The Weierstrass preparation theorem

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

$\displaystyle g = g' u''$

and

$\displaystyle g = g' + r,$

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. $\Box$

Climbing Mount Bourbaki