# How come the genus of algebraic curve can be any natural number?

One the one hand, any topological surface has a complex structure, so for any natural number $$g$$, there exists a complex curve with genus $$g$$. On the other hand, we have a Chow's theorem, which says that any complex analytic variety is algebraic. This means that there exists a non-singular algebraic curve with genus $$g$$. But algebraic curves have a well-defined degree $$d$$ which connect to the notion of genus using the formula $$g=(d-1)(d-2)/2$$. This suggests that $$g$$ cannot be any natural number.

Obviously this argument has a flaw, but where?

The formula you gives concerns plane algebraic curves, but, in general a curve cannot embed in $$\bf CP^2$$ but in $$\bf CP^3$$.
• Is there a formula for the genus of a degree $d$ curve in $\mathbb{C}\mathbb{P}^3$ ? – Mihail Apr 20 at 4:47
• @Mihail one way to produce a curve of any genus in $\Bbb P^3$ is as a divisor on $\Bbb P^1\times\Bbb P^1\subset \Bbb P^3$: a divisor of type $(a,b)$ on this surface has genus $(a-1)(b-1)$ by adjunction. – KReiser Apr 20 at 4:55
• @Mihail, there exists smooth rational (= genus 0) curves of any degree in $CP^3$. – Thomas Apr 20 at 9:51