# 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

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Do you know Riemann Roch? – Soarer Mar 3 '11 at 19:12
checking it out right now – Leon Mar 3 '11 at 19:58
...and having a hard time with it. R.-R. theorem: for a divisor $D$ and a canonical divisor $W$ (whatever that is), there holds the equation $l(D)-l(W-D)=deg(D)+1-g$, where g is the arithmetic genus of the curve (in my case $1$) and $l(D):=dim_\mathbb{F}L(D)$, $L(D):=\{f\!\in\!\mathbb{F}(\mathcal{C});\;D+(f)\geq0\}\cup\{0\}$. So in my case, $l(D)=l(W-D)$. How can this help? – Leon Mar 4 '11 at 0:40
1. Put $D = 0$, you get $l(W) = 1$. Put $D = W$, you get $deg(W) = 0$. 2. Show that if degree of $D$ (sum of coefficients) is smaller than 0, $l(D) = 0$. 3. Given any divisor $D$ of degree 0, apply RR to $D+A_0$, which is of degree 1. Then $deg(W-(D+A_0)) < 0$, so RR tells you that $l(D+A_0) = 1$. Say a nonzero element of $L(D+A_0)$ is $f$. Then $(f) + D + A_0 \ge 0$ and is of degree 1. What does this tell you? – Soarer Mar 4 '11 at 1:42
So the question is why $(f)$ is of degree 0. The simplest way of proving it is to study the degree of the pullback of divisor under a morphism of varieties, then notice that $(f)$ is the pullback of divisor of $(0) - (\infty)$ in $\mathbb{P}^1$, under the map $f$. For a rigorous proof, check out Silverman's "Arithmetic of Elliptic Curves" Chapter 2, or maybe Hartshorne's book, section 2.6. – Soarer Mar 4 '11 at 6:10

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.

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