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We have to prove that if two divisors $D, D'$ are linearly equivalent in a smooth cubic curve $X$ in $\mathbb{P}^2_\mathbb{C}$, then there exists two curves $C, C' $ such that $D-D' =C\cdot X-C'\cdot X$ where this last terms are the intersection divisors. Any hint?

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    Although you can figure this out from context, where these things are living is somewhat important to the question. We have $X\hookrightarrow \mathbb{P}^2$ a degree 3 embedding. The two "curves" $C,C'\subset \mathbb{P}^2$ and the intersection is taken in the projective plane as well.2012-11-28
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    yes, sorry for the misunderstanding2012-11-28
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    The naive idea probably is this one: $D-D^\prime =(f)$, where $f$ is a rational function on $X$. $f$ is the restriction of a rational function $F$ on $\mathbb{P}^2$, hence $F=A/B$ for some homogenous polynomials $A,B\in\mathbb{C}[X,Y,Z]$ of the same degree. $A$ and $B$ define curves, which are the candidates for $C$ and $C^\prime$.2012-11-28
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    ok that's probably it, thank you!2012-11-28

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