Let X be a curve over $F\_{p^2}$ defined by equation $X^{p+1}-Y^p-Y$. 
In general ideal $I$ of $F(X)$ is should not be principal. But for some $\alpha$ $I^\alpha$ becomes principal. Do you know how I calculate this $\alpha$?



What is the Picard group of Hermitian curve
The zeta function of $X/\mathbb{F}\_{p^2}$ is $P(t)/((1-t)(1-p^2t))$, where $P(t)=(1+pt)^{2g}, g=p(p-1)/2$ is the genus of $X$. Finally, from the general theory of curves over finite fields, the class number of $X$ is $\alpha=P(1)$ which in this case is $\alpha = (p+1)^{2g}$.


Edit: Actually the class group is $(\mathbb{Z}/(p+1))^{2g}$ so the $p+1$-st power of every ideal is principal.


