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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 Shende Jun 16 '13 at 15:53

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up vote 2 down vote accepted

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.

All these things are very classical and can be found in any book of introduction to Algebraic geometry.

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Thank you for the answer! Can you suggest me a good book to look? – user35018 Jun 16 '13 at 14:56
If you are beginner, look at the book of Miles Reid "undegraduate algebraic geometry". And then Hartshorne or Shafarevich for example. – Jérémy Blanc Jun 16 '13 at 16:41

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