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Let $r^3=d\ne 0$ and let $\zeta$ be a primitive cube root of unity. The nine inflection points of a plane cubic on $X^3+Y^3+d{Z}^3=0$ areThe Hessian criterion for flexes of a plane cubic gives the Hessian determinant $216dXYZ$, so every flex lies on one of the three coordinate lines. Their intersections give precisely the displayed points. With $(1:-1:0)$ as identity, these are exactly the three-torsion points are flexes of a plane cubic configurations: the tangent divisor at a flex is $3P$, giving $3(P-O)=0$.

Over $\mathbb{Q}$ only the first point with $j=0$ is rational on $Z=0$. The other two coordinate lines each give one rational point precisely when $d$ is a rational cube. Consequently the rational three-torsion subgroup is $\mathbb{Z}/3\mathbb{Z}$ when $d$ is a cube and is trivial otherwise.

The Hessian matrix describes how the slope of a function changes in different directions. For a real-valued function $f$ with continuous second partial derivatives on an open subset of ${\mathbb{R}}^n$, it isThe symmetry of second derivatives makes $H_f$ a symmetric matrix under these hypotheses. It is also the Jacobian matrix of the gradient: near $x$, a small displacement $h$ changes $\nabla f$ by $H_f(x)h+o(\Vert h\Vert )$. Thus the gradient measures the current slope, while the Hessian measures its local variation. ^[ref]

For example, take the quadratic formIts gradient and Hessian matrix areThe only critical point is the origin. Completing the square gives $q=(x+y{)}^2+2y^2$, so the origin is a strict global minimum. The eigenvalues $4-2\sqrt{2}$ and $4+2\sqrt{2}$ are both positive, recovering the same conclusion from the positive definiteness of the Hessian. Its eigenvectors identify the rotated directions in which the quadratic form has its smallest and largest bending.

For a fixed vector $v$, restriction to a straight line gives the directional interpretationThis is the quadratic form associated with the Hessian matrix. For a unit eigenvector $v$, it equals the corresponding eigenvalue. Opposite signs therefore mean that some directions bend upwards and others downwards. The second-derivative test turns this information into a classification when the gradient vanishes.

The same object is useful for computation. Newton's method in optimization chooses a displacement $h$ by solving $H_f(x)h=-\nabla f(x)$ when the Hessian matrix is an invertible matrix; this makes the gradient of the local quadratic model vanish. For the displayed $q$, the equation gives $h=-(x,y{)}^T$, so one exact step reaches the minimum from any starting point. For a general function, a Hessian with opposite-sign eigenvalues or a large condition number need not give a useful minimizing step, and the local minimum need not be a global minimum.

The name honors Otto Hesse. ^[ref] An older geometric use is the Hessian criterion for flexes of a plane cubic, connecting a determinant of second partial derivatives with inflection points of an algebraic curve. Hesse's 1844 paper Über die Wendepuncte der Curven dritter Ordnung is an early source on that geometric problem. ^[ref]

The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

Put $F=x^3+y^3+13z^3-2xyz$. The Hessian criterion for flexes of a plane cubic applies because the projective cubic is smooth. To check smoothness, a singular point with all coordinates nonzero would satisfy $3x^2=2yz$, $3y^2=2xz$, $39z^2=2xy$. Multiplication would force $351=8$, a contradiction. If one coordinate is zero, the three equations force all coordinates to vanish, which is not a projective point.

For completeness, the Hessian criterion for flexes of a plane cubic follows locally by putting a smooth point at $(0:0:1)$ and its tangent at $y=0$. After normalization the homogeneous cubic has terms $y{z}^2+a{x}^{2}z+bxyz+c{y}^{2}z$ and terms cubic in $x,y$. Its tangent has contact order three precisely when $a=0$. Its Hessian matrix at the point has determinant $-8a$, giving exactly the same condition.

For the present cubic the Hessian determinant isOn $F=0$ this is $2744xyz=8\cdot 343xyz$. Thus the plane cubic flexes are exactly the points on the cubic with $xyz=0$. Let $c=\sqrt[3]{13}$ and let $\omega$ be a primitive cube root of unity. The complete list over the algebraic closure isThese nine points are distinct, and the Hessian calculation proves there are no others. They are the flexes of a Hesse cubic. The only rational one is $O^{\prime }=(1:-1:0)$: the other two points with $z=0$ require a nonrational cube root of unity, and a rational point in either remaining triple would give a rational cube root of $-13$. By prime valuations, $13$ is not a rational cube.