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A flex of a smooth plane cubic is a point whose tangent line has intersection multiplicity three with the cubic. Over an algebraically closed field, choose a flex $O$ as the elliptic curve identity. A line cuts out a divisor linearly equivalent to $3O$, so $P$ is a flex exactly when $3P=O$: the tangent divisor is then $3P$. Thus the flexes are precisely the geometric 3-torsion points.

If the characteristic is not three, prime-to-characteristic geometric torsion gives nine distinct flexes. In characteristic three, the scheme-theoretic fibre $E[3]$ still has length nine, but need not have nine distinct points: an ordinary elliptic curve has three geometric 3-torsion points, and a supersingular elliptic curve has only the identity. A flex chosen over the algebraic closure need not be rational over the original field.

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.