Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathcal{C}$


share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.