Proving an Isomorphism regarding Group law for Elliptic Curves (independent of neutral element) - Fulton Exercise 5.36

Question

This is Fulton's (Algebraic Curves) Exercise 5.36 for reference. If we consider $C$ an irreducible cubic, and $O,O'$ simple (smooth) points on $C$ we have that each gives rise to a group operation $+,+'$ on the set of simple points $C^{\circ}$. Let $ Q= \phi(O,O')$. We define $A+B = \phi(O,\phi(A,B))$ and similar for $+'$. We assume that $+,+' $ are commutative and associative.

Now we need to show that the map $\alpha :(C^{\circ},+,O) \to (C^{\circ},+',O')$ by $\alpha(P) = \phi(Q,P)$ is a group isomorphism. I'm not getting the homomorphism part at all, bijectivity is fine.

This question is similar : Proving a group isomorphism from $(S,+)$ to $(S,+')$ but it doesn't define the isomorphism the same way, here $F(P)= \phi(O,\phi(P,O'))$ where $\alpha(P)=\phi(P,\phi(O,O'))$.

What I have is $\alpha(A)+'\alpha(B) = \phi(O',\phi(\phi(Q,A),\phi(Q,B))$ and $\alpha(A+B) = \phi(Q,\phi(O,\phi(A,B))$ But I do not see equality at all. Even just hints are appreciated. Thanks!

Answer 1

I will denote by $\mathcal C$ the given curve. Lines through two points $A,B$ are denoted by $\mathcal L_{A,B}$.

In order to have a more compositional notation, it is useful to denote as in loc. cit the point $\phi(A,B)$ by $A*B$. Then the composition star has the following properties for simple points $A,B,C,D$:

• commutativity, $A*B=B*A$,
• idempotency (for $A$), $A*(A*B)=B$,
• $3\times 3$-match, $(A*B)*(C*D)=(A*C)*(B*D)$.

The last property is best seen in a picture:

This is Proposition 3, page 124 in Fulton's book, applied for the two intersections $\mathcal C\cdot\mathcal C_1$ and $\mathcal C\cdot\mathcal C_2$, where $\mathcal C_1,\mathcal C_2$ are the reducible cubics (each product/union of three lines): $$ \begin{aligned} \mathcal C_1 &=\mathcal L_{AB}\ \mathcal L_{CD}\ \mathcal L_{A*C\ B*D}\ ,\\ \mathcal C_2 &=\mathcal L_{AC}\ \mathcal L_{BD}\ \mathcal L_{A*B\ C*D}\ . \end{aligned} $$ Now we can work: $$ \begin{aligned} \alpha(A)+'\alpha(B) &=O'*(\alpha(A)*\alpha(B)) &&\text{definition of $+'$} \\ &=O'*((Q*A)*(Q*B)) &&\text{definition of $\alpha$} \\ &=O'*((Q*Q)*(A*B)) && 3\times 3 \text{ inside the right operator} \\ &=(Q*O)*((Q*Q)*(A*B)) &&\text{definition of $Q=O*O'$, idempotency for $O$} \\ &=(Q*(Q*Q))*(O*(A*B)) && 3\times 3 \\ &=Q*(O*(A*B)) &&\text{by idempotency for $Q$, $Q*(Q*Q)=Q$} \\ &=Q*(A+B) &&\text{by definition of $+$} \\ &=\alpha(A+B)&&\text{by definition of $\alpha$ .} \end{aligned} $$