\section{Translation directions of singular Hessians}
\label{sec:rank-three-apex}

Ranks of polynomial and rational matrices are taken over the rational function field in the source coordinates. We write \(X=(X_1,\ldots,X_m)^{\mathsf T}\), \(\partial_j=\partial/\partial X_j\), and \(e_j\) for the \(j\)-th standard coordinate vector. Subscripts on functions denote partial derivatives. The dot product is the complex bilinear pairing \(a\cdot b=a^{\mathsf T}b\). Closures of algebraic images are taken in the Zariski topology. A translation direction of an algebraic set \(Y\subseteq\mathbb C^m\) is a constant vector \(d\) such that \(Y+\tau d=Y\) for every \(\tau\in\mathbb C\).

\begin{theorem}\label{thm:rational-symmetrizer}
Let \(V\colon\mathbb C^3\dashrightarrow\mathbb C^3\) be rational with nilpotent \(DV\). Suppose \(H=\operatorname{Hess} f\) for a polynomial \(f\in\mathbb C[X_1,X_2,X_3]\), \(\det H\ne0\) as a polynomial, and \(H\,DV\) is symmetric. Then the image of \(V\) is contained in an affine plane.
\end{theorem}

\begin{proof}
If \(DV\) has rank zero, \(V\) is constant. Suppose its rank is one. By the characteristic-zero rank criterion the image closure is a curve. At a point where \(V\) is defined and its differential has rank one, a line in a direction outside the kernel has nonconstant restriction. Such a restriction is dominant onto the image curve. Its function field thus embeds in a one-variable rational function field. By L\"uroth's theorem the image function field is generated by one rational function \(t=p/q\) of the source coordinates, with coprime polynomials \(p,q\) and \(q\ne0\). Write \(V=A(t)\), where \(A(T)\) is a vector of rational functions in an indeterminate \(T\).

If the image is contained in no affine plane, the three components of \(A'(T)\) are linearly independent over \(\mathbb C\). The trace identity is \(A'(t)\cdot\nabla t=0\). It implies
\[
 A'(T)\cdot\nabla(p-Tq)=0
 \quad\text{in }\mathbb C(T)[X_1,X_2,X_3].
\]
Indeed \(q\nabla t=\nabla(p-Tq)|_{T=t}\), where the gradient on the right holds \(T\) fixed. After clearing the denominators of \(A'(T)\), the resulting polynomial vanishes at \(T=p/q\), so it is divisible by \(p-Tq\) over the rational function field in the source coordinates. This divisor is primitive as a polynomial in \(T\) over the source polynomial ring, since \(p,q\) are coprime. Gauss's lemma gives divisibility in the polynomial ring. Differentiation lowers the total degree in the source variables, and multiplication by a polynomial in \(T\) alone does not raise it. The differentiated expression therefore has smaller source degree than \(p-Tq\) and must vanish.

Let \(M(T)\) be the matrix whose rows are the coefficient vectors of the source monomials in \(\nabla(p-Tq)\). Its entries are affine in \(T\), and its rank is at most two. If its rank is two, the cross product of two independent rows gives a polynomial kernel vector of degree at most two. The independence of the components of \(A'\) implies independence of the three constant coefficient vectors of this polynomial vector. Conjugating \(V\) by the linear map that sends these three coefficient vectors to the standard coordinate vectors makes the kernel vector \((1,T,T^2)^{\mathsf T}\). In coordinates \(x,y,z\) the coefficient identities are
\[
 p_x=0,\qquad p_y=q_x,\qquad p_z=q_y,\qquad q_z=0.
\]
The four identities give \(p=P(y)+B(z)\), \(q=xP'(y)+C(y)\), and \(B'(z)=xP''(y)+C'(y)\). Comparison of the coefficient of \(x\) gives \(P''=0\), and the remaining equality makes \(B'\) and \(C'\) constant. Thus \(p=ay+bz+c\), \(q=ax+by+d\), and \(\nabla(p-Tq)\) is constant in the source variables. This contradicts rank two for \(M\).

The matrix \(M\) has rank one, since rank zero would make \(p,q\) and hence \(t\) constant. Its row space defines a nonzero linear form \(L(T)\) in the source variables over \(\mathbb C(T)\). Derivatives of \(p-Tq\) along the kernel of \(L(T)\) vanish, so \(p-Tq\) is a polynomial in \(L(T)\). If its source degree is \(D\ge2\), its top homogeneous part is \(c(T)L(T)^D\) with \(c(T)\ne0\). This part is affine in \(T\), so its second derivative in \(T\) is zero. Dividing that derivative by \(L^{D-2}\) and reducing modulo \(L\) gives \(D(D-1)c(T)(L')^2=0\bmod L\). Since a nonzero linear form is prime, \(L'\) is proportional to \(L\), and all ratios of nonzero coefficients of \(L\) have zero \(T\)-derivative. They are therefore constant. Then \(p,q\) are both polynomials in a single constant linear form, obtained by evaluating \(p-Tq\) at two general constants \(T\). The same conclusion holds for \(D=1\) if their linear parts are proportional. In either case the identity \(A'(T)\cdot\nabla(p-Tq)=0\) gives a fixed nonzero linear relation on \(A'\). Thus \(p,q\) are affine with independent linear parts. Use \(x=p,\ y=q\) as the first two new source coordinates and conjugate \(V\) by the same affine change. The Jacobian changes by conjugation and the Hessian by inverse congruence, which preserve nilpotence and Hessian symmetry.

Now \(t=x/y\), and \(\nabla t=(1,-t,0)^{\mathsf T}/y\) in the trace identity gives \(A_1'=tA_2'\). Independence gives \(A_2'\ne0\). Put \(k=A_3'/A_2'\) and \(F(t,y,z)=f(ty,y,z)\). Symmetry of the outer product \(H\,DV=HA'(t)\nabla t^{\mathsf T}\) makes \(HA'(t)\) proportional to \(\nabla t\). Thus \(H(t,1,k)^{\mathsf T}\) is proportional to \((1,-t,0)^{\mathsf T}\). Contracting with \((t,1,0)\) and \((0,0,1)\) gives, by the chain rule at fixed \(t\),
\[
 \partial_y(F_y+kF_z)=\partial_z(F_y+kF_z)=0.
\]
In particular \(F_y+kF_z\) depends only on \(t\).

The coefficient of \(y^jz^\ell\) in \(F\) has \(t\)-degree at most \(j\), since substituting \(x=ty\) sends \(x^ay^bz^\ell\) to \(t^ay^{a+b}z^\ell\). Since \(\det H\ne0\), we have \(F\ne0\). Write \(F=\sum_{\ell=0}^N F_\ell(t,y)z^\ell\), with \(N=\deg_z F\). For \(N\ge1\), the coefficient of \(z^N\) in \(F_y+kF_z\) is \(\partial_y F_N=0\). The degree restriction at \(j=0\) makes \(F_N=c\) a nonzero constant. For \(N\ge2\), the coefficient of \(z^{N-1}\) is \(\partial_y F_{N-1}+Nc\,k=0\). Hence the coefficient of \(yz^{N-1}\) in \(F\) is \(-Nc\,k\), so \(k=at+b\) for constants \(a,b\). Then \(A_3'=aA_1'+bA_2'\), contradicting independence. For \(N\le1\), the only possible term involving \(z\) is a constant multiple of \(z\), and the condition on \(F_y+kF_z\) makes \(F_0\) affine in \(y\). The same degree restriction gives \(F=(at+b)y+cz+d\) with constant coefficients, where \(c=0\) if \(N=0\). This makes \(f\) affine, contrary to \(\det H\ne0\).

Suppose now that \(N=DV\) has rank two. At a point where \(N\) has rank two and \(H\) is invertible, use a constant conjugation to make \(N\) the Jordan block \(J\) with \(J_{12}=J_{23}=1\) and all other entries zero. The identity \(HJ=J^{\mathsf T}H\) gives
\[
 H=\begin{pmatrix}0&0&a\\0&a&b\\a&b&c\end{pmatrix},\qquad a\ne0.
\]
A further constant change preserving \(J\) makes \(H\) the antidiagonal identity at this point. Write \(H=R(aI+bJ+cJ^2)\), where \(I\) is the identity and \(R\) the antidiagonal identity. Put \(U=(bJ+cJ^2)/a\), so \(U^3=0\). The change given by the congruence matrix \(a^{-1/2}(I-U/2+3U^2/8)\) has this effect, since the square of \(I-U/2+3U^2/8\) multiplied by \(I+U\) is \(I\), and \(RJ=J^{\mathsf T}R\). Write \(K_{i,j\ell}=\partial_j\partial_\ell V_i\) and \(T_{ij\ell}=\partial_i\partial_j\partial_\ell f\), evaluated at this point. At the chosen point the derivatives of \(\operatorname{tr}(N^3),\operatorname{tr}(N^2),\operatorname{tr}(N)\), divided by \(3,2,1\), respectively, give
\[
 K_{3,1\ell}=0,\qquad K_{2,1\ell}+K_{3,2\ell}=0,\qquad
 K_{1,1\ell}+K_{2,2\ell}+K_{3,3\ell}=0.
\]
The \((1,2)\) and \((1,3)\) symmetry equations for \(\partial_\ell(HN)=(\partial_\ell H)J+R\,\partial_\ell N\) at the chosen point give
\[
 T_{11\ell}+K_{3,2\ell}=K_{2,1\ell},
 \qquad T_{12\ell}+K_{3,3\ell}=K_{1,1\ell}.
\]
The second trace equation makes the first identity \(T_{11\ell}=-2K_{3,2\ell}\). The first trace equation gives \(K_{3,31}=K_{3,13}=0\). Thus \(T_{112}=T_{121}\) gives \(K_{1,11}=-2K_{3,22}\). The second trace equation gives \(K_{2,21}=K_{2,12}=-K_{3,22}\). Substituting into the third at \(\ell=1\) yields \(K_{1,11}=K_{3,22}\). In particular \(K_{3,22}=0\). Let \(\omega\) be a local nonzero one-form annihilating \(\operatorname{im}N\), with value \(dX_3\) at this point. Differentiating \(\omega N=0\) gives
\[
 \partial_\ell\omega_1=-K_{3,2\ell},\qquad
 \partial_\ell\omega_2=-K_{3,3\ell}.
\]
The coefficient of \(dX_1\wedge dX_2\wedge dX_3\) in \(\omega\wedge d\omega\) at the point is \(-K_{3,31}+K_{3,22}=0\). Thus the rank-two distribution \(\operatorname{im}N\) is involutive on the regular locus.

Assume the image of \(V\) is not planar. Locally write \(V=(u,v,w(u,v))\), where \(u,v\) are two algebraically independent components of \(V\) and \(w\) is algebraic over \(\mathbb C(u,v)\). Let \(p=w_u\), \(q=w_v\); derivatives in \(u,v\) extend uniquely to the image function field in characteristic zero. Write \(M\) for the \(3\times2\) matrix with columns \((1,0,p)^{\mathsf T},(0,1,q)^{\mathsf T}\), and let \(J_0=D(u,v)\). Since \(N=MJ_0\) is nilpotent of rank two in dimension three, \(N^2\) has rank one. The matrix \(M\) is injective and \(J_0\) is surjective over the source function field. Hence \(N^2=M(J_0M)J_0\) gives \(\operatorname{rank}(J_0M)=1\), and \(N^3=0\) gives \((J_0M)^2=0\). A nonzero nilpotent two-by-two matrix has a nonzero off-diagonal entry. Exchange the first two source and target coordinates if needed to put that entry in position \((1,2)\). Call it \(b\) and write the \((1,1)\) entry as \(-b\beta\). Trace and determinant zero give
\[
 J_0M=b\begin{pmatrix}-\beta&1\\-\beta^2&\beta\end{pmatrix},
 \qquad b\ne0.
\]
The columns of \(M\) define vector fields \(L_1=\partial_1+p\partial_3\) and \(L_2=\partial_2+q\partial_3\). Their bracket is \((L_1(q)-L_2(p))\partial_3\). The two columns of \(J_0M\) give \((L_1(u),L_1(v))=(-b\beta,-b\beta^2)\) and \((L_2(u),L_2(v))=(b,b\beta)\). The chain rule then gives
\[
 [L_1,L_2]=-b(w_{uu}+2\beta w_{uv}+\beta^2w_{vv})\partial_3.
\]
Since \(\partial_3\notin\operatorname{im}N\), involutivity makes the coefficient vanish. If all three second derivatives vanished, \(w_u,w_v\) would have zero differential and would be constant. Indeed, if an element of the image field has both partial derivatives zero, differentiating its monic minimal polynomial over \(\mathbb C(u,v)\) gives a relation of smaller degree for each partial derivation. Minimality forces both derivatives of every coefficient to vanish, so the coefficients are in \(\mathbb C\) and the element itself is in \(\mathbb C\). Subtracting the corresponding linear function of \(u,v\) from \(w\) would then give a constant. This contradicts nonplanarity. Hence \(\beta\) is algebraic over the image function field. On each component of the geometric generic fiber, \(p,q,\beta\) are constant. Here the geometric generic fiber means the fiber after extending the image function field to its algebraic closure. Its tangent line is \(\ker N\). Since \(N^2=M(J_0M)J_0\) has rank one and image equal to \(\ker N\), that kernel is generated by
\[
 (1,\beta,p+q\beta)^{\mathsf T}.
\]
Two independent linear forms vanishing on this vector are constant on the component, because their differentials vanish there in characteristic zero. The closure of the component is therefore an affine line.

Let \(B\) be the closure of the image of the rational map sending a source point \(X\) to the projective closure of \(X+\ker N(X)\) in the Grassmannian of lines in \(\mathbb P^3\). The fibers of this assignment are contained in their assigned lines and contain a dense open subset of those lines at general points, by the description of the geometric generic fibers of \(V\). Hence \(B\) is an irreducible surface. On a nonempty open subset of \(B\) the restrictions of \(V\) to the lines are defined and constant, and the lines meet the rank-two regular locus. To see openness after discarding a proper closed set, trivialize the family of lines on a Grassmannian chart. For each component of \(V\), constancy on a line is the proportionality of the coefficient vectors of the numerator and denominator as polynomials in the line parameter, with denominator not identically zero. The proportionality identities hold at the generic line. The omitted subfamily has dimension at most one, so its incidence has dimension at most two. At a general point of the affine chart each remaining line through the point has tangent direction \(\ker N\) there. Thus there is exactly one family line through a general point.

Over a resolution of \(B\), the incidence space is a smooth \(\mathbb P^1\)-bundle. Its evaluation map to \(\mathbb P^3\) is proper and birational. The ramification divisor is the divisor of the determinant of the differential, a section of the canonical line bundle of the incidence space tensored with the pullback of the inverse canonical line bundle of \(\mathbb P^3\). On a general incidence fiber these canonical bundles have degrees \(-2\) and \(-4\), respectively, so the ramification divisor has degree two there. Over the generic point of a divisor of \(\mathbb P^3\), the evaluation map has zero-dimensional fiber. Otherwise the inverse image of the divisor would have dimension at least three, contradicting dominance and irreducibility of the incidence space. By upper semicontinuity of fiber dimension, the map is proper and quasi-finite over an open neighborhood of that generic point, hence finite there. A finite birational morphism to a normal variety is an isomorphism. The image of each ramification component therefore has dimension at most one. If one component maps to a point, its intersection with each incidence fiber contains at most one point because evaluation embeds each fiber as its corresponding line. Thus the two-dimensional component dominates the two-dimensional base, and all general family lines pass through its image. Otherwise the focal locus, defined as the scheme-theoretic image of the ramification divisor, has one-dimensional support.

De Poi's classification \cite[Theorem~0.1]{depoi} gives three possibilities with a one-dimensional focal locus. The lines are the secants of a twisted cubic, the pencils in the planes through a fixed line with their centers on that line, or the joins of a fixed line and a curve. We consider these families and the families through one point. In each case \(\ker N\) is orthogonal to \(\operatorname{im}N\) for the bilinear pairing \(a^{\mathsf T}Hb\), since \(N\) is self-adjoint for that pairing. Also \(\ker N=\operatorname{im}N^2\subseteq\operatorname{im}N\). In particular \(d^{\mathsf T}Hd=0\) for every \(d\in\ker N\).

\emph{Two generically finite endpoints.}
For a secant family or a join family whose two endpoint loci are not contained in the plane at infinity, write
\(X=(1-\lambda)A(s)+\lambda B(t)\), \(r=B(t)-A(s)\),
and \(V(X)=W(s,t)\), where \(s,t\) are local parameters on the endpoint curves. The derivatives of \(X\) with respect to \(s,t,\lambda\) are \((1-\lambda)A',\lambda B',r\), so dominance gives generic independence of \(A',B',r\). In this frame the columns of \(N\) are the coordinates of \(W_s/(1-\lambda),W_t/\lambda,0\). Its trace is the sum of the \(A'\) coefficient of \(W_s\) divided by \(1-\lambda\) and the \(B'\) coefficient of \(W_t\) divided by \(\lambda\). Multiplying by \(\lambda(1-\lambda)\) shows that these coefficients both vanish, since they depend only on \(s,t\). Thus \(W_s-\alpha B'\) and \(W_t-\gamma A'\) are proportional to \(r\), with \(\alpha,\gamma\) depending on \(s,t\). The sum of the three principal two-by-two minors is \(-\alpha\gamma/(\lambda(1-\lambda))\), which vanishes by nilpotence. Since \(W_s,W_t\) are independent generically, exactly one of \(\alpha,\gamma\) is nonzero generically. Take \(\alpha=0\), so \(\operatorname{im}N=\operatorname{span}(A',r)\). Let \(n\) be a normal to this plane with respect to the usual bilinear dot product. Then \(Hr\) is proportional to \(n\). Integrating along the line and using polynomiality at the finite endpoint gives
\[
 \nabla f(X)=\nabla f(A)+n\,R(\lambda),\qquad
 f(X)=f(A)+\lambda\nabla f(A)\cdot r.
\]
Here \(R\) may depend on \(s,t\). The second identity follows by integrating \(\partial_\lambda f=\nabla f(X)\cdot r=\nabla f(A)\cdot r\). Differentiating it in \(t\) at fixed \(s,\lambda\) gives \(\lambda(\nabla f(X)-\nabla f(A))\cdot B'=0\). Since \(n\) annihilates precisely \(\operatorname{span}(A',r)\), we have \(n\cdot B'\ne0\). It follows that \(R=0\). This makes the gradient on a dominant family depend only on \(s\), contradicting its rank three. For \(\gamma=0\) the same calculation uses the finite endpoint \(B(t)\) and differentiation in \(s\).

\emph{Lines through a point.}
If the point \(c\) is finite, the direction \(X-c\) is \(H\)-isotropic. The identity \((X-c)^{\mathsf T}H(X)(X-c)=0\) multiplies the degree-\(j\) part of the expansion of \(f\) about \(c\) by \(j(j-1)\). Thus \(f\) is affine, a contradiction.

For a point at infinity choose the common line direction to be \(e_3\). Then \(V=(A(x,y),B(x,y),C(x,y))\), and \(H\)-isotropy gives \(f=zP(x,y)+Q(x,y)\). The polynomial \(P\) is nonconstant since \(H\) is nonsingular. The two-dimensional Jacobian of \((A,B)\) is nilpotent. If \((A,B)\) is constant the desired relation follows. Otherwise it has rank one and \((A,B)=(a(t),b(t))\) for a nonconstant rational function \(t\) by L\"uroth's theorem as above. Let \(r=(a'(t),b'(t))^{\mathsf T}\). The planar Jacobian is \(r(\nabla t)^{\mathsf T}\). Its trace is \((\nabla t)^{\mathsf T}r=0\). Since \(He_3=(P_x,P_y,0)^{\mathsf T}\) annihilates \(\operatorname{im}N\), also \((\nabla P)^{\mathsf T}r=0\). Therefore \(dt\wedge dP=0\), as \(r\) is nonzero. By the characteristic-zero rank criterion \(t\) is algebraic over \(\mathbb C(P)\). On each component of a general level of \(P\), \(t\) is constant by algebraicity. The tangent direction is then the constant direction \(r\). A linear form annihilating \(r\) has zero differential on the component, so the component is dense in an affine line. Since \(P\) is polynomial, each such line lies in its level. Lines in distinct levels cannot meet, so they are parallel. Comparing with a line in a different level shows that the direction is common to all components of general levels. A constant linear form annihilating this direction therefore annihilates \((a',b')\), giving a constant affine relation between \(A\) and \(B\).

\emph{Pencils with centers on a fixed line.}
For a fixed line not contained in the plane at infinity, a general center is finite, with the constant center case already treated. The restriction of \(f\) to each general pencil plane is affine along each line through its center by \(H\)-isotropy. Expansion at the center makes \(f\) affine on the entire plane. Expanding instead at a fixed finite point common to these planes shows every homogeneous part of degree at least two vanishes on their union, which is dense. This contradicts nonsingularity.

If the fixed line is at infinity, write the parallel planes and their pencils as
\[
 X=(\lambda,q+a(t)\lambda,t),\qquad V(X)=W(t,q),
\]
where \(t\) specifies the plane, \(q\) specifies its affine line, and \(a(t)\) its common slope. A constant \(a\) gives parallel lines. Otherwise \(q=y-a(z)x\) and \(t=z\) give
\[
 \operatorname{tr}N=-aW_{1,q}+W_{2,q}+W_{3,t}-\lambda a'W_{3,q}.
\]
The coefficient of \(\lambda\) vanishes, so \(a'\ne0\) implies \(W_{3,q}=0\). Now the third component of \(V\) depends only on \(t=z\), and its \(z\)-derivative is an eigenvalue of \(N\) by the shape of the bottom row. Nilpotence makes that derivative zero, so \(V_3\) is constant.

\emph{Joins with an endpoint locus at infinity.}
If the fixed line is at infinity and the curve has finite general point, the lines through each such point have all directions in a fixed two-dimensional vector plane. By \(H\)-isotropy, the restriction of \(f\) to the corresponding affine plane is affine. As these planes sweep a dense set, \(f=\alpha(z)x+\beta(z)y+\gamma(z)\) in suitable affine coordinates. Its Hessian is singular.

Suppose the fixed line has a finite point and the curve lies at infinity. Put the affine part of the fixed line on the \(z\)-axis. Choose the first coordinate so that it is nonzero at the general point of the infinite curve. The infinite endpoint then has coordinates \([1:t:q(t):0]\), and the finite endpoint is \((0,0,s)\). The general join has affine coordinates \(X=(\lambda,t\lambda,s+\lambda q(t))\). The parameter \(t\) is nonconstant, since otherwise the joins would all lie in the plane \(y=tx\). The function \(q(t)\) may be algebraic over \(\mathbb C(t)\). Write \(V(X)=W(s,t)=(a,ta+b,c)\) with \(a,b,c\) functions of \(s,t\). The two parameter differentials are \(dt=(dy-t\,dx)/\lambda\) and \(ds=dz-q\,dx-\lambda q' dt\). The trace of \(N\) is \((a+b_t)/\lambda+c_s-q a_s-q'b_s\), so \(b_t=-a\) and \(c_s=q a_s+q'b_s\). If \(b_s=0\) identically, then \(a_s=c_s=0\), contrary to rank two. The Jacobian \(N\) is the product of the matrix with columns \(W_s,W_t\) and the matrix with rows \(ds,dt\). Applying \(ds\) and \(dt\) to these columns gives the reversed product
\[
 \begin{pmatrix}
 c_s-q a_s-q'b_s & c_t-q a_t-q'(a+b_t)\\
 b_s/\lambda & (a+b_t)/\lambda
 \end{pmatrix}
 =
 \begin{pmatrix}
 0 & c_t-q a_t\\
 b_s/\lambda & 0
 \end{pmatrix}.
\]
The reversed product is nilpotent, since its fourth power has \(N^3=0\) between the same two factors. Its determinant thus vanishes. As \(b_s\ne0\), this gives \(c_t=q a_t\). Differentiating \(c_s=q a_s+q'b_s\) in \(t\) and \(c_t=q a_t\) in \(s\) yields \(q'a_s+q''b_s+q'b_{st}=0\). The relation \(b_t=-a\) cancels the first and third terms, so \(q''=0\). Differentiating the monic minimal polynomial of \(q'\) over \(\mathbb C(t)\) forces its coefficients to have zero derivative, by minimality. Hence \(q'\) is constant, and the same argument for \(q-tq'\) shows that \(q\) is affine in \(t\). The covector \((tq'-q,-q',1)\) is therefore constant and nonzero. Its products with \(W_s,W_t\) are \(c_s-q a_s-q'b_s\) and \(c_t-q a_t-q'(a+b_t)\), respectively, both zero. It gives a constant affine relation on \(V\). If both endpoint loci lie at infinity, the joins remain in that plane and cannot dominate the affine chart.
\end{proof}

\begin{lemma}\label{lem:characteristic-slice}
Let \(h\in\mathbb C[X_1,X_2,X_3,w]\) have Hessian rank three, and suppose \(A(X)=\operatorname{Hess}_X h(X,0)\) is nonsingular over \(\mathbb C(X_1,X_2,X_3)\), where \(\operatorname{Hess}_X\) differentiates in \(X_1,X_2,X_3\) alone. Set
\[
 v(X)=A(X)^{-1}\nabla_X h_w(X,0).
\]
Then
\[
 \nabla_X h(X-wv(X),w)=\nabla_X h(X,0).
\]
Moreover \(Dv\) is nilpotent and \(A\,Dv\) is symmetric.
\end{lemma}

\begin{proof}
The gradient-image closure is an irreducible hypersurface, because the dimension of the image of a polynomial map in characteristic zero equals the generic rank of its differential. Let \(R\) be an irreducible defining polynomial of this hypersurface and put \(Q=\nabla R(\nabla h)\). The relation \(R(\nabla h)=0\) implies \(\operatorname{Hess}(h)Q=0\). For \(Z=(X_1,X_2,X_3,w)\), the derivation \(D_Q=\sum_{i=1}^4 Q_i\,\partial/\partial Z_i\) annihilates every component of \(\nabla h\). By the chain rule it also annihilates every component of \(Q\). Inductively, for any polynomial \(g\) and integer \(j\ge1\),
\[
 D_Q^j g=\sum_{i_1,\ldots,i_j=1}^4
   Q_{i_1}\cdots Q_{i_j}\,
   \frac{\partial^j g}{\partial Z_{i_1}\cdots\partial Z_{i_j}},
\]
since \(D_Q\) acts only on the derivative of \(g\) in each summand. The coefficient of \(\tau^j\) in the polynomial \(g(Z+\tau Q(Z))\) is therefore \(D_Q^j g/j!\). Taking \(g\) to be each component of \(\nabla h\) gives \(\nabla h(Z+\tau Q(Z))=\nabla h(Z)\).

The slice \(w=0\) maps densely into the hypersurface, since the first three gradient coordinates have Jacobian \(A\) there. Thus \(Q(X,0)\ne0\) generically. On the slice the first three Hessian rows are \((A\ \nabla_X h_w(X,0))\), so their kernel is spanned by \((-v,1)\). The fourth row leaves this kernel unchanged because the full Hessian has rank three there. Hence \(Q(X,0)=Q_4(X,0)(-v(X),1)\), where \(Q_4\) is its fourth component. Substitution of \(\tau=w/Q_4(X,0)\) gives the claimed characteristic identity. Differentiating it in \(X\) yields
\[
 \operatorname{Hess}_X h(X-wv(X),w)(I-wDv)=A(X),
\]
where \(I\) is the three-by-three identity. Both matrices on the left have entries in \(\mathbb C(X)[w]\). Their determinants have nonzero product independent of \(w\), so each determinant is constant in \(w\). At \(w=0\) the second determinant is one, giving \(\det(I-wDv)=1\). Thus the characteristic polynomial of \(Dv\) is \(T^3\), and Cayley--Hamilton gives \((Dv)^3=0\). Finally, differentiating \(Av=\nabla_X h_w(X,0)\) in \(X\) gives \(A Dv\) as the difference of \(\operatorname{Hess}_X h_w(X,0)\) and the matrix with \((i,j)\) entry \(\sum_{k=1}^3 v_k\partial_j A_{ik}\). The latter is symmetric because \(\partial_j A_{ik}=\partial_i A_{jk}\).
\end{proof}

\begin{theorem}\label{thm:r4}
If \(h\in\mathbb C[X_1,X_2,X_3,X_4]\) has Hessian rank three, then the closure of its gradient image has a nonzero translation direction.
\end{theorem}

\begin{proof}
Choose a constant linear change and a constant slice as in Lemma~\ref{lem:characteristic-slice}. Such a choice exists because a symmetric rank-three matrix at a regular point restricts nondegenerately to a three-plane. Translate the remaining coordinate so the slice is \(w=0\). Apply Theorem~\ref{thm:rational-symmetrizer} to \(v\), using the Hessian of \(h(X,0)\) as the symmetrizing matrix. There are a constant \(b\in\mathbb C^3\setminus\{0\}\) and \(d\in\mathbb C\) with \(b\cdot v=d\).

As in Lemma~\ref{lem:characteristic-slice}, let \(R\) be an irreducible equation of the gradient-image closure and \(Q=\nabla R(\nabla h)\). On the slice \(w=0\), the kernel direction \((-v,1)\) gives \((b,d)\cdot Q=0\). The slice has dense image in \(R=0\), so the derivative of \(R\) in direction \((b,d)\) vanishes on \(R=0\). Irreducibility makes that derivative divisible by \(R\), and the degree bound forces it to vanish identically. Thus \(R(Y+\tau(b,d))=R(Y)\) for every target point \(Y\) and scalar \(\tau\). An affine source change \(Z=SZ'+a\) sends the gradient image by the invertible target map \(Y\mapsto S^{\mathsf T}Y\), so existence of a nonzero translation direction also holds in the original coordinates.
\end{proof}

\begin{corollary}\label{cor:debondt-problem-one}
Let \(h\in\mathbb C[X_1,X_2,X_3,X_4]\) have identically zero Hessian determinant. There exist a nonzero polynomial \(R\in\mathbb C[Y_1,Y_2,Y_3,Y_4]\) and a nonzero \(d\in\mathbb C^4\) such that \(R(\nabla h)=0\) and \(D_dR=0\), where \(D_dR=\sum_{i=1}^4 d_i\partial R/\partial Y_i\). Thus Problem~1 of \cite[Section~6]{quasi} has an affirmative answer.
\end{corollary}

\begin{proof}
For rank three, take an irreducible equation \(R\) of the gradient-image closure and the nonzero translation direction \(d\) from Theorem~\ref{thm:r4}. Translation invariance implies that \(D_dR\) vanishes on \(R=0\), hence is divisible by \(R\). Its degree forces \(D_dR=0\). If the rank is at most two, the Jacobian of the first three gradient coordinates has rank at most two, so those coordinates are algebraically dependent in characteristic zero. A nonzero relation \(R\) in \(Y_1,Y_2,Y_3\) then satisfies \(D_dR=0\) for \(d=(0,0,0,1)^{\mathsf T}\).
\end{proof}