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.

  • $\begingroup$ You meant to write $h(0)\neq0$. $\endgroup$ Mar 3, 2015 at 20:06
  • $\begingroup$ The branch locus is the set of $z$'s which make the discriminant of $g$ vanish. $\endgroup$ Mar 3, 2015 at 20:07

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.