Can one indicate to me the Weierstrass factorization theorem in several variables (real or complex). In one complex variable the result is well known. Thank you in advance.
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1$\begingroup$ My inclination is that any extension of the theorem would have to be drastically different. The issue is that in one dimension, zeros of an entire function are isolated, whereas in higher dimensions they need not be. $\endgroup$– Christopher A. WongCommented Nov 6, 2012 at 8:32
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1$\begingroup$ Could you have in mind the Weierstrass preparation theorem (en.wikipedia.org/wiki/Weierstrass_preparation_theorem)? Unlike the single complex variable version, this result is local. $\endgroup$– Igor KhavkineCommented Nov 6, 2012 at 11:18
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Before addressing this question, think of the case of polynomials. In one dimension, Weierstrass theorem for polynomials says that there exists a polynomial with prescribed zeros. In several dimensions, zeros are never isolated. So what does it mean "prescribed zeros"? Zeros of a polynomial form an "algebraic set". But an algebraic set is defined as... zero set of a polynomial. We obtain a tautology.
However there is something which can be called a "generalization of Weierstrass theorem". This is called the "solution of the Second Cousin Problem". See any book on several complex variables for exact formulation.
For example, the Weierstrass theorem has the following consequence: every meromorphic function in the whole plane is a ratio of two entire functions. (Here a meromorphic function is defined as a function which is LOCALLY (in a neighborhod of every point) can be represented as a ratio of two analytic functions in a neighborhood of this point. So the statement is not a tautology.)
The same statement is true in any dimension and it is proved using the solution of the second Cousin problem. In this sense the second Cousin problem is the generalization of the Weierstrass theorem to several variables.
answered Nov 6, 2012 at 13:03
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1$\begingroup$ nice answer. in this vein perhaps one may say that the weierstrass theorem says that any "subset" (with multiplicities) of C which is locally the zero locus of an analytic function is also globally the zero locus of an analytic function. in this form it is also true in C^n. $\endgroup$ Commented Nov 7, 2012 at 1:56