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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 \(z' \in V'\)


    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.


    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.

    This page titled 6.2: Weierstrass Preparation and Division Theorems is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform.