Consider $E$ be an elliptic curve and $k$ a field. I read that one way to show that $E(k)$ has an abelian group structure can be derived using Riemann Roch. Could anybody explain how it concretely provides the desired result?

My considerations:

We have a group of divisors $Div(E)$ where the divisors are formal sums $\sum_{P \in E(k)} n_P (P)$ with $n_P \in \mathbb{Z}$ and the principal divisors $div(f) = \sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.

The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.

We can define canonically a map $E(k) \to Cl(E), p \to (P)-(O)$ where $O$ is the special point (=neutral element).

Obviously it suffice to show that every divisor $D$ obtained from intersection of a line $L$ with $E$ has three points (counted we multiplicies). Or in language of divisors: For $D:= L \cap E$ we have to show that $deg(D)=3$, so $D= (P) + (Q) +(R) + div(f)$ for some principal divisor $div(f)$. This would settle $P+Q=-R$.

But I have some problems to derive it with Riemann Roch:

The RR-formula is:

$$l(D)-l(K-D) = deg(D) +g-1$$

Since $E$ elliptic $g=1$ so it suffice to show $l(D)-l(K-D)=3$.

And here I stuck. I know that $l(D) := dim_kH^0(D, \mathcal{O}_D)$ but this doesn't help me. Futhermore what to do with $l(K-D)$?

Remark: I know that there are a lot of other ways to derive the group law but the point of this question is to derive it using Riemann Roch.

Background on my question: @Awenshi's comment in https://mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group


The idea is to prove that the map that you defined above (called the Abel-Jacobi map) $$\begin{align*} J \colon E(k) &\to \mathrm{Pic}^0(E)\\ P &\mapsto (P)-(O) \end{align*}$$

is a bijection, where $\mathrm{Pic}^0(E)$ is the subgroup of $\mathrm{Pic}(E)$ of elements of degree $0$. Then the group law of $E(k)$ will be the one that makes the map above an isomorphism of groups.

Remarks: In this context $\mathrm{Pic}(E)$ is just another notation for $\mathrm{Cl}(E)$. As $k$ may not be algebraically closed we have to recall that the degree of $D=\sum_{p\in E}n_p(P)$ is given by $\deg(D)=\sum_{p\in E}n_p[k(p):k]$ so we have $P\in E(k)$ if and only if $deg(P)=1$.

  • Injectivity of the map came from the fact that if $(P)-(Q)=\text{div}(f)$ then by replacing $D=(Q)$ on Riemann-Roch we get $h^0(Q)=2-g=1$ and hence $f\in H^0(E,Q)$ must be constant.

    Over an algebraically closed field you can also proceed without R-R as here.

  • Surjectivity came from the fact that if $D\in \mathrm{Div}^0(E)$ then by R-R we have $h^0(D+(O))=1$ so if $f\in H^0(E,D+(O))$ is not constant we have $\text{div}(f)=-D-(O)+(P)$ for some $P$. As $\deg(D)=\deg(\mathrm{div}(f))=0$ we get $\deg(P)=\deg(O)=1$ hence $P\in E(k)$ and then $$P\mapsto (P)-(O)\sim D$$

Now to prove that this group law $\oplus$ coincides with the geometric group law (the one defined with lines) when $E$ is given by a Weierstrass equation $$E:Y^2Z=4X^3-aX^2-bZ^3$$ it's enough to notice that $P\oplus Q\oplus R = O$ $\iff$ $P,Q,R$ are coolinear $\iff$ there is a degree one homogeneous polynomial $F(X,Y,Z)$ with $V(F)=\{P,Q,R\}$ $\iff$ $\mathrm{div}(\frac{F}{Z})=(P)+(Q)+(R)-3(O)$ (notice that the intersection multiplicity between $V(Z)$ and $E$ is $3$, hence the $3(O)$ term) $\iff$ $P+Q+R=O$ with the addition descrived above. Notice that all the above its true when the set $\{P,Q,R\}$ degenerates with tangencies.

  • $\begingroup$ So RR is only used to show bijectivity of the map? And the fact that all divisors of the shape $D = L \cap E$ have $deg(D)=3$ follows from Bezout /intersection multiplicity argument? And so it can't be deduced directly with RR? One remark to your argument: How do you deduce $h^0(Q)=2-g=1$ in the "injectivity" part. $\endgroup$ – KarlPeter Jan 14 at 2:22
  • $\begingroup$ With this approach (that I think is the most classic one) RR is only used to prove the bijectivity. Also you can use Bezout to prove that $L\cap E$ has degree 3 but you can also give an easy direct proof of Bezout when one curve is a line. The fact that $h^0(Q)=1$ came from $h^0(Q)=h^0(W-Q)+\deg(Q)+1-g$ and $h^0(W-Q)=0$ because $\deg(W-Q)=-1<0$ (recall that $\deg(W)=2g-2$ in general). $\endgroup$ – yamete kudasai Jan 14 at 2:31
  • $\begingroup$ bwt: Do you see a quick argument that for the canonic divisor of ell curve $W= K_E$ we have $W = \mathcal{O}_E$? RR says $h^0(W) = h^0(\mathcal{O}_E)+deg(W) +g-1= h^0(\mathcal{O}_E)=1$ since elliptic. I don't see wy this already imply $W = \mathcal{O}_E$. $\endgroup$ – KarlPeter Jan 14 at 2:41

Do you know that for an elliptic curve $K$ is the zero divisor? That is, $K=\mathcal{O}_E$? If you knew this, RR implies $l(D)\geq 3$ first. Thus $D$ is an effective divisor of degree 3 and thus $l(K-D)=l(-D)=0$, since negative degree divisor can not be effective. Then, you have $l(D)=3$, which is what you want.

I do not understand your statement `so it suffices to show $l(D)-l(K-D)=3$', but isn't that what RR says, since $\deg D=3$?

  • $\begingroup$ Ah yes, of course we have $K=\mathcal{O}_E$. I think this follows from $K_C = \mathcal{O}(3-deg(C))$ for all nice enough curves $C \subset \mathbb{P}^2$. Then $K$ is exactly the structure sheaf. This reduces RR to $l(D)= deg(D)$. Now to your question: I think that $deg(D) =3$ is exactly the statement that has to be shown. So I don't understand how do you conclude $l(D) =3$. $\endgroup$ – KarlPeter Jan 14 at 1:39

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