I have a curve but I dont know how find it genus.

Question

$y^{2}+y=\frac{\alpha}{x^{2}+x}$ is a curve on a finite field with even characteristic. I dont know any thing about this curve. About genus, pole and ...

Answer 1

We get the genus using the formula from Stichtenoth's book. More precisely, Proposition 3.7.8 on page 127. It describes the form the Riemann-Hurwitz genus formula takes when we are dealing with an Artin-Schreier extension of function fields (which is the case here).

Anyway, here $F=\Bbb{F}_q(x,y)$ is an Artin-Schreier extension of $K=\Bbb{F}_q(x)$, and the rational function field has genus zero. We need to calculate the different $\operatorname{Diff}(F/K)$.

From the partial fraction decomposition $$ \frac{\alpha}{x^2+x}=\frac{\alpha}x+\frac{\alpha}{x+1} $$ it follows that the only places of $K$ where we have ramification are the two poles of degree one: $P_1$ corresponding to $x=0$ and $P_2$ corresponding to $x=1$. Both those poles are simple, so the parameters $m_{P_1},m_{P_2}$ (see the statement of Prop. 3.7.8) are both equal to $1$. This means that the different is $$ \operatorname{Diff}(F/K)=2\cdot P_1+2\cdot P_2, $$ and Riemann-Hurwitz says that, part d) of Prop. 3.7.8, $$ g(F)=g(K)+\frac12\left(-2+2\deg P_1+2\deg P_2\right)=0+\frac22=1. $$