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NOTATION: Let $\mathcal{C}=\mathcal{V}(F)\subseteq\mathbb{P}^2$ be a curve of degree $3\!=\!deg(F)$ with no singularities and let $A_0\!\!\in\!\mathcal{C}$ be fixed. Let $Div(\mathcal{C})$ denote the group of divisors on $\mathcal{C}$, i.e. the set of all formal sums $\{\sum\limits_{P\in\mathcal{C}} n_PP\,|\; n_P\!\in\!\mathbb{Z}, \text{only finitely many } n_P \text{ are not zero}\},$ let $\mathbb{F}(\mathcal{C})$ be $\{\text{rational functions from }\mathcal{C}\text{ to }\mathbb{F}\}$, i.e. the field of fractions of $\mathbb{F}[x_0\!:\!x_1\!:\!x_2]/I(\mathcal{C})$. Let $\psi:\mathbb{F}(\mathcal{C})\setminus\{0\}\rightarrow Div(\mathcal{C})$ denote the mapping, that sends each rational function $f$ to the principal divisor $(f)=\sum_{P\in\mathcal{C}}\mu_P(f,F)P$ where $\mu_P(f,F)$ is the intersection multiplicity of curves $\mathcal{V}(f),\mathcal{V}(F)$ in $P$. Then $Cl(\mathcal{C})$ denotes the group of divisor classes on $\mathcal{C}$, i.e. $Div(\mathcal{C})/im(\psi)$. So any two divisors $D_1$ and $D_2$ are equivalent, $D_1\sim D_2$, iff $D_1-D_2=(f)$ for some $f\in\mathbb{F}(\mathcal{C})$.

QUESTION: Define $\varphi:\mathcal{C}\rightarrow Cl^0(\mathcal{C})\!=\!\{\text{divisor classes on }\mathcal{C}\text{ of degree }0\}$ as a mapping, that sends each $A$ to the divisor class of $A-A_0$. How can I prove that $\varphi$ is surjective?

WHAT IS ALREADY KNOWN: on a smooth cubic curve $\mathcal{C}$ for $P,Q,R,S\in\mathcal{C}$:

  • $P\sim Q\Leftrightarrow P=Q$
  • $P+Q\sim R+S \;\;\Longleftrightarrow\;\;$ the line through $P,Q$ intersects the line through $R,S$ on $\mathcal{C}$

help

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    anyway, you are a lifesaver man. You can't imagine how much you helped, I'm really grateful. Oh, and write "solved" below as an answer, so I can make a checkmark that It's solved. kind regards2011-03-04

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Given what is already known in the original question, you don't need Riemann--Roch to prove surjectivity. (The content of Riemann--Roch is already encoded in the given facts.)

Rather, given two points $P$ and $Q$, draw a line through them, which meets $\mathcal C$ in a third point $R$. Now draw a line through $R$ and $A_0$, which meets $\mathcal C$ in a third point $S$. From the given facts, we find that $P + Q \sim A_0 + S$.

Now if $D$ is a divisor of degree zero, write $D = D_+ - D_-$, with all the coefficients of $D_+$ and $D_-$ being positive. Repeatedly applying the procedure of the preceding paragraph, we may write $D_+ \sim A_+ + n A_0$ for some point $A_+$, and $D_- \sim A_- + n A_0$ for some point $A_-$. (We get the same number $n$ in both cases because $D$ has degree zero by assumption.)

Thus $D \sim A_+ - A_-.$

Now an evident variation on the preceding construction shows that if we have points $P$ and $S$, we may find a point $Q$ such that $P + Q \sim A_0 + S.$ (Draw the line through $A_0$ and $S$, which meets $\mathcal C$ in a third point $R$. Now draw the line through $R$ and $P$, which meets $\mathcal C$ in a third point $Q$, which is the desired point.)

In particular, we may find a point $A$ so that $A_- + A \sim A_0 + A_+$. Thus $D \sim A - A_0$, as required.