# the genus of a nonsingular projective curve is a birational invariant [closed]

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(d-1)(d-2), where d is the degree of the curve. Do i just can see it from this?

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## closed as off topic by Qiaochu Yuan, Dan Petersen, Karl Schwede, Felipe Voloch, Vivek ShendeJun 16 '13 at 15:53

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Two non-singular 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 $(d-1)(d-2)/2$ as you said. If it is smooth, the arithmetic and geometric genus coincide. Otherwise, you can blow-up one singular point, and the arithmetic genus decreases by $m(m-1)/2$, where $m$ is the multiplicity at the point blown-up. The process has an end, since the arithmetic genus is non-negative. So you can compute the geometric genus as $(d-1)(d-2)/2-\sum m_i(m_i-1)/2$ where the sum runs over all singular points that you see when doing the desingularisation.