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!

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.
