Is true that two birational projectives nonsingular curves have the same genus? I know that for a non singular projective curve the genus is given by the formula 1/2(d1)(d2), where d is the degree of the curve. Do i just can see it from this?
closed as off topic by Qiaochu Yuan, Dan Petersen, Karl Schwede, Felipe Voloch, Vivek Shende Jun 16 '13 at 15:53Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question. 


Two nonsingular projective curves are birational if and only if they are isomorphic (a birational map between smooth projective varieties has indeterminacy points in codimension $\ge 2$). The geometric genus of an irreducible projective curve $C$ is the arithmetic genus of its desingularisation $\tilde{C}$. Moreover, $C$ is birational to $\tilde{C}$. By the above observation, the desingularisation $\tilde{C}$ is unique up to isomorphism, and the geometric genus is then well defined. The arithmetic genus of an irreducible curve of degree $d$ in $\mathbb{P}^2$ is $(d1)(d2)/2$ as you said. If it is smooth, the arithmetic and geometric genus coincide. Otherwise, you can blowup one singular point, and the arithmetic genus decreases by $m(m1)/2$, where $m$ is the multiplicity at the point blownup. The process has an end, since the arithmetic genus is nonnegative. So you can compute the geometric genus as $(d1)(d2)/2\sum m_i(m_i1)/2$ where the sum runs over all singular points that you see when doing the desingularisation. All these things are very classical and can be found in any book of introduction to Algebraic geometry. 

