\section{The quadratic case}\label{sec:quadratic}

We prove the initial degree needed for the induction.
The only previously established finite-dimensional input is the theorem
of Fr\"oberg and L\"ofwall \cite[Theorem~6.2]{FL02}: the Ideal Generation
Conjecture holds for quadrics through general points in projective
dimension at most \(20\). We use it only when \(n\leq12\).

\begin{theorem}\label{thm:quadratic-column}
For every \(n\geq2\) and \(n\leq p<N_2(n)\), the multiplication map
\[
 I_Z(2)\otimes k[x_1,\ldots,x_n]_1\longrightarrow I_Z(3)
\]
has maximal rank for a general set of \(p\) points in
\(\mathbf P^{n-1}\).
\end{theorem}

We use two elementary constructions, according to the residue of \(n\)
modulo three. Throughout this section \(E(n,2)=n(n+1)/6\).

\subsection{Cones over a trace}

Write \(S=T[s_1,\ldots,s_c]\), where \(T=k[y_1,\ldots,y_m]\),
\(n=m+c\), and put \(P=(s_1,\ldots,s_c)\).
Let \(Z_2\subset\mathbf P^{m-1}=V(P)\) be a trace set, and let \(Z_1\)
consist of points \((\sigma_z;\mu_z y_z)\), where \(\sigma_z\ne0\),
\(\mu_z\ne0\), and \([y_z]\in Z_2\). Repeated projected supports are
allowed. Set
\[
 K=P_2\cap I_{Z_1}(2),\qquad
 V=I_{Z_2,T}(2),\qquad M=P_3\cap I_{Z_1}(3).
\]
Assume that the evaluations on \(P_2\) and \(T_2\) have full row rank.
Every pure trace form vanishing on \(Z_2\) also vanishes on \(Z_1\).
Separating the pure trace part of a polynomial therefore gives
\[
 I_{Z_1\cup Z_2}(j)=I_{Z_2,T}(j)\oplus
                 \bigl(P_j\cap I_{Z_1}(j)\bigr),\qquad j=2,3.
\]
In these decompositions multiplication is the direct sum of trace
multiplication and
\begin{equation}\label{eq:quad-middle}
 \mu_M:(K\otimes S_1)\oplus
        \bigl(V\otimes\langle s_1,\ldots,s_c\rangle\bigr)
       \longrightarrow M.
\end{equation}

It suffices to construct a trace configuration for which the second
map has the desired rank. Indeed the space of marked cone configurations,
with a fixed assignment of projected supports to trace labels, is
irreducible and dominates the space of ordered trace configurations.
Full evaluation rank and the desired rank of \(\mu_M\) are open
conditions. Their nonempty open set consequently meets the pullback of
the nonempty open set where the inductive trace theorem holds. This
argument allows the trace used in an explicit calculation to be special.

\subsection{A two-normal interpolation construction}

Let \(c=2\), write the normal coordinates as \(s,t\), and split the
trace variables into \(y_1,\ldots,y_h,z_1,\ldots,z_l\), where \(h\geq2\)
is even, \(l\geq2\), and \(m=h+l\). Put
\(\pi_a=e_{2a-1}+e_{2a}\).
Choose \(N_2(h)\) base points imposing independent conditions on quadrics in
\(\mathbf P\langle y_1,\ldots,y_h\rangle\), including the coordinate
points and all \(\pi_a\). For example, the coordinate points and the
points \(e_i+e_j\), \(i<j\), have this property. In each layer add
\[
 e_{z_j},\qquad \pi_a+e_{z_j}\quad(1\leq a\leq h/2).
\]
Call the resulting set \(Z_f\); thus
\[
 p_f=|Z_f|=N_2(h)+l(h/2+1).
\]
The designated cone supports are the \(h\) base coordinate points and
the two points \(e_{z_j},\pi_1+e_{z_j}\) in each layer. There are
\(p_1=h+2l=2m-h\) supports.

Over \(e_i\) choose normal direction \((c_i,-1)\), with the \(c_i\)
distinct and nonzero. In each layer choose two independent normal
directions. Choose nonzero cone heights. The mixed evaluation equations
have independent rows: the base equations use separate coordinates,
and each new layer adds an invertible two-by-two system. Hence
\[
 K\longrightarrow \operatorname{Sym}^2\langle s,t\rangle
 \quad\hbox{is onto},\qquad
 K_{00}:=\ker(K\longrightarrow\operatorname{Sym}^2\langle s,t\rangle)
 \quad\hbox{has dimension }h.
 \tag{\(\ast\)}
 \label{eq:quad-qmap}
\]
Write a basis of \(K_{00}\) as
\[
 H_i=s\kappa_i^1+t\kappa_i^2,\qquad
 \kappa_i=(\kappa_i^1,\kappa_i^2)
          =y_i(1,c_i)+\text{layer terms}.
\]
Since the noncoordinate layer supports use only \(\pi_1\), the layer
terms vanish for \(i\geq3\).

Let
\[
 \mathcal A=\{(A,B)\in T_2^{\oplus2}:
                \sigma_z\cdot(A,B)(y_z)=0
                \text{ at every designated support}\}.
\]
Put
\[
 \mathcal U=\sum_{i=1}^h\kappa_iT_1,\qquad I_f=I_{Z_f,T}(2).
\]
We claim that
\begin{equation}\label{eq:quad-flag}
 \mathcal A=\mathcal U\oplus I_f^{\oplus2},\qquad
 \dim\mathcal U=hm.
\end{equation}
On the base, the map
\[
 (\ell_i)\longmapsto
 \left(\sum y_i\ell_i,\sum c_i y_i\ell_i\right)
\]
is injective: the coefficients of \(y_i^2\) vanish, and the two equations
for \(y_iy_j\) force both coefficients to vanish because \(c_i\ne c_j\).
Its image is exactly the space satisfying the base support equations,
by dimension.

Inductively add a layer. Restriction, setting the new variable \(z_j\)
equal to zero, identifies the new mixed residual space with the old one.
At the layer points the allowed values of
\(\sum\lambda_i\kappa_i\) have total dimension \(h\), and evaluation is
injective. Here the two designated supports have one-dimensional
fibres cut out by their normal equations, and the \(h/2-1\) remaining
points have unrestricted two-dimensional fibres.
In fact, vanishing at \(e_{z_j}\) first kills the layer
correction; vanishing at every \(\pi_a+e_{z_j}\) then gives
\[
 \lambda_{2a-1}+\lambda_{2a}
 =c_{2a-1}\lambda_{2a-1}+c_{2a}\lambda_{2a}=0,
\]
so all \(\lambda_i\) vanish. This evaluation is therefore bijective.
After lifting a decomposition from the preceding layers, its error is
divisible by \(z_j\); the preceding bijectivity supplies the required
mixed correction. The layer points impose independent scalar conditions
because their underlying vectors
\(e_{z_j},\pi_1,\ldots,\pi_{h/2}\) are independent.
This proves the sum assertion. Injectivity of
\(K_{00}\otimes T_1\to T_2^2\) follows by restriction and induction:
the remaining coefficient forms are multiples of \(z_j\), and linear
independence of the \(\kappa_i\) kills them.
Finally,
\[
 hm+2(N_2(m)-p_f)=2N_2(m)-p_1=\dim\mathcal A,
\]
so the sum is direct.

Choose lifts
\[
 K_1=s^2+sU_1+tV_1,\quad
 K_2=st+sU_2+tV_2,\quad
 K_3=t^2+sU_3+tV_3
\]
in \(K\). They may be normalized so that the \(U_j\) contain only
layer variables. Define
\[
 \mathcal W=\{(A,B):sA+tB\in K S_1\},\qquad
 \mathcal W'=\mathcal W\cap I_f^{\oplus2},
 \qquad
 \Omega=\{(k_s,k_t)\in K^2:s\mathfrak q(k_s)+t\mathfrak q(k_t)=0\},
\]
where \(\mathfrak q\) takes the pure normal quadratic part.
Since \(\mathfrak q\) is onto, \(\dim\Omega=2h+2\).
If \(k_s=Q_s+su_s+tv_s\) and \(k_t=Q_t+su_t+tv_t\), the polynomial
\begin{equation}\label{eq:quad-elimination}
 -K_1u_s-K_2(v_s+u_t)-K_3v_t+sk_s+tk_t
\end{equation}
has only normal degree one. Let \(\phi(k_s,k_t)\) be its coefficient
pair, and let \(\pi:\mathcal A\to I_f^{\oplus2}\) be the projection in
\eqref{eq:quad-flag}. Eliminating normal degrees three and two shows
that
\[
 \mathcal W=\mathcal U+\phi(\Omega),\qquad
 \mathcal W'=\pi\phi(\Omega),\qquad \dim\mathcal W'\leq2h+2.
\]

\begin{lemma}\label{lem:quad-pair-interpolation}
The cone parameters can be chosen so that \(\dim\mathcal W'=2h+2\)
and evaluation \(\mathcal W'\longrightarrow(k^2)^{h+1}\) is an
isomorphism at the \(h+1\)
points
\begin{equation}\label{eq:quad-extra}
 e_{y_1}+e_{z_1},\quad e_{y_1}+e_{z_2},\quad
 e_{z_1}+e_{z_2},\quad
 e_{y_{2a-1}}+e_{z_j}\quad(2\leq a\leq h/2,\ j=1,2).
\end{equation}
These points impose independent scalar conditions on \(I_f\).
\end{lemma}

\begin{proof}
First restrict to \(y_1,y_2,z_1,z_2,s,t\).
The seven trace points in this subspace have quadratic ideal
\[
 I_{f,c}=\langle
 (y_1-y_2)z_1,\ (y_1-y_2)z_2,\ z_1z_2\rangle.
\]
Restriction commutes with the projection \(\pi\):
it maps \(\mathcal U\) onto its core counterpart, while the restriction
of a form in \(I_f\) vanishes on the core flag. It also maps \(K\) and
\(\Omega\) onto the corresponding core spaces.

We use the following direct calculation. If \(N\) points are linearly
independent in \(\mathbf P^{N-1}\), with pairwise distinct normal
directions \(s:t\), then
\[
 ((s,t)_2\cap I_Z)\otimes S_1
       \longrightarrow (s,t)_3\cap I_Z
\]
is bijective. Put the points at the coordinate vectors, and write
\(s=\sum a_ix_i\), \(t=\sum b_ix_i\).
The quadrics \(w_i=x_i(b_is-a_it)\) span the residual space, with the
single relation \(\sum w_i=0\).
If \(\sum\ell_iw_i=0\), the coefficient of \(x_i^2x_j\) shows that
the \(i\)-th coefficients of \(\ell_i\) and \(\ell_j\) agree, since
\(b_ia_j-a_ib_j\ne0\). Thus all \(\ell_i\) agree, and the relation is
the indicated redundant-generator relation. The dimensions of source
and target are both \(N(N-1)\).

For the six core cone points use, in order, the trace supports
\[
 Y_1=e_{y_1},\quad Y_2=e_{y_2},\quad Y_3=e_{z_1},\quad
 Y_4=e_{y_1}+e_{y_2}+e_{z_1},\quad
 Y_5=e_{z_2},\quad Y_6=e_{y_1}+e_{y_2}+e_{z_2}.
\]
Take normal directions
\[
 (1,-1),(2,-1),(1,1),(1,2),(1,3),(1,4)
\]
and normal heights
\((\theta_1,\ldots,\theta_6)=(1,1,1,2,1,3)\), respectively.
The actual six projective points are \(P_i=(\theta_i\sigma_i;Y_i)\),
where \(\sigma_i\) denotes the corresponding displayed direction.
Eliminating four support-coordinate columns in their coordinate
matrix leaves determinant \(-17\); the six points are independent.
The preceding calculation implies
\(\mathcal W'_c=(I_{f,c})^{\oplus2}\). The first three points in
\eqref{eq:quad-extra} evaluate the displayed basis of \(I_{f,c}\)
as the identity matrix.

For an untouched pair \(i=2a-1,i'=2a\), \(a\geq2\), put
\(\beta_i=(1,c_i)\).
The projected products
\[
 sH_i-y_iK_1-c_iy_iK_2,\qquad
 tH_i-y_iK_2-c_iy_iK_3
\]
give elements
\begin{equation}\label{eq:quad-four-columns}
 \beta_{i'}(y_i-y_{i'})L_{i,s}(z),\quad
 \beta_{i'}(y_i-y_{i'})L_{i,t}(z),
\end{equation}
and the analogous two elements with \(i,i'\) interchanged.
Here the projection can be calculated by first subtracting the
base-quadratic terms in \(\mathcal U\), and then subtracting
linear combinations of \(\kappa_i z_j,\kappa_{i'}z_j\).
If a remaining coefficient in the \(j\)-th layer is
\(a\beta_i+\gamma\beta_{i'}\), this last subtraction replaces
\(y_i z_j(a\beta_i+\gamma\beta_{i'})\) by
\(\gamma\beta_{i'}(y_i-y_{i'})z_j\), which proves the asserted form.

We check that the two forms for each fixed index are independent on
\(\langle z_1,z_2\rangle\). Let the normal directions of layer \(j\)
be \(\sigma_\nu=(a_\nu,b_\nu)\), \(\nu=1,2\), and let its independently
variable normal heights be \(\theta_1,\theta_2\). Write
\(M=(\sigma_1;\sigma_2)\) and \(d_\nu=M^{-1}e_\nu\).
Solving the support equations for the three lifts and differentiating
with respect to \(\theta_\nu\), the two coefficients of
\((L_{i,s},L_{i,t})\) in layer \(j\) have Jacobian columns
\[
 \lambda_i(d_\nu)(a_\nu+c_ib_\nu)
                   \binom{a_\nu}{b_\nu},\qquad
 \lambda_i(v)=\frac{\det(\beta_i,v)}
                    {\det(\beta_i,\beta_{i'})}.
\]
Both scalar factors are nonzero if
\(\sigma_\nu\cdot\beta_i\ne0\) for both \(\nu\).
The Jacobian is then invertible. Thus general heights in the first
two layers make both forms independent; the finitely many such
conditions hold simultaneously. They are compatible with the core
rank condition, since all are nonempty opens in the same irreducible
parameter space.

For each pair, \eqref{eq:quad-four-columns} and its two companions
span a four-dimensional space \(\mathcal V_a\). These spaces are
independent, have zero core restriction, and have total dimension
\(2h-4\). Together with the six-dimensional core quotient they reach
the upper bound \(2h+2\). At the last two points assigned to pair \(a\)
in \eqref{eq:quad-extra}, its four columns give an invertible block:
the two directions \(\beta_i,\beta_{i'}\) are independent, as are the
two layer forms in each direction. The other pair spaces vanish there
and all pair spaces vanish at the core points. The evaluation matrix
is block triangular with invertible diagonal blocks.
For scalar independence use the three displayed core quadrics, followed
by \((y_i-y_{i'})z_1,(y_i-y_{i'})z_2\); their evaluation matrix is the
identity.
\end{proof}

Take \(q\) additional trace points, first among
\eqref{eq:quad-extra} and then general if necessary.
As long as these impose independent scalar conditions on \(I_f\),
their vector evaluation on \(\mathcal W'\) has maximal rank.
Normal-degree elimination in \eqref{eq:quad-elimination} identifies
the kernel and cokernel of \(\mu_M\) with those of this evaluation.
For clarity, \(K\otimes S_1\to KS_1\) is injective:
a relation first gives an element of \(\Omega\); its projection in
\(\mathcal W'\) vanishes, so that element is zero because
\(\Omega\to\mathcal W'\) is an isomorphism; then
\eqref{eq:quad-flag} kills the remaining mixed coefficients.
The pure normal parts of degrees two and three are covered by the
three lifts and their normal multiples. The remaining cokernel is
precisely the evaluation cokernel. Consequently
\begin{equation}\label{eq:quad-middle-ranks}
 \dim\ker\mu_M=\max\{2h+2-2q,0\},\qquad
 \dim\operatorname{coker}\mu_M=\max\{2q-2h-2,0\}.
\end{equation}

For \(n=3\nu+2\geq8\), choose
\[
 h=2\nu-2,\quad l=\nu+2,\quad q=h+1,\quad
 \dim K=2\nu+1,\quad \dim V=E(3\nu,2).
\]
The trace count is \(p_f+q=\nu(3\nu+1)\), and
\(\dim K+\dim V=E(n,2)\). Thus \(\mu_M\) is bijective.
For \(n=3\nu+1\geq7\), choose
\[
 h=2\nu-2,\quad l=\nu+1,\quad q=\nu-1,\quad
 \dim V=E(3\nu-1,2).
\]
Now \(p_f+q=\nu(3\nu-1)\), and
\(\dim K+\dim V=\lceil E(n,2)\rceil\).
Equation~\eqref{eq:quad-middle-ranks} gives surjectivity.
In both cases the trace endpoint is integral, so the cone splitting
and the open-set argument prove the required full endpoint from the
quadratic theorem in \(n-2\) variables.

\subsection{The injective endpoint in the remaining residue class}

Keep the latter parameters, but choose the \(q=h/2\) extra trace points
\[
 E_0=e_{y_1}+e_{z_1},\qquad
 E_a=e_{y_{2a-1}}+e_{z_1}\quad(2\leq a\leq h/2).
\]
An element of \(\Omega\) has a unique expression
\[
 k_s=\sum\alpha_iH_i+aK_2+bK_3,\qquad
 k_t=\sum\beta_iH_i-aK_1-bK_2.
\]
Write \(w_\omega=\pi\phi(\omega)\). The old middle kernel is
\(\{\omega:w_\omega(E_a)=0\text{ for every }a\}\).
Add the point \(e_s\) to \(Z_1\).
Since \(K_1(e_s)=1\) and the other basis quadrics vanish there,
the new residual space omits \(K_1\). The new middle kernel is therefore
the subspace of the old kernel on which the coefficient of \(K_1\)
in the relation is zero. Writing
\(H_i=su_i+tv_i\), this coefficient is
\begin{equation}\label{eq:quad-contraction}
 C(\omega)=-\sum_i\alpha_i u_i-a(t+U_2)-bU_3.
\end{equation}
The corrections in \(\mathcal U\) and \(V\otimes\langle s,t\rangle\)
do not alter it.

Here is an explicit core calculation for the parameters used in the
preceding proof:
\[
\begin{aligned}
u_1&=y_1+3z_1+15z_2,&v_1&=y_1-3z_1-5z_2,\\
u_2&=y_2+5z_1+27z_2,&v_2&=2y_2-5z_1-9z_2,\\
U_2&=-4z_1-12z_2,&V_2&=-y_1-2y_2+3z_1+3z_2,\\
U_3&=10z_1+132z_2,&V_3&=y_1+y_2-11z_1-47z_2.
\end{aligned}
\]
The coefficients of \(y_1,y_2,t,z_1\) in \(C=0\) force
\(\alpha_1=\alpha_2=a=b=0\).
The remaining two core directions, \((0,H_1),(0,H_2)\), have projected
values \((30,60),(75,75)\) at \(E_0\), with determinant \(-2250\).
One can check these values without solving the full core system.
Set \(x=y_1,y=y_2,z=z_1,z_2=0\).
The coefficient pair for \((0,H_i)\) is
\(-u_i(U_2,V_2)-v_i(U_3,V_3)\).
Subtracting
\(\kappa_1(-y-41z)+\kappa_2(x+33z)\) for \(i=1\), and
\(\kappa_1(y-70z)+\kappa_2(-x+56z)\) for \(i=2\), leaves respectively
\(30(1,2)(x-y)z\) and \(75(1,1)(x-y)z\).

Thus the joint core map
\(\omega\mapsto(C(\omega),w_\omega(E_0))\) is injective.
For each untouched pair, the two projected directions
\((0,H_i),(0,H_{i'})\) have the forms
\[
 \beta_{i'}(y_i-y_{i'})L_{i,t},\qquad
 \beta_i(y_{i'}-y_i)L_{i',t}.
\]
The Jacobian calculation above allows both \(z_1\)-coefficients to
be nonzero simultaneously with the core condition.
If \(C(\omega)=0\), its \(y_i\)-coefficient first gives
\(\alpha_i=0\) for every \(i\geq3\).
Restriction to the core, together with \(w_\omega(E_0)=0\), then
kills all core parameters. Finally the evaluation at \(E_a\) kills
\(\beta_i,\beta_{i'}\) for its pair, since the displayed two vectors
are independent. Hence the new middle map is injective.

The new residual dimension is \(2\nu\), and
\[
 2\nu+E(3\nu-1,2)=\frac{3\nu(\nu+1)}2=\lfloor E(n,2)\rfloor.
\]
Scalar independence persists because \(K_1\) separates the added point
from all previous points. Pure trace forms vanish at \(e_s\), so the
cone splitting remains valid. Keeping \(e_s\) fixed, the same marked
cone family still dominates the trace configuration space; intersecting
the middle-rank open with the trace-isomorphism open proves the full
injective endpoint.


\subsection{The quadratic step in three normal directions}
\label{subsec:quadratic-three-normal}

We prove the remaining quadratic induction step.  Throughout this
subsection, $h\geq 8$ and $h\equiv2\pmod3$, and we put
\[
\begin{gathered}
 m=h+4,\qquad n=h+7,\qquad
 T=k[y_1,\ldots,y_h,z_1,\ldots,z_4],\\
 S=T[s_1,s_2,s_3],\qquad P=(s_1,s_2,s_3).
\end{gathered}
\]
The construction below gives a set $Z_1$ of $2h+12$ points outside the
trace $\mathbf P^{m-1}$ and a set $Z_2$ of $m(m+1)/3$ trace points, with
all projections of $Z_1$ contained in $Z_2$, such that
\begin{equation}\label{eq:c3-middle}
 \bigl(K\otimes S_1\bigr)\oplus
 \bigl(V\otimes\langle s_1,s_2,s_3\rangle\bigr)
 \longrightarrow P_3\cap I_{Z_1}(3)
\end{equation}
is an isomorphism.  Here $K=P_2\cap I_{Z_1}(2)$ and $V=I_{Z_2,T}(2)$.
Their dimensions will be $h+6=n-1$ and $m(m+1)/6$, respectively.
Consequently, quadratic ideal generation in $m=n-3$ variables implies
quadratic ideal generation in $n$ variables: the two dimensions add to
$n(n+1)/6$, and the trace multiplication is an isomorphism at its
integral critical count.

\subsubsection*{The cone family and its coefficient map}
Choose distinct nonzero parameters $t_i$, and a primitive cube root
$\omega$.  Set
\begin{align*}
 v_i&=(1,t_i,t_i^2),& \ell_i(s)&=s\mathbin{\cdot}v_i,\\
 \sigma_i^+&=(\omega^2t_i^2,\omega t_i,1),&
 \sigma_i^-&=(\omega t_i^2,\omega^2t_i,1),\\
 \theta_i^+&=-\omega^2,&\theta_i^-&=\omega.
\end{align*}
Over the trace coordinate point $e_i$ place the two cone points
$(\theta_i^+\sigma_i^+;e_i)$ and
$(\theta_i^-\sigma_i^-;e_i)$.  The vectors $\sigma_i^\pm$ are
independent and annihilate $v_i$.  Moreover,
\begin{equation}\label{eq:c3-gram}
 \theta_i^+\sigma_i^+-\theta_i^-\sigma_i^-=(t_i^2,-2t_i,1),
 \qquad
 (\sigma_i^+\mathbin{\cdot}v_j)
 (\sigma_i^-\mathbin{\cdot}v_j)
 =(t_i-t_j)^2(t_i^2+t_it_j+t_j^2).
\end{equation}
We choose the $t_i$ so that the last expression is nonzero whenever
$i\ne j$.

Partition the first six indices into two triples, leave $\{7,8\}$ as
a pair, and partition any remaining indices into triples.  Write
$\pi_B=\sum_{i\in B}e_i$ for a part $B$.  In each layer $z_j$, put
trace points at $e_{z_j}$ and at
$\pi_B+\zeta_{j,B}e_{z_j}$ for all parts $B$, where the $\zeta_{j,B}$
are nonzero.  Lift the first three of these points, namely $e_{z_j}$
and the points belonging to the first two triples, to cone points
whose three normal-coordinate vectors are independent.  There are
three cone points and $(h+4)/3$ trace points in each layer.
In addition to the $h$ trace coordinate points, choose
\[
 f_h=\frac{h^2+2h-8}{3}
\]
free trace points in the base $z=0$, and four further trace points.
The latter four will remain in the fixed subspace on
$y_1,\ldots,y_8,z_1,\ldots,z_4$.  The total trace count is
\[
 h+f_h+4\frac{h+4}{3}+4=\frac{(h+4)(h+5)}3.
\]

The mixed equations at the cone points are independent.  Thus
$K\to k[s_1,s_2,s_3]_2$, given by restriction to $y=z=0$, is onto,
and its kernel has basis
\[
 H_i=\ell_i(s)y_i+\text{terms involving the }z_j.
\]
Only $H_1,\ldots,H_6$ have layer corrections.  For each normal
quadratic $Q$, choose a lift $K_Q\in K$, linearly in $Q$.  At each
base point its mixed coefficient $A_i(Q)\in k^3$ satisfies
\begin{equation}\label{eq:c3-lift}
 \sigma_i^\pm\mathbin{\cdot}A_i(Q)
   =-\theta_i^\pm Q(\sigma_i^\pm).
\end{equation}
The three equations in a layer then determine its coefficient
uniquely.  These choices can be made compatibly whenever further base
indices are added.

Put $U=\langle s_1,s_2,s_3\rangle$ and
\[
 \Omega_h=\ker\bigl(U\otimes K\longrightarrow k[s_1,s_2,s_3]_3\bigr).
\]
Since restriction $K\to k[s]_2$ is onto, $\dim\Omega_h=3h+8$.
For $\eta=(k_1,k_2,k_3)\in\Omega_h$, write
\[
 \sum_{a=1}^3s_ak_a=\sum_b x_bR_b(s),
 \qquad x_b\in\{y_1,\ldots,y_h,z_1,\ldots,z_4\},
\]
and subtract $\sum_bx_bK_{R_b}$.  The result has normal degree one,
so it is $s\mathbin{\cdot}P_h(\eta)$ for a vector-valued quadric
$P_h(\eta)$ on the trace.  Let $\Phi_h$ denote the vector-valued
quadrics arising from $K_{00}\otimes T_1$, where
$K_{00}=\langle H_1,\ldots,H_h\rangle$.

At a trace point supporting a cone point with normal vector $a$, the
value of such a vector quadric satisfies $a\mathbin{\cdot}F=0$.
At $e_i$ there are two such conditions.  Denote the direct sum of
these allowed value spaces, with unrestricted $k^3$ at all other
trace points, by $A(Z_2)$.  Consider the square map
\begin{equation}\label{eq:c3-Dh}
 D_h:\Omega_h\oplus(K_{00}\otimes T_1)\longrightarrow A(Z_2),
 \qquad (\eta,\phi)\longmapsto
       \operatorname{ev}_{Z_2}\bigl(P_h(\eta)+\phi\bigr).
\end{equation}
Indeed, both dimensions equal
\[
 3h+8+h(h+4)=3|Z_2|-(2h+12).
\]
The kernels of $D_h$ and \eqref{eq:c3-middle} are naturally isomorphic.
In particular, one map is invertible exactly when the other is.  To see this,
start with a relation in \eqref{eq:c3-middle}.  Its normal-degree-three
part gives an element of $\Omega_h$; its normal-degree-two part
uniquely determines the coefficients of the $K_QT_1$ terms.  The
remaining relation has normal degree one and says that
$P_h(\eta)+\phi$ belongs to $V\otimes U$, or equivalently has zero
values on $Z_2$.  Conversely, a pair in the kernel of $D_h$ uniquely supplies the remaining
$V\otimes U$ coefficients and hence an original relation.  The dimensions of
\eqref{eq:c3-middle} agree.  This argument is ordinary coefficient
elimination and uses no ideal-generation assertion for the chosen
special trace.

\subsubsection*{Enlargement by three base variables}
We next show that invertibility of $D_h$ implies invertibility of
$D_{h+3}$ for a suitable enlargement.  Keep every old point fixed, add
three new base indices $B$, their coordinate trace points and their
two cone points each, and choose their $t_b$ generally.  All old lifts
restrict to their original values, while
$H_b=\ell_b(s)y_b$ for $b\in B$.  The kernel of restriction on the
source of \eqref{eq:c3-Dh} is the direct sum of
\begin{enumerate}
\item the nine vectors $U\otimes\langle H_b:b\in B\rangle$;
\item the $6h+9$ products consisting of old $H_i$ times new $y_b$,
and new $H_b$ times arbitrary old or new base variables;
\item the twelve products $H_bz_j$.
\end{enumerate}
Every vector in this kernel has zero value at all old trace points.
The first two groups, restricted to the base, form a vector-quadric
space of dimension $6h+18$, as we now verify.

On the base, the products $H_iy_j$ have coefficient $v_i$ on $y_iy_j$.
They are independent: diagonal coefficients involve $v_i$, and an
edge $ij$ involves the independent vectors $v_i,v_j$.  Write
$\lambda_{ij}=v_i\times v_j$,
$\sigma_i^+\times\sigma_i^-=c_iv_i$, and
$L_i=(t_i^2,-2t_i,1)$.  Modulo the product space, measure an edge
coefficient $b_{ij}$ by $-\lambda_{ij}\mathbin{\cdot}b_{ij}$.
Put $q_{ij}=(t_i-t_j)^2(t_i^2+t_it_j+t_j^2)$.
The vector identity
\[
 \lambda_{ij}=
 \frac{\sigma_j^+(\sigma_j^-\mathbin{\cdot}v_i)
       -\sigma_j^-(\sigma_j^+\mathbin{\cdot}v_i)}{c_j}
\]
and \eqref{eq:c3-lift} give
\[
 \lambda_{ij}\mathbin{\cdot}
 A_j\bigl((s\mathbin{\cdot}u_i)\ell_i\bigr)
 =-\frac{q_{ij}}{c_j}\,L_j\mathbin{\cdot}u_i.
\]
Consequently the edge coordinate of the sum of the products
\[
 (s\mathbin{\cdot}u_i)H_i-y_iK_{(s\mathbin{\cdot}u_i)\ell_i}
\]
is
\[
 -\frac{q_{ij}}{c_j}L_j\mathbin{\cdot}u_i
 +\frac{q_{ij}}{c_i}L_i\mathbin{\cdot}u_j.
\]
Multiplication of this row by $c_ic_j/q_{ij}$ and the invertible
change of variables $w_i=c_i u_i$ therefore give
\begin{equation}\label{eq:c3-motion}
 L_i\mathbin{\cdot}w_j-L_j\mathbin{\cdot}w_i.
\end{equation}
Choose three old indices as roots.  With all old $w_i=0$, the nine
root--new edge coefficients determine the three new vectors $w_b$,
since the three root vectors $L_i$ are independent.  Hence the nine
new motions are independent modulo the product space.

It remains to interpolate this space by actual points.  After removing
its three new diagonal columns, its coefficient description has two
columns spanning $\langle v_i,v_j\rangle$ on every edge between an old
nonroot and a new vertex and on the three new--new edges; on each of
the nine root--new edges it has three columns spanning $k^3$.  The
latter columns may also have coefficients on the former edges.  The
number of two-column edges is $3h-6$, so the off-diagonal dimension is
\[
 2(3h-6)+27=6h+15=3(2h+5).
\]

Here is an explicit interpolation argument.  Assign positive integer
exponents $a_i$ close to $1,\ldots,h-3$ on old nonroots, exponents
close to $M,M+1,M+2$ on new vertices, and much larger exponents on the
three roots.  Choose $M$ large, then perturb and clear denominators
so that all relevant edge sums are distinct.  The old-nonroot--new
edges occur in matching groups of sizes
$1,2,3,\ldots,3,2,1$; no three consecutive edges form a star.  Their
number is divisible by three, and the final three new--new edges
form a triangle.  By a general choice of the three new conic points,
the normals of every such triple of chords are independent.  This
choice is possible even with the old conic points fixed: two chords
with a common endpoint meet there, and a third chord not containing
that endpoint does not pass through it; for three disjoint chords,
varying a new endpoint prevents persistent concurrence.

Evaluate at $2h+5$ points
\[
 y(\tau_j)=(\tau_j^{a_1},\ldots,\tau_j^{a_{h+3}}),
 \qquad \tau_j=T^{b_j},\qquad 0<b_1<\cdots<b_{2h+5}.
\]
In the determinant, the maximal power of $T$ groups each three
successive two-column edges into two successive point blocks.
The corresponding six-by-six leading coefficient is, up to a
nonzero factor, the determinant of the three chord normals.
For completeness, if the three coefficient planes have bases
$(A_1,A_2)$, $(B_1,B_2)$, $(C_1,C_2)$, this coefficient is
\[
 \det(A_1,A_2,B_1)\det(B_2,C_1,C_2)
 -\det(A_1,A_2,B_2)\det(B_1,C_1,C_2),
\]
which equals the determinant of their plane normals up to sign.
Each root--new edge subsequently contributes an invertible
three-by-three block.  All extra coefficients of those columns have
strictly smaller edge degrees and therefore cannot cancel this
leading term.  Adding the three new coordinate points handles the
three diagonal columns.  We have obtained an invertible
$(6h+18)$-square evaluation block at $2h+8$ new base points.

Finally add, in each layer, the point
$\pi_B+\zeta_je_{z_j}$.  On the twelve columns $H_bz_j$, the four new
point blocks are
\[
 \zeta_j[\,v_b:b\in B\,],\qquad 1\le j\le4,
\]
which are invertible.  These twelve columns vanish on all new base
points.  Ordering old rows first, then new base rows, then new layer
rows, the enlarged map is therefore
\[
 \begin{pmatrix}D_h&0&0\\ *&A&0\\ *&*&B\end{pmatrix},
 \qquad \det A\ne0,\quad\det B\ne0.
\]
This proves the enlargement.  Notice that
$f_{h+3}-f_h=2h+5$, exactly the number of new free base points used.
Scalar quadratic independence can be retained simultaneously:
quadrics involving the three new base variables have dimension
$3h+6\ge2h+8$ and impose independent conditions at the new coordinate
and general base points.  The four new layer rows are distinguished
by the monomials $y_bz_j$ for a fixed new index $b$.  Their nonempty
scalar-rank opens intersect the nonempty determinant opens just
constructed.

\subsubsection*{An explicit initial configuration}
\label{subsec:c3-initial}
We give the finite initial calculation in full reproducible form.
Take $h=8$, $t_i=i$, and work first over $\mathbf Q(\omega)$.
The eight coordinate trace points and sixteen base cone points are
as above.  The twenty-four free base trace points, in the coordinates
$(y_1,\ldots,y_8)$, are the rows of Table~\ref{tab:c3-base}; their four
$z$ coordinates are zero.  The four layer records are in
Table~\ref{tab:c3-layers}: the trace points of layer $j$ are
\[
 e_{z_j},\quad
 (e_1+e_2+e_3)+\zeta_{j,1}e_{z_j},\quad
 (e_4+e_5+e_6)+\zeta_{j,2}e_{z_j},\quad
 (e_7+e_8)+\zeta_{j,3}e_{z_j},
\]
and the first three are lifted using the three listed normal vectors.
The four final trace points are the rows of
Table~\ref{tab:c3-final}.  All entries in these tables are rational
integers, not algebraic parameters.

\begin{table}[htbp]\centering\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{rrrrrrrr}
138&583&868&822&783&65&262&121\\
508&780&461&484&668&389&808&215\\
97&500&30&915&856&400&444&623\\
781&786&3&713&457&273&739&822\\
235&606&968&105&924&326&32&23\\
27&666&555&10&962&903&391&703\\
222&993&433&744&30&541&228&783\\
449&962&508&567&239&354&237&694\\
225&780&471&976&297&949&23&427\\
858&939&570&945&658&103&191&645\\
742&881&304&124&761&341&918&739\\
997&729&513&959&991&433&520&850\\
933&687&195&311&291&602&997&904\\
512&867&964&518&403&604&874&36\\
492&249&762&817&414&425&681&178\\
376&562&904&720&795&691&756&384\\
89&450&680&521&111&798&168&534\\
861&403&380&502&751&31&481&45\\
316&721&869&1007&630&608&593&404\\
663&175&173&515&233&1006&13&790\\
205&553&943&881&562&238&415&527\\
353&976&868&592&362&471&932&276\\
676&562&624&981&747&6&393&803\\
878&841&978&908&961&759&525&829
\end{tabular}
\caption{Free base points for the three-normal initial configuration.}
\label{tab:c3-base}
\end{table}

\begin{table}[htbp]\centering\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{c|c|ccc}
$j$&$(\zeta_{j,1},\zeta_{j,2},\zeta_{j,3})$&
$a_{j,0}$&$a_{j,1}$&$a_{j,2}$\\\hline
1&$(133,532,797)$&$(575,211,437)$&$(973,58,493)$&$(891,374,584)$\\
2&$(568,205,964)$&$(517,424,497)$&$(833,366,425)$&$(355,2,552)$\\
3&$(554,639,806)$&$(628,340,470)$&$(615,29,824)$&$(236,651,182)$\\
4&$(564,599,186)$&$(882,94,818)$&$(565,817,872)$&$(837,954,262)$
\end{tabular}
\caption{Layer parameters and normal-coordinate vectors.}
\label{tab:c3-layers}
\end{table}

\begin{table}[htbp]\centering\small
\setlength{\tabcolsep}{3pt}
\begin{tabular}{rrrrrrrrrrrr}
34&862&967&690&73&86&889&18&464&15&773&774\\
288&256&276&113&817&640&190&353&298&72&172&164\\
262&541&975&173&673&280&664&729&302&466&720&330\\
509&486&117&25&320&396&352&432&816&193&265&112
\end{tabular}
\caption{The final four trace points, with the $y$ coordinates first.}
\label{tab:c3-final}
\end{table}

Here are precise instructions and intermediate values for the
calculation, so that it requires no further data.  Reduce modulo
$1009$ and set $\omega=374$; then
$374^2+374+1\equiv0\pmod{1009}$.  Order variables as
$(s_1,s_2,s_3,y_1,\ldots,y_8,z_1,\ldots,z_4)$ and monomials
$x_{i_1}\cdots x_{i_d}$ lexicographically by their nondecreasing
index tuples.  Form the evaluation matrices $E_K$ on the 42 quadratic
monomials in $P_2$ at the 28 cone points, and $E_V$ on the 78 trace
quadratic monomials at the 52 trace points.  Order points as in their
construction: base coordinates or base cone pairs first, followed by
free base points where applicable, successive layers, and final
points.  Row reduction, scanning columns from left to right and using
the first available nonzero pivot, gives
\[
 \operatorname{rank}E_K=28,\qquad
 \operatorname{rank}E_V=52.
\]
The determinants on their pivot columns are respectively $948$ and
$604$ modulo $1009$.  The four layer-normal determinants are
$832,164,963,296$.

For each evaluation matrix, use its reduced row-echelon form to make
a kernel basis: in increasing order of free columns, set that free
coordinate to one and all other free coordinates to zero.  Multiply
each of the 14 resulting $K$ basis vectors by each of the 15 variables,
then multiply each of the 26 resulting $V$ basis vectors by
$s_1,s_2,s_3$.  The resulting cubic coefficient matrix has 680 rows
and 288 columns.  Forward elimination with the same first-pivot rule
has 288 pivots.  If the original row labels are carried along during
elimination, the square submatrix on those pivot rows, in pivot order,
has determinant
\[
 280\pmod{1009}.
\]
These assertions involve only the displayed integer tables, the
sixteen explicit algebraic cone points, and addition, multiplication
and inversion in $\mathbf F_{1009}$.

Every column of this cubic matrix lies in $P_3\cap I_{Z_1}(3)$.
Evaluation of $P_3$ on $Z_1$ has rank 28, since multiplication by
$s_1$, which is nonzero at all cone points, carries the full-rank
quadratic evaluation to cubic evaluation.  Therefore
\[
 \dim(P_3\cap I_{Z_1}(3))=(680-364)-28=288,
\]
and the initial middle map is an isomorphism.  The two nonzero
evaluation minors ensure that the kernel bases are defined over a
localization of $\mathbf Z[\omega]$ at the prime
$(1009,\omega-374)$.  Thus the nonzero reduced cubic determinant
proves that its characteristic-zero determinant is nonzero as well.
This uses the displayed algebraic formulas for the first sixteen
cone points; the other coordinates are the literal integers in the
tables.  The initial $D_8$ is consequently invertible.

The enlargement now proves invertibility for every $h=8+3a$.
Finally, the marked cone family used here is irreducible and dominates
the space of ordered trace configurations: normal coordinates vary
independently over each designated trace label.  Middle-map
invertibility and scalar independence are nonempty open conditions
by the construction.  If quadratic ideal generation holds in
$m=n-3$ variables, trace bijectivity is another nonempty open in this
same family.  Intersecting these opens and applying the exact cone
splitting proves the quadratic step for every $n\equiv0\pmod3$,
$n\ge15$.

\begin{proof}[Proof of Theorem~\ref{thm:quadratic-column}]
The theorem of Fr\"oberg and L\"ofwall \cite[Theorem~6.2]{FL02}
supplies the cases \(n\leq12\).
Assume \(n>12\), and induct on \(n\).
If \(n\equiv2\pmod3\), the two-normal construction gives the integral
endpoint from the theorem in \(n-2\) variables.
If \(n\equiv1\pmod3\), that construction and its one-point enlargement
give the surjective and injective endpoints, respectively, again from
\(n-2\) variables.
If \(n\equiv0\pmod3\), then \(n\geq15\), and the three-normal
construction gives the integral endpoint from \(n-3\) variables.
Every predecessor is smaller than \(n\). The reduction to critical
point counts therefore proves the assertion for every
\(n\leq p<N_2(n)\).
\end{proof}