# $n$-sheeted branched covering

Michael Artin's algebra

let $f(x,y)$ be an irreducible polynomial in $\mathbb{C}[x,y]$ which has degree $n>0$ in the variable $y$. The Riemann surface of $f(x,y)$ is an $n$-sheeted branched covering of the plane.

The description in the text book for the $n$-sheeted branch covering is too complex for me to understand. can someone explain this Corollary in not that abstract way to me , and give me some hints in terms of understanding these concepts ?

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This is not algebra but geometry – specifically complex geometry. How familiar are you with this subject? – Zhen Lin Oct 18 '12 at 11:16
I am not farmilar with that, but it did occur in the book called Algebra. section Ring - Algebra geometry. – zinking Oct 18 '12 at 12:04
@zinking You need to be familiar with covering spaces to fully understand this. – user38268 Oct 18 '12 at 12:17
I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 19 '12 at 6:51

I assume that by "the Riemann surface of $f$" you mean the set $V = \{(x,y)\in \mathbb{C}^2 : f(x,y) = 0\}$. This set $V$ is not exactly a Riemann surface, since it will probably have some singularities, but it will be a one (complex) dimensional analytic space.

Let $\pi\colon V\to \mathbb{C}$ be the projection $\pi(x,y) = x$. This will be the $n$-fold branched covering map. Intuitively, you want to think of this as saying that the preimage $\pi^{-1}(x_0)$ of most points $x_0\in \mathbb{C}$ consists of $n$ points in $V$. Suppose I write out $f$ as $$f(x,y) = p_n(x)y^n + p_{n-1}(x)y^{n-1} + \cdots + p_0(x),$$ where the $p_i(x)$ are polynomial is $x$, and, by assumption, $p_n(x)\not\equiv 0$. Then, if I've fixed $x_0\in \mathbb{C}$, the preimage $\pi^{-1}(x_0)$ consists of those points $(x_0,y)\in \mathbb{C}^2$ for which $p_n(x_0)y^n + \cdots + p_0(x_0) = 0$. If $p_n(x_0)\neq 0$, the degree of this polynomial is $n$, and hence there are exactly $n$ such values of $y$, at least when counted with multiplicities. For most values of $x_0$, there will even by exactly $n$ such values of $y$. In fact, you can show the following

There exists a finite set $S\subset\mathbb{C}$ such that for every $x_0\in \mathbb{C}\smallsetminus S$, the preimage $\pi^{-1}(x_0)$ consists of exactly $n$ points, and moreover, the map $\pi\colon V\smallsetminus \pi^{-1}(S)\to \mathbb{C}\smallsetminus S$ is a degree $n$ covering map.

Intuitively, the term "branched cover" refers to a map that is an actual covering map except over some bad points that will have less than the $n$ preimage points. It's easier to picture than to define, so here's a picture I've taken from wikipedia.

This is a schematic picture of a degree $3$ branched cover. You should think of $V$ as being $X$, and $\mathbb{C}$ as being $Y$, and the map $\pi$ as being the map $f$ from the picture (the projection downward). This map is an actual degree $3$ covering map except for at the points marked with blue dots; these would be the set $S$.

I should also say something about the irreducibilty assumption on your polynomial $f$. If you didn't have this assumption, it would be possible for, say $x$ to divide $f(x,y)$. If this were the case, then $\pi^{-1}(0)$ would be the set $\{(0,y) : y\in \mathbb{C}\}$, an infinite set. If this were to happen, then $\pi\colon V\to \mathbb{C}$ would not be considered a branched cover: the preimage $\pi^{-1}(x_0)$ of a point $x_0\in \mathbb{C}$ is allowed to have fewer than $n$ elements, but not more. The irreducibility hypothesis is there to rule out things like this.

I hope that sheds some light one your question.

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