1. Weierstrass Polynomials
Let f (z, w) be holomorphic function in a neighborhood of C2 with f (0, 0) = 0. Suppose f is not identically zero on the w-axis. Consider the Taylor expansion of f (0, w) in a neighborhood of w = 0 in C. There is d > 0 and ad 6= 0 so that f (0, w) = adwd+ · · · in
|w| < R for some R > 0. The number R can be chosen so that f (0, w) is nonzero when 0 < |w| ≤ R/2. Choose δ > 0 so that |f (0, w)| ≥ δ for |w| = r = R/2. By continuity of f (z, w), we can choose > 0 so that
|f (z, w) − f (0, w)| < δ 2, for all |z| < and |w| = r. Hence
|f (z, w) − f (0, w)| < |f (0, w)|
for all |z| < and |w| = r. By Rouche’s theorem, for any given |z| < , f (z, w) and f (0, w) has the same number of zeros in |w| < r, said b1(z), · · · , bd(z), i.e. f (z, bi(z)) = 0 for 1 ≤ i ≤ d. Moreover, for any given |z| < ,
|f (z, w)| ≥ δ 2 > 0 on |w| = r. For any |z| < ,
b1(z)j + · · · + bd(z)j = 1 2π√
−1 Z
|w|=r
ζj
∂f
∂w(z, ζ) f (z, ζ) dζ.
By Lebesgue dominated convergence theorem and f (z, w) being holomorphic, we have
∂
∂z Z
|w|=r
ζj
∂f
∂w(z, ζ) f (z, ζ) dζ =
Z
|w|=r
∂
∂z
∂f
∂w(z, ζ) ζj f (z, ζ)dζ
= Z
|w|=r
∂
∂w
∂f
∂z(z, ζ) ζj f (z, ζ)dζ
= 0.
This implies that b1(z)j+ · · · + bd(z)j are holomorphic functions in |z| < . Let σj(z) be the j-th elementary symmetric function in b1(z), · · · , bd(z), i.e.
σj(z) = X
1≤i1<···<ij≤d
bi1(z) · · · bij(z), 1 ≤ j ≤ d.
Using the following Lemma, we obtain σj(z) is holomorphic in |z| < .
Lemma 1.1. Let ej(x1, · · · , xn) ∈ Z[x1, · · · , xn] be the j-th elementary symmetric polyno- mial in x1, · · · , xn and pk be the k-th power sum of x1, · · · , xn. Then ej can be expressed in terms of polynomial in {p1, · · · , pn} and pk can be expressed in terms of polynomial in {e1, · · · , en}.
Since the power sums of b1(z), · · · , bd(z) are holomorphic functions in |z| < , σj(z) are holomorphic function in |z| < . Moreover, since w = 0 is the only zero of f (0, w) in |w| < r, we find σj(0) = 0 for 1 < j < d. Define
g(z, w) = wd− σ1(z)wd−1+ · · · + (−1)dσd(z).
1
2
Definition 1.1. A Weierstrass polynomial is a polynomial in w of the form wd+ a1(z)wd+ · · · + ad(z),
where ai(z) are holomorphic functions in a neighborhood of z = 0 with ai(0) = 0.
Let us define
h(z, w) = f (z, w) g(z, w)
for |z| < and |w| < r. For each fixed |z| < , f (z, w) and g(z, w) has the same zeros in
|w| < r. Hence h(z, w) has only removable singularities in |w| < r for all |z| < . Hence h(z, w) can be extended to a holomorphic function in variable w with |w| < r for all |z| < given. By Cauchy integral formula, for any |z| < given, we have
h(z, w) = 1 2π√
−1 Z
|ζ|=r
f (z, ζ) g(z, ζ)
1 ζ − wdζ.
By Lebesgue dominated convergence theorem and the fact that f (z, w) and g(z, w) are holomorphic in z,
∂
∂z Z
|ζ|=r
f (z, ζ) g(z, ζ)
1
ζ − wdζ = Z
|ζ|=r
∂
∂z f (z, ζ) g(z, ζ)
1 ζ − wdζ
= Z
|ζ|=r
fz(z, ζ)g(z, ζ) − f (z, w)gz(z, w) (g(z, ζ))2
1 ζ − wdζ
= 0
This shows that h(z, w) is also holomorphic for |z| < . Notice that h(0, w) = f (0, w)/g(0, w) = ad+ ad+1w + · · · . Hence h(0, 0) = ad6= 0. We proved that
Theorem 1.1. If f (z, w) is holomorphic in a neighborhood of (0, 0) in C2 and is not identically zero on the w-axis, then there is > 0 and r > 0 so that in |z| < , |w| < r, one has
f (z, w) = h(z, w)g(z, w) where g(z, w) is a Weierstrass polynomial in w and h(0, 0) 6= 0.
Suppose f (z, w) = h1(z, w)g1(z, w) is another decomposition of f (z, w) with g1(z, w) a Weierstrass polynomial and h1(0, 0) 6= 0. In a neighborhood of (0, 0), h1(z, w) and h2(z, w) are both nonzero. For any fixed z such that h1(z, w) and h2(z, w) are non zeros when
|w| < r, the zeros of g1(z, w) and g2(z, w) coincide with the zeros b1(z), · · · , bd(z) of f (z, w).
Hence the degree d polynomials g1(z, w) and g2(z, w) in w share the same zeros when |z| < is given. For any given |z| < , g2(z, w) = c(z)g1(z, w) for some c(z) ∈ C. Since both the leading coefficient of g1(z, w) and g2(z, w) are 1, c(z) = 1 and hence g1(z, w) = g2(z, w) when |z| < is given. In a open neighborhood of (0, 0) where g1(z, w) and g2(z, w) are nonzero, h1(z, w) = h2(z, w). Then h1(z, w) = h2(z, w) in |z| < and |w| < r by coincidence principle. We prove that:
Corollary 1.1. Let f (z, w) be as above. If f (z, w) = hi(z, w)gi(z, w) for i = 1, 2, where hi(0, 0) = 0 for i = 1, 2 and gi(z, w) are Weierstrass polynomial, then h1(z, w) = h2(z, w) and g1(z, w) = g2(z, w).
