every divisor of degree $0$ on a smooth cubic curve $\mathcal{C}\subseteq\mathbb{P}^2$ is equivalent to $A-A_0$ for a fixed $A_0\in\!\mathcal{C}$ 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
 A: 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.
