 Solution

DefineIf $D$ is a divisor class of degree zero, Riemann--Roch gives $\ell (D+Q)=1$, because $K_E\sim 0$ and the divisor $-D-Q$ has negative degree. A nonzero section therefore yieldsfor some point $P$, proving that $\alpha$ is surjective. If $\alpha (P)=\alpha (P^{\prime })$, then $P-P^{\prime }$ is a principal divisor. For $P\ne P^{\prime }$ its defining function would have one simple pole and give a degree-one map $E\to {\mathbb{P}}^1$, impossible for a genus-one curve. Thus $P=P^{\prime }$. The map and its inverse are regular, so this proves the Abel-Jacobi map of a genus-one curve is an isomorphism.

Transport the addition of the divisor class group through $\alpha$: define $P⊞R$ byThen $Q$ is the identity, and this is the chord-and-tangent group law on the plane cubic.

For the displayed equation, inversion is $(x,y)\mapsto (x,-y)$. Thus $P⊞P=Q$ exactly when $P=-P$. Besides the point at infinity $Q$, this requires $y=0$. Therefore the complete set of two-torsion points is