Talk:Weierstrass preparation theorem

Equivalence of Weierstrass Preparation and Weierstrass Division?

I have just removed the following sentences, about the Weierstrass division theorem:

This is equivalent to the preparation theorem, since the Weierstrass factorization of f may be obtained by applying the division theorem for g = z^N for the least N that gives an h not zero at the origin; the desired Weierstrass polynomial is then z^N + j/h. For the other direction, we use the preparation theorem on g, and the normal polynomial remainder theorem on the resulting Weierstrass polynomial.

As far as I can tell, the arguments given here don't make sense. In the first direction (division implies preparation), Weierstrass dividing f by z^N yields f/h = z^N + j/h, but there's no reason why this should be a Weierstrass polynomial. For example, in two variables, if f = z + z² + w, then N = 1 and the Weierstrass division is

f = z*(1+z) + w,

so h = 1 + z and j = z. Then f/h = z + w/(1+z), which is not a Weierstrass polynomial.

In the second direction (preparation implies division), the argument again doesn't make sense: applying Weierstrass preparation to g is pointless, because g is already a Weierstrass polynomial in the division theorem (in the version stated here). In the case where g is a Weierstrass polynomial, Weierstrass division does not follow easily from the polynomial remainder theorem.

In Grauert and Remmert's book Coherent Analytic Sheaves (in section 2.1.2, on page 42) I was able to find the statement that Weierstrass Preparation and Weierstrass Division are equivalent. They give a proof of preparation from division, but note that they are using a stronger version of division, where g is not assumed to be a Weierstrass polynomial, but rather an arbitrary thing of order N. Then one obtains Weierstrass Preparation from this by dividing zⁿ by g, not vice versa (which was what the deleted text stated).

Grauert and Remmert don't explain how one can derive Weierstrass Division from Weierstrass Preparation, and google searches don't turn up much of anything. In Grauert and Remmert's book Analytische Stellenalgebren, there is some discussion of this (on page 43, paragraph 3 of the supplement to section 4 of chapter 1). I can't exactly understand what's going on (I don't know enough German). But it seems like they are getting Weierstrass division as a corollary of the proof of Weierstrass preparation by Stickelberger, and this uses properties of Laurent series and functions on annuli. At any rate, the argument for Weierstrass division given there seems to be equally complicated as the argument for Weierstrass preparation, and using the same techniques, so it's not clear that one can prove Weierstrass division any faster by having Weierstrass preparation as a black box.

If someone knows what's actually going on, feel free to add it to the article. But I don't know, and what I deleted from the article didn't seem reasonable, so there. 67.188.231.203 (talk) 01:21, 18 June 2014 (UTC)