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I am reading Griffiths and Harris' Principles of Algebraic Geometry but I am having trouble making sense of a statement following the Weierstrass Preparation Theorem (p.9 in my edition).

The Weierstrass Preparation Theorem says the following.

Theorem (Weierstrass Preparation Theorem). If $f$ is holomorphic around the origin in $\mathbb{C}^n$ and is not identically zero on the $w$-axis, then in some neighbourhood of the origin $f$ can be written uniquely as $$f=g\cdot h$$ *where $g$ is a Weierstrass polynomial of degree $d$ in $w$, i.e., $g$ if of the form*$$g(z,w)=w^d+a_1(z)w^{d-1} + ... + a_d(z)$$ and $h(0)\neq 0$.

I understand the theorem and its proof, but then the authors conclude the following which I am not being able to understand.

"(Therefore) The zero locus of an analytic function $f(z_1,...,z_{n-1},w)$, not vanishing identically on the $w$-axis, projects locally onto the hyperplane ($w=0$) as a finite-sheeted cover branched over the zero locus of an analytic function."

Could someone help me elaborate a little bit further so that I can understand the meaning of this last paragraph?

I understand that as a consequence of the theorem, the zero locus of $f$ is locally (for most choices of coordinate systems) the same as the zero locus of $g$. This is evident since $h$ does not vanish around zero. So my intuition says that the sheets will be related to the zero loci of the functions $a_i$. But I am missing the geometric picture to see the branched cover.

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  • $\begingroup$ You meant to write $h(0)\neq0$. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '15 at 20:06
  • $\begingroup$ The branch locus is the set of $z$'s which make the discriminant of $g$ vanish. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '15 at 20:07
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I think I have a rough solution to my question now. We may write $f$ as $$ f=h\cdot g=h\cdot (w-b_1(z))...(w-b_d(z)) $$ where the $b_i(z)$ are the "roots" of $g$. They depend on $z$ since the coefficients of $g$ depend on $z$. It follows that the zero locus of $f$ contains the hypersurfaces $w=b_i(z)$ for all $i=1,...,d$.

What suggests now the idea of projecting into the plane ($w=0$) is that we can regard each hypersurface as a function of $w$ in coordinates $(z,w)$. So they look like surface in this coordinates.

A way to think geometrically in the case $z\in\mathbb{C}$ goes as follows. If I draw the real parts of each function in the plane $(Re(z),Re(w))$ what we will see is lines representing a function, and this lines intersect only where $g$ has multiple roots. This lines form a branched covering of the horizontal axis which is $w=0$.

The discriminant function of a polynomial $g$ is precisely a holomorphic function whose zero locus is the set of multiple roots. Therefore the conclusion that the zero locus forms a branch cover when projecting to $(w=0)$ is now evident, the branch locus of this projection is the zero locus of the determinant of $g$.

I now think the conclusion of the statement is trivial, but what I needed to "see" is that the roots can be seen as functions $w=b_i(z)$ ,that $g$ has multiple roots exactly where its discriminant vanishes (I did not know this) and that this function is analytic.

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