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Consider the bijective correspondance between compact Riemann surfaces and smooth irreducible complex projective algebraic curves.

1) Do we take only plane curves or not.

2) If yes then every compact Riemann surface will correspond to a curve $$C=\{[x:y:z]\in P^2(\mathbb C)\, |\, f(x,y,z)=0\}$$ where $f$ is a homogeneous polynomial with coefficients in $\mathbb C$. Is this correct and i'm also asking : when precising complex or real projective curve are we making precise points in $ P^2(\mathbb C)$, in $ P^2(\mathbb R)$ or making precise the coefficients of the polynomial $f$ being in $\mathbb C$ or $\mathbb R$?

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The answer to (1) is no. The equivalence is for all compact Riemann surfaces and all smooth projective curves. – user18119 Nov 26 '12 at 12:53

1) I don't understand. Not all smooth projective curves are plane curves. In fact,the genus of a smooth projective plane curve of degree d is $(d-1)(d-2)/2$.

Moreover, there is not just a bijection of sets, but an equivalence of categories and even more....

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Yes i know that there are skew projective curves, for example the twisted cubics. but my question is that : are non plane projective curves concerned in the equivalence or not – palio Nov 26 '12 at 11:44

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