First, thanks to all who kindly contributed advice on the previous post.

A long on-and-off project of mine has been to learn several complex variables. My latest attempt started a few days back, though the commencement of the fall semester may derail it. As is now customary for myself, I have started writing a set of notes that I intend to make grow reasonably large, though right now it is less than thirty pages (even with parts of my AG notes on coherence copied and pasted in). I shall post here an excerpt on the Weierstrass preparation theorem, which will assume only the definition of a holomorphic function in several variables).

Let {U} be an open subset of {\mathbb{C}^n}; we write {H(U)} for the ring of holomorphic functions {U \rightarrow \mathbb{C}}. If {V \subset U} is open, we have a homomorphism of rings

\displaystyle  H(U) \rightarrow H(V)

so that the association {U \rightarrow H(U)} is a presheaf. It is immediate that it is in fact a sheaf of rings.

The stalk at {z_0} will be denoted {\mathcal{O}_{z_0}}. By definition, an element of {\mathcal{O}_{z_0}} is a germ of a holomorphic function at {z_0}, i.e. a pair {(f, U)} for {U} a neighborhood of {z_0} and {f: U \rightarrow \mathbb{C}} holomorphic. Two germs {(f,U), (g,V)} are considered equivalent if and only if they coincide on a neighborhood {W \subset U \cap V} of {z_0} (and hence on all of {U \cap V}).

Definition 1 The stalk {\mathcal{O}_{z_0}} is called the local ring at {z_0}.

It turns out that the ring {\mathcal{O}_{z_0}} has many convenient properties: it is noetherian and, though not complete, henselian. First, however, we note a reduction: any two rings {\mathcal{O}_{z_0}} and {\mathcal{O}_{z_1}} are clearly isomorphic by translating {z \rightarrow z-z_0 +z_1}. So, in the future, we shall focus on the ring {\mathcal{O} = \mathcal{O}_{0}}.

First, we remark:

Proposition 2 The stalk {\mathcal{O}_{z_0}} is a local ring whose maximal ideal is the set of germs that vanish at {z_0}; the residue field is isomorphic to {\mathbb{C}}.

Proposition 3 {\mathcal{O}} is isomorphic to the ring of convergent power series (i.e. convergent in some neighborhood of the origin) in {n} variables.

Proof: This is clear from the definition of holomorphicity: we can always get a unique local expansion of a holomorphic function in a power series. \Box

Suppose {f \in \mathcal{O}}. We can write

\displaystyle  f(z_1, \dots, z_n) = \sum_{k=0}^{\infty} z_1^k f_k(z_2, \dots, z_n)

where the {f_k} are functions in the variables {z_2, \dots, z_n}. Suppose the {f_k}‘s do not all vanish at zero; this is equivalent to saying that {f} is not identically zero along the {z_1}-axis.

Definition 4 If {f} is not identically zero along the {z_1}-axis, we say that {f} is regular in {z_1}. By a change of coordinates, this can always be arranged, since {f} cannot vanish on every line through the origin (unless it is identically zero). In fact, the collection of lines through the origin on which {f} vanishes is of measure zero, so almost any choice of coordinates will do. It follows that if {f_1, \dots} is a sequence of holomorphic functions, we can make one choice of coordinates such that all are regular in {z_1}.

Let {f} be regular in {z_1}. Suppose we have {f_m(0) \neq 0} while {f_k(0)=0} for {k<m}. Then we will say that {f} is of order {m} in {z_1}. For instance, consider the ring {\mathcal{O}'}, the local ring at zero in {\mathbb{C}^{n-1}}, i.e., in the variables {z_2, \dots, z_{n}} only, which obviously injects into {\mathcal{O}}. Suppose {g_1, \dots, g_{m-1} \in \mathcal{O}'} are non-units, i.e. vanish at the origin. Then

\displaystyle  z_1^m + z_1^{m-1} g_{m-1} + \dots + g_1 \ \ \ \ \ (1)

is regular of order {m} in {z_1}.

Definition 5 Any expression of the form is called a Weierstrass polynomial.

The next result states that this is essentially the only way.

Theorem 6 (Weierstrass preparation theorem) Let {f \in \mathcal{O}} be regular of order {m} in {z_1}. Then there is a unit {u \in \mathcal{O}} and a Weierstrass polynomial {g} of degree {m} such that {f = ug}. This representation is unique.

We shall prove this result using the Weierstrass division theorem, for which we shall give an algebraic proof.

Theorem 7 (Weierstrass division theorem) Let {g \in \mathcal{O}} be regular in {z_1} of degree {m} and let {f \in \mathcal{O}}. Then there is a unique expression {f = qg + r}, where {q \in \mathcal{O}} and {r \in \mathcal{O}'[z_1]} has degree {\leq m-1}.

Proof: We shall define the operator {\tau: \mathcal{O} \rightarrow \mathcal{O}} as follows. Let {\sum c_k(z_2, \dots, z_n) z_1^k} be a convergent power series representing an element of {\mathcal{O}}. We map this by {\tau} to

\displaystyle  \sum_{k \geq m} c_k(z_2, \dots, z_n) z_1^{k-m},

which is a “truncation” of the original power series in the {z_1} variable.

In particular, we are trying to solve the equation

\displaystyle  \tau(gq) = \tau(f)

in {q}. For convenience, we write {A = \tau(f)}, and {T: \mathcal{O} \rightarrow \mathcal{O}} to be the operator {q \rightarrow \tau(gq)}. So we are trying to solve the equation

\displaystyle  Tq = A,

i.e. to invert {T}. If we show that {T} is invertible, then we will be done.

Now we can write {g = g_0 z_1^m + s}, where {s} consists of terms of smaller degree in {z_1}. Moreover, the coefficients in the powers of {z_1} of {s} all vanish at {z_2 =z_3 = \dots = z_n = 0}. So the operator {T} decomposes into a sum {T_1 + T_2} where

\displaystyle  T_1(q) = \tau(g_0 q), \ T_2(q) = \tau(sq).

I claim that {T_1} is invertible and {T_2} is “close to” nilpotent, so we will be able to use a geometric series to invert {T}.

First, {T_1} is just multiplication by {g_0}! This is because {g_0} is just {z_1^m} times something in {z_2, \dots, z_n}; the application of {\tau} removes the {z_1^m}. Since {g_0} is invertible in {\mathcal{O}'}, {T_1} is invertible.

In fact, {T_2} is not nilpotent. But it is close to being that. For this, we will need to define a set of norms on various subspaces of {\mathcal{O}}. Let {\rho = (\rho_1, \dots, \rho_n) \in \mathbb{R}_+^{n}} be an {n}-tuple of positive numbers. For a formal power series {\sum c_{i_1\dots i_n} z_1^{i_1} \dots z_{n}^{i_n}}, define the {\rho}-norm to be

\displaystyle \left \lVert { \sum c_{i_1\dots i_n} z_1^{i_1} \dots z_{n}^{i_n}}\right \rVert_{\rho} = \sum |c_{i_1 \dots i_n}| \rho_1^{i_1} \dots \rho_n^{i_n}

With a slight abuse of notation, we let {\mathcal{O}_{\rho}} be the subring of {\mathcal{O}} such that the {\rho}-norm is finite.

Lemma 8 With respect to the {\rho}-norm, {\mathcal{O}_{\rho}} is a Banach algebra.

Proof: It is easy to see that the {\rho}-norm is indeed a norm and (with a little algebra) that it is submultiplicative. To see completeness, note that if there is a Cauchy sequence of formal power series {\sum c^{(v)}_{i_1 \dots i_n} z_1^{i_1} \dots z_n^{i_n}}, then the {c^{(v)}_{i_1 \dots i_n}} are each Cauchy and have limits. The pointwise limit of the individual {c^{(m)}_{i_1 \dots i_n}} is then the limit of the sequence of the power series. In fact, this is nothing more than the completeness of {L^1} with respect to the measure induced by {\rho}. \Box

Note also that {\bigcup_{\rho} \mathcal{O}_{\rho} = \mathcal{O}}. In fact, if { f \in \mathcal{O}} and {\rho} is close to zero, then {\left \lVert {f }\right \rVert_{\rho}} is close to {\left \lvert {f(0) }\right \rvert}. To show that {T_2} is near-nilpotent (more precisely, a contraction), we shall compute its norm as an operator on the algebras {\mathcal{O}_{\rho}}.

Lemma 9 Let {f \in \mathcal{O}_\rho}. Then {\tau(f) \in \mathcal{O}_{\rho}} and\displaystyle \left \lVert { \tau(f) }\right \rVert_{\rho} \leq \rho_1^{-m} \left \lVert { f }\right \rVert_{\rho}.

In particular, the operator norm of {\tau: \mathcal{O}_{\rho} \rightarrow \mathcal{O}_{\rho}} is at most {\rho_1^{-m}}.

Proof: Indeed, this is evident, because {\tau} cuts down the {z_1}-exponent by {m} and leaves everything else unchanged. \Box

Now we are ready to obtain estimates on norms of {T_1^{-1}, T_2}, and thus show that {(T_1 + T_2)^{-1}} exists. We shall use {\left \lVert {\cdot }\right \rVert_{\rho}} to denote the operator norm {\mathcal{O}_{\rho} \rightarrow \mathcal{O}_\rho}.

Lemma 10 {\left \lVert {T_1^{-1} }\right \rVert_{\rho} = \left \lVert {g_0^{-1} }\right \rVert_\rho}.

Proof: This is immediate since {T_1^{-1}} is multiplication by {g_0^{-1}}. Note that multiplication by some element {x} in a Banach algebra is a continuous operator with norm {\left \lVert { x }\right \rVert} by submultiplicativty of the norm and the fact that the unit element has norm 1. \Box

Lemma 11 {\left \lVert {T_2 }\right \rVert_\rho \leq \rho_1^{-m} \left \lVert { s }\right \rVert_\rho}.

Proof: This follows because {T_2} is the composition of {\tau} (which has norm {\leq \rho_1^{-m}}) and multiplication by {s} (which has norm {\left \lVert { s}\right \rVert_\rho} because {\mathcal{O}_\rho} is a Banach algebra). \Box

We now have the required estimates, and will be able to complete the proof that {T} is invertible (which will prove the division theorem). More precisely, we shall prove that given an {n}-tuple in {\mathbb{R}_+^n}, there is a smaller {\rho} such that {T} is invertible on {\mathcal{O}_{\rho}}. This evidently implies that {T} is invertible on {\mathcal{O}}.

First, choose {\rho' = (\rho'_1, \dots , \rho'_n)} as our initial {n}-tuple; we will construct a smaller {\rho \in \mathbb{R}_+^n}. We will leave {\rho'_1} the same and shrink {\rho'_2, \dots, \rho'_n}. Now {\left \lVert { s }\right \rVert_\rho} is an infinite sum where each term has one of {\rho_1, \dots, \rho_n} occurring to a nonzero power, since each coefficient of {z_1^k} in {s} vanishes at zero. So choose {\rho_2, \dots, \rho_n} smaller than the counterparts {\rho'_2, \dots, \rho'_n} such that (with {\rho = (\rho'_1, \rho_2, \dots, \rho_n)})

\displaystyle  \left \lVert { s }\right \rVert_\rho < \left \lVert {g_0^{-1} }\right \rVert_\rho^{-1}.

Indeed, as {\rho_2, \dots, \rho_n} tend to zero (and {\rho_1 = \rho'_1} remains fixed), the norm {\left \lVert {s }\right \rVert_\rho} tends to zero, while the norm {\left \lVert {g_0^{-1} }\right \rVert_\rho} remains bounded below by {\left \lvert {g_0(0)^{-1} }\right \rvert}.

Now, {\left \lVert {T_2 }\right \rVert_{\rho} < \left \lVert { T_1^{-1} }\right \rVert_\rho^{-1}}. It follows that

\displaystyle  T= T_1 + T_2 = I + T_2 T_1^{-1} ,

and the latter differs from {I} by at most norm one. Thus it is invertible on {\mathcal{O}_\rho}, by elementary facts about Banach algebras (basically, the geometric series). It follows that {T} must be invertible on {\mathcal{O}}, and we are done for existence and uniqueness. \Box

The result has an immediate corollary.

Corollary 12 Let {g} be regular in {z_1} of degree {m}. Then the {\mathcal{O}'}-module {\mathcal{O}/\mathcal{O} g} is free of rank {m}; the map sends each {f \in \mathcal{O}} (modulo {g\mathcal{O}}) to the {m-1} coefficients of its remainder.

We now return to the Weierstrass preparation theorem and will give a proof. Proof: Suppose {f} is regular of order {m} in {z_1}. We can, by the division theorem, write

\displaystyle  z_1^m = qf + r

where {r \in \mathcal{O}'[z_1]} is of degree {\leq m-1}. If {q(0)=0}, then {qf} will be unable to contribute the term {z_1^m} because {f(\cdot, 0, \dots, 0)} is divisible by {z_1^m}; note that {r} has too small a degree. So {q(0) \neq 0} and {q} is a unit. It follows that

\displaystyle  f = (z_1^m - r)q^{-1},

which is the required representation. Indeed, the coefficients of {z_1^k} in {r} must all vanish at {z_2 = \dots = z_n =0} since this is the case for {f}.

We now show uniqueness . Suppose

\displaystyle  gu = g'u'

for {g,g'} Weierstrass polynomials and {u, u'} units. We have to show that {g = g'}. The question reduces to showing that two Weierstrass polynomials that differ by a unit in {\mathcal{O}} are equal.

So suppose {g,g'} are Weierstrass polynomials that differ by the unit {u''}. Then we have that

\displaystyle  g = g' u''


\displaystyle  g = g' + r,

where {r \in \mathcal{O}'[z_1]} has degree {\leq m-1}. The uniqueness of the division theorem now implies that {u'' =1} and {r = 0}, since Weierstrass polynomials are regular. \Box