I have a curve but I dont know how find it genus. $y^{2}+y=\frac{\alpha}{x^{2}+x}$ is a curve on a finite field with even characteristic. I dont know any thing about this curve. About genus, pole and ...
 A: 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,
$$
g(F)=g(K)+\frac12\left(-2+2\deg P_1+2\deg P_2\right)=0+\frac22=1.
$$
