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.


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).

So additional question is:

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.

  • $\begingroup$ Are you using the projective closure in the Zariski topology? $\endgroup$
    – user40276
    Feb 2, 2014 at 17:32
  • $\begingroup$ 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). $\endgroup$ Feb 2, 2014 at 17:43
  • $\begingroup$ 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. $\endgroup$
    – user40276
    Feb 2, 2014 at 17:46
  • $\begingroup$ 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. $\endgroup$ Feb 2, 2014 at 18:44
  • $\begingroup$ 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. $\endgroup$
    – user40276
    Feb 2, 2014 at 18:51

1 Answer 1


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.

  • $\begingroup$ 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). $\endgroup$ Feb 2, 2014 at 19:48
  • $\begingroup$ 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) $\endgroup$ Feb 2, 2014 at 20:43
  • $\begingroup$ (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 $\bar V$. $\endgroup$ Feb 3, 2014 at 0:07
  • $\begingroup$ Dear Pavel, it's my pleasure ( and note that I have made a correction to my previous comment). $\endgroup$ Feb 3, 2014 at 0:07

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