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For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.

A divisor on an algebraic curve is a finite formal integer sum $D={\sum }_p{n}_{p}p$. Its degree is ${\sum }_p{n}_p$. The Riemann-Roch space $L(D)$ consists of zero and the meromorphic functions $h$ with $(h)+D\ge 0$, and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K$ is a canonical divisor, of degree $2g-2$. In genus one, $\mathrm{deg}(K-D)=-n<0$ for $\mathrm{deg}D=n>0$. A negative-degree divisor on an algebraic curve has no nonzero section, since a principal divisor has degree zero and an effective divisor has nonnegative degree. HenceA nonsingular plane cubic has genus one by the genus-degree formula. Therefore $\ell (p+q)=2$, and there is a nonconstant $f\in L(p+q)$. It has at most simple poles at $p,q$ and no other poles. It cannot have no poles, since a holomorphic function on a compact Riemann surface is constant. It cannot have just one simple pole either, by the allowed fact, or by $\ell (p)=\ell (q)=1$. Thus $f$ has simple poles at both $p$ and $q$, and no others.

Let $F$ define the nonsingular plane cubic and let $H_F=\det \mathrm{Hess}F$ be its Hessian curve of a plane cubic equation. The Hessian criterion for a flex applies in the allowed field characteristic. Here is the local calculation, so no characteristic-zero argument is needed.

Move any point $P$ to $[0:1:0]$ and its tangent line to $Z=0$. Smoothness gives a nonzero coefficient $g$ of $Y^{2}Z$, while the coefficients of $Y^3$ and $X{Y}^2$ vanish. ThusThe intersection multiplicity with the tangent line is at least three precisely when $b=0$, which is precisely $H_F(P)=0$ because the field characteristic is not two. The tangent cannot be a component of a nonsingular plane cubic.

If $H_F$ vanishes identically, every point is a flex by this calculation. Otherwise it is a degree-three homogeneous polynomial. Its zero locus meets the cubic by Bézout's theorem; if they share a component, that also supplies a point of intersection. Every such point is smooth and satisfies the criterion, so a flex exists.

The genus-degree formula givesbecause the nonsingular plane cubic is already its own normalization of an algebraic curve. If it were birational to the projective line, the two smooth projective curves would have isomorphic function fields. The uniqueness of the finite normalization model of a one-variable function field would then make them isomorphic. But the projective line has genus of a smooth projective curve zero. Therefore the cubic is not rational. This obstruction works in the stated positive characteristics as well as in characteristic zero.