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We take $d\in {\mathbb{Q}}^{\times }$, as required for this rational elliptic curve. The tangent at $O=(1:-1:0)$ is $X+Y=0$. Its intersection with the curve has $d{Z}^3=0$, so it meets the curve three times at $O$: the identity is an inflection point of a plane cubic.

For $P,Q$, let their chord meet the diagonal cubic elliptic curve at the third point $R$, using the tangent if $P=Q$ and counting intersection multiplicities. Then join $R$ to $O$ and take the third intersection $S$. Define $P+Q=S$. Equivalently, three collinear intersection points satisfy $P+Q+R=O$.

Here inversion has the particularly simple formIndeed $P$, its displayed inverse, and $O$ lie on the same line, since swapping $X,Y$ preserves both the curve and $X+Y$. Thus addition is the chord or tangent operation followed by swapping the first two coordinates. The rule includes limiting cases involving $O$ and repeated intersections. Its associativity follows from the elliptic curve group law from Riemann-Roch: the identification $P\mapsto [P-O]$ with the degree-zero Picard group turns the line-intersection Weil divisor relation into addition of divisor classes. Since a line cuts a Weil divisor linearly equivalent to $3O$, that relation is exactly the chord rule just described.

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.

Embed $E$ as a smooth plane cubic with $O_E$ an inflection point. A line through $P,Q$, using the tangent when $P=Q$, has a third intersection $S$ counted with multiplicity. The chord-and-tangent group law defines $P+Q=-S$, where $-S$ is the third point on the line through $S$ and $O_E$.

The divisor cut out by the first line isand the line through $S$ and $O_E$ gives $(S)+(-S)-2(O_E)$. Their quotient therefore showsUnder the bijection from part (a), the chord-and-tangent operation is exactly addition in the abelian group ${\mathrm{Pic}}^0(E)$. It is consequently associative and commutative, has identity $O_E$, and has the geometrically defined point $-P$ as inverse. Thus it makes $E$ an abelian group.

The Hessian determinant of $u^3+v^3+w^3$ is a nonzero scalar multiple of $uvw$. Hence the inflection points areThere are nine of them. Since $O_C=(1:-1:0)$ is an inflection point, these are exactly $C[3]$. They are all defined overso $C(K)[3]\cong (\mathbb{Z}/3\mathbb{Z}{)}^2$.