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Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

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