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For an elliptic curve over a number field, fix the absolute logarithmic $x$-height $h_x(P)=h([x(P):1])$, with $h_x(O)=0$. The normalization used here isThe factor $1/2$ makes this the Néron–Tate height associated with the degree-one divisor $(O)$. Some sources omit $1/2$ and obtain twice this function; formulas and the regulator of an elliptic curve must use a consistent convention.^[ref]

The estimate $h_x(2P)=4h_x(P)+O(1)$ makes successive terms differ by a summable geometric series, proving convergence and $\hat{h}(P)=h_x(P)/2+O(1)$. The resulting quadratic form satisfies $\hat{h}([m]P)=m^2\hat{h}(P)$ and the height parallelogram identity. It is nonnegative and vanishes exactly on torsion points of an elliptic curve; bounded-height rational points over a fixed number field form a finite set. For an isogeny $\varphi :E\to E^{\prime }$, the corresponding normalized heights satisfy ${\hat{h}}_{E^{\prime }}(\varphi P)=\mathrm{deg}(\varphi ){\hat{h}}_E(P)$.

Let $E:y^2=x^3+a_2{x}^2+a_{4}x+a_6$ be a smooth Weierstrass equation of an elliptic curve over a field of characteristic different from two, with identity $O$ at infinity. For affine points $P,Q$ with $x(P)\ne x(Q)$, putFor $P=Q$ with $y(P)\ne 0$, use the tangent slope $\lambda =(3x(P{)}^2+2a_{2}x(P)+a_4)/(2y(P))$. In either case,Substitute the secant or tangent equation into the cubic: the sum of its three intersection $x$-coordinates is ${\lambda }^2-a_2$. Reflecting the third intersection in $y=0$ gives the sum.^[ref]

The omitted cases belong to the elliptic curve group law on the full projective curve: $P+O=P$ and $P+(-P)=O$, where $-P=(x(P),-y(P))$. In particular, a point with $y(P)=0$ satisfies $2P=O$; its vertical tangent cannot be handled by dividing by $2y(P)$.

Let $E/K$ be an elliptic curve with $\mathrm{char}K\ne 3$. Its geometric 3-torsion is a two-dimensional vector space over ${\mathbb{F}}_3$. Choosing a basis gives the continuous Galois representationHere $G_K$ is the absolute Galois group, acting on torsion coordinates in a separable closure. Its fixed vectors are exactly $E(K)[3]$. Thus a nonzero fixed vector means a rational point of order three; a stable one-dimensional vector subspace need only give a Galois-stable cyclic subgroup, not a rational generator.

The Weil pairing identifies ${\bigwedge }^{2}E[3]$ with ${\mu }_3$, so $\det {\rho }_{E,3}$ is the mod-three cyclotomic character. In particular, the image lies in ${\mathrm{SL}}_2({\mathbb{F}}_3)$ if $K$ contains all third roots of unity.^[ref]

In characteristic three, geometric 3-torsion is not ${\mathbb{F}}_3^2$. The group scheme $E[3]$ and its nonreduced structure require a different formulation.