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Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

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

The intersection with a line $\ell$ is a hyperplane section:The canonical bundle of a smooth projective hypersurface formula, obtained from the adjunction formula, gives for a smooth degree-four curve in ${\mathbb{P}}^2$Thus ${\mathcal{O}}_C(D)\cong K_C$, so the divisor class of $D$ is the canonical divisor class. Therefore $D$ is a canonical divisor. Its degree is also $4=2g-2$, consistently with $g=3$.

For a smooth projective curve $C$ of geometric genus one, the Riemann-Roch theorem saysThe canonical divisor has degree zero and is principal because a nonzero regular differential has no zeros. Hence $K_C\sim 0$, so equivalentlyIn particular, $\ell (D)=\mathrm{deg}D$ when $\mathrm{deg}D>0$.

The group ${\mathrm{Pic}}^0(E)$ is the group of degree-zero divisor classes on $E$, with addition induced by addition of divisors. ConsiderFor surjectivity, let $D$ have degree zero. Since $\mathrm{deg}(D+(O_E))=1$, Riemann--Roch gives $\ell (D+(O_E))=1$. A nonzero element of this space makes $D+(O_E)$ linearly equivalent to an effective divisor of degree one, necessarily $(P)$ for some point $P$. Thus $[D]=[(P)-(O_E)]$.

For injectivity, suppose $(P)-(O_E)$ is a principal divisor. If $P\ne O_E$, its defining function would be nonconstant and would have at most one simple pole, whereas Riemann--Roch gives $\ell ((O_E))=1$, so every such function is constant. Therefore $P=O_E$, and $\iota$ is a bijection.

For a degree-$g$ divisor class $D$ with $h^0({\mathcal{O}}_C(D))=1$, let $D_0$ be its unique effective representative. Under ${\varphi }_D(P)=[D-P]$, evaluation of the unique global section gives ${\varphi }_D^{\ast }\Theta =D_0$, including multiplicities. If $h^0({\mathcal{O}}_C(D))\ge 2$, every evaluation kernel is nonzero and the entire curve lies in the theta divisor. These restrictions also show that $h^0(\mathcal{O}(\Theta ))=1$: after pulling back to a dominant product of curves, the ratio of any section to the defining section of the theta divisor is constant on each general coordinate curve, hence constant everywhere.