Projective closure of an algebraic curve as a compactification of Riemann surface

Assume $f \in \mathbb{C}[x,y]$ a polynomial such that the affine algebraic curve $X=V(f)$ has no singular points. Then there is a natural structure of non-compact Riemann surface on $X$, which can be made into compact Riemann surface by adding several (finitely many) points.

Question:

Is this compactification the same thing as taking projective closure of the curve $X$? If so, how does one generally define the holomorphic maps in the neighborhoods of the "points at infinity"?

Up until now I have thought so. However, I came across the following example (I will further assume that the projective closure is indeed the compactification):

Consider a polynomial

$$f(x,y)=x^ 2-g(y),$$

where $g(y)$ is a complex polynomial of an even degree $k, \; k>2$ and, for simplicity's sake, leading coefficient $1$. Assume further that $g$ has $k$ distinct roots. Say I want to compute the genus of the compactification of $V(f)$.

Then the projective closure of $V(f)$ is $V_{proj}(f^{*}),$ where

$$f^{*}(x,y,z)=x^2z^{k-2}-y^k-(\text{other monomials of }g\text{ multiplied by some nonzero power of }z)$$

Now I want to compute the points at infinity, this leads to the equation $y^k=0,$ hence $y=0$ and thus, there is only one such point: $(1:0:0)$.

However, consider the holomorphic map $\pi: V_{proj}(f^*) \rightarrow \mathbb{S}$ defined by $\pi(x:y:1)=y, \pi(1:0:0)=\infty$. Then it is easy to compute that the degree of $\pi$ is $2$ and that $b(\pi)=k+1$ (where $b(\pi):= \sum_{P \in V_{proj}(f^*)}(e_P-1)$ and $e_P$ denotes the ramification index at the point $P$). So by Riemann-Hurwitz formula I get

$$g(V_{proj}(f^*))=1+(g(\mathbb{S})-1)\deg \pi +\frac{1}{2}b(\pi)=\frac{k+1}{2}-1,$$

which is not an integer. (Note that if tha considered curve had two points at infinity, the number $b(\pi)$ would be even and everything would work fine).

If the compactification can really be obtained via the projective closure, where is the mistake in the previous example?

Thanks in advance for any help.

• Are you using the projective closure in the Zariski topology? Feb 2 '14 at 17:32
• Yes, i.e. it is the smallest projective algebraic set containing the set $X$ (i.e. its image under the inclusion $i:(x,y)\mapsto (x:y:1)$). In the case $X=V(f)$, it can be computed as $V_{proj}(f^*),$ where $f^*$ denotes the homogenization of $f$ (which is what I'm doing). Feb 2 '14 at 17:43
• So I don't think this is true, but as far as I know you can embed into the projective space as a manifold and them work with the usual topology. Feb 2 '14 at 17:46
• The thing is, I cannot see a reason why the projective closure should not be the compactification, provided that it is non-singular as well. More explicitly, assume in the previous example that $g(y)=y^4-y$. A theorem says that a non-singular projective algebraic curve is a compact Riemann surface. Then V(f^*) is non singular, hence a compact Riemann surface which contains $V(f)$ (or its image under the embedding) as a subspace, so it should be its one-point compactification with a structure of Riemann surface. I do not see what is wrong with that line of thought. Feb 2 '14 at 18:44
• I've never seen a theorem saying this, but the way you're working it's not clear what topology you're using in each part and GAGA correspondence is pretty subtle. Anyway, the question is very interesting. Feb 2 '14 at 18:51

The compactification (= completion) $\bar X$ of a smooth affine irreducible algebraic curve $X\subset \mathbb A^2(\mathbb C)$ is the closure of $X$ in $\mathbb P^2(\mathbb C)$ .
Strangely but pleasantly the closure is the same in the Zariski or the transcendental topology of $\mathbb P^2(\mathbb C)$.
That closure is however in general non-smooth (more about that below) and is thus not the Riemann surface associated to $X$.
However there is a canonical way to obtain that Riemann surface:
Take the normalization $\nu:Y \to \bar X$ of $\bar X$. You obtain a normal irreducible complete algebraic curve $Y$ and the good news is that in dimension one normality is equivalent to smoothness.
So the required Riemann surface compactifying $X$ is just the complex manifold associated to the algebraic curve $Y$ .

A complement
That the compactification $\bar X$ is not smooth in general is easy to check on simple examples, as in Pavel's question.
But there is a more theoretical reason:
A smooth projective curve of degree $d$ in $\mathbb P^2$ has genus $g=\frac{(d-1)(d-2)}{2}$.
The integers of the form $\frac{(d-1)(d-2)}{2}$ are quite scarce in $\mathbb N$ whereas any integer is the genus $g$ of some complete smooth curve (for example, one lying on $\mathbb P^1(\mathbb C)\times \mathbb P^1(\mathbb C)$).
So most compact Riemann surfaces cannot be embedded in $\mathbb P^2 (\mathbb C)$ at all: this is one reason why the compactifcation in $\mathbb P^2$ of an affine plane smooth curve cannot in general be its associated compact Riemann surface.

• Thank you for your answer. So do I understand it correctly that the projective closure of $X$ is its compactification provided that the projective closure is non-singular? If that is the case, what about the example $V(f)$, where $f=x^ 2-y^4+y$? The curve, as well as its projective closure, should be non-singular. However, the closure contains only one additional point, whereas the compactification should allegedly contain two additional points (as mentioned in an answer here). Feb 2 '14 at 19:48
• Dear Pavel: the completion $\bar V\subset \mathbb P^2$ of your affine curve $V$ is given by the homogeneous equation $X^2Z^2-Y^4+YZ^3=0$. It has one point at infinity, namely $P=(X:Y:Z)=(1:0:0)$. At that point you can take affine coordinates $u=Y/X, v=Z/X$ and $\bar V$ has its intersection with the affine plane $X\neq0$ given by the equation $v^2-u^4+uv^3=0$. The point $P$ has coordinates $u=v=0$ and the curve $\overline V$ is thus singular at $P$. (to be continued) Feb 2 '14 at 20:43
• (continued) The normalization $\nu:Y→\bar V$ gives the required compactification $Y$. That smooth curve $Y$ is an elliptic curve and has two points $P_1,P_2$ more than $V$. More precisely $\nu^{-1}(P)=\{P_1,P_2\}$. However those points $P_1,P_2$ are now smooth on Y , whereas P was singular on $\bar V$. Feb 3 '14 at 0:07
• Dear Pavel, it's my pleasure ( and note that I have made a correction to my previous comment). Feb 3 '14 at 0:07