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 be an open subset of ; we write for the ring of holomorphic functions . If is open, we have a homomorphism of rings

so that the association is a presheaf. It is immediate that it is in fact a sheaf of rings.

The stalk at will be denoted . By definition, an element of is a **germ** of a holomorphic function at , i.e. a pair for a neighborhood of and holomorphic. Two germs are considered equivalent if and only if they coincide on a neighborhood of (and hence on all of ).

Definition 1The stalk is called thelocal ringat .

It turns out that the ring has many convenient properties: it is noetherian and, though not complete, henselian. First, however, we note a reduction: any two rings and are clearly isomorphic by translating . So, in the future, we shall focus on the ring .

First, we remark:

Proposition 2The stalk is a local ring whose maximal ideal is the set of germs that vanish at ; the residue field is isomorphic to .

Proposition 3is isomorphic to the ring of convergent power series (i.e. convergent in some neighborhood of the origin) in 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.

Suppose . We can write

where the are functions in the variables . Suppose the ‘s do not all vanish at zero; this is equivalent to saying that is not identically zero along the -axis.

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

Let be regular in . Suppose we have while for . Then we will say that is of **order in **. For instance, consider the ring , the local ring at zero in , i.e., in the variables only, which obviously injects into . Suppose are non-units, i.e. vanish at the origin. Then

is regular of order in .

Definition 5Any 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 be regular of order in . Then there is a unit and a Weierstrass polynomial of degree such that . 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 be regular in of degree and let . Then there is a unique expression , where and has degree .

*Proof:* We shall define the operator as follows. Let be a convergent power series representing an element of . We map this by to

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

In particular, we are trying to solve the equation

in . For convenience, we write , and to be the operator . So we are trying to solve the equation

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

Now we can write , where consists of terms of smaller degree in . Moreover, the coefficients in the powers of of all vanish at . So the operator decomposes into a sum where

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

First, is just multiplication by ! This is because is just times something in ; the application of removes the . Since is invertible in , is invertible.

In fact, 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 . Let be an -tuple of positive numbers. For a formal power series , define the **-norm** to be

With a slight abuse of notation, we let be the subring of such that the -norm is finite.

Lemma 8With respect to the -norm, is a Banach algebra.

*Proof:* It is easy to see that the -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 , then the are each Cauchy and have limits. The pointwise limit of the individual is then the limit of the sequence of the power series. In fact, this is nothing more than the completeness of with respect to the measure induced by .

Note also that . In fact, if and is close to zero, then is close to . To show that is near-nilpotent (more precisely, a contraction), we shall compute its norm as an operator on the algebras .

Lemma 9Let . Then and

In particular, the operator norm of is at most .

*Proof:* Indeed, this is evident, because cuts down the -exponent by and leaves everything else unchanged.

Now we are ready to obtain estimates on norms of , and thus show that exists. We shall use to denote the operator norm .

Lemma 10.

*Proof:* This is immediate since is multiplication by . Note that multiplication by some element in a Banach algebra is a continuous operator with norm by submultiplicativty of the norm and the fact that the unit element has norm 1.

Lemma 11.

*Proof:* This follows because is the composition of (which has norm ) and multiplication by (which has norm because is a Banach algebra).

We now have the required estimates, and will be able to complete the proof that is invertible (which will prove the division theorem). More precisely, we shall prove that given an -tuple in , there is a smaller such that is invertible on . This evidently implies that is invertible on .

First, choose as our initial -tuple; we will construct a smaller . We will leave the same and shrink . Now is an infinite sum where each term has one of occurring to a nonzero power, since each coefficient of in vanishes at zero. So choose smaller than the counterparts such that (with )

Indeed, as tend to zero (and remains fixed), the norm tends to zero, while the norm remains bounded below by .

Now, . It follows that

and the latter differs from by at most norm one. Thus it is invertible on , by elementary facts about Banach algebras (basically, the geometric series). It follows that must be invertible on , and we are done for existence *and* uniqueness.

The result has an immediate corollary.

Corollary 12Let be regular in of degree . Then the -module is free of rank ; the map sends each (modulo ) to the coefficients of its remainder.

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

where is of degree . If , then will be unable to contribute the term because is divisible by ; note that has too small a degree. So and is a unit. It follows that

which is the required representation. Indeed, the coefficients of in must all vanish at since this is the case for .

We now show uniqueness . Suppose

for Weierstrass polynomials and units. We have to show that . The question reduces to showing that two Weierstrass polynomials that differ by a unit in are equal.

So suppose are Weierstrass polynomials that differ by the unit . Then we have that

and

where has degree . The uniqueness of the division theorem now implies that and , since Weierstrass polynomials are regular.

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