\section{The highest homogeneous part}
\label{sec:ordinary-top}

\begin{proposition}\label{prop:ordinary-top}
Let \(f\in\mathbb C[x_1,x_2,x_3,x_4]\) have degree at least three and nonzero constant Hessian determinant. Then either \(\nabla f\) is a polynomial automorphism or, in constant linear coordinates \(x,y,z,w\), the polynomial has the form \(R(x,y)+zA(x,y)+wB(x,y)\) with \(\det D(A,B)\in\mathbb C^\times\).
\end{proposition}
\begin{proof}
Subtract the affine part of \(f\). Restoring it translates the gradient and, in the affine-in-\(z,w\) form, adds constants to \(A,B\) and an affine polynomial to \(R\). For an invertible linear change \(L\), the identity \(\nabla(f\circ L)=L^{\mathsf T}(\nabla f)\circ L\) shows that gradient invertibility is invariant under the change.

Write \(D\geq3\) for the total degree and \(f_D\) for the sum of the monomials of degree \(D\). The degree-\(4(D-2)\) component of \(\det\Hess f\) is \(\det\Hess f_D\), hence it vanishes. Some second derivative of \(f_D\) is nonzero because \(f_D\) has degree \(D\geq3\) in characteristic zero. By \cite[Corollary~7.3(iii)]{debondt}, a constant linear change makes \(f_D\) depend on one, two or three coordinates, according to its Hessian rank, with a nonsingular Hessian in those coordinates.

Use positive rational weights and denote the weight of a variable \(t\) by \(w_t\). Each path stops at the first wall or at \(W=0\), whichever comes first. At \(W=0\), including a tie, Lemma~\ref{lem:endpoint} gives a polynomial inverse. A constant vector in the Hessian kernel of a leading form gives a constant directional derivative. This derivative is zero because the affine terms were removed. Conversely, for a rank-three leading form, such a nonzero vector gives a linear relation on its gradient. The image closure then has dimension three and lies in a hyperplane, so it is that hyperplane and its apex space has dimension three.

By Theorem~\ref{thm:r4}, a rank-three leading form has a nonzero apex space. One, two or three apices means that this space has the indicated dimension. With two apices, neither affine fiber coefficient is constant and the two are not proportional, since either case gives a constant kernel. By Lemma~\ref{lem:lift} the coefficient pair has zero Jacobian. Its coefficients are thus nonzero constant multiples of positive powers of a common weighted homogeneous polynomial, as shown in Section~\ref{sec:weights}. Equal coefficient degrees would give proportional coefficients, so the fiber weights are distinct.

We use the following weight change when the face is \(G(x,y)+c_1x^pz+c_2x^qw\) with \(p>q\geq1\) and \(c_1c_2\neq0\). Suppose \(x\) has minimum weight and the largest exponent \(m\) of \(y\) in \(G\) is at least two. Increase the weight of \(y\) at speed one and those of \(z,w\) at speed \(m\), fixing the weight of \(x\). The maximal slope among the monomials of the face is \(m\). The monomials with that slope are
\[
 c_0 x^\ell y^m+c_1 x^p z+c_2 x^q w,
\]
where \(c_0\neq0\) and \(\ell\) is a nonnegative integer, since homogeneity fixes the \(x\)-exponent of a term with \(y\)-exponent \(m\). There are finitely many monomials outside the face, all with strict weight gaps. A sufficiently small positive step therefore selects the displayed form, keeps \(W>0\), and makes the other weights strictly greater than that of \(x\). After dividing the weights by that of \(x\), Lemma~\ref{lem:unary} applies.

\medskip\noindent\textbf{Highest part of rank three.}\par

Write \(f_D=h(x,y,z)\) with nonsingular ternary Hessian. Keep the weights of \(x,y,z\) equal to one and raise that of \(w\). There is a monomial involving \(w\), since otherwise \(\det\Hess f=0\). If the exponent of \(w\) in such a monomial is one, the monomial has another factor of weight at least one and reaches weight \(D\) by \(w\)-weight \(D-1\). For exponent at least two it reaches weight \(D\) by \(w\)-weight \(D/2\). Thus the first new face \(P\) is reached at some \(\alpha\in(1,D-1]\). Since \(P|_{w=0}=h\), the ternary Hessian minor has nonzero restriction at \(w=0\). Also \(W=4D-2(3+\alpha)\geq2D-4>0\), so \(\det\Hess P=0\). Thus \(P\) has Hessian rank three.

The apex space splits between the ternary block and the \(w\) axis. A single apex on the \(w\) axis would make the complementary Hessian have rank at most one, contradicting the restriction to \(h\). Thus a single apex has minimum weight and Lemma~\ref{lem:unique-min} applies. If two apices include the \(w\) axis, their complementary source plane lies in the ternary block. Then \(h\) is affine in two coordinates of that block. The two affine coefficients are homogeneous of degree \(D-1\) in the one remaining coordinate, hence proportional. This would give a constant kernel for the ternary Hessian. So the two apices lie in the ternary block. Align them with \(x,y\) and write
\[
 P=G(x,y)+z A(x,y)+w B(x,y).
\]
If the common polynomial of \(A,B\) is not a polynomial in a constant linear form, apply Lemma~\ref{lem:cotangent-propagation}. Otherwise, a constant change of the equal-weight coordinates \(x,y\) makes both coefficients powers of \(x\). If \(G_{yy}=0\), then \(P\) is jointly affine in \(y,z,w\). Its Hessian vanishes on their principal block, and adjoining the \(x\) row and column gives rank at most two. Thus \(G_{yy}\neq0\), and the weight change keeping the two power coefficients applies. Finally three apices would give a constant kernel line. The ternary Hessian of \(h\) excludes a line in its block, and \(P_w\neq0\) excludes the \(w\) axis.

\medskip\noindent\textbf{Highest part of rank two.}\par

Write \(f_D=h(s,t)\) with nonsingular binary Hessian. Keep the weights of \(s,t\) equal to one and increase those of \(z,u\) together. A monomial of total degree at least two involving \(z\) or \(u\) meets weight \(D\) by the common weight \(D-1\), by the same exponent bound as above. Such a monomial exists by the unit Hessian condition. At \(D-1\) the determinant weight is zero, so assume the first new face \(P\) occurs at a common weight \(\beta\in(1,D-1)\). The face has rank two or three and \(P(s,t,0,0)=h(s,t)\).

\paragraph{Rank two at the first face.}
A two-dimensional constant kernel would lie entirely in the \(z,u\) block by the old binary Hessian. The kernel would be the whole \(z,u\) block, so \(P\) would be independent of \(z,u\), contrary to the new terms at the wall. In the one-apex case the apex lies in the \(s,t\) block, since an apex in the \(z,u\) block would have the nonsingular binary Hessian in its complementary space. Align the apex with \(s\). The normal form is affine in \(t,z,u\). Its \(t\) coefficient is a nonzero multiple of \(s^{D-1}\), by the binary Hessian. Its \(z,u\) coefficients are proportional because they have the same weighted degree in \(s\). A constant change in \(z,u\) gives
\begin{equation}
 P=a s^D+b t s^{D-1}+c z s^{D-\beta},\qquad bc\neq0.
 \label{eq:rank-two-binary-top-face}
\end{equation}
Here \(D-\beta\) is a positive integer. The variable \(u\) is unused.

\paragraph{Rank three at the first face.}
If there is a single apex, it belongs to the \(s,t\) block, since the complementary rank bound excludes the \(z,u\) block. Apply Lemma~\ref{lem:unique-min}. If there are two apices, they cannot span the \(z,u\) block, since that would give \(P\) affine in \(s,t\). They cannot span the \(s,t\) block either: its two affine coefficients in \(z,u\) would have equal weighted degree. Aligning the mixed apex plane with \(s,z\) gives
\[
 P=G(s,z)+t A(s,z)+u B(s,z).
\]
Here \(A(s,0)\) is a nonzero multiple of \(s^{D-1}\) by the binary Hessian. The coefficients have degrees \(D-1>D-\beta\). If their common polynomial belongs to \(\mathbb C[L]\) for a constant linear form \(L\), weighted homogeneity and its nonzero restriction to \(z=0\) make it a power of \(s\) up to a nonzero scalar. In this case \(G_{zz}\neq0\), since otherwise \(P\) is jointly affine in \(t,z,u\) and has Hessian rank at most two. Apply the weight change keeping the two power coefficients. If the common polynomial is not of this type, apply Lemma~\ref{lem:cotangent-propagation}. If there are three apices, the constant kernel line lies in the \(z,u\) block. Align the kernel with \(u\), so that \(P=H(s,t,z)\) with nonsingular ternary Hessian.

\paragraph{Increasing the unused weight.}
For either \(H(s,t,z)\) or \eqref{eq:rank-two-binary-top-face}, increase only the weight of \(u\), starting at \(\beta\). Its first new face \(Q\) occurs at a weight \(\gamma\in(\beta,D-1]\), by the exponent bound and the unit Hessian condition. The determinant weight is
\[
 4D-2(2+\beta+\gamma)\geq 2(D-\beta-1)>0.
\]
The terms independent of \(u\) retain their weights, so \(Q|_{u=0}\) is the preceding face. Some monomial involving \(u\) reaches the face at the wall, and therefore \(Q_u\neq0\).

If this restriction is \(H\), the rank remains three and the \(u\) cofactor is nonzero. Thus the apex space has zero \(u\) component, since the \(u\) axis is its own weight block and the cofactor excludes that axis. Three apices are impossible because a constant kernel line would have to be the \(u\) axis. A unique apex in the \(s,t\) block is handled by Lemma~\ref{lem:unique-min}. A unique \(z\) apex would give a single nonlinear phase over \(\mathbb C(z)\) in \(s,t,u\). At least one of the \(s,t\) phase coefficients is nonzero. Otherwise \(Q|_{u=0}=H\) is affine in \(s,t\), which makes \(\Hess h\) zero. Since \(w_s=w_t=1\), any ratio of two nonzero \(s,t\) phase coefficients is a rational function of \(z\) of weight zero and therefore constant, by the phase coefficient formula in Section~\ref{sec:weights}. After normalizing one such coefficient, the phase is \(L(s,t)+a(z)u\), with \(L\) a constant linear form. Choose a second constant linear form \(M\) so that \(L,M\) are coordinates on the \(s,t\) plane. Setting \(u=0\) gives
\[
 H=\Phi(z,L)+C(z)M
\]
over \(\mathbb C(z)\). Here \(\Phi(z,L)\) and \(C(z)\) are polynomials, since they are respectively the restriction at \(M=0\) and the derivative with respect to \(M\) of the polynomial \(H\). The weighted degree \(D-1>0\) of \(C\) gives \(C(0)=0\). Hence \(h=\Phi(0,L)\) has Hessian rank at most one, a contradiction.

With two apices in the \(H\) case, their plane is the \(s,t\) block or is spanned by \(z\) and one line in that block. For the first plane the two affine coefficients depend on \(s,t\) and have respective degrees \(D-\beta>D-\gamma>0\). A common polynomial in a constant linear form can be made a power of \(s\) by a constant base change. Then the pure base term has a nonzero \(tt\) derivative, because \(H\) has nonsingular Hessian. For the second plane align the base line with \(s\). The affine coefficients in \(t,u\) have degrees \(D-1>D-\gamma>0\), and the coefficient of \(t\) restricts to a nonzero multiple of \(s^{D-1}\) at \(z=0\). A common polynomial in a constant linear form is therefore a power of \(s\). The pure base term now has a nonzero \(zz\) derivative by the ternary Hessian of \(H\). The weight change keeping the two power coefficients applies in both cases. A common polynomial that is not a polynomial in a constant linear form gives Lemma~\ref{lem:cotangent-propagation}.

It remains to handle \eqref{eq:rank-two-binary-top-face} at this new face. Write \(p=D-1\) and \(q=D-\beta\). The old mixed Hessian minors on \((s,t)\) and \((s,z)\) are respectively
\[
 -(bp s^{p-1})^2,\qquad -(cq s^{q-1})^2.
\]
They survive by restriction to \(u=0\). The old face has no constant kernel on its \(s,t,z\) variables: its \(s\) derivative contains nonzero \(t,z\) terms, and the other derivatives are nonzero multiples of distinct powers of \(s\). Thus a constant kernel of \(Q\), decomposed into the new weight blocks, could only be the \(u\) axis. This excludes the rank-two case with two apices and the rank-three case with three apices.

For a single apex at either rank, the \(z,u\) axes are excluded by the \((s,t)\) minor. Indeed, the Hessian on the complementary space has rank at most one at rank three, and is zero at rank two. For an apex \(d_s e_s+d_t e_t\) in the base block, where \(e_s,e_t,e_z\) are coordinate vectors, its annihilator contains \(d_t e_s-d_s e_t\) and \(e_z\). The old second derivative in the \(z\) direction is zero and the mixed derivative on these directions is \(d_tcq s^{q-1}\). The restricted minor is therefore \(-(d_tcq s^{q-1})^2\), so \(d_t=0\). If the rank is three, apply Lemma~\ref{lem:unique-min}. If the rank is two, the normal form gives
\[
 Q=a s^D+b t s^p+c z s^q+d u s^{D-\gamma},\qquad d\neq0.
\]
The exponent \(D-\gamma\) is a positive integer smaller than \(q\). A small common increase of the three weights other than that of \(s\) selects the three linear partners, whose slopes are all one, while \(a s^D\) has slope zero. Lemma~\ref{lem:three-partners} applies.

Suppose \(Q\) has rank three and two apices. Their plane splits between the \(s,t\) block and the \(z,u\) axes, and must contain the \(s\) axis. Indeed, the plane spanned by \(z,u\) is excluded by the \((s,t)\) minor. For a plane spanned by \(d_s e_s+d_t e_t\) and \(e_u\), the preceding restricted minor forces \(d_t=0\). For a plane spanned by \(d_s e_s+d_t e_t\) and \(e_z\), the second derivative of the old face along \(d_t e_s-d_s e_t\) has \(t\) coefficient \(d_t^2 bp(p-1)s^{p-2}\). This must vanish by joint affineness in the complementary plane, so \(d_t=0\). The restrictions used here are derivatives in the old variables and commute with setting \(u=0\).

If the apex plane is the \(s,t\) block, write \(Q=G(s,t)+c s^q z+u B(s,t)\), where the old restriction fixes \(G(s,t)=a s^D+b t s^p\). The zero Jacobian of the coefficient pair gives \(c q s^{q-1}B_t=0\), so \(B\) depends only on \(s\). This makes \(Q\) affine in \(t,z,u\), with Hessian rank at most two. For the \(s,z\) plane, write \(Q=G(s,z)+b s^p t+u B(s,z)\), where \(G(s,z)=a s^D+c z s^q\). Now \(b p s^{p-1}B_z=0\), so the same rank bound holds. For the \(s,u\) plane the coefficients of \(t,z\) restrict at \(u=0\) to \(b s^p,c s^q\). If their common polynomial is a polynomial in a constant linear form, it is a power of \(s\), and the pure base term must have \(u\)-degree at least two by rank three. The unary weight change applies. If their common polynomial does not belong to \(\mathbb C[L]\) for any constant linear form \(L\), Lemma~\ref{lem:cotangent-propagation} applies.

\medskip\noindent\textbf{Highest part of rank one.}\par

Write \(f_D=a_0 s^D\) with \(a_0\neq0\). Keep \(s\) at weight one and initially increase the other three weights together. We first examine any rank-three face \(P\) of weight \(D\) containing \(a_0 s^D\), with all other weights greater than one and positive determinant weight.

A single apex of minimum weight is handled by Lemma~\ref{lem:unique-min}. If instead the unique apex is a heavier coordinate \(t\) after a constant change within its weight block, the complementary normal form over \(\mathbb C(t)\) has a phase with a nonzero \(s\) coefficient because \(P\) contains \(a_0s^D\). Normalize the phase to \(s+a(t)u+b(t)v\). Setting \(u=v=0\) shows that its highest power is \(D\) and has coefficient \(a_0\): a larger exponent of \(s\) would exceed the weight \(D\). The coefficients of \(s^{D-1}u,s^{D-1}v\) are \(D a_0 a(t),D a_0 b(t)\). They are polynomials in \(t\) of weights \(1-w_u,1-w_v<0\). Since \(w_t>0\), both vanish. Then \(P=G(s,t)+u A(t)+v B(t)\). The gradient image closure lies in the product of the \(s,t\) target plane with the image closure of \(t\mapsto(A(t),B(t))\). The latter is irreducible of dimension at most one. Since the gradient image closure has dimension three, it equals the product. This gives at least two apices, excluding the heavy single-apex case.

A plane of two apices contains the \(s\) axis: by weight invariance, otherwise it lies in the heavy block and its complementary space contains \(s\), forcing \(P_{ss}=0\). Align the plane with \(s,t\) and write
\[
 P=G(s,t)+u A(s,t)+v B(s,t).
\]
If the common polynomial of \(A,B\) is not a polynomial in a constant linear form, apply Lemma~\ref{lem:cotangent-propagation}. Otherwise the unequal weights of \(s,t\) make the common polynomial a power of one of them. A power of \(s\) gives the weight change above, since \(G_{tt}=0\) would make the face affine in \(t,u,v\) and force rank at most two. For a power of \(t\), order the fibers so the coefficients are \(c_1t^p,c_2t^q\) with \(p>q\geq1\). Increase the weight of \(s\) at speed one and those of \(u,v\) at speed \(D\), leaving the \(t\) weight fixed. In \(G(s,t)\) only \(a_0s^D\) has \(s\)-exponent \(D\), and none has a higher exponent because the old weight of \(s\) is one and the old weight of \(t\) is positive. Thus the slope-\(D\) terms are exactly \(a_0s^D+c_1t^pu+c_2t^qv\). Divide all weights by the \(t\) weight. The new \(s\) weight \(\beta\) is less than one for a sufficiently small step. Since the selected terms have weight \(D\beta\), the two fiber weights are \(D\beta-p,D\beta-q>0\). In particular \(0<q<p<D\) and \(p/D<\beta<1\), so Lemma~\ref{lem:low-base} applies with \(t\) as its weight-one variable.

If the common weight of the other three variables is \(\beta\), the determinant weight along the initial path is \(4D-2(1+3\beta)\). Thus the endpoint occurs at \(\beta=(2D-1)/3\) if there is no earlier wall. At an earlier wall \(P_{ss}\neq0\) because of the term \(a_0s^D\). In rank one, \cite[Corollary~7.3(i),(ii)]{debondt} gives a three-dimensional constant kernel. This kernel splits along the weight blocks and cannot contain \(s\), so it is the whole heavy block. That contradicts the terms involving heavy variables at a wall. In rank two with a two-dimensional constant kernel, the same weight argument places that kernel in the heavy block. A linear change within the block puts the kernel on two coordinate axes, so \(P\) is independent of those variables. In rank two with a single apex, the apex is \(s\), since any heavy apex gives \(P_{ss}=0\) by the complementary affine form. Then the three heavy variables occur affinely, with coefficients proportional to the same power of \(s\) by their equal weights. A linear change within their block makes \(P\) independent of two of them. Rank three with one or two apices was treated above.

If the rank-three face has three apices, its constant kernel lies in the heavy block by the \(s^D\) term. A linear change within that block makes the form \(H(s,t,u)\), with nonsingular ternary Hessian and no dependence on \(v\), whose weight is maximal. More generally, whenever such a ternary form occurs, raise only the weight of \(v\). The determinant weight decreases at speed two because the weight \(D\) of the leading form is fixed. At a wall before \(W=0\), the cofactor complementary to \(v\) restricts at \(v=0\) to the nonzero determinant of the ternary Hessian. Thus the new Hessian has rank three. Its constant kernel, if nonzero, would have to be the \(v\) axis by weight invariance and the nonsingular ternary restriction. This is impossible since \(v\) occurs. The form therefore has one or two apices.

For the rank-two face with two unused variables, write \(H(s,t)\) with nonsingular binary Hessian. The unused \(u,v\) have equal weights at least as large as that of \(t\). Increase both weights at unit speed. The determinant weight decreases at speed four. At a wall before \(W=0\), the restriction at \(u=v=0\) is \(H\), so the new Hessian has rank at least two. Any constant kernel splits by the new weights and is contained in the \(u,v\) block, by the nonsingular binary Hessian of \(H\). A two-dimensional kernel would be the entire block, contrary to the new terms. A rank-three face with three apices has its kernel in the \(u,v\) block by the binary Hessian, and a linear change in that block gives the ternary case just treated. A rank-three face with fewer apices was handled above. Finally a rank-two face with one apex has apex \(s\), since the \(s^D\) term excludes all heavy apices. Its \(u,v\) coefficients are proportional by their equal weight. A constant linear change in \(u,v\) gives the face
\[
 a_0 s^D+b t s^p+c u s^q,\qquad bc\neq0,\quad p>q\geq1.
\]
The restriction \(H(s,t)=a_0s^D+bts^p\) has Hessian determinant \(-b^2p^2s^{2p-2}\neq0\), so \(b\neq0\) and \(p\geq1\). A monomial involving \(u,v\) reaches the face, so \(c\neq0\). The exponents \(p=D-w_t\) and \(q=D-w_u\) are integers, with \(p>q\) since \(w_u>w_t\), and \(q>0\) because the affine part was removed. The weight of \(v\) equals that of \(u\).

Raise only the weight of \(v\). The determinant weight decreases at speed two. At a wall before \(W=0\) the four weights are distinct. The old face has only \(v\) as a constant kernel direction, since its \(s\) derivative has nonzero \(t,u\) terms and its \(t,u\) derivatives are independent powers of \(s\). The new face involves \(v\), so no constant kernel can survive. Its Hessian rank is two or three by the surviving \((s,t)\) minor and the positive determinant weight. Rank three gives one of the cases already handled. Rank two gives the unique apex \(s\), again by the \(s^D\) term. The new \(v\) coefficient is a nonzero multiple of \(s^r\) with \(r=D-w_v\). Homogeneity and the absence of affine terms give \(r\in\mathbb Z_{>0}\), and \(w_v>w_u\) gives \(r<q\). A small common increase of the three weights other than that of \(s\) now selects the three linear partners, all with slope one, while \(a_0s^D\) has slope zero. Apply Lemma~\ref{lem:three-partners}.
\end{proof}