Genus computation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:25:02Z http://mathoverflow.net/feeds/question/112725 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112725/genus-computation Genus computation LMN 2012-11-17T22:09:11Z 2012-11-18T03:58:50Z <p>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.)</p> <p>There are two ways I can think of, but neither is very clean. I'm wondering if this can (or cannot?) be avoided.</p> <p>$\textbf{More detail:}$</p> <p>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.</p> <p>Note that the image of $\phi|_D$ is the complement of the point $(1,0) \in \mathbb{P}^1$.</p> <p>$\textbf{Method 1}$: 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!</p> <p>$\textbf{Method 2}$: Let $P = (0,1,0)$. Then, compute the invariant $\delta_P$, where <code>$\delta_P = \textrm{length}( \tilde{\mathcal{O}}_{D,P}/ \mathcal{O}_{D,P})$</code>. 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 $\delta_P$?</p> <p>What's the easiest way to compute the genus?</p> http://mathoverflow.net/questions/112725/genus-computation/112726#112726 Answer by Will Sawin for Genus computation Will Sawin 2012-11-17T22:17:38Z 2012-11-18T03:58:50Z <p>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.</p> <p>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!</p> <p>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 $n/2-1$.</p> <p>This should work in odd characteristic as well.</p> <p>In characteristic $0$, you can also prove it by computing the fundamental group of $\mathbb P^1$ minus $n$ points.</p> http://mathoverflow.net/questions/112725/genus-computation/112727#112727 Answer by Qing Liu for Genus computation Qing Liu 2012-11-17T22:19:01Z 2012-11-17T22:19:01Z <p>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. </p> <p>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 $$(\frac{y}{x^d})^2=\frac{f(x)}{x^{2d}}\in k[1/x].$$ </p>