Weierstrass preparation and division

Intuition Beginner

In one complex variable, a holomorphic function near a point looks like a power of the variable times a non-vanishing factor: if $f$ has a zero of order $k$ at the origin, then $f(z)=z^k\cdot u(z)$ with $u$ a unit. This is the whole local story — zeros are isolated, and counting them is counting the exponent $k$.

In several variables the zero set is no longer a scatter of points; it is a hypersurface, a curve, a whole web of vanishing. You want a normal form that turns a general holomorphic function near the origin into something you can manipulate like a polynomial in one chosen variable. The Weierstrass preparation theorem provides exactly that: after a harmless rotation of coordinates, every holomorphic germ factors as a unit times a monic polynomial in the last variable whose coefficients are holomorphic in the others.

The payoff is that questions about a complicated analytic object become questions about an honest polynomial, where you can divide, count roots, and compute degrees.

Visual Beginner

Picture the zero set of a holomorphic function of two variables near the origin: a curve passing through $0$ in ${\mathbb{C}}^2$. Slice the picture along the last coordinate axis. The preparation theorem says that, looking down each such slice, the function vanishes at exactly $k$ points counted with multiplicity, and these $k$ points are the roots of a monic degree-$k$ polynomial whose coefficients vary holomorphically as you move the slice.

Worked example Beginner

Take the function of two variables $f(z_1,z_2)=z_2^2-z_1$ near the origin of ${\mathbb{C}}^2$. Fix a small value of $z_1$ and ask where $f$ vanishes as a function of $z_2$. The answer is $z_2=+\sqrt{z_1}$ and $z_2=-\sqrt{z_1}$, two points that come together as $z_1$ goes to $0$.

So along each vertical slice the function has exactly $2$ roots in $z_2$. The number $2$ is the order of vanishing of $f$ at the origin in the $z_2$-direction, because setting $z_1=0$ gives $f(0,z_2)=z_2^2$, a power of $z_2$ with exponent $2$.

Here $f$ is already monic of degree $2$ in $z_2$, with coefficients $0$ and $-z_1$ that are holomorphic in $z_1$ and vanish at $z_1=0$. The unit factor is just $1$. This is the simplest shape a Weierstrass polynomial can take.

What this tells us: the order of vanishing in the chosen direction is the degree of the polynomial you get, and the polynomial records the roots slice by slice.

Check your understanding Beginner

Formal definition Intermediate+

Let ${\mathcal{O}}_n=\mathbb{C}\{z_1,\dots ,z_n\}$ denote the ring of germs at the origin of holomorphic functions on ${\mathbb{C}}^n$, equivalently the ring of convergent power series in $n$ variables. It is a local ring with maximal ideal ${\mathfrak{m}}_n=\{f:f(0)=0\}$ and residue field $\mathbb{C}$; the units are the germs with $f(0)\ne 0$. Write ${\mathcal{O}}_{n-1}=\mathbb{C}\{z_1,\dots ,z_{n-1}\}$ and use $z^{\prime }=(z_1,\dots ,z_{n-1})$, $z=(z^{\prime },z_n)$.

A germ $f\in {\mathcal{O}}_n$ is regular of order $k$ in $z_n$ when the restriction $f(0,z_n)$ is a non-zero germ in ${\mathcal{O}}_1$ vanishing to order exactly $k$, that is $f(0,z_n)=c{z}_n^k+(\text{higher order})$ with $c\ne 0$. Equivalently, ${\partial }^{j}f/\partial z_n^j(0)=0$ for $j<k$ and ${\partial }^{k}f/\partial z_n^k(0)\ne 0$.

A Weierstrass polynomial of degree $k$ in $z_n$ is an element $$ W(z', z_n) = z_n^k + a_{k-1}(z') z_n^{k-1} + \cdots + a_1(z') z_n + a_0(z'), $$ monic in $z_n$, with coefficients $a_j\in {\mathcal{O}}_{n-1}$ satisfying $a_j(0)=0$ for all $j$. The last condition says $W(0,z_n)=z_n^k$, so a Weierstrass polynomial is itself regular of order equal to its degree.

Notation. The boundary-of-vanishing data sits in ${\mathfrak{m}}_{n-1}\subset {\mathcal{O}}_{n-1}$: each coefficient $a_j$ lies in the maximal ideal of the smaller ring. The symbol ${\mathcal{O}}_n^{\times }$ denotes the unit group.

Counterexamples to common slips

• The factorisation $f=u\cdot W$ requires $f$ to be regular in $z_n$. The germ $f(z_1,z_2)=z_1$ is not regular in $z_2$ (its restriction $f(0,z_2)=0$ vanishes identically), and indeed it admits no Weierstrass factorisation in $z_2$ — a coordinate change is mandatory first.
• A Weierstrass polynomial need not be irreducible: $W=z_2^2-z_1^2=(z_2-z_1)(z_2+z_1)$ factors into two Weierstrass polynomials of degree $1$.
• "Monic polynomial in $z_n$ with holomorphic coefficients" is weaker than "Weierstrass polynomial". The condition $a_j(0)=0$ is essential; $z_n^2+1$ is monic but not Weierstrass, and it is a unit in ${\mathcal{O}}_2$.

Key theorem with proof Intermediate+

Theorem (Weierstrass preparation, Weierstrass 1860s). Let $f\in {\mathcal{O}}_n$ be regular of order $k$ in $z_n$. Then there exist a unique unit $u\in {\mathcal{O}}_n^{\times }$ and a unique Weierstrass polynomial $W\in {\mathcal{O}}_{n-1}[z_n]$ of degree $k$ with $$ f = u \cdot W. $$

Proof. The argument runs through the Cauchy integral in the single variable $z_n$, with the remaining variables as holomorphic parameters ^{[Krantz Ch. 1]}.

Because $f$ is regular of order $k$, the one-variable function $z_n\mapsto f(0,z_n)$ has a zero of order $k$ at $0$ and no other zero in a small disc $|z_n|\le r$. By continuity in the parameters $z^{\prime }$, choose $r>0$ and a polydisc neighbourhood $|z^{\prime }|\le s$ small enough that for each fixed $z^{\prime }$ with $|z^{\prime }|\le s$ the function $z_n\mapsto f(z^{\prime },z_n)$ has no zero on the circle $|z_n|=r$ and exactly $k$ zeros (counted with multiplicity) inside it. The argument principle applied in $z_n$ gives the count $$ N(z') = \frac{1}{2\pi i} \oint_{|z_n| = r} \frac{\partial f / \partial z_n}{f}(z', z_n), dz_n = k, $$ constant in $z^{\prime }$ since it is a continuous integer-valued function.

Let $w_1(z^{\prime }),\dots ,w_k(z^{\prime })$ be those $k$ zeros. The power sums $$ p_m(z') = \sum_{j=1}^{k} w_j(z')^m = \frac{1}{2\pi i} \oint_{|z_n| = r} z_n^m , \frac{\partial f / \partial z_n}{f}(z', z_n), dz_n $$ are holomorphic in $z^{\prime }$, being parameter integrals of holomorphic integrands. By Newton's identities the elementary symmetric functions ${\sigma }_1(z^{\prime }),\dots ,{\sigma }_k(z^{\prime })$ are polynomials in the $p_m$ with rational coefficients, hence also holomorphic in $z^{\prime }$. Set $$ W(z', z_n) = \prod_{j=1}^{k}\bigl(z_n - w_j(z')\bigr) = z_n^k - \sigma_1(z') z_n^{k-1} + \cdots + (-1)^k \sigma_k(z'). $$ The coefficients $a_{k-j}=(-1{)}^j{\sigma }_j$ lie in ${\mathcal{O}}_{n-1}$. At $z^{\prime }=0$ all $k$ zeros collapse to $w_j(0)=0$, so ${\sigma }_j(0)=0$ and $W(0,z_n)=z_n^k$; thus $W$ is a Weierstrass polynomial of degree $k$.

Define $u=f/W$ on the punctured neighbourhood where $W\ne 0$. For each fixed $z^{\prime }$ the functions $f(z^{\prime },\cdot )$ and $W(z^{\prime },\cdot )$ have the same $k$ zeros with the same multiplicities inside $|z_n|<r$, so the quotient extends holomorphically across them (removable singularities in one variable), and it has no zero in the disc. Holomorphic dependence on $z^{\prime }$ follows from the same Cauchy-integral representation; the extended $u$ is holomorphic on the full polydisc and satisfies $u(0,0)=f(0,z_n)/z_n^k{|}_{z_n\to 0}\ne 0$, so $u$ is a unit. Hence $f=u\cdot W$.

For uniqueness, suppose $f=u_1{W}_1=u_2{W}_2$ with both $W_i$ Weierstrass of degree $k$. For each fixed $z^{\prime }$ the polynomials $W_1(z^{\prime },\cdot )$ and $W_2(z^{\prime },\cdot )$ are monic of degree $k$ with the identical zero set (the zeros of $f(z^{\prime },\cdot )$ inside the disc, with multiplicity), so $W_1=W_2$ as polynomials in $z_n$; then $u_1=u_2$. This is the factorisation, and its uniqueness. $\square$

Bridge. The preparation theorem builds toward the division theorem stated next, and the same Cauchy-integral-in-$z_n$ machinery appears again in the inductive proof that ${\mathcal{O}}_n$ is Noetherian and factorial. The foundational reason a several-variable local ring behaves like a polynomial ring is exactly this: preparation identifies the germ $f$ with $u\cdot W$, and dividing by $f$ in ${\mathcal{O}}_n$ is the same as dividing by the Weierstrass polynomial $W$ in ${\mathcal{O}}_{n-1}[z_n]$ — putting these together, the analytic question generalises the elementary polynomial division algorithm one variable at a time. The central insight is that finiteness over ${\mathcal{O}}_{n-1}$, not over $\mathbb{C}$, is what makes the induction close, and the bridge is the degree bound on remainders that division supplies. This is exactly the structure exploited in 04.06.02 when coherence of ${\mathcal{O}}_n$ is established.

Exercises Intermediate+

Advanced results Master

The division theorem is the analytic core; preparation is its specialisation to $g$ regular. Stated in full: for $g\in {\mathcal{O}}_n$ and a Weierstrass polynomial $W$ of degree $k$ in $z_n$, there are unique $q\in {\mathcal{O}}_n$ and $r\in {\mathcal{O}}_{n-1}[z_n]$ with ${\mathrm{deg}}_{z_n}r<k$ and $g=qW+r$ ^{[Hörmander §6.1]}. The quotient and remainder are given by Cauchy integrals over a circle $|z_n|=\rho$ enclosing the $k$ zeros of $W(z^{\prime },\cdot )$: $$ q(z) = \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{g(z', \zeta)}{W(z', \zeta)}, \frac{d\zeta}{\zeta - z_n}, \qquad r(z', z_n) = \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{g(z', \zeta)}{W(z', \zeta)}, \frac{W(z', \zeta) - W(z', z_n)}{\zeta - z_n}, d\zeta. $$ The kernel $(W(z^{\prime },\zeta )-W(z^{\prime },z_n))/(\zeta -z_n)$ is a polynomial in $z_n$ of degree $k-1$ with coefficients holomorphic in $(z^{\prime },\zeta )$, which is exactly why $r$ has $z_n$-degree at most $k-1$. Preparation is recovered by dividing $z_n^k$ by $f$ (after reduction) or, equivalently, the symmetric-function construction given in the Intermediate proof.

The local ring is Noetherian. ${\mathcal{O}}_0=\mathbb{C}$ is a field, hence Noetherian. Assume ${\mathcal{O}}_{n-1}$ is Noetherian. Given a non-zero ideal $I\subset {\mathcal{O}}_n$, choose $0\ne f\in I$; after a coordinate rotation $f$ is regular in $z_n$, and preparation replaces $f$ by a Weierstrass polynomial $W\in I$ of degree $k$. Division by $W$ identifies ${\mathcal{O}}_n/(W)$ with the free ${\mathcal{O}}_{n-1}$-module ${⨁}_{j=0}^{k-1}{\mathcal{O}}_{n-1}{z}_n^j$ of rank $k$, which is Noetherian as a module over the Noetherian ring ${\mathcal{O}}_{n-1}$. The image $\bar{I}=I/(W)$ is a submodule, hence finitely generated; lifting generators and adjoining $W$ gives a finite generating set for $I$. By induction ${\mathcal{O}}_n$ is Noetherian (Rückert basis theorem) ^{[Grauert–Remmert §1.2]}.

The local ring is factorial. With ${\mathcal{O}}_{n-1}$ a unique factorisation domain by induction, Gauss's lemma makes ${\mathcal{O}}_{n-1}[z_n]$ a unique factorisation domain. Preparation shows every germ of ${\mathcal{O}}_n$ is, up to a unit, a Weierstrass polynomial, and the factorisation of a Weierstrass polynomial in ${\mathcal{O}}_n$ agrees with its factorisation in ${\mathcal{O}}_{n-1}[z_n]$ (the argument of Exercise 7). Thus factorisation descends from the polynomial ring to ${\mathcal{O}}_n$, and ${\mathcal{O}}_n$ is a unique factorisation domain. The base case ${\mathcal{O}}_0=\mathbb{C}$ is a field.

Henselian structure. The division theorem makes ${\mathcal{O}}_n$ a Henselian local ring with respect to the variable $z_n$: a factorisation of the reduction $W(0,z_n)=z_n^k$ into coprime monic factors lifts uniquely to a factorisation of $W$ over ${\mathcal{O}}_{n-1}$. This is the analytic Hensel's lemma, and it is what couples the algebra of ${\mathcal{O}}_n$ to the geometry of analytic germs: irreducible factors of $W$ correspond to the irreducible branches of the hypersurface $\{f=0\}$ through the origin.

Comparison with the formal and algebraic cases. The same statements hold verbatim for the formal power series ring $\mathbb{C}[[z_1,\dots ,z_n]]$ and for the ring of algebraic power series; the proofs differ only in how convergence is tracked. The convergent (analytic) case carries the extra content that the Cauchy integrals converge on a common polydisc, so the symmetric functions and quotients are genuinely holomorphic, not merely formal. This is the distinction that keeps the analytic Weierstrass theorems outside the reach of Mathlib's existing formal-power-series preparation lemma.

Synthesis. Weierstrass preparation is the foundational reason the local analytic geometry of ${\mathbb{C}}^n$ is governed by polynomial algebra one variable at a time. The preparation theorem identifies an arbitrary germ with a unit times a Weierstrass polynomial; division then identifies the quotient ${\mathcal{O}}_n/(W)$ with a finite free module over ${\mathcal{O}}_{n-1}$, and this is exactly the finiteness that drives the induction. Putting these together, the Noetherian and factorial properties of ${\mathcal{O}}_n$ are not separate facts but a single inductive consequence: each is dual to a statement about ${\mathcal{O}}_{n-1}[z_n]$ transported through preparation. The central insight is that the residue circle $|z_n|=\rho$ does double duty — counting zeros (preparation) and producing remainders of bounded degree (division) — so the analytic and algebraic structures are the same structure read two ways; this pattern recurs whenever the local geometry of an analytic set is reduced to a branched cover of a coordinate subspace, and it builds toward the coherence of the structure sheaf ${\mathcal{O}}_{{\mathbb{C}}^n}$ developed in 04.06.02.

Full proof set Master

Proposition (Weierstrass division theorem). Let $W\in {\mathcal{O}}_{n-1}[z_n]$ be a Weierstrass polynomial of degree $k$ and let $g\in {\mathcal{O}}_n$. Then there exist unique $q\in {\mathcal{O}}_n$ and $r\in {\mathcal{O}}_{n-1}[z_n]$ with ${\mathrm{deg}}_{z_n}r<k$ such that $g=qW+r$.

Proof. Choose $\rho >0$ and a polydisc $|z^{\prime }|\le s$ so small that for each $z^{\prime }$ with $|z^{\prime }|\le s$ all $k$ zeros of $W(z^{\prime },\cdot )$ lie strictly inside $|z_n|<\rho$ and $g$ is holomorphic on the closed polydisc $|z^{\prime }|\le s,|z_n|\le \rho$; this is possible because $W(0,z_n)=z_n^k$ has its only zero at $0$, and zeros vary continuously. Define $$ q(z', z_n) = \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{g(z', \zeta)}{W(z', \zeta)}, \frac{d\zeta}{\zeta - z_n}, \qquad |z_n| < \rho. $$ The integrand is holomorphic in $(z^{\prime },z_n)$ for $|z_n|<\rho$ because $W(z^{\prime },\zeta )\ne 0$ on the contour, so $q\in {\mathcal{O}}_n$. Set $r=g-qW$. Then $$ r(z', z_n) = \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{g(z', \zeta)}{W(z', \zeta)}\bigl(W(z', \zeta) - W(z', z_n)\bigr)\frac{d\zeta}{\zeta - z_n}, $$ using $g(z^{\prime },z_n)=\frac{1}{2\pi i}\oint g(z^{\prime },\zeta )d\zeta /(\zeta -z_n)$ (Cauchy in $z_n$) and the definition of $q$. The difference quotient $$ \frac{W(z', \zeta) - W(z', z_n)}{\zeta - z_n} = \sum_{j=0}^{k-1} c_j(z', \zeta), z_n^{,j} $$ is a polynomial in $z_n$ of degree $k-1$ with $c_j$ holomorphic, since $W$ is monic of degree $k$ in its last slot. Therefore $r$ is a polynomial in $z_n$ of degree $\le k-1$ with coefficients in ${\mathcal{O}}_{n-1}$.

For uniqueness, suppose $qW+r=q^{\prime }W+r^{\prime }$ with both remainders of $z_n$-degree $<k$. Then $(q-q^{\prime })W=r^{\prime }-r$. For each fixed $z^{\prime }$ the left side, if $q\ne q^{\prime }$, vanishes at the $k$ zeros of $W(z^{\prime },\cdot )$ counted with multiplicity, forcing the right side — a polynomial of degree $<k$ — to have $\ge k$ roots, hence to vanish identically. So $r=r^{\prime }$, and then $(q-q^{\prime })W=0$ with $W\ne 0$ gives $q=q^{\prime }$. $\square$

Proposition (preparation from division). Every germ $f\in {\mathcal{O}}_n$ regular of order $k$ in $z_n$ factors as $f=uW$ with $u$ a unit and $W$ a Weierstrass polynomial of degree $k$, uniquely.

Proof. Apply division of $g=z_n^k$ by... rather, apply the division proposition with the roles arranged as follows. The construction of $W$ from the symmetric functions of the zeros of $f(z^{\prime },\cdot )$, given in the Intermediate proof, yields a Weierstrass polynomial of degree $k$ whose zeros (with multiplicity) coincide with those of $f(z^{\prime },\cdot )$ inside $|z_n|<\rho$. Set $u=f/W$. On the contour $|z_n|=\rho$ neither $f$ nor $W$ vanishes, and inside they share zeros with equal multiplicities, so $u$ extends to a holomorphic function with no zero on the polydisc; thus $u\in {\mathcal{O}}_n^{\times }$ and $f=uW$. Uniqueness: if $f=u_1{W}_1=u_2{W}_2$, then $W_1$ and $W_2$ have, for each $z^{\prime }$, the same monic factorisation of degree $k$ determined by the shared zero set, so $W_1=W_2$ and $u_1=u_2$. $\square$

Proposition (Rückert basis theorem). ${\mathcal{O}}_n$ is Noetherian.

Proof. Induct on $n$, with ${\mathcal{O}}_0=\mathbb{C}$ a field. Assume ${\mathcal{O}}_{n-1}$ Noetherian. Let $I\subseteq {\mathcal{O}}_n$ be an ideal; if $I=0$ it is finitely generated, so take $0\ne f\in I$. After a linear change of coordinates $f$ is regular of some order $k$ in $z_n$ (a non-zero germ does not vanish on every line through $0$), and preparation gives a Weierstrass polynomial $W\in I$ of degree $k$, associate to $f$. By the division proposition, ${\mathcal{O}}_n/(W)\cong {⨁}_{j=0}^{k-1}{\mathcal{O}}_{n-1}{z}_n^j$ as ${\mathcal{O}}_{n-1}$-modules, a finitely generated module over the Noetherian ring ${\mathcal{O}}_{n-1}$, hence Noetherian. The image of $I$ in ${\mathcal{O}}_n/(W)$ is a submodule, so finitely generated; pulling back generators and adjoining $W$ produces a finite generating set for $I$. $\square$

Proposition. ${\mathcal{O}}_n$ is a unique factorisation domain.

Proof. Induct on $n$, base case ${\mathcal{O}}_0=\mathbb{C}$. Assume ${\mathcal{O}}_{n-1}$ factorial; by Gauss's lemma ${\mathcal{O}}_{n-1}[z_n]$ is factorial. Let $f\in {\mathcal{O}}_n$ be non-zero and non-unit; after rotation it is regular in $z_n$, and preparation gives $f=uW$ with $W$ Weierstrass. It suffices to factor $W$ uniquely. Factor $W=W_1\cdots W_m$ into irreducibles in ${\mathcal{O}}_{n-1}[z_n]$; each $W_i$ may be taken Weierstrass (a monic factor of a Weierstrass polynomial restricts at $z^{\prime }=0$ to a power of $z_n$, hence has vanishing lower coefficients). By the argument of Exercise 7 any analytic factorisation of $W$ in ${\mathcal{O}}_n$ refines to a polynomial factorisation, so the $W_i$ are irreducible in ${\mathcal{O}}_n$ as well, and uniqueness of the polynomial factorisation gives uniqueness in ${\mathcal{O}}_n$ up to units. $\square$

Connections Master

• Holomorphic functions of several variables 06.07.01. The germ ring ${\mathcal{O}}_n=\mathbb{C}\{z_1,\dots ,z_n\}$ whose elements the preparation theorem factors is built directly on the multi-variable power-series and Cauchy theory introduced there; regularity in $z_n$ is a statement about the one-variable restriction studied in that unit.

• Hartogs phenomenon 06.07.02. The automatic-extension phenomena of $n\ge 2$ and the local normal form are two faces of the same several-variable rigidity: where Hartogs forbids isolated singularities, preparation supplies the polynomial model that makes the codimension-one zero sets tractable. Both rely on the parameter-Cauchy-integral technique.

• Analytic continuation 06.01.04. The Cauchy-integral-with-parameters argument that produces the holomorphic coefficients $a_j(z^{\prime })$ is a continuation statement: the symmetric functions of the zeros extend holomorphically across the polydisc, the same principle that governs continuation of one-variable germs.

• Meromorphic functions 06.01.05. The division theorem is the local source of the structure of meromorphic germs: a quotient $g/f$ in ${\mathcal{O}}_n$ becomes, after preparation, a quotient by a Weierstrass polynomial, exhibiting the field of fractions of ${\mathcal{O}}_n$ and the local theory of poles along a hypersurface.

• Coherent sheaves 04.06.02. The Noetherian and factorial properties proved here are the local input to the Oka coherence theorem: ${\mathcal{O}}_{{\mathbb{C}}^n}$ is a coherent sheaf of rings precisely because each stalk ${\mathcal{O}}_n$ is Noetherian, and the division theorem furnishes the finite free resolutions used to verify coherence.

Historical & philosophical context Master

Karl Weierstrass developed the preparation theorem (the Vorbereitungssatz) in his Berlin lectures of the 1860s, in the course of constructing the theory of analytic functions of several variables on a foundation of convergent power series rather than geometry ^{[Weierstrass 1895]}. The theorem first appeared in print in his collected works (Mathematische Werke, Berlin 1895), where it served as the technical instrument for studying the local structure of analytic sets and the algebraic dependence of analytic functions. Weierstrass's motivation was the inversion of the local problem of one-variable function theory: where in one variable a holomorphic function is locally $z^k$ times a unit, he sought the corresponding normal form in $n$ variables, and found it in the factorisation by a monic polynomial in one distinguished variable.

The algebraic consequences — that ${\mathcal{O}}_n$ is Noetherian and a unique factorisation domain — were drawn out in the twentieth century as the local foundations of analytic geometry were systematised. Rückert's 1933 work and the subsequent development by Behnke, Thullen, and the Oka–Cartan school placed preparation and division at the base of the theory of coherent analytic sheaves (Gunning–Rossi 1965 ^{[Gunning–Rossi]}; Grauert–Remmert ^{[Grauert–Remmert §1.2]}). The same theorems hold for formal and algebraic power series, and the analytic case is distinguished by the convergence bookkeeping in the Cauchy integrals — a point Hörmander makes precise in §6.1 of his text ^{[Hörmander §6.1]}. Krantz presents the theorem twice: as elementary local structure in Ch. 1 and again in the integral-formula chapters ^{[Krantz Ch. 1]}.

Bibliography Master

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}

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}

@article{Ruckert1933,
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}