ and $\eta $ disappearing after the exterior product with $\mathrm d\xi \wedge \mathrm d\eta $; so $\partial \mu /\partial t_ l=0$ as distributions. Testing this against functions $\chi (\xi ,\eta )\psi (t)$ with $\chi =1$ near $0$, we find that all distributional derivatives $\partial \sigma /\partial t_ l$ vanish. On a smaller ball in the $t$-variables, the convolutions of $\sigma $ with a smoothing kernel therefore have zero gradient and are constant; integrating them against a fixed test function of integral one, we see that these constants converge as the kernel shrinks, and that $\sigma $ is a constant multiple $c$ of Lebesgue measure. Hence $R=c[\{ \zeta =0\} ]$ near $y$, with $c$ real. (If $n=1$, $\sigma $ is a measure on a point and the same conclusion holds.)

Step 3: conclusion. Let $D=Z_\alpha $ be an irreducible component of codimension one, the closure of the connected component $S_\alpha $ of $A_{\mathrm{reg}}$. By Step 2, near each point of $S_\alpha $ we have $R=c[D]$ for a real constant $c$, and this constant is locally constant on the connected manifold $S_\alpha $; we call it $c_ D$. By corollary 9.3.16 the sum $\sum _ Dc_ D[D]$ is locally finite, and by lemma 9.3.19 it is a closed real current with measure coefficients. The difference $R'=R-\sum _ Dc_ D[D]$ vanishes near every point of $A_{\mathrm{reg}}$: near a point of $S_\alpha $ with $Z_\alpha $ of codimension one by Step 2, since the other components do not meet $S_\alpha $; and near a point of $A_{\mathrm{reg}}$ where $A$ has codimension at least two by Step 1, since no component of codimension one passes through such a point. Thus the support of $R'$ lies in $A_{\mathrm{sing}}$.

We show that $A_{\mathrm{sing}}$ has codimension at least two; then $R'=0$ by Step 1. A branch of $A$ of dimension $n$ at a point $y$ would be a subgerm of the germ $(M,y)$, which is irreducible of dimension $n$ by lemma 9.3.12, hence equal to it by proposition 9.3.9; then $A$ would contain a neighborhood of $y$, which is impossible because $A$ is nowhere dense. So the branches of $A$ have dimension at most $n-1$, and $A_{\mathrm{sing}}$, which has smaller dimension than each branch (corollary 9.3.13(1)), has codimension at least two. The coefficients are unique, because near a point of $S_\alpha $ the current $R$ determines $c_ D$; and if $R\geq 0$, then $c_ D\geq 0$, because $[D]$ is positive and nonzero near such a point. $\square$