The local ring of holomorphic germs

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Every local question about holomorphic functions of several variables becomes a question about the ring \(\mathcal O_n\) of convergent power series. This lesson establishes the algebraic structure of that ring: it is a regular local ring of dimension \(n\) and a unique factorization domain, coprimality of two germs persists at nearby points, and quotients by ideals containing Weierstrass polynomials are finite modules over a ring of convergent power series in fewer variables. The tool throughout is the Weierstrass preparation and division theorem, which turns a germ into a polynomial in one variable whose coefficients are germs in the others.

We use Holomorphic functions of several variables and the proof of the Weierstrass preparation and division theorem in Complex analytic spaces and analytification, Section 2, together with its Lemma 2.1: \(\mathcal O_n\) is Noetherian and its completion is the formal power series ring. From commutative algebra we use Krull's height theorem Dimension theory of Noetherian local rings, Theorem 3.1 and the facts that a polynomial ring over a unique factorization domain is again one and that factorial domains are integrally closed, with the proofs cited where they are used.

Basic references are [Demailly] and [Lebl SCV].

1. The ring of germs

Let \(\mathcal O_n=\mathbf C\{z_1,\ldots,z_n\}\) be the ring of germs at \(0\) of holomorphic functions, equivalently of power series converging on some polydisc about \(0\). A germ with nonzero constant term is a unit, because its reciprocal is holomorphic near \(0\); the other germs form the unique maximal ideal

\[ \mathfrak m_n=\{f\in\mathcal O_n:\ f(0)=0\}=(z_1,\ldots,z_n). \tag{1.1} \]

The last equality holds because a germ with \(f(0)=0\) can be written \(f=\sum_j z_jf_j\), grouping the terms of its power series by the first variable that occurs in them. For \(k\leq n\) we regard \(\mathcal O_k=\mathbf C\{z_1,\ldots,z_k\}\) as the subring of germs independent of \(z_{k+1},\ldots,z_n\). The ring \(\mathcal O_n\) is an integral domain: if \(fg=0\) with \(f\neq0\), then \(f\) does not vanish on some nonempty open set near \(0\), so \(g\) vanishes there and \(g=0\) by the identity theorem Holomorphic functions of several variables, Theorem 2.3.

Proposition 1.1. \(\mathcal O_n\) is a regular local ring of Krull dimension \(n\). The classes of \(z_1,\ldots,z_n\) form a basis of \(\mathfrak m_n/\mathfrak m_n^2\).

Proof. Taylor expansion identifies \(\mathfrak m_n/\mathfrak m_n^2\) with the space of linear forms, so \(\dim_{\mathbf C}\mathfrak m_n/\mathfrak m_n^2=n\). The ideals \(\mathfrak p_k=(z_1,\ldots,z_k)\), \(0\leq k\leq n\), are prime, since \(\mathcal O_n/\mathfrak p_k\cong\mathbf C\{z_{k+1},\ldots,z_n\}\) is a domain (a germ lies in \(\mathfrak p_k\) exactly when it vanishes on \(z_1=\cdots=z_k=0\), by the grouping used for (1.1)). The chain \(\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_n\) shows \(\dim\mathcal O_n\geq n\). By Krull's height theorem, the maximal ideal, generated by \(n\) elements, has height at most \(n\); so \(\dim\mathcal O_n=n\), which equals the number of generators of \(\mathfrak m_n\). This is the definition of a regular local ring. \(\square\)

2. Weierstrass polynomials

Write \(z=(z',z_n)\) with \(z'=(z_1,\ldots,z_{n-1})\). A Weierstrass polynomial of degree \(s\) in \(z_n\) is a monic polynomial

\[ P(z',z_n)=z_n^s+a_1(z')z_n^{s-1}+\cdots+a_s(z'),\qquad a_j\in\mathcal O_{n-1},\ a_j(0)=0, \tag{2.1} \]

equivalently a monic element of \(\mathcal O_{n-1}[z_n]\) with \(P(0,z_n)=z_n^s\). A germ \(g\in\mathcal O_n\) is regular of order \(s\) in \(z_n\) if \(g(0,z_n)=z_n^sv(z_n)\) with \(v(0)\neq0\).

Theorem 2.1 (Weierstrass preparation and division). Let \(g\in\mathcal O_n\) be regular of order \(s\) in \(z_n\).

  1. There are a unique unit \(u\in\mathcal O_n\) and a unique Weierstrass polynomial \(P\) of degree \(s\) with \(g=uP\).
  2. Every \(f\in\mathcal O_n\) can be written uniquely as \(f=qg+R\) with \(q\in\mathcal O_n\) and \(R\in\mathcal O_{n-1}[z_n]\) of degree less than \(s\) in \(z_n\).

Proof. This is proved in Complex analytic spaces and analytification, Section 2, by C. L. Siegel's contour integral argument. \(\square\)

By Holomorphic functions of several variables, Lemma 4.1, every nonzero germ becomes regular in \(z_n\) after a linear change of coordinates, and the admissible directions form a dense open set. We record the order of the coefficients.

Proposition 2.2. Let \(g\in\mathcal O_n\) vanish to order exactly \(m\) at \(0\), with lowest homogeneous part \(g_m\), and suppose \(g_m(0,\ldots,0,1)\neq0\). Then \(g\) is regular of order \(m\) in \(z_n\), and its Weierstrass polynomial \(z_n^m+\sum_ka_k(z')z_n^{m-k}\) has \(a_k\) vanishing to order at least \(k\) at \(0\).

Proof. \(g(0,z_n)=g_m(0,\ldots,0,1)z_n^m+O(z_n^{m+1})\), so \(g\) is regular of order \(m\). Write \(g=uP\). Since \(u(0)\neq0\), \(P\) also vanishes to order exactly \(m\). Suppose some \(a_k\) vanished to order \(\ell_k<k\), and let \(D=\min_k(\ell_k+m-k)<m\), the minimum over such \(k\). The homogeneous part of degree \(D\) of \(P\) is \(\sum_k(a_k)_{D-m+k}z_n^{m-k}\), where \((a_k)_j\) is the homogeneous part of degree \(j\) of \(a_k\). The terms carry distinct powers of \(z_n\), and at least one of them is nonzero, so this part is nonzero. This contradicts that \(P\) vanishes to order \(m\). \(\square\)

Lemma 2.3. Let \(P\in\mathcal O_{n-1}[z_n]\) be a Weierstrass polynomial and \(F\in\mathcal O_{n-1}[z_n]\). If \(P\) divides \(F\) in \(\mathcal O_n\), then \(P\) divides \(F\) in \(\mathcal O_{n-1}[z_n]\).

Proof. Since \(P\) is monic, Euclidean division in \(\mathcal O_{n-1}[z_n]\) gives \(F=PQ+R\) with \(\deg R<\deg P\). If also \(F=Ph\) with \(h\in\mathcal O_n\), the uniqueness in Theorem 2.1(2), applied to \(f=F\) and \(g=P\), gives \(h=Q\) and \(R=0\). \(\square\)

Lemma 2.4. Let \(P\) be a Weierstrass polynomial of degree \(s\).

  1. If \(P=P_1\cdots P_k\) with \(P_j\in\mathcal O_{n-1}[z_n]\) of positive degree, then each leading coefficient is a unit of \(\mathcal O_{n-1}\), and after dividing each \(P_j\) by its leading coefficient (and \(P_1\) by the inverse product), every \(P_j\) is a Weierstrass polynomial.
  2. \(P\) is irreducible in \(\mathcal O_n\) if and only if it is irreducible in \(\mathcal O_{n-1}[z_n]\).

Proof. (1) The product of the leading coefficients is \(1\), so each is a unit. Normalize so that each \(P_j\) is monic. Setting \(z'=0\) gives \(z_n^s=\prod_jP_j(0,z_n)\) in \(\mathbf C[z_n]\); by unique factorization there, each \(P_j(0,z_n)\) is a power of \(z_n\), so each \(P_j\) is a Weierstrass polynomial.

(2) Suppose \(P=g_1g_2\) with nonunits \(g_1,g_2\in\mathcal O_n\). Then \(z_n^s=g_1(0,z_n)g_2(0,z_n)\), so each \(g_i\) is regular of some order \(s_i\), and \(s_i>0\) because \(g_i(0)=0\); also \(s_1+s_2=s\). Write \(g_i=u_iP_i\) by Theorem 2.1. Then \(P\) divides the monic polynomial \(P_1P_2\) of the same degree in \(\mathcal O_n\), hence in \(\mathcal O_{n-1}[z_n]\) by Lemma 2.3, so \(P=P_1P_2\) is reducible in \(\mathcal O_{n-1}[z_n]\). Conversely a factorization in \(\mathcal O_{n-1}[z_n]\) into factors of positive degree gives, by (1), a factorization into Weierstrass polynomials of positive degree, which are nonunits of \(\mathcal O_n\). A factor of degree zero dividing a monic polynomial is a unit of \(\mathcal O_{n-1}\). \(\square\)

3. Unique factorization

Theorem 3.1. \(\mathcal O_n\) is a unique factorization domain.

Proof. Induction on \(n\); \(\mathcal O_0=\mathbf C\) is a field. Let \(n\geq1\) and assume \(\mathcal O_{n-1}\) is factorial. Then \(\mathcal O_{n-1}[z_n]\) is factorial Stacks, Tag 0BC1. Every nonzero nonunit \(f\in\mathcal O_n\) is, after a linear change of coordinates, a unit times a Weierstrass polynomial; a factorization of that polynomial into irreducible elements of \(\mathcal O_{n-1}[z_n]\) of maximal length consists, by Lemma 2.4, of Weierstrass polynomials irreducible in \(\mathcal O_n\). So factorizations exist. For uniqueness it suffices that an irreducible \(g\) dividing \(f_1f_2\) divides \(f_1\) or \(f_2\). After one linear change of coordinates that makes \(g,f_1,f_2\) all regular in \(z_n\) (Lemma 4.1 of the previous lesson, applied to the product \(gf_1f_2\)), we may replace them by their Weierstrass polynomials. By Lemmas 2.3 and 2.4(2), \(g\) is irreducible in \(\mathcal O_{n-1}[z_n]\) and divides \(f_1f_2\) there. In the factorial ring \(\mathcal O_{n-1}[z_n]\) the irreducible element \(g\) is prime, so it divides \(f_1\) or \(f_2\). \(\square\)

A second proof: \(\mathcal O_n\) is regular by Proposition 1.1, and every regular local ring is factorial Stacks, Tag 0AG0. Since factorial domains are integrally closed Stacks, Tag 0AFV, \(\mathcal O_n\) is integrally closed in its field of fractions; this is used in Lesson 4.

Germs at points other than \(0\) are treated by translation: \(\mathcal O_{\mathbf C^n,x}\cong\mathcal O_n\). The next result compares the structure at nearby points. For a polydisc \(\Delta'\subset\mathbf C^{n-1}\) and \(x'\in\Delta'\), a monic polynomial \(F\in\mathcal O(\Delta')[z_n]\) defines a germ of \(\mathcal O_{n-1,x'}[z_n]\) at every point.

Lemma 3.2 (splitting at a nearby point). Let \(F\in\mathcal O(\Delta')[z_n]\) be monic of degree \(\mu\), let \(x=(x',x_n)\) with \(x'\in\Delta'\), and let \(\mu'\) be the order of vanishing of \(z_n\mapsto F(x',z_n)\) at \(x_n\). Then \(F=f'f''\) in \(\mathcal O_{\mathbf C^{n-1},x'}[z_n]\), where \(f'\) is a Weierstrass polynomial of degree \(\mu'\) in \(z_n-x_n\) at \(x'\) and \(f''\) is monic of degree \(\mu-\mu'\) with \(f''(x)\neq0\).

Proof. If \(\mu'=0\), take \(f'=1\). Otherwise the germ of \(F\) at \(x\) is regular of order \(\mu'\) in \(z_n-x_n\), so Theorem 2.1 at \(x\) gives \(F=uf'\) with \(u\) a unit of \(\mathcal O_{\mathbf C^n,x}\) and \(f'\) a Weierstrass polynomial in \(z_n-x_n\). By Lemma 2.3 at \(x\), \(F=f'f''\) with \(f''\in\mathcal O_{\mathbf C^{n-1},x'}[z_n]\), and comparing leading terms \(f''\) is monic of degree \(\mu-\mu'\). The uniqueness of the quotient in Theorem 2.1(2) gives \(f''=u\) as germs at \(x\), so \(f''(x)\neq0\). \(\square\)

Lemma 3.3 (Gauss). Let \(A\) be a unique factorization domain with field of fractions \(K\), and let \(P,Q\in A[T]\). If \(P\) and \(Q\) have no common factor of positive degree in \(A[T]\), they have none in \(K[T]\).

Proof. Let \(D\in K[T]\) of positive degree divide \(P\) and \(Q\) in \(K[T]\). Multiplying \(D\) by an element of \(K\) we may assume \(D\in A[T]\) with coefficients having no common prime factor, since gcds exist in \(A\). Write \(P=DE\) with \(E\in K[T]\), and choose \(c\in A\setminus\{0\}\) with \(cE\in A[T]\), so that \(cP=D\cdot cE\). If a prime \(\pi\) of \(A\) divides \(c\), reduce modulo \(\pi\): in \((A/\pi)[T]\), whose coefficient ring is a domain, \(\overline D\cdot\overline{cE}=0\) and \(\overline D\neq0\), so \(\overline{cE}=0\); that is, \(\pi\) divides every coefficient of \(cE\), and we may replace \(c\) by \(c/\pi\). After finitely many steps \(c\) is a unit, so \(E\in A[T]\) and \(D\) divides \(P\) in \(A[T]\). The same holds for \(Q\), so \(D\) is a common factor of positive degree in \(A[T]\). \(\square\)

Theorem 3.4 (coprimality persists). Let \(f,g\) be holomorphic near \(a\in\mathbf C^n\) with coprime germs at \(a\). Then their germs at every point of some neighbourhood of \(a\) are coprime.

Proof. Take \(a=0\). If \(f(0)\neq0\) or \(g(0)\neq0\), one of them is a unit near \(0\). Otherwise, after a linear change of coordinates, \(f\) and \(g\) are units times Weierstrass polynomials \(P,Q\) of positive degree, which are coprime in \(\mathcal O_n\). They are also coprime in \(\mathcal O_{n-1}[z_n]\): a common factor of positive degree would, by Lemma 2.4, be a common nonunit factor in \(\mathcal O_n\), and a common factor of degree zero divides a monic polynomial, hence is a unit. Let \(K\) be the field of fractions of \(\mathcal O_{n-1}\). By Lemma 3.3, applied with \(A=\mathcal O_{n-1}\), they are also coprime in the principal ideal domain \(K[z_n]\), and clearing denominators in a Bézout identity gives

\[ AP+BQ=c,\qquad A,B\in\mathcal O_{n-1}[z_n],\quad 0\neq c\in\mathcal O_{n-1}. \tag{3.1} \]

Write \(f=uP\) and \(g=vQ\) with units \(u,v\). Choose a polydisc \(\Delta'\times\Delta_n\) about \(0\) on which \(u,v,P,Q,A,B,c\) are holomorphic and \(u,v\) have no zeros, with \(\Delta'\) connected. At every point of the polydisc the germs of \(f,g\) are then associates of those of \(P,Q\). Since \(c\neq0\), its germ at every \(x'\in\Delta'\) is nonzero by the identity theorem.

Let \(x\in\Delta'\times\Delta_n\) and suppose a nonunit \(h\in\mathcal O_{\mathbf C^n,x}\) divides the germs of \(P\) and \(Q\) at \(x\). By (3.1), \(h\) divides \(c\). Split \(P=f'f''\) at \(x\) by Lemma 3.2; since \(f''\) is a unit at \(x\), \(h\) divides \(f'\). Restricting to the line \(z'=x'\), \(h(x',z_n)\) divides \(f'(x',z_n)=(z_n-x_n)^{\mu'}\), and \(h(x)=0\); so \(h\) is regular of some order \(k\geq1\) in \(z_n-x_n\), and \(h=vh'\) with \(v\) a unit and \(h'\) a Weierstrass polynomial of degree \(k\) in \(z_n-x_n\). Now \(h'\) divides \(c\), so by Lemma 2.3 at \(x\), \(c=h'q\) in \(\mathcal O_{\mathbf C^{n-1},x'}[z_n]\). Comparing degrees in \(z_n\), \(q=0\), so the germ of \(c\) at \(x'\) is zero, a contradiction. \(\square\)

4. Finiteness

Theorem 4.1. Let \(P\in\mathcal O_{n-1}[z_n]\) be a Weierstrass polynomial of degree \(s\).

  1. \(\mathcal O_n/(P)\) is a free \(\mathcal O_{n-1}\)-module with basis the classes of \(1,z_n,\ldots,z_n^{s-1}\).
  2. If an ideal \(I\subset\mathcal O_n\) contains \(P\), then \(\mathcal O_n/I\) is generated as an \(\mathcal O_{n-1}\)-module by the classes of \(1,z_n,\ldots,z_n^{s-1}\), and \(\mathcal O_{n-1}\to\mathcal O_n/I\) is injective if and only if \(I\cap\mathcal O_{n-1}=0\).

Proof. (1) By Theorem 2.1(2) every \(f\) is \(qP+R\) with \(R=\sum_{j<s}c_jz_n^j\), \(c_j\in\mathcal O_{n-1}\), and \((c_j)\) is unique. (2) The generators come from (1), and the kernel of \(\mathcal O_{n-1}\to\mathcal O_n/I\) is \(I\cap\mathcal O_{n-1}\). \(\square\)

Corollary 4.2. Let \(0\leq d<n\) and let \(I\subset\mathcal O_n\) be an ideal containing, for each \(k=d+1,\ldots,n\), a Weierstrass polynomial \(P_k\in\mathcal O_{k-1}[z_k]\) of degree \(s_k\). Then \(\mathcal O_n/I\) is generated as an \(\mathcal O_d\)-module by the classes of the monomials \(z_{d+1}^{\alpha_{d+1}}\cdots z_n^{\alpha_n}\) with \(\alpha_k<s_k\). In particular \(\mathcal O_n/I\) is a finite \(\mathcal O_d\)-module and every element of \(\mathcal O_n/I\) is integral over \(\mathcal O_d\).

Proof. Divide \(f\in\mathcal O_n\) by \(P_n\): \(f=P_nq_n+R_n\) with \(R_n\in\mathcal O_{n-1}[z_n]\) of degree less than \(s_n\). Divide each coefficient of \(R_n\), an element of \(\mathcal O_{n-1}\), by \(P_{n-1}\), and continue down to \(P_{d+1}\). The result is \(f\equiv R\) modulo \((P_{d+1},\ldots,P_n)\subset I\), with \(R\in\mathcal O_d[z_{d+1},\ldots,z_n]\) of degree less than \(s_k\) in each \(z_k\). A finite module over a ring consists of integral elements, by the determinant trick. \(\square\)

5. Exercises

Exercise 5.1. Show that \(g=z_2-z_2^2-z_1\in\mathcal O_2\) is regular of order \(1\) in \(z_2\), find its factorization \(g=uP\) explicitly, and verify Proposition 2.2.

Solution. \(g(0,z_2)=z_2(1-z_2)\), so \(g\) is regular of order \(1\). The Weierstrass polynomial has degree \(1\), so \(P=z_2-\varphi(z_1)\), where \(\varphi\) is the root of \(g(z_1,\cdot)\) near \(0\): \(\varphi-\varphi^2=z_1\), that is \(\varphi(z_1)=\tfrac12\bigl(1-\sqrt{1-4z_1}\bigr)=\sum_{k\geq0}C_kz_1^{k+1}\) with the Catalan numbers \(C_k\). Then \((z_2-\varphi)(1-\varphi-z_2)=z_2-z_2^2-\varphi+\varphi^2=g\), so \(u=1-\varphi(z_1)-z_2\), a unit since \(u(0)=1\). The lowest homogeneous part of \(g\) is \(z_2-z_1\), nonzero at \((0,1)\), and the coefficient \(a_1=-\varphi\) vanishes to order \(1\), as Proposition 2.2 predicts.

Exercise 5.2. Show that \(z_1z_2\) is not regular in \(z_2\), and find a linear change of coordinates after which it is.

Solution. \(g(0,z_2)=0\), so it is not regular in \(z_2\). Put \(z_1=w_1+w_2\), \(z_2=w_2\). Then \(g=(w_1+w_2)w_2\) and \(g(0,w_2)=w_2^2\), so \(g\) is regular of order \(2\) in \(w_2\). Its lowest homogeneous part \(z_1z_2\) is nonzero at the new axis vector \((1,1)\), as Proposition 2.2 requires.

Exercise 5.3. Show that \(\mathcal O_n\) is not a principal ideal domain for \(n\geq2\), although it is factorial.

Solution. If \(\mathfrak m_n=(f)\), then \(\mathfrak m_n/\mathfrak m_n^2\) would be spanned by the class of \(f\), so it would have dimension at most \(1\), contradicting Proposition 1.1.

Exercise 5.4. Let \(f=z_1\) and \(g=z_1+z_2^2\) in \(\mathcal O_2\). Show directly that their germs are coprime at every point of \(\mathbf C^2\).

Solution. At \(0\): \(f\) is irreducible, being of degree one, and it does not divide \(g\), since \(g(0,z_2)=z_2^2\neq0\). At a point with \(z_1\neq0\), \(f\) is a unit; at a point \((0,b)\) with \(b\neq0\), \(g(0,b)=b^2\neq0\), so \(g\) is a unit. Note that \(f\) and \(g\) have a common zero only at \(0\), yet the ideal \((f,g)=(z_1,z_2^2)\) is not the maximal ideal: coprimality concerns common factors, not common zeros.

Exercise 5.5. Let \(I=(z_2^2-z_1^3,\ z_3-z_1z_2)\subset\mathcal O_3\). Show that \(\mathcal O_3/I\) is a free \(\mathcal O_1\)-module of rank \(2\) with basis \(1,z_2\).

Solution. \(P_3=z_3-z_1z_2\) is a Weierstrass polynomial of degree \(1\) in \(z_3\) over \(\mathcal O_2\) and \(P_2=z_2^2-z_1^3\) one of degree \(2\) in \(z_2\) over \(\mathcal O_1\). By Corollary 4.2, \(1\) and \(z_2\) generate \(\mathcal O_3/I\) over \(\mathcal O_1\). Eliminating \(z_3\), \(\mathcal O_3/I\cong\mathcal O_2/(P_2)\), which is free with basis \(1,z_2\) by Theorem 4.1.

References