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

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.
