6.2: Weierstrass Preparation and Division Theorems

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.