# 6.2: Weierstrass Preparation and Division Theorems

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Suppose $$f$$ is (a germ of) a holomorphic function at a point $$p \in \mathbb{C}^n$$. Write $f(z) = \sum_{k=0}^\infty f_k(z-p),$ where $$f_k$$ is a homogeneous polynomial of degree $$k$$, that is, $$f_k(tz) = t^k f_k(z)$$.

## Definition: Order of Vanishing

Let $$p \in \mathbb{C}^n$$ and $$f$$ be a function holomorphic in a neighborhood of $$p$$. If $$f$$ is not identically zero, define $\text{ord}_p f \overset{\text{def}}{=} \min \bigl\{ k \in \mathbb{N}_0 : f_k \not\equiv 0 \bigr\} .$ If $$f \equiv 0$$, then define $$\text{ord}_p f = \infty$$. The number $$\text{ord}_p f$$ is called the order of vanishing of $$f$$ at $$p$$.

In other words, the order of vanishing of $$f$$ at $$p$$ is $$k$$, whenever all partial derivatives of order less than $$k$$ vanish at $$p$$, and there exists at least one derivative of order $$k$$ that does not vanish at $$p$$.

In one complex variable, a holomorphic function $$f$$ with $$\text{ord}_0 f = k$$ can be written (locally) as $$f(z) = z^k u(z)$$ for a nonvanishing $$u$$. In several variables, there is a similar theorem, or in fact a pair of theorems, the so-called Weierstrass preparation and division theorems.

## Definition: Weierstrauss Polynomial

Let $$U \subset \mathbb{C}^{n-1}$$ be open, and let $$z' \in \mathbb{C}^{n-1}$$ denote the coordinates. Suppose a polynomial $$P \in \mathcal{O}(U)[z_n]$$ is monic of degree $$k \geq 0$$, that is, $P(z',z_n) = z_n^k + \sum_{j=0}^{k-1} c_j(z') \, z_n^j ,$ where $$c_j$$ are holomorphic functions defined on $$U$$, such that $$c_j(0) = 0$$ for all $$j$$. Then $$P$$ is called a Weierstrauss polynomial of degree $$k$$. If the $$c_j$$ are germs in $$\mathcal{O}_0 = {}_{n-1}\mathcal{O}_0$$, then $$P \in \mathcal{O}_0[z_n]$$ and $$P$$ is a germ of a Weierstrauss polynomial.

The definition (and the theorem that follows) still holds for $$n=1$$. If you read the definition carefully, you will find that if $$n=1$$, then the only Weierstrass polynomial of degree $$k$$ is $$z^k$$. Note that for any $$n$$, if $$k=0$$, then $$P = 1$$.

The purpose of this section is to show that every holomorphic function in $$\mathcal{O}_0$$ is up to a unit and a possible small rotation a Weierstrass polynomial, which carries the zeros of $$f$$. Consequently the algebraic and geometric properties of $${}_n\mathcal{O}_0$$ can be understood via algebraic and geometric properties of $${}_{n-1}\mathcal{O}_0[z_n]$$.

## Theorem $$\PageIndex{1}$$: Weierstrass Preparation Theorem

Suppose $$f \in \mathcal{O}(U)$$ for an open $$U \subset \mathbb{C}^{n-1} \times \mathbb{C}$$, where $$0 \in U$$, and $$f(0)=0$$. Suppose $$z_n \mapsto f(0,z_n)$$ is not identically zero near the origin and its order of vanishing at the origin is $$k \geq 1$$.

Then there exists an open polydisc $$V = V' \times D \subset \mathbb{C}^{n-1} \times \mathbb{C}$$ with $$0 \in V \subset U$$, a unique $$u \in \mathcal{O}(V)$$, $$u(z) \not=0$$ for all $$z \in V$$, and a unique Weierstrass polynomial $$P$$ of degree $$k$$ with coefficients holomorphic in $$V'$$ such that $f(z',z_n) = u(z',z_n) \, P(z',z_n) ,$ and such that all $$k$$ zeros (counting multiplicity) of $$z_n \mapsto P(z',z_n)$$ lie in $$D$$ for all $$z' \in V'$$

Proof

There exists a small disc $$D \subset \mathbb{C}$$ centered at zero such that $$\{0\} \times \overline{D} \subset U$$ and such that $$f(0,z_n) \not= 0$$ for $$z_n \in \overline{D} \setminus \{ 0 \}$$. By continuity of $$f$$, there is a small polydisc $$V = V' \times D$$ such that $$\overline{V} \subset U$$ and $$f$$ is not zero on $$V' \times \partial D$$. We will consider the zeros of $$z_n \mapsto f(z',z_n)$$ for $$z'$$ near zero. See Figure $$\PageIndex{1}$$. Figure $$\PageIndex{1}$$

By the one-variable argument principle (Theorem B.25) the number of zeros (with multiplicity) of $$z_n \mapsto f(z',z_n)$$ in $$D$$ is $\frac{1}{2\pi i} \int_{\partial D} \frac{\frac{\partial f}{\partial z_n} (z',\zeta)}{f(z',\zeta)} ~d\zeta .$ As $$f(z',\zeta)$$ does not vanish when $$z' \in V'$$ and $$\zeta \in \partial D$$, the expression above is a continuous integer-valued function of $$z' \in V'$$. The expression is equal to $$k$$ when $$z'=0$$, and so it is equal to $$k$$ for all $$z' \in V'$$. Write the zeros of $$z_n \mapsto f(z',z_n)$$ as $$\alpha_1(z'),\ldots,\alpha_k(z')$$, including multiplicity. The zeros are not ordered in any particular way. Pick some ordering for every $$z'$$. Write $P(z',z_n) = \prod_{j=1}^k \bigl(z_n-\alpha_j(z')\bigr) = z_n^k + c_{k-1}(z') \, z_n^{k-1} + \cdots + c_0 (z') .$ For a fixed $$z'$$, $$P$$ is uniquely defined as the ordering of zeros does not matter in its definition (see exercise below). It is clear that $$u$$ and $$P$$ are unique if they exist (that is, if they exist as holomorphic functions).

The functions $$\alpha_j$$ are not even continuous in general (see Example $$\PageIndex{1}$$). However, we will prove that the functions $$c_j$$ are holomorphic. The functions $$c_j$$ are (up to sign) the elementary symmetric functions of $$\alpha_1,\ldots,\alpha_k$$ (see below). It is a standard theorem in algebra (see Exercise $$\PageIndex{1}$$) that the elementary symmetric functions are polynomials in the so-called power sum functions in the $$\alpha_j$$s: $s_m(z') = \sum_{j=1}^k \alpha_j{(z')}^m , \qquad m = 1,\ldots,k.$ Therefore, if we show that the power sums $$s_m$$ are holomorphic, then $$c_\ell$$ are holomorphic.

A refinement of the argument principle (see also Theorem B.25) says: If $$h$$ and $$g$$ are holomorphic functions on a disc $$D$$, continuous on $$\overline{D}$$, such that $$g$$ has no zeros on $$\partial D$$, and $$\alpha_1,\ldots,\alpha_k$$ are the zeros of $$g$$ in $$D$$, then $\frac{1}{2 \pi i} \int_{\partial D} h(\zeta) \frac{g'(\zeta)}{g(\zeta)} ~d\zeta = \sum_{j=1}^k h(\alpha_j) .$

The formula above with $$h(\zeta) = \zeta^m$$ and $$g(\zeta)=f(z',\zeta)$$ says that $s_m(z') = \sum_{j=1}^k \alpha_j{(z')}^m = \frac{1}{2\pi i} \int_{\partial D} \zeta^m \frac{\frac{\partial f}{\partial \zeta} (z',\zeta)}{f(z',\zeta)} ~d\zeta .$ The function $$s_m$$ is clearly continuous, and if we differentiate under the integral with $$\frac{\partial}{\partial\bar{z}_\ell}$$ for $$\ell=1,\ldots,{n-1}$$, we find that $$s_m$$ is holomorphic.

Finally, we wish to show that $$P$$ divides $$f$$ as claimed. For each fixed $$z'$$, one variable theory says that $$z_n \mapsto \frac{f(z',z_n)}{P(z',z_n)}$$ has only removable singularities, and in fact, it has no zeros as we defined $$P$$ to exactly cancel them all out. The Cauchy formula on $$\frac{f}{P}$$ then says that the function $u(z',z_n) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(z',\zeta)}{P(z',\zeta)(\zeta-z_n)} \, d\zeta$ is equal to $$\frac{f(z',z_n)}{P(z',z_n)}$$. The function $$u$$ is clearly continuous and holomorphic in $$z_n$$ for each fixed $$z'$$. By differentiating under the integral, we find that it is also holomorphic in $$z'$$.

## Example $$\PageIndex{1}$$

A useful example to keep in mind is $$f(z_1,z_2) = z_2^2 - z_1$$, a Weierstrass polynomial in $$z_2$$ of degree $$k=2$$. So $$z' = z_1$$. For all $$z_1$$ except the origin there are two zeros, $$\pm \sqrt{z_1}$$. Call one of them $$\alpha_1(z_1)$$ and one of them $$\alpha_2(z_1)$$. Recall there is no continuous choice of a square root that works for all $$z_1$$, so no matter how you choose, $$\alpha_1$$ and $$\alpha_2$$ will not be continuous. At the origin there is only one zero of order two, so $$\alpha_1(0) = \alpha_2(0) = 0$$. On the other hand the symmetric functions $$c_1(z_1) = - \alpha_1(z_1) - \alpha_2(z_1) = 0$$ and $$c_0(z_1') = \alpha_1(z_1)\alpha_2(z_1) = -z_1$$ are holomorphic.

The $$k$$ depends on the coordinates chosen. If we do a linear change of coordinates and consider $$g(z_1,z_2) = -f(z_2,z_1)$$, then $$g(z_1,z_2) = z_1^2 - z_2$$, which is a Weierstrass polynomial in $$z_2$$ of degree $$k=1$$. After the change, there is only one zero, $$\alpha_1(z_1) = z_1^2$$, and so $$c_0(z_1) = -z_1^2$$.

A function $$f(z_1,\ldots,z_n)$$ is symmetric if $$f = f \circ p$$ for all permutations of the variables $$p$$. The elementary symmetric functions of $$\alpha_1,\ldots,\alpha_k$$ are the coefficients $$\sigma_j$$ of the polynomial $\prod_{j=1}^k \bigl(t+\alpha_j\bigr) = t^k + \sigma_{1} \, t^{k-1} + \cdots + \sigma_{k-2} \, t^2 + \sigma_{k-1} \, t + \sigma_k .$ In other words: \begin{align}\begin{aligned} \sigma_{1} & = \alpha_1 + \alpha_2 + \cdots + \alpha_k, \\ \sigma_{2} & = \alpha_1 \alpha_2 \, + \, \alpha_1 \alpha_3 \, + \, \cdots \, + \, \alpha_{k-1} \alpha_k , \\ & \;\: \smash{\vdots} \\ \sigma_{k-1} & = \alpha_2 \alpha_3 \cdots \alpha_{k} \, + \, \alpha_1 \alpha_3 \alpha_4 \cdots \alpha_{k} \, + \, \cdots \,+\, \alpha_1 \alpha_2 \cdots \alpha_{k-1}, \\ \sigma_k & = \alpha_1 \alpha_2 \cdots \alpha_k.\end{aligned}\end{align} So for example when $$k=2$$, then $$\sigma_2 = \alpha_1\alpha_2$$ and $$\sigma_1 = \alpha_1 + \alpha_2$$. The function $$\sigma_1$$ happens to already be a power sum. We can write $$\sigma_2$$ as a polynomial in the power sums: $\sigma_2 = \frac{1}{2} \left( {\bigl(\alpha_1 + \alpha_2\bigr)}^2 - \bigl(\alpha_1^2 + \alpha_2^2\bigr) \right) .$

## Exercise $$\PageIndex{1}$$

Show that elementary symmetric functions are polynomials in the power sums.

## Exercise $$\PageIndex{2}$$

Prove the fundamental theorem of symmetric polynomials: Every symmetric polynomial can be written as a polynomial in the elementary symmetric functions. Use the following procedure. Using double induction, suppose the theorem is true if the number of variables is less than $$k$$, and the theorem is true in $$k$$ variables for degree less than $$d$$. Consider a symmetric $$P(z_1,\ldots,z_k)$$ of degree $$d$$. Write $$P(z_1,\ldots,z_{k-1},0)$$ by induction hypothesis as a polynomial in the elementary symmetric functions of one less variable. Use the same coefficients, but plug in the elementary symmetric functions of $$k$$ variables except the symmetric polynomial in $$k$$ variables of degree $$k$$, that is except the $$z_1\cdots z_k$$. You will obtain a symmetric function $$L(z_1,\ldots,z_k)$$ and you need to show $$L(z_1,\ldots,z_{k-1},0) = P(z_1,\ldots,z_{k-1},0)$$. Now use symmetry to prove that $P(z_1,\ldots,z_k) = L(z_1,\ldots,z_k) + z_1\cdots z_k Q(z_1,\ldots,z_k) .$ Then note that $$Q$$ has lower degree and finish by induction.

## Exercise $$\PageIndex{3}$$

Extend the previous exercise to power series. Suppose $$f(z_1,\ldots,z_k)$$ is a convergent symmetric power series at 0, show that $$f$$ can be written as a convergent power series in the elementary symmetric functions.

## Exercise $$\PageIndex{4}$$

Suppose $$P(z',z_n)$$ is a Weierstrass polynomial of degree $$k$$, and write the zeros as $$\alpha_1(z'), \ldots, \alpha_k(z')$$. These are not holomorphic functions, but suppose that $$f$$ is a symmetric convergent power series at the origin in $$k$$ variables. Show that $$f\bigl(\alpha_1(z'), \ldots, \alpha_k(z')\bigr)$$ is a holomorphic function of $$z'$$ near the origin.

The hypotheses of the preparation theorem are not an obstacle. If a holomorphic function $$f$$ is such that $$z_n \mapsto f(0,z_n)$$ vanishes identically, then we can make a small linear change of coordinates $$L$$ ($$L$$ can be a matrix arbitrarily close to the identity) such that $$f \circ L$$ satisfies the hypotheses of the theorem. For example, $$f(z_1,z_2,z_3) = z_1z_3+z_2z_3$$ does not satisfy the hypotheses of the theorem as $$f(0,0,z_3) \equiv 0$$. But for an arbitrarily small $$\epsilon \not= 0$$, replacing $$z_2$$ with $$z_2 + \epsilon z_3$$ leads to $$\tilde{f}(z_1,z_2,z_3) = f(z_1,z_2+\epsilon z_3,z_3) = z_1z_3+z_2z_3 + \epsilon z_3^2$$, and $$\tilde{f}(0,0,z_3) = \epsilon z_3^2$$. Thence $$\tilde{f}$$ satisfies the hypotheses of the theorem.

## Exercise $$\PageIndex{5}$$

Prove the fact above about the existence of $$L$$ arbitrarily close to the identity.

## Exercise $$\PageIndex{6}$$

Prove that a monic polynomial $$P(\zeta)$$ of one variable is uniquely determined by its zeros up to multiplicity. That is, suppose $$P$$ and $$Q$$ are two monic polynomials with the same zeros up to multiplicity, then $$P=Q$$. That proves the uniqueness of the Weierstrass polynomial.

## Exercise $$\PageIndex{7}$$

Suppose $$D \subset \mathbb{C}$$ is a bounded domain, $$0 \in D$$, $$U' \subset \mathbb{C}^{n-1}$$ is a domain, $$0 \in U'$$, and $$P \in \mathcal{O}(U')[z_n]$$ is a Weierstrass polynomial such that $$P(z',z_n)$$ is not zero on $$U' \times \partial D$$. Then for any $$z' \in U$$, all zeros of $$z_n \mapsto P(z',z_n)$$ are in $$D$$.

## Exercise $$\PageIndex{8}$$

Let $$D \subset \mathbb{C}$$ be a bounded domain, and $$U' \subset \mathbb{C}^{n-1}$$ a domain. Suppose $$f$$ is a continuous function on $$U' \times \overline{D}$$ holomorphic on $$U' \times D$$, where $$f$$ is zero on at least one point of $$U' \times D$$, and $$f$$ is never zero on $$U' \times \partial D$$. Prove that $$z_n \mapsto f(z',z_n)$$ has at least one zero in $$D$$ for every $$z' \in U'$$.

The order of vanishing of $$f$$ at the origin is a lower bound on the number $$k$$ in the theorem. The order of vanishing for a certain variable may be larger than this lower bound. If $$f(z_1,z_2) = z_1^2 + z_2^3$$, then the $$k$$ we get is 3, but $$\text{ord}_0 f = 2$$. We can make a small linear change of coordinates to ensure $$k = \text{ord}_0 f$$. With the $$f$$ as above, $$f(z_1 + \epsilon z_2,z_2)$$ gives $$k = 2$$ as expected.

The Weierstrass preparation theorem is a generalization of the implicit function theorem. When $$k=1$$ in the theorem, then we obtain the Weierstrass polynomial $$z_n + c_0(z')$$. That is, the zero set of $$f$$ is a graph of the holomorphic function $$-c_0$$. Therefore, the Weierstrass theorem is a generalization of the implicit function theorem to the case when $$\frac{\partial f}{\partial z_n}$$ is zero. We can still “solve” for $$z_n$$, but we find $$k$$ solutions given as the zeros of the obtained Weierstrass polynomial.

There is an obvious statement of the preparation theorem for germs.

## Exercise $$\PageIndex{9}$$

State and prove a germ version of the preparation theorem.

## Theorem $$\PageIndex{2}$$

Weierstrass Division Theorem

Suppose $$f$$ is holomorphic near the origin, and suppose $$P$$ is a Weierstrass polynomial of degree $$k \geq 1$$ in $$z_n$$. Then there exists a neighborhood $$V$$ of the origin and unique $$q,r \in \mathcal{O}(V)$$, where $$r$$ is a polynomial in $$z_n$$ of degree less than $$k$$, and on $$V$$, $f = qP + r .$

Note that $$r$$ need not be a Weierstrass polynomial; it need not be monic nor do the coefficients need to vanish at the origin. It is simply a polynomial in $$z_n$$ with coefficients that are holomorphic functions of the first $$n-1$$ variables.

Proof

Uniqueness is left as an exercise. Consider a connected neighborhood $$V = V' \times D$$ of the origin, where $$D$$ is a disc, $$f$$ and $$P$$ are continuous in $$V' \times \overline{D}$$, and $$P$$ is not zero on $$V' \times \partial D$$. Let $q(z',z_n) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(z',\zeta)}{P(z',\zeta)(\zeta-z_n)} ~d\zeta .$ As $$P$$ is not zero on $$V' \times \partial D$$, the function $$q$$ is holomorphic in $$V$$ (differentiate under the integral). If $$P$$ did divide $$f$$, then $$q$$ would really be $$\frac{f}{P}$$. But if $$P$$ does not divide $$f$$, then the Cauchy integral formula does not apply and $$q$$ is not equal to $$\frac{f}{P}$$. Interestingly the expression does give the quotient in the division with remainder.

Write $$f$$ using the Cauchy integral formula in $$z_n$$ and subtract $$qP$$ to obtain $$r$$: $r(z',z_n) = f(z',z_n) - q(z',z_n)P(z',z_n) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(z',\zeta)P(z',\zeta) - f(z',\zeta)P(z',z_n)}{P(z',\zeta)(\zeta-z_n)} ~d\zeta .$ We need to show $$r$$ is a polynomial in $$z_n$$ of degree less than $$k$$. In the expression inside the integral, the numerator is of the form $$\sum_j h_j(z',\zeta)(\zeta^j-z_n^j)$$ and is therefore divisible by $$(\zeta-z_n)$$. The numerator is a polynomial of degree $$k$$ in $$z_n$$. After dividing by $$(\zeta-z_n)$$, the integrand becomes a polynomial in $$z_n$$ of degree $$k-1$$. Use linearity of the integral to integrate the coefficients of the polynomial. Each coefficient is a holomorphic function in $$V'$$ and the proof is finished. Some coefficients may have integrated to zero, so we can only say that $$r$$ is a polynomial of degree $$k-1$$ or less.

For example, let $$f(z,w) = e^z + z^4 e^w + z w^2 e^w + zw$$ and $$P(z,w) = w^2 + z^3$$. Then $$P$$ is a Weierstrass polynomial in $$w$$ of degree $$k=2$$. A bit of computation shows $\frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{e^z + z^4 e^{\zeta} + z {\zeta}^2 e^{\zeta} + z \zeta}{(\zeta^2+z^3)(\zeta-w)} d\zeta = z e^w , \quad \text{so} \quad f(z,w) = \underbrace{\bigl( ze^w \bigr)}_{q} \underbrace{\bigl( w^2 + z^3 \bigr)}_{P} + \underbrace{z w + e^z}_{r} .$ Notice that $$r$$ is a polynomial of degree 1 in $$w$$, but it is neither monic, nor do coefficients vanish at 0.

## Exercise $$\PageIndex{10}$$

Prove the uniqueness part of the theorem.

## Exercise $$\PageIndex{11}$$

State and prove a germ version of the division theorem.

The Weierstrass division theorem is a generalization of the division algorithm for polynomials with coefficients in a field, such as the complex numbers: If $$f(\zeta)$$ is a polynomial, and $$P(\zeta)$$ is a nonzero polynomial of degree $$k$$, then there exist polynomials $$q(\zeta)$$ and $$r(\zeta)$$ with degree of $$r$$ less than $$k$$ such that $$f = qP + r$$. If the coefficients are in a commutative ring, we can divide as long as $$P$$ is monic. The Weierstrass division theorem says that in the case of the ring $$\mathcal{O}_p$$, we can divide by a monic $$P \in {}_{n-1}\mathcal{O}_p[z_n]$$, even if $$f$$ is a holomorphic function (a “polynomial of infinite degree”) as long as $$f(0,z_n)$$ has finite order.

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