Rational embedded irreducible curves in a complex surface. Given a complex surface $X$ and an embedded irreducible compact curve $C$ with its arithmetical genus $g(C) = 0$, how can one show that $C$ is non-singular ?
Thanks for your answers!

Given an irreducible  complete curve $C$ and its normalization $\tilde C$ you have the following relation between their arithmetic genera:$$ p_a(C)=p_a(\tilde C)+\sum_{P\in Sing(C)}dim_\mathbb C (\mathcal O_{C,\tilde P}/\mathcal O_{C, P})$$
where for a singular point $P\in Sing (C)$ the ring $\mathcal O_{C,\tilde P}$ is the integral closure of the ring $\mathcal O_{C, P}$.
It is then clear that $ p_a(C)=0$ forces $ p_a(\tilde C)=0$ and also forces all $P$ to  actually  not be singular at all, since all $\mathcal O_{C,\tilde P}=\mathcal O_{C, P}$ . The curve $C$ was smooth after all and isomorphic to $\mathbb P^1$.   
Note that the embedding of $C$ in some  surface $X$ is irrelevant. 
