Some photos I have taken are too blurry, so I am also following some nicely organized notes by Aleksander Horawa. Of course, this blog is all about my personal learning notes, so they are not identical to the actual lectures.
- Given a complex analytic space, the local ring at any point is Noetherian.
- Let $X$ be a complex variety. Given any $\mathbb{C}$-point $x$ of $X,$ the local map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is faithfully flat.
Our goal is to prove the first for the case of complex manifolds and the second for the case of smooth complex varieties. (In this posting, we only deal with the first.)
Ring of convergent power series. We write $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ to mean the $\mathbb{C}$-subalgebra of $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ that consists of convergent power series, meaning $$f(z_{1}, \dots, z_{n}) = \sum_{d_{1} \geq 0} \cdots \sum_{d_{n} \geq 0} a_{d_{1}, \dots, d_{n}}z_{1}^{d_{1}} \cdots z_{n}^{d_{n}}$$ such that there exits some $\epsilon > 0$ such that $f(z_{1}, \dots, z_{n})$ converges uniformly and absolutely for $|z_{j}| < \epsilon$ for all $1 \leq j \leq n.$
Remark. Note that $f(z_{1}, \dots, z_{n})$ given above is convergent if and only if there is $\epsilon > 0$ such that the sequence $(|a_{d_{1}, \dots, d_{n}}|\epsilon^{d_{1} + \cdots + d_{n}})_{d_{1}, \dots, d_{n} \geq 0}$ is bounded. By Cauchy-Hadamard, this is also equivalent to $$\limsup_{d_{1} + \cdots + d_{n} \rightarrow \infty}|a_{d_{1}, \dots, d_{n}}|^{1/(d_{1} + \cdots + d_{n})} < \infty,$$ although I cannot quite comprehend this just yet.
The map $\mathscr{O}_{\mathbb{C}^{n}, \boldsymbol{0}} \rightarrow \mathbb{C}[[z_{1}, \dots, z_{n}]]$ given by $$f(z_{1}, \dots, z_{n}) \mapsto \sum_{\boldsymbol{d} \in (\mathbb{Z}_{\geq 0})^{n}}\frac{1}{\boldsymbol{d}!} \frac{\partial^{|\boldsymbol{d}|}f}{\partial \boldsymbol{z}^{\boldsymbol{d}}}(\boldsymbol{0}) \boldsymbol{z}^{\boldsymbol{d}}$$ is injective $\mathbb{C}$-algebra map, and its image is $\mathbb{C}\{z_{1}, \dots, z_{n}\}.$ Hence, for any complex manifold $X$ and a point $x \in X,$ we have $$\mathscr{O}_{X,x} \simeq \mathscr{O}_{\mathbb{C}^{n}, \boldsymbol{0}} \simeq \mathbb{C}\{z_{1}, \dots, z_{n}\}.$$ Thus, to show local rings are Noetherian, it is enough to show:
Theorem. The $\mathbb{C}$-algebra $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is Noetherian.
This will be a direct consequence of another theorem, called "Weierstrass Preparation". For this, we define that an element in $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Weierstrass polynomial in $z_{n}$ if it can be written as the following form: $$z_{n}^{d} + a_{1}(z_{1}, \dots, z_{n-1})z_{n}^{d-1} + \cdots + a_{d}(z_{1}, \dots, z_{n-1}) \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ such that $a_{j}(0, \dots, 0) = 0$ for $1 \leq j \leq d.$
Weierstrass Preparation Theorem. For $f(z_{1}, \dots, z_{n}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that $f(0, \dots, 0, z_{n}) \neq 0 \in \mathbb{C}\{z_{n}\},$ there are unique $g(\boldsymbol{z}), h(\boldsymbol{z}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that
Theorem. The $\mathbb{C}$-algebra $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is Noetherian.
This will be a direct consequence of another theorem, called "Weierstrass Preparation". For this, we define that an element in $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Weierstrass polynomial in $z_{n}$ if it can be written as the following form: $$z_{n}^{d} + a_{1}(z_{1}, \dots, z_{n-1})z_{n}^{d-1} + \cdots + a_{d}(z_{1}, \dots, z_{n-1}) \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ such that $a_{j}(0, \dots, 0) = 0$ for $1 \leq j \leq d.$
Weierstrass Preparation Theorem. For $f(z_{1}, \dots, z_{n}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that $f(0, \dots, 0, z_{n}) \neq 0 \in \mathbb{C}\{z_{n}\},$ there are unique $g(\boldsymbol{z}), h(\boldsymbol{z}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that
- $h(0, \dots, 0) \neq 0,$
- $g(\boldsymbol{z})$ is a Weierstrass polynomial in $z_{n},$ and
- $f(\boldsymbol{z}) = g(\boldsymbol{z})h(\boldsymbol{z}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}.$
Remark. Note that Weierstrass Preparation basically says that if a convergent power series has a monomial consisting of a single variable, then it is divisible by a Weierstrass polynomial (in somewhat unique fashion). For the case $n = 1,$ this is already interesting: any nonzero convergent power series $f(z)$ over $\mathbb{C}$ can be written as $$f(z) = z^{d} h(z)$$ with $h(0) \neq 0.$ This shows that zeros of $f \in \Gamma(X, \mathscr{O}_{X})$ do not accumulate in $X$ for any complex manifold $X$ of dimension $1.$
We first apply Weierstrass Preparation to prove Theorem.
Proof of Theorem. We proceed by induction on $n \geq 0.$ For $n = 0,$ we have $\mathbb{C},$ a field, which is Noetherian. Now, let $n \geq 1,$ and suppose that the result is true for any convergent power series $\mathbb{C}$-algebras with the number of variables $< n.$ Fix any nonzero ideal $I \subset \mathbb{C}\{z_{1}, \dots, z_{n}\}$ and a nonzero element $f(z_{1}, \dots, z_{n}) \in I.$ We may apply a $\mathbb{C}$-linear map to the variables $z_{1}, \dots, z_{n}$ to ensure that $f(z_{1}, \dots, z_{n})$ has a monomial that looks like $z_{n}^{d}$ for some $d \in \mathbb{Z}_{\geq 0}.$ Indeed, if there is any other $z_{j}^{d},$ we may permute the indices. If there is no monomial consisting of a single variable, consider any term, which must look like $c z_{i_{1}}^{d_{1}} \cdots z_{i_{r}}^{d_{r}}$ for some $c \neq 0.$ Permuting indices and dividing by $c,$ we may assume that the term we have is $z_{n}^{d_{1}} \cdots z_{n-r}^{d_{r}},$ and we just need to consider the invertible $\mathbb{C}$-linear map given by $z_{j} \mapsto z_{n}$ for $1 \leq j < n$ and $z_{n} \mapsto z_{n}.$ Thus, we are in the situation where $f(0, \dots, 0, z_{n}) \neq 0.$ Now, take any generating set $S$ of $I$ containing $f(\boldsymbol{z}),$ which is our modified version. For any $g(\boldsymbol{z}) \in S,$ if $g(0, \dots, 0, z_{n}) = 0,$ then replace $g$ with $g + f$ in $S$ so that we may assume $g(0, \dots, 0, z_{n}) \neq 0.$ This constructs another generating set for $I$ because any element in $I$ can be written as a linear sum of finitely many generators over the base ring. Hence, after this modification, we may assume that $S$ is a generating set for $I$ such that for any $f \in S,$ we have $$f(0, \dots, 0, z_{n}) \neq 0.$$ But then now note that Weierstrass Preparation says that we have $f \in I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$ This shows that $$S \subset I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$$ By the induction hypothesis $\mathbb{C}\{z_{1}, \dots, z_{n-1}\}$ is Noetherian, so by Hilbert basis, $\mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$ is Noetherian as well. This implies that $I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$ is finitely generated, so $(S) = I$ must be too. $\Box.$
Proof of Weierstrass Preparation. We will use the following residue theorem: consider any nonempty open $U \subset \mathbb{C}$ and fix distinct $p_{1}, \dots, p_{r} \in D,$ an open disc such that $\bar{D} \subset U.$ Given any $f \in \Gamma(U \smallsetminus \{p_{1}, \dots, p_{r}\}, \mathscr{O}_{\mathbb{C}}),$ we have $$\frac{1}{2\pi i}\int_{\partial D}f(z)dz = \sum_{j=1}^{r}\mathrm{Res}_{p_{j}}(f).$$ The proof of this fact is in any usual complex analysis text (e.g., Ch. 3 Corollary 2.2 of Stein and Shakarchi). We will apply this in the following specific situation.
Lemma. Let $U \subset \mathbb{C}$ be nonempty and open and say $f \in \Gamma(U, \mathscr{O}_{\mathbb{C}}).$ Say we have an open disk $D$ in $\mathbb{C}$ such that $\bar{D} \subset U$ and $f$ has no roots on $.$ For any $j \in \mathbb{Z}_{\geq 0},$ we have $$\int_{\partial D} z^{j} \frac{f'(z)}{f(z)} dz = \lambda_{1}^{j} + \cdots + \lambda_{m}^{j},$$ where $\lambda_{1}, \dots, \lambda_{m}$ are the roots of $f(z)$ in $D$ counted with multiplicity.
Proof. If there are no roots of $f$ in $D,$ then it has no roots in $\bar{D}$ due to the hypothesis. Since $\bar{D}$ is closed in $\mathbb{C},$ this means that there is an open subset $W$ of $\mathbb{C}$ containing $\bar{D}$ such that $f$ takes no roots in $W.$ On $W,$ the expression $z^{j} f'(z)/f(z)$ defines a holomorphic function in $z \in W,$ so by Cauchy's theorem the integral in question is evaluated to be $0.$
Now, suppose that we have some roots of $f$ in $D.$ Let $\lambda$ be any one of them. We may write $$f(z) = (z - \lambda)^{m}h(z),$$ where $m \in \mathbb{Z}_{\geq 1}$ and $h(z)$ is a holomorphic function defined in a small enough open disc centered at $\lambda$ such that $h(\lambda) \neq 0.$ Then $$\frac{f'(z)}{f(z)} = \frac{m(z - \lambda)^{m-1}h(z) + (z - \lambda)^{m}h'(z)}{(z - \lambda)^{m}h(z)} = \frac{m}{z - \lambda} + \frac{h'(z)}{h(z)},$$ so $$\begin{align*}\mathrm{Res}_{\lambda}\left(z^{j}\frac{f'(z)}{f(z)}\right) &= \mathrm{Res}_{\lambda}\left(\frac{mz^{j}}{z - \lambda}\right) \\ &= \lim_{z \rightarrow \lambda} (z - \lambda) \left(\frac{mz^{j}}{z - \lambda}\right) \\ &= m\lambda^{j}.\end{align*}$$ Since $m$ is the multiplicity of the root $\lambda.$ Applying the residue theorem, this finishes the proof. $\Box$
We are now ready to prove Weierstrass Preparation.
Proof of Weierstrass Preparation. Fix an open neighborhood $U$ of $(0, \dots, 0) \in \mathbb{C}.$ Since $f(0, \dots, 0, z_{n}) \neq 0,$ we know that zeros of $f(0, \dots, 0, z_{n})$ in $z_{n}$ stay outside a small disc centered at the origin. Hence, we may take $\epsilon_{n} > 0$ such that $f(0, \dots, 0, z_{n})$ has no zeros in $z_{n}$ for $0 < |z_{n}| \leq \epsilon_{n}.$ Since $z_{n} \in \mathbb{C}$ with $|z_{n}| = \epsilon_{n}$ defines a compact set in $\mathbb{C},$ we may find $\epsilon > 0$ such that whenever $|z_{1}|, \dots, |z_{n-1}| < \epsilon$ and $|z_{n}| = \epsilon_{n},$ we have $f(z_{1}, \dots, z_{n-1}, z_{n}) \neq 0.$ We may also assume that $\epsilon > 0$ is small enough so that $$D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0) \subset U.$$ Fix any $\boldsymbol{z} = (z_{1}, \dots, z_{n-1}) \in D_{\epsilon}(0)^{n-1},$ and let $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m_{\boldsymbol{z}}}(\boldsymbol{z})$ be the roots of $f(\boldsymbol{z}, z_{n})$ in $z_{n} \in D_{\epsilon_{n}}(0),$ counted with multiplicities. (The can only be finitely many zeros because otherwise, they will accrue in $\bar{D},$ which will contradict that $f(\boldsymbol{z}, z_{n}) \neq 0.$) Applying Lemma, for any $j \in \mathbb{Z}_{\geq 0},$ we get $$\sum_{l=1}^{m_{\boldsymbol{z}}}\lambda_{l}(\boldsymbol{z})^{j} = \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} w^{j}\frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw,$$ The upshot is that the right-hand side is a holomorphic function in $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1}.$ Taking $j = 0,$ we have $$ \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} \frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw = m_{\boldsymbol{z}},$$ but since the right-hand side is always integer no matter where we vary $\boldsymbol{z}$ in the connected set $D_{\epsilon}(0)^{n-1},$ it must be the same integer regardless of the choice of $\boldsymbol{z}.$ Hence, we may write $m = m_{\boldsymbol{z}}.$
Denote by $\sigma_{1}(\boldsymbol{z}), \dots, \sigma_{n}(\boldsymbol{z})$ the elementary symmetric polynomials in $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$ meaning, we have $$\begin{align*}g(z_{1}, \dots, z_{n}) &:= z_{n}^{m} - \sigma_{1}(\boldsymbol{z})z_{n}^{m-1} + \cdots + (-1)^{m}\sigma_{m}(\boldsymbol{z}) \\ &= (z_{n} - \lambda_{1}(\boldsymbol{z})) \cdots (z_{n} - \lambda_{m}(\boldsymbol{z})).\end{align*}$$ By definition this is a Weierstrass polynomial in $z_{n}.$ Note that $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ defines a holomorphic function on $$(D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0)) \smallsetminus \{(\boldsymbol{z}, z_{n}) : z_{n} \neq \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z})\}.$$ For every fixed $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ the fraction $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ extends to a unique holomorphic function in $z_{n} \in D_{\epsilon_{n}}(0),$ since the numerator can cancel out all the roots of the denominator. Calling this function $h(\boldsymbol{z}, z_{n}),$ we check that $h(0, \dots, 0) \neq 0$ (since the roots of the numerator is precisely given by those of the denominator) and since $h(\boldsymbol{z}, z_{n}) = f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n}),$ fixing $z_{n},$ we see that $h(\boldsymbol{z}, z_{n})$ is holomorphic in $\boldsymbol{z}$ as well. This gives us the existence of the desired factorization: $$f(z_{1}, \dots, z_{n}) = g(z_{1}, \dots, z_{n})h(z_{1}, \dots, z_{n}).$$ For the uniqueness, suppose that we have another factorization $$f(z_{1}, \dots, z_{n}) = g'(z_{1}, \dots, z_{n})h'(z_{1}, \dots, z_{n}),$$ where $$g'(z_{1}, \dots, z_{n}) = z_{n}^{m'} + \cdots \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ with $h'$ holomorphic such that $h'(0, \dots, 0) \neq 0.$ Then $$z_{n}^{m}h(0, \dots, 0, z_{n}) = f(0, \cdots, 0, z_{n}) = z_{n}^{m'}h'(0, \dots, 0, z_{n}),$$ and both sides must have the same number of roots in $z_{n}$ counting with multiplicities, so $m' = m.$ Fix $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ where we may have to make $\epsilon > 0$ smaller. Then $g'(\boldsymbol{z}, z_{n})$ vanishes at $$z_{n} = \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$$ counting with multiplicities. Hence, we must have $g(\boldsymbol{z}, z_{n}) = g'(\boldsymbol{z}, z_{n}),$ and $h = h'$ follows from here. $\Box$
Proof of Weierstrass Preparation. We will use the following residue theorem: consider any nonempty open $U \subset \mathbb{C}$ and fix distinct $p_{1}, \dots, p_{r} \in D,$ an open disc such that $\bar{D} \subset U.$ Given any $f \in \Gamma(U \smallsetminus \{p_{1}, \dots, p_{r}\}, \mathscr{O}_{\mathbb{C}}),$ we have $$\frac{1}{2\pi i}\int_{\partial D}f(z)dz = \sum_{j=1}^{r}\mathrm{Res}_{p_{j}}(f).$$ The proof of this fact is in any usual complex analysis text (e.g., Ch. 3 Corollary 2.2 of Stein and Shakarchi). We will apply this in the following specific situation.
Lemma. Let $U \subset \mathbb{C}$ be nonempty and open and say $f \in \Gamma(U, \mathscr{O}_{\mathbb{C}}).$ Say we have an open disk $D$ in $\mathbb{C}$ such that $\bar{D} \subset U$ and $f$ has no roots on $.$ For any $j \in \mathbb{Z}_{\geq 0},$ we have $$\int_{\partial D} z^{j} \frac{f'(z)}{f(z)} dz = \lambda_{1}^{j} + \cdots + \lambda_{m}^{j},$$ where $\lambda_{1}, \dots, \lambda_{m}$ are the roots of $f(z)$ in $D$ counted with multiplicity.
Proof. If there are no roots of $f$ in $D,$ then it has no roots in $\bar{D}$ due to the hypothesis. Since $\bar{D}$ is closed in $\mathbb{C},$ this means that there is an open subset $W$ of $\mathbb{C}$ containing $\bar{D}$ such that $f$ takes no roots in $W.$ On $W,$ the expression $z^{j} f'(z)/f(z)$ defines a holomorphic function in $z \in W,$ so by Cauchy's theorem the integral in question is evaluated to be $0.$
Now, suppose that we have some roots of $f$ in $D.$ Let $\lambda$ be any one of them. We may write $$f(z) = (z - \lambda)^{m}h(z),$$ where $m \in \mathbb{Z}_{\geq 1}$ and $h(z)$ is a holomorphic function defined in a small enough open disc centered at $\lambda$ such that $h(\lambda) \neq 0.$ Then $$\frac{f'(z)}{f(z)} = \frac{m(z - \lambda)^{m-1}h(z) + (z - \lambda)^{m}h'(z)}{(z - \lambda)^{m}h(z)} = \frac{m}{z - \lambda} + \frac{h'(z)}{h(z)},$$ so $$\begin{align*}\mathrm{Res}_{\lambda}\left(z^{j}\frac{f'(z)}{f(z)}\right) &= \mathrm{Res}_{\lambda}\left(\frac{mz^{j}}{z - \lambda}\right) \\ &= \lim_{z \rightarrow \lambda} (z - \lambda) \left(\frac{mz^{j}}{z - \lambda}\right) \\ &= m\lambda^{j}.\end{align*}$$ Since $m$ is the multiplicity of the root $\lambda.$ Applying the residue theorem, this finishes the proof. $\Box$
We are now ready to prove Weierstrass Preparation.
Proof of Weierstrass Preparation. Fix an open neighborhood $U$ of $(0, \dots, 0) \in \mathbb{C}.$ Since $f(0, \dots, 0, z_{n}) \neq 0,$ we know that zeros of $f(0, \dots, 0, z_{n})$ in $z_{n}$ stay outside a small disc centered at the origin. Hence, we may take $\epsilon_{n} > 0$ such that $f(0, \dots, 0, z_{n})$ has no zeros in $z_{n}$ for $0 < |z_{n}| \leq \epsilon_{n}.$ Since $z_{n} \in \mathbb{C}$ with $|z_{n}| = \epsilon_{n}$ defines a compact set in $\mathbb{C},$ we may find $\epsilon > 0$ such that whenever $|z_{1}|, \dots, |z_{n-1}| < \epsilon$ and $|z_{n}| = \epsilon_{n},$ we have $f(z_{1}, \dots, z_{n-1}, z_{n}) \neq 0.$ We may also assume that $\epsilon > 0$ is small enough so that $$D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0) \subset U.$$ Fix any $\boldsymbol{z} = (z_{1}, \dots, z_{n-1}) \in D_{\epsilon}(0)^{n-1},$ and let $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m_{\boldsymbol{z}}}(\boldsymbol{z})$ be the roots of $f(\boldsymbol{z}, z_{n})$ in $z_{n} \in D_{\epsilon_{n}}(0),$ counted with multiplicities. (The can only be finitely many zeros because otherwise, they will accrue in $\bar{D},$ which will contradict that $f(\boldsymbol{z}, z_{n}) \neq 0.$) Applying Lemma, for any $j \in \mathbb{Z}_{\geq 0},$ we get $$\sum_{l=1}^{m_{\boldsymbol{z}}}\lambda_{l}(\boldsymbol{z})^{j} = \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} w^{j}\frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw,$$ The upshot is that the right-hand side is a holomorphic function in $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1}.$ Taking $j = 0,$ we have $$ \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} \frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw = m_{\boldsymbol{z}},$$ but since the right-hand side is always integer no matter where we vary $\boldsymbol{z}$ in the connected set $D_{\epsilon}(0)^{n-1},$ it must be the same integer regardless of the choice of $\boldsymbol{z}.$ Hence, we may write $m = m_{\boldsymbol{z}}.$
Denote by $\sigma_{1}(\boldsymbol{z}), \dots, \sigma_{n}(\boldsymbol{z})$ the elementary symmetric polynomials in $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$ meaning, we have $$\begin{align*}g(z_{1}, \dots, z_{n}) &:= z_{n}^{m} - \sigma_{1}(\boldsymbol{z})z_{n}^{m-1} + \cdots + (-1)^{m}\sigma_{m}(\boldsymbol{z}) \\ &= (z_{n} - \lambda_{1}(\boldsymbol{z})) \cdots (z_{n} - \lambda_{m}(\boldsymbol{z})).\end{align*}$$ By definition this is a Weierstrass polynomial in $z_{n}.$ Note that $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ defines a holomorphic function on $$(D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0)) \smallsetminus \{(\boldsymbol{z}, z_{n}) : z_{n} \neq \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z})\}.$$ For every fixed $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ the fraction $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ extends to a unique holomorphic function in $z_{n} \in D_{\epsilon_{n}}(0),$ since the numerator can cancel out all the roots of the denominator. Calling this function $h(\boldsymbol{z}, z_{n}),$ we check that $h(0, \dots, 0) \neq 0$ (since the roots of the numerator is precisely given by those of the denominator) and since $h(\boldsymbol{z}, z_{n}) = f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n}),$ fixing $z_{n},$ we see that $h(\boldsymbol{z}, z_{n})$ is holomorphic in $\boldsymbol{z}$ as well. This gives us the existence of the desired factorization: $$f(z_{1}, \dots, z_{n}) = g(z_{1}, \dots, z_{n})h(z_{1}, \dots, z_{n}).$$ For the uniqueness, suppose that we have another factorization $$f(z_{1}, \dots, z_{n}) = g'(z_{1}, \dots, z_{n})h'(z_{1}, \dots, z_{n}),$$ where $$g'(z_{1}, \dots, z_{n}) = z_{n}^{m'} + \cdots \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ with $h'$ holomorphic such that $h'(0, \dots, 0) \neq 0.$ Then $$z_{n}^{m}h(0, \dots, 0, z_{n}) = f(0, \cdots, 0, z_{n}) = z_{n}^{m'}h'(0, \dots, 0, z_{n}),$$ and both sides must have the same number of roots in $z_{n}$ counting with multiplicities, so $m' = m.$ Fix $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ where we may have to make $\epsilon > 0$ smaller. Then $g'(\boldsymbol{z}, z_{n})$ vanishes at $$z_{n} = \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$$ counting with multiplicities. Hence, we must have $g(\boldsymbol{z}, z_{n}) = g'(\boldsymbol{z}, z_{n}),$ and $h = h'$ follows from here. $\Box$
No comments:
Post a Comment