MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the smartest way to compute the genus of a hyperelliptic curve $C: y^2 = f(x)$ (with $f$ a separable polynomial of degree $n> 3$ over a field $k = \bar{k}$ of characteristic $0$ (prob. characteristic unequal $2$ is enough). (Just to be precise, I am referring to the unique nonsingular curve proper over $k$ defined by this equation.)

There are two ways I can think of, but neither is very clean. I'm wondering if this can (or cannot?) be avoided.

$\textbf{More detail:}$

We have the equation $z^{n-2}y^2 = z^n f(x/z) \subset \mathbb{P}^2$. Just to be precise, let's call this singular planar model of $C$ by $D$. The rational map from $\phi: \mathbb{P}^2 \rightarrow \mathbb{P}^1: (x,y,z) \mapsto (x,z)$ is of degree $2$ on $D$, and well-defined on the curve $D$ (except at the point $(x,y,z) = (0,1,0)$). Computing the ramification (outside this point, which is the unique singular point of $D$) is trivial, the entire issue is really just to determine the ramification at the (blow-up) of this point.

Note that the image of $\phi|_D$ is the complement of the point $(1,0) \in \mathbb{P}^1$.

$\textbf{Method 1}$: We can blow up the unique singular point $(0,1,0) \in D$ enough times, find the number of points of the strict transform, and that settles the issue. Even if the degree of $f$, $n$ is $6$ this seems like a lot of work!

$\textbf{Method 2}$: Let $P = (0,1,0)$. Then, compute the invariant $\delta_P$, where $\delta_P = \textrm{length}( \tilde{\mathcal{O}}_{D,P}/ \mathcal{O}_{D,P})$. Then, $p_a(D) - \delta_P(D) = p_a(C)$. It's of course easy to compute the arithmetic genus $p_a$ of $D$, since this we can change to a non singular element of the relevant linear system, and then compute the genus of a ns planar curve. But, is it easy to compute $\delta_P$?

What's the easiest way to compute the genus?

share|cite|improve this question

I would use the double cover $C\to \mathbb P^1$ induced by the rational function $x$ and compute the genus with Riemann-Hurwitz as you do.

Over the finite part $\mathrm{Spec} k[x]$, the cover is given by $k[x][y]/(y^2-f(x))$. Over the part containning $x=+\infty$, the cover is given by $k[1/x][y/x^d]$, where $d=[(\deg f+1)/2]$, with the equation $$(\frac{y}{x^d})^2=\frac{f(x)}{x^{2d}}\in k[1/x].$$

share|cite|improve this answer
topologically think of the complex plane plus a point at infinity as a sphere. then cut slits between successive pairs of zeroes of f, obtaining [n/2] slits (if n is odd cut one from the last zero to the point at infinity]. Then the branched double cover represented by the curve is obtained by joining two copies of such spheres along the edges of their corresponding slits. This gives a topological surface with [n/2] tubes, i.e. [n/2 - 1] holes. so the genus is [n/2 -1]. – roy smith Nov 17 '12 at 22:37
Thanks for your comment Roy!, I'm still thinking about it :) @Qing, Thanks! I didn't think one could make this all so explicit, so succinctly. – LMN Nov 17 '12 at 22:56
@Roy, Nice argument! I hate to ask, but why is the branched double cover represented by the (singular) curve obtained in this way? I would be very happy if you could give me a reference, so I can learn this point of view. – LMN Nov 17 '12 at 23:01
I think he only wants to be changing by $x^{d/2}$, so that the curve is nonsingular. – Will Sawin Nov 17 '12 at 23:34
LMN, you are apparently asking for the genus of the normalization of the singular plane curve, i.e. analytically that is the genus of the unique riemann surface that maps birationally onto the singular curve. The construction given does that. does that answer the question? – roy smith Nov 18 '12 at 2:34

It is easy to compute the ramification at that point, because of a general fact about covers of degree $2$. Each point is either ramified ($e_p=1$) or unramified, and the the total number of ramified points is even.

Probably the easiest way to see this is from Riemann-Hurwitz, which I presume you're already using to compute the genus. If there are an odd number of ramification points, the formula for the genus that you get is not a whole number!

Thus, if $n$ is odd, the point at $\infty$ is ramified in the blowup. If $n$ is even, it is unramified. So $g$ is the ceiling of $n/2-1$.

This should work in odd characteristic as well.

In characteristic $0$, you can also prove it by computing the fundamental group of $\mathbb P^1 $ minus $n$ points.

share|cite|improve this answer
Great, Thanks Will! – LMN Nov 17 '12 at 22:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.