Projective curve covered by two affine pieces

Question

A non-singular projective curve $X$ is covered by two affine pieces (with respect to different embeddings) which are affine plane curves with equations $y^2 = f (x)$ and $v^2 = g(u)$ respectively, with f a square-free polynomial of even degree 2n and $u = 1/x, v = y/x^n$ in $K(X)$. Determine the polynomial g(u) and show that the canonical class on X has degree $2n − 4$. Why can we not just say that $X$ is the projective plane curve associated to the affine curve $y^2 = f (x)$?

First course in algebraic geometry, $k$ is closed. I'd appreciate if someone could check my solution here, I found this question conceptually very difficult. I don't really know how to picture whats going on. I think $y^2=f(x)$ is supposed to describe the 'finite' points of the curve and $v^2=g(u)$ describes the behaviour at infinity, and this distinguishes it from usual homogenization. It turns out (below), this only corresponds to two points at infinity, $(u,v)=(0,\pm 1)$. Yet the result implies the curve is genus $n-1$, which is very non-intuitive.

$g(u)=u^{2n}f(1/u)$ -- algebra omitted.

I take $\omega = dx$. $x-x_0$ is a local parameter at points $(x_0,y_0)$ with $y_0\neq 0$ and by construction this has no zeros or poles. When $y_0=0$, $y$ is a local parameter and we write $$\omega = \frac{2y}{f'}dy$$. We know that $f'$ is a unit at these points so we can ignore it. There are $2n$ distinct points of the form $(x_0,0)$, so we have counted $2n$ zeros of $\omega$ so far.

Now, we need to look at infinity. If $u\neq 0$, then the map between embeddings tells us we have already looked at that part, so the remaining points to look at are when $u=0$. At these points, $v^2=g(0)=1$ so we need to look at $(0,\pm 1)$ and $u$ is a local parameter here (just as $x-x_0$ was), so writing $$\omega = \frac{-1}{u^2} du$$ we see there is a pole order 2 at each of these points.

Answer 1

Yes, your work is correct. You can compare with my answer here, which contains basically the same calculations as yours.

Regarding the unusual definition of the curve: if one takes the affine plane curve $y^2 = f(x)$ and just tries homogenizing it, the resulting projective plane curve given by $Y^2 Z^{2n-2} = Z^{2n} f(X/Z)$ will have a singular point at infinity. To get a nonsingular model, one typically considers the closure in weighted projective space $\mathbb{P}(1,n,1)$, where instead of the usual action of $k^\times$ defining projective space, we take the weighted action $$ \lambda \cdot (U, V, W) = (\lambda U, \lambda^n V, \lambda W) \, . $$

The transition functions between the open set where $W \neq 0$ and the open set where $U \neq 0$ are then given by the formulas you wrote above. (I compute these transitions functions when $n=2$ in my answer here.)

So the upshot is that this curve is not the projective plane curve obtained by taking the closure in $\mathbb{P}^2$, as this would have a singular point. It is instead the closure in $\mathbb{P}(1,n,1)$.

Let $g$ be the genus of the curve. Here's an alternative way to see that we must have $g = n-1$. Consider the projection map $\pi: X \to \mathbb{P}^1$ given by $(x,y) \mapsto x$ on the first affine patch, or \begin{align*} \pi: X &\to \mathbb{P}^1\\ [U : V : W] &\mapsto [U : W] \end{align*} in weighted projective space. Generically, the fiber of $\pi$ above a point $x_0$ consists of the $2$ points $\{(x_0, \sqrt{f(x_0)}), (x_0, -\sqrt{f(x_0)})\}$, so $\deg(\pi) = 2$. The points $x_0$ above which $\pi$ is ramified are those such that the fiber has size $1$, which occurs exactly when $f(x_0) = 0$, i.e., $x_0$ is one of the $2n$ roots of $f$. (You showed that there are $2$ points at infinity, so $\pi$ is unramified above $\infty = [1:0] \in \mathbb{P}^1$.) At these points the ramification index is $2$, so applying the Riemann--Hurwitz formula, we have \begin{align*} 2g - 2 = 2(2 \cdot 0 - 2) + \sum_{i=1}^{2n} (2-1) = 2n-4 \, . \end{align*} Thus $g = n-1$, as you found by computing the degreeof the canonical class.