# Derive Group Law on Elliptic Curve with Riemann Roch

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.

• 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. – KarlPeter Jan 14 at 2:22
• 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). – yamete kudasai Jan 14 at 2:31
• 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 $W = \mathcal{O}_E$. – 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 $$\deg D=3$$?

• 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 $l(D) =3$. – KarlPeter Jan 14 at 1:39