Compactness of Algebraic Curves over $\mathbb C^2$

Question

I was reading through Kirwan's Complex Algebraic Curves and I've been stuck on the following exercise: Given a (non-constant) polynomial $P(x,y)$, show that the curve in $\mathbb C^2$ defined by $P(x,y)=0$ is not compact.

My attempt at a solution: If we can show that there are only finitely many $a\in \mathbb C$ such that there is no $b \in C$ with $P(a,b)=0$, I think we're done. Once we've shown this, we can consider any sequence of complex numbers $x_n$ with $|x|=n$. For all but finitely many $x_n$, there is then some $y_n$ with $P(x_n,y_n)=0$. This implies then that the curve defined by $P$ isn't bounded, and thus is not compact.

I'm unsure as to my proof of this step though, and would love some input on my sketch. Fix $a_0$, and assume $P(a_0,y)$, viewed as a polynomial in $\mathbb C[y]$, has no roots. Clearly then $P(a_0,y)=c_0$ for some $c_0 \in \mathbb C$. Then $a_0$ is a root of $P(x,y)-c_0$, so we can write $P(x,y)-c_0=(x-a_0)\cdot f_0(x,y)$ for some polynomial $f_0(x,y)$. Now, fix $a_1$ distinct from $a_0$ and assume $P(a_1,y)$ is some constant $c_1$. Then $P(a_1,y)=c_1$, so $P(a_1,y)-c_0=c_1-c_0$ and then $(a_1-a_0)\cdot f_0(a_1,y)=c_1-c_0$. Thus $a_1$ is a root of $f_0(x,y)-\frac{c_1-c_0}{a_1-a_0}$, so we can write $f_0(x,y)-\frac{c_1-c_0}{a_1-a_0}=(x-a_1)\cdot f_1(x,y)$, for some polymomial $f_1$ of degree strictly less than $f_0$, and thus $P(x,y)=(x-a_0)(x-a_1)f_1(x,y)+(x-a_1)\frac{c_1-c_0}{a_1-a_0}+c_0$. If we continue in this fashion, we obtain a sequence of polynomials $f_i$ of strictly decreasing degree. This must terminate in finitely many steps, so there can only be finitely many distinct $a_i$.

Does this look like the right idea? I feel like I'm probably over-complicating this step, so if anyone has a more elegant solution I'd love to hear it!

Answer 1

Hint: Pick $y_0$ to be arbitrarily far from the origin. Then, $P(x,y_0)$ is a polynomial in $\mathbb{C}[x]$, and so necessarily has a root.

EDIT: I was trying to be leading and not give it away, but I see how this could be misinterpreted by Georges's response. Write

$$P(x,y)=f_0(y)+f_1(y)x+f_2(y)x^2+\cdots+ f_m(y)x^m$$

Suppose that $P(x,y_0)$ is constant. This implies necessarily that $f_1(y_0)=\cdots=f_m(y_0)=0$. In particular, we have two cases:

1. All the $f_i$, $i>1$, are zero, in which case $P(x,y)=f_0(y)$ so then, choose any root of $f_0(y)$ and tack any $x$ on. In this case. $Z(P)$ contains an entire line, and so clearly can't be compact.

2. One of the $f_{i_0}$, $i_0>1$, is not zero. Then we know that there are only finitely many (bound by the degree of $f_{i_0}$) points $y_0$ for which $P(x,y_0)$ is constant. In particular, there exists an unbounded sequence $y_n$ such that $P(x,y_n)$ is non-constant, so there exists a corresponding $x_n$ such that $P(x_n,y_n)=0$. Thus, $(x_n,y_n)$ is an unbounded sequence in $P(x,y)=0$, and so $Z(P)$ is not compact.

EDIT: It is interesting to note, since you put this under the algebraic geometry heading, that there is, in some cases, a very trivial solution to this. Clearly we can reduce ourselves to showing that $Z(P)$ is not compact when $P$ is irreducible, for any irreducible component of $Z(P)$ would also necessarily be compact. Then, if $P$ is non-singular, you know that $Z(P)$ is a compact Riemann surface, and the only compact Riemann surfaces in $\mathbb{C}^2$ are points. Since non-constant polynomials (in two variables over an algebraically closed field) vanish infinitely often, this would be a contradiction.

You might be able to fix this to work everywhere, by considering the normalization of any non-singular curve.

Answer 2

Let $P(z, w) = A_0(z)w^n + A_1(z)w^{n-1} + \cdots + A_n(z)$, where $A_i(z)$ is a polynomial of $z$ and $A_0(z) \neq 0$.

Suppose $n = 0$. Since $P(z, w) = A_0(z)$ is not constant, $A_0(z)$ has a root $c$. Then for every complex number $d$, $P(c, d) = 0$. Hence the curve $P(z, w) = 0$ is not bounded. Hence it is not compact.

Suppose $n \ge 1$. Since $A_0(z)$ has only finitely many roots, there exists a complex number $c$ such that $A_0(c) \neq 0$ and $|c|$ is arbitrarily large. Since $A_0(c) \neq 0$ and $n \ge 1$, there exists a complex number $d$ such that $P(c, d) = 0$. Since $|c|$ is arbitrarily large, the curve $P(z, w) = 0$ is not bounded. Hence it is not compact.

Answer 3

Just another way of seeing this quickly, at least intuitively (details are routine).

Write a polynomial $P(x,y)$ as $P(x,y)=P_0+P_1+\dots+P_d$ where each $P_i$ is homogeneous of degree $i$ and where $P_d$ is not zero ($d$ is in fact the degree of $P$). Since $P$ is not constant, we have $d\ge 1$. Since $\mathbb{C}$ is algebraically closed, we can decompose $P_d$ and write it as a polynomial of degree $1$ (fundamental theorem of Algebra). Hence, the zeroes of $P_d$ are union of $d$ lines (maybe some can coincide).

When $(x,y)\in\mathbb{C}^2$ is far from the origin, the value of the polynomial $P_d$ is "almost" the value of $P$, since the other terms are smaller. Hence, the zeroes of $P_d$ are asymptotical lines of your curve (this is indeed true and can be checked with homogeneous coordinates for instance). Hence, the curve is not compact.

The same argument shows that over $\mathbb{R}$, curves of odd degree are not compact and curves of even degree are compact if and only if the leading homogeneous polynomial has no real root.

Answer 4

Here's an elegant proof (sketch) my professor gave if we assume $P$ is non-singular (i.e., $P$ has non-vanishing gradient on $Z(P)$). By the implicit function theorem, $Z(P)$ is locally a graph of holomorphic functions, which allows us to give $Z(P)$ the structure of a Riemann surface. Then one can check that either $(z, w) \mapsto z$ or $(z, w) \mapsto w$ is a non-constant holomorphic map on $Z(P)$. But compact Riemann surfaces admit no holomorphic maps, which implies $Z(P)$ is non-compact.