For a holomorphic function of one variable, we can consider the decomposition where is an -order zero of and is holomorphic. We want to obtain a similar description of holomorphic functions of several variables near a zero.
To find such decomposition, we need to figure out a special kind of holomorphic functions which share the same zeros with the original holomorphic functions. Consider the Weierstrass polynomials (in , of degree ), which have the form where are holomorphic functions in on a neighborhood of such that
We see that for any fixed such that the above Weierstrass polynomial is defined, has exactly zeros (with multiplicities) as a holomorphic function in .
Lemma. Suppose is a holomorphic function on a neighborhood of , and is a -order zero of the function in . Then there exists a polydisc and such that for each , has exactly zeros (with multiplicities) on as a holomorphic function in .
Proof: There exists such that is the only zero of on . Then there exists such that for any and .
Let be the number of zeros (with multipicities) of on . The argument principle implies that and hence is holomorphic in on by a lemma from the previous post. Thus must be constant, i.e.,
Theorem (Weierstrass preparation theorem). Suppose is a holomorphic function on a neighborhood of , such that and is a nonzero holomorphic function in . Then there exists a polydisc around , a unique Weierstrass polynomial in and a holomorphic function on such that
Proof: Take the same polydisc and the same as the proof of the lemma. Suppose is a -order zero of . By the lemma, we can let be the zeros of on for each .
By residue theorem we have and hence the left hand side is holomorphic in . It follows that the function defined by is a Weierstrass polynomial in a neighborhood of , such that and have exactly the same zeros on this neighborhood.
For each , there is a unique such that is holomorphic in on and that . We can see from the Cauchy integral formula in and a lemma from the previous post that is also holomorphic in , and hence holomorphic in .
The uniqueness of is clear from the fact that and must have the same zeros. The uniqueness of then follows from the identity theorem.
For each holomorphic function on an open subset , we denote the zero set of by . The Weierstrass preparation actually gives a description of in a neighborhood of a zero. The construction of the holomorphic in the theorem can be modified to a proof of the following Riemann extension theorem.
Theorem (Riemann extension theorem). Suppose is a nonzero holomorphic function on a connected open set and is holomorphic and bounded. Then can be extended to a holomorphic function .
Another Weierstrass theorem is concerned with the Euclid algorithm of holomorphic functions.
Theorem (Weierstrass division theorem). Suppose is a holomorphic function on a neighborhood of and is a Weierstrass polynomial in of degree defined around . Then there exists a sufficiently small neighborhood of , a uniquely determined holomorphic function on and a uniquely determined Weierstrass polynomial in of degree less than , such that
Proof: The unique statement can be proved in a similar way to the unique statement of an ordinary Euclid algorithm. Now we show the existence. Taking a sufficient small neighborhood we may define such that is holomorphic for and . It remains to show that is a Weierstrass polynomial with degree less than the degree of .
Suppose Then we can write where are determined by and holomorphic in and . It follows that where are holomorphic in .
For the preceding theorems, we often need to take smaller and smaller neighborhoods. In convenience, we introduce the concept of germs of functions.
Suppose is a fixed point. Let be the set of holomorphic functions defined in some neighborhood of . Define an equivalence relation on by that if and only if and agree with each other on a sufficiently small neighborhood of . An equivalence class of with respect to this equivalence relation is called a germ of holomorphic functions near , and the collection of all such germs is denoted by .
We can see that has a natural ring structure with identity given by the constant function . It can be verified that the units in are the germs given by the functions such that . This imples that is a local ring, whose unique maximal ideal consisting of the germs given by the functions vanishing at . We may use the same notation for a function and the germ it determines for short.
Now we can formulate Weierstrass theorems using the language of germs.
Theorem (Weierstrass preparation theorem). If satisfies that and that is nontrivial, then can be uniquely decomposed as , where is a Weierstrass polynomial and is a unit in .
Theorem (Weierstrass division theorem). Let and let be a Weierstrass polynomial of degree . Then there exist uniquely determined and a Weierstrass polynomial such that .