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

We prove the degree-three case by adding two variables at a time.
The construction used below retains all three normal degrees of a
cubic.  This is essential: the relations in the two upper normal
degrees contribute to the final interpolation problem.

\begin{theorem}\label{thm:cubic}
For every \(n\geq2\), general points in \(\mathbf P^{n-1}\) satisfy
the ideal generation conjecture in initial degree three.
\end{theorem}

\subsection{Reduction to a cone construction}
The cases \(n\leq11\) follow from
Gesmundo--Kayser--Telen~\cite[Theorem~5.1]{GKT25}.
Their table uses projective dimension: the two largest cases needed
here are \(165\) points in \(\mathbf P^9\) and \(214,215\) points in
\(\mathbf P^{10}\), within the respective bounds \(207\) and \(267\).
The elementary case of \(\mathbf P^1\) can also be treated directly.

Observe that
\[
 E(m,3)=\frac{m(m+1)(m+2)}{24}
\]
is integral exactly when \(m\not\equiv1,3,5\pmod8\).
Indeed, divisibility by \(3\) is automatic.  If \(m\) is even, the
two even factors have product divisible by \(8\); if \(m\) is odd,
divisibility by \(8\) is equivalent to \(8\mid m+1\).
Consequently Proposition~\ref{prop:integral-transfer} applies from \(n-1\)
to \(n\) unless \(n\equiv2,4,6\pmod8\).  Its residual hypothesis is
quadratic injectivity at
\[
 \left\lceil\frac{n(n+1)}8\right\rceil.
\]
This is supplied by Theorem~\ref{thm:quadratic-column}, since for \(n\geq10\)
\[
 \left\lceil\frac{n(n+1)}8\right\rceil
 \leq\left\lfloor\frac{n(n+1)}6\right\rfloor.
\]
It remains to provide a two-variable transfer for every even
\(n\geq12\).

Write \(n=m+2\), put \(T=k[y_0,\ldots,y_{m-1}]\), and set
\[
 S=T[s,t],\qquad U=\langle s,t\rangle,\qquad P=(s,t).
\]
Choose \(p_2=3N_3(m)/4\) points \(Z_2\) in
\(\Lambda=\{s=t=0\}\).  A point attached to \(y\in Z_2\) has a
representative
\[
 z=(\sigma,\mu y),\qquad \sigma\in k^2\setminus\{0\},\quad
 \mu\neq0.
\]
Let \(Z_1\) consist of \(p_1=3(m+2)^2/4\) such points, and set
\[
 K=P_3\cap I_{Z_1}(3),\quad V=I_{Z_2,T}(3),\quad
 M=P_4\cap I_{Z_1}(4).
\]
Every member of \(V\), viewed as a polynomial in \(S\), vanishes on
the attached points.  Separating the pure \(y\)-part of a polynomial
therefore gives
\[
 I_{Z_1\cup Z_2}(3)=V\oplus K.
\]
Multiplication splits into the trace multiplication and the map
\begin{equation}\label{eq:cubic-middle}
 K\otimes S_1\ \oplus\ V\otimes U\longrightarrow M .
\end{equation}
When the scalar evaluations have full rank,
\[
 \dim K=\frac{(m+2)^2}{4},\qquad \dim V=\frac{N_3(m)}4.
\]
Thus \eqref{eq:cubic-middle} is square.  Its size, in terms of
\(n=m+2\), is \((4n^3-3n^2+2n)/12\).

The attachment pattern specifies an irreducible parameter space:
the ordered points of \(Z_2\) are free, and their normal directions
and heights are free.  This space dominates the ordered trace
configuration space.  Hence one full-rank instance of
\eqref{eq:cubic-middle} gives a nonempty open set that meets the
pullback of the inductive trace open set.  We may therefore use
specially supported trace points to construct the middle map,
without requiring that special trace to satisfy ideal generation.

\subsection{Eliminating the upper normal degrees}
Write a residual cubic as
\[
 F=C+\sum_c B_cy_c+A,\qquad
 C\in\operatorname{Sym}^3U,\quad
 B_c\in\operatorname{Sym}^2U,\quad A\in U\otimes T_2.
\]
Suppose the projection of \(K\) onto its \(C,B\) coefficients is
onto.  Let \(K_{00}\subset U\otimes T_2\) be its kernel, and choose
linear lifts
\[
 K_R=R+a(R),\qquad K_{Q,c}=Qy_c+a(Q,c)
\]
for \(R\in\operatorname{Sym}^3U\) and
\(Q\in\operatorname{Sym}^2U\).
Define
\[
 \Omega=\{(k_s,k_t)\in K^2:sC(k_s)+tC(k_t)=0\},
 \qquad
 \mathcal X=\Omega\oplus
  \bigl(\operatorname{Sym}^2U\otimes\textstyle\bigwedge^2T_1\bigr)
  \oplus(K_{00}\otimes T_1).
\]

\begin{lemma}\label{lem:cubic-elimination}
There is a linear map \(\Phi:\mathcal X\longrightarrow U\otimes T_3\)
with the following properties.  Its image is
\((K S_1)\cap(U\otimes T_3)\).  At every attached point
\((\sigma,\mu y)\), it satisfies \(\sigma\cdot\Phi(y)=0\).
If
\[
 \mathcal A(Z_2)=
 \{(v_y)\in(k^2)^{Z_2}:
           \sigma\cdot v_y=0\text{ for every point attached to }y\},
\]
then the kernel of \eqref{eq:cubic-middle} is naturally the kernel of
\begin{equation}\label{eq:cubic-D}
 D:\mathcal X\longrightarrow\mathcal A(Z_2),
 \qquad D=\operatorname{ev}_{Z_2}\Phi.
\end{equation}
With at most two independent normal directions above each support,
\(\dim\mathcal A(Z_2)=2p_2-p_1\).
\end{lemma}

\begin{proof}
Consider a tensor
\[
 k_s\otimes s+k_t\otimes t+\sum_b k_b\otimes y_b
       +V_s\otimes s+V_t\otimes t
\]
in the source of \eqref{eq:cubic-middle}.
Its normal-degree-four equation says
\((k_s,k_t)\in\Omega\).  The normal-degree-three equation uniquely
sets
\[
 C(k_b)=-sB_b(k_s)-tB_b(k_t)=:R_b.
\]
Write \(B(k_b)=\sum_cQ_{bc}y_c\).  The normal-degree-two equation
determines the symmetric part of the array \(Q_{bc}\) through
\[
 \sum_{b,c}Q_{bc}y_by_c=-sA(k_s)-tA(k_t).
\]
Its antisymmetric part is arbitrary and is precisely an element of
\(\operatorname{Sym}^2U\otimes\bigwedge^2T_1\).
Thus
\[
 k_b=K_{R_b}+\sum_cK_{Q_{bc},c}+\kappa_b,
       \qquad \kappa_b\in K_{00}.
\]
These equations define \(\Phi=\sum_bA(k_b)y_b\).
The remaining equation is
\(\Phi+sV_s+tV_t=0\).  It has a solution with \(V_s,V_t\in V\)
exactly when \(\Phi\) vanishes on \(Z_2\); then that solution is
unique.  Every preceding elimination was reversible.
Moreover, \(\Phi\in KS_1\) vanishes at each attached point; since it
has normal degree one, this is equivalent to
\(\sigma\cdot\Phi(y)=0\).
Finally each support contributes two scalar coordinates minus the
number of its independent normal directions.
\end{proof}

\subsection{The supported configurations and their lifts}
Let \(H\) be a graph on the \(m\) trace coordinates.  Include in
\(Z_2\) the vertices \(e_i\), the points
\(\pi_{ij}=e_i+\lambda_{ij}e_j\) for \(i<j\), and
\[
 \frac{m(m+1)(m-2)}8
\]
additional free points.  Attach two points above every vertex;
above \(\pi_{ij}\), attach one point if \(ij\in H\) and two otherwise.
Directions at a double support are independent.  The required
number of attached points is equivalent to
\begin{equation}\label{eq:cubic-edge-count}
 |H|=\frac{m^2-8m-12}{4}.
\end{equation}
If the direction at a single support is \(\sigma_{ij}\), write
\(\ell_{ij}=\sigma_{ij,t}s-\sigma_{ij,s}t\).

\begin{lemma}\label{lem:cubic-supported}
For these configurations,
\[
 K_{00}=\bigoplus_{ij\in H}k\,\ell_{ij}y_iy_j,
 \qquad \dim K=4+3m+|H|=\frac{(m+2)^2}{4},
\]
and the projection onto the \(C,B\) coefficients is onto.
The lifts can be chosen with \(a(Q,c)=y_cL_c(Q)\), where
\(L_c(Q)\) is a vector-valued linear form.
\end{lemma}

\begin{proof}
For a member of \(U\otimes T_2\), the two vertex equations kill the
coefficient of each \(y_i^2\).  A double-edge support then kills the
coefficient of \(y_iy_j\), whereas a single-edge support restricts it
to \(k\ell_{ij}\).  This proves the formula for \(K_{00}\).
For arbitrary \(C,B\), first solve the two equations for each
coefficient of \(y_i^2\); then solve the one or two equations for
each coefficient of \(y_iy_j\).  The same independent normal
directions make each solve possible.

Here are the explicit equations used below.  Write
\(L_c(Q)=\sum_j\alpha_{cj}(Q)y_j\).
For a cone point \((\sigma,\mu e_c)\), and for a point above the
edge joining \(c\) and \(j\), respectively, they are
\begin{equation}\label{eq:cubic-local-lifts}
\begin{aligned}
 \sigma\cdot\alpha_{cc}(Q)&=-Q(\sigma)/\mu,\\
 \sigma\cdot\alpha_{cj}(Q)&=
 \begin{cases}
 -\bigl(Q(\sigma)/\mu+\sigma\cdot\alpha_{cc}(Q)\bigr)/\lambda_{cj},
       &c<j,\\
 -Q(\sigma)/\mu-\lambda_{jc}\sigma\cdot\alpha_{cc}(Q),
       &j<c .
 \end{cases}
\end{aligned}
\end{equation}
At a single support choose any fixed linear solution of its one
equation.  Likewise the coefficients \(\beta_{ij}(R)\) of \(a(R)\)
are obtained from
\[
 \sigma\cdot\beta_{ii}(R)=-R(\sigma)/\mu^2,\qquad
 \sigma\cdot
  \bigl(\beta_{ii}+\lambda_{ij}\beta_{ij}
                   +\lambda_{ij}^2\beta_{jj}\bigr)
     =-R(\sigma)/\mu^2.
\]
These formulas follow by evaluating the cubic at its cone point.
In particular \(L_c(Q)\) uses only the data at \(c\) and its
incident edges, not the vertex data at their other endpoints.
\end{proof}

We use the following graph for every even \(m\geq10\).
Start with
\[
 A_0=\{0,1,2,3,4\},\qquad B_0=\{5,6,7,8,9\},\qquad
 H_0=\{17,28\}.
\]
Successively add \(b_j=8+2j\) to \(B\) and \(a_j=9+2j\) to \(A\),
with
\begin{equation}\label{eq:cubic-periodic-graph}
 N_H(b_j)=A_{\mathrm{old}}\setminus\{0\},\qquad
 N_H(a_j)=B_{\mathrm{old}}\setminus\{5,6\}.
\end{equation}
There is no edge \(a_jb_j\) in \(H\).  The graph remains bipartite,
and each step adds \(m-3\) edges, which preserves
\eqref{eq:cubic-edge-count}.  Thus triangles never occur.
For general directions, multiplication
\(K_{00}\otimes T_1\to U\otimes T_3\) is injective: a cubic monomial
is supported by at most two graph edges, whose normal linear forms
are independent when both occur.

\subsection{Adding a pair of variables}
At a step from \(m\) to \(m+2\), denote the new variables by
\(w_1=b_{j+1}\) and \(w_2=a_{j+1}\), and retain every old point.
Put \(N_1=A_{\mathrm{old}}\setminus\{0\}\) and
\(N_2=B_{\mathrm{old}}\setminus\{5,6\}\).
Restriction \(w_1=w_2=0\) maps the new space \(\mathcal X\) onto
the old one.  This follows either from the elimination equations,
using compatible local lifts, or directly by lifting its
\(\Omega\), antisymmetric, and \(K_{00}\) coordinates separately.
Let \(\mathcal X_{\mathrm{new}}\) be its kernel.
All products from this kernel vanish on the old trace space, so in
compatible bases
\begin{equation}\label{eq:cubic-triangular}
 D_{m+2}=
 \begin{pmatrix}D_m&0\\ *&D_{\mathrm{new}}\end{pmatrix},
 \qquad
 \dim\mathcal X_{\mathrm{new}}
       =\frac{3m^2}{2}+3m-3.
\end{equation}
The latter count follows by taking differences in
\[
 \dim\mathcal X=(2\dim K-5)+3\binom m2+m|H|.
\]

We describe explicit coordinates on \(\mathcal X_{\mathrm{new}}\).
There are \(2m-3\) coordinates whose products have no term linear
in \(w_1,w_2\): they are
\[
 \ell_{iw_b}y_iw_b\otimes w_{b'},\quad
       i\in N_b,\ b,b'\in\{1,2\},
 \qquad Q\otimes(w_1\wedge w_2),\quad
       Q\in\operatorname{Sym}^2U .
\]
The remaining coordinates number \(3m^2/2+m\).  Their linear
parts are pure in \(w_1\) or \(w_2\).
For \(w=w_b\), after invertible column changes they are the following
vector-valued quadrics on the old trace space:
\begin{equation}\label{eq:cubic-linear-families}
\begin{array}{ll}
 \ell_{ij}y_iy_j,&ij\in H,\\
 \ell_{iw}y_iy_c,&i\in N_b,\quad 0\leq c<m,\\
 y_i\bigl(L_w(Q)-L_i(Q)\bigr)|_{w=0},
       &0\leq i<m,\quad Q\in\{s^2,st,t^2\},\\
 y_iL_w(u\ell_{iw})|_{w=0},
       &i\in N_b,\quad u\in\{s,t\},\\
 \displaystyle a(uQ)|_{w=0}+
       \sum_jy_jL_w\bigl(u\alpha_{wj}(Q)\bigr)|_{w=0},
       &u\in\{s,t\},\quad Q\in\{s^2,st,t^2\}.
\end{array}
\end{equation}
Here \(w=0\) means both new coordinates are set to zero.
To check the list, substitute the local lifts into
Lemma~\ref{lem:cubic-elimination}.  The first two families come
from \(K_{00}\otimes T_1\), and the third from
\(Q\otimes(y_i\wedge w)\).
For the fourth, the product associated to
\((\ell_{iw}y_iw,0)\in\Omega\) has linear part
\(-y_i(L_i(u\ell_{iw})+L_w(u\ell_{iw}))/2\).
Adding half of the corresponding third-family column gives the
fourth family, up to sign.  The same substitution for
\((K_{Q,w},0)\) gives the last family after adding half of the
appropriate third-family columns.
This also accounts for every coordinate: their total, together
with the \(2m-3\) higher coordinates, equals
\(\dim\mathcal X_{\mathrm{new}}\).

Choose
\[
 \frac{3m^2}{4}+\frac m2
\]
of the new free trace points in the form
\(p+\varepsilon(c_1w_1+c_2w_2)\), with \(p\) in the old space.
Keep another \(m/2\) new free points away from the old space.
There are also \(m-3\) new single-edge supports, each contributing
one allowed evaluation coordinate.
Let \(B_{11}\) evaluate \eqref{eq:cubic-linear-families} at the
first group, with weights \(c_1,c_2\), and let \(B_{22}\) evaluate
the higher coordinates at the single-edge supports and the second
group.  Dividing the first \(3m^2/2+m\) evaluation rows by
\(\varepsilon\) gives at \(\varepsilon=0\)
\[
 \begin{pmatrix}B_{11}&0\\ *&B_{22}\end{pmatrix}.
\]
Thus
\begin{equation}\label{eq:cubic-leading}
 \det D_{\mathrm{new}}(\varepsilon)
  =\varepsilon^{3m^2/2+m}\det B_{11}\det B_{22}
       +O\bigl(\varepsilon^{3m^2/2+m+1}\bigr).
\end{equation}
It remains to make these two fixed-type blocks nonsingular.

\subsection{The two evaluation blocks}
First consider \(B_{22}\).
At the support of \(iw_b\), the same-index column
\(\ell_{iw_b}y_iw_b\otimes w_b\) has a nonzero scalar evaluation,
whereas all the other higher columns vanish there.
After these \(m-3\) pivots, pair the remaining \(m-3\) cross
columns, leaving one, with index \(i_0=1\), unpaired.
For each pair use a point supported on its two old vertices and
\(w_1,w_2\).  Its two rows give the two independent normal lines.
Two final points supported on \(\{i_0,w_1,w_2\}\) see the leftover
cross column and the three columns
\[
 w_1w_2\bigl(L_{w_2}(Q)-L_{w_1}(Q)\bigr),
       \qquad Q=s^2,st,t^2.
\]
The resulting last block is \(4\times4\).
Proposition~\ref{prop:cubic-certificates} provides parameters for
which it is nonsingular.  Its coefficients depend only on the
two new vertices and their incident edges, so this is the same
nonzero polynomial at every step.

For \(B_{11}\), call the first two families in
\eqref{eq:cubic-linear-families} the local columns.
A face is a vertex or an edge of the old coordinate simplex.
On a face \(F\), the number of local columns in direction \(b\) is
\[
 t_b(F)=\mathbf1_{\{F\in H\}}+|F\cap N_b|\leq2.
\]
Call \(F\) full or half according as this number is two or one.
The inequality follows from bipartiteness: three local columns
on an edge would give a triangle after adding \(w_b\).
Call the third and fourth families the star columns, and the last
family the six remaining columns.

Take the following rows on the ten-coordinate initial pool:
\begin{equation}\label{eq:cubic-root-rows}
\begin{aligned}
 &\text{one point on every pool face with }t_b(F)>0,
       \text{ in pure direction }w_b;\\
 O_1={}&\{\{5\},\{6\},\{7\},\{8\},\{9\},\{0\},
                          \{7,8\},\{7,9\},\{8,9\}\},\\
 O_2={}&\{\{0\},\{1\},\{2\},\{3\},\{4\},\{5\},\{6\},
                          \{1,2\},\{1,3\},\{2,3\}\}.
\end{aligned}
\end{equation}
Use one additional point on each listed optional face in its
specified pure direction.  The two initial-pool blocks have sizes
\(86\) and \(74\), respectively.

For an old pair \((a_j,b_j)\), assign a face involving two old
pairs to the earlier pair.  Use one pure-direction point on every
full face assigned to \(j\), and thirteen points with general
direction weights on
\[
 S_j=\{a_j,b_j,0,5,6\}.
\]
In direction \(w_1\), the half faces involving the pair are
\[
 \{a_j\},\ \{a_j,b_j\},\ \{a_j,0\},\ \{a_j,5\},\ \{a_j,6\};
\]
in direction \(w_2\), they are
\[
 \{b_j\},\ \{a_j,b_j\},\ \{b_j,0\},\ \{b_j,5\},\ \{b_j,6\}.
\]
There are no full faces of the pair inside \(S_j\).
There are sixteen star columns for the pair: five from \(a_j\)
and three from \(b_j\) in the first direction, and three and five
in the second.  Consequently the thirteen points give a square
\(26\times26\) block consisting of these sixteen star columns
and the ten half-face columns.

These rows and columns factor the determinant of \(B_{11}\).
Indeed, every pair column has a factor \(y_{a_j}\) or \(y_{b_j}\).
Pool rows annihilate all pair columns, and the rows assigned to
a later pair annihilate earlier pair columns.  Within a pair
block, each full-face row sees precisely its two local columns;
the thirteen points on \(S_j\) annihilate these full-face columns.
Therefore, up to nonzero coordinate factors and a sign,
\begin{equation}\label{eq:cubic-factorization}
 \det B_{11}
  =\det R\prod_j
      \left(\det E_j\prod_{\substack{F\text{ full}\\
                                  F\text{ assigned to }j}}
                         \det(\ell_F,\ell'_F)\right).
\end{equation}
Here \(R\) is the \(160\times160\) initial-pool block and every
\(E_j\) is the same \(26\times26\) polynomial construction, with
its coordinate names changed.  Their nonsingularity for some
parameters is established in
Proposition~\ref{prop:cubic-certificates}.

\subsection{Compatible parameters and conclusion}
We must ensure that the successive nonvanishing conditions can be
imposed on one configuration.  The locality in
\eqref{eq:cubic-local-lifts} gives this directly.
The determinant of \(R\) uses only the old pool data and the new
vertices and their edges to the pool.
The determinant of \(E_j\) uses the vertex data at \(a_j,b_j\),
their edges to \(0,5,6\), and the new data on these five coordinates.
In particular it does not use the vertex-cone data at \(0,5,6\).
The \(4\times4\) determinant for \(B_{22}\) likewise uses only
new vertex and incident-edge data.

A nonzero polynomial \(f(x,y)\) remains a nonzero polynomial in
\(y\) for \(x\) in a nonempty open set: exclude the common zero
set of its coefficients.  At the initial level \(m=10\), impose
both the nonvanishing of the middle map and this coefficient-open
condition for \(R\); both are nonempty by
Proposition~\ref{prop:cubic-certificates}.
For every subsequent pair, retain the coefficient-open condition
for its future \(26\times26\) block.
When adding the next pair, each existing block is a nonzero
polynomial in the shared new parameters and its own evaluation
points.  Their finite product is nonzero.
The future condition for the new pair remains a nonempty open
condition with the pool fixed, because its old-side parameters
exclude the vertex data at the three fixed roots.
Intersect these conditions with the nonzero normal determinants
and the nonzero \(4\times4\) determinant.
The parameter space is irreducible, so the intersection is
nonempty.

Equations \eqref{eq:cubic-leading} and
\eqref{eq:cubic-triangular} now give an invertible map \(D\) at
every even level \(m\geq10\).
This also proves scalar independence of the trace if it has not
already been imposed.  Indeed, Lemma~\ref{lem:cubic-elimination}
gives injectivity of \eqref{eq:cubic-middle}; if
\(\dim V>N_3(m)/4\), its source would be larger than its target,
contrary to that injectivity.  Scalar independence of the cone
points in degree three follows from
Lemma~\ref{lem:cubic-supported}, and independence in degree four
follows by multiplication by a linear form nonzero at the points.
Thus the middle map is an isomorphism with the asserted dimensions.

Finally restore a general trace in the irreducible attachment
family and apply induction in \(n\).
The two-variable construction handles every even exceptional
level \(n\geq12\); the integral hyperplane transfer handles every
other level.  Together with the published cases \(n\leq11\) and
the quadratic residual theorem, this proves the two cubic critical
counts for all \(n\).  Point-count monotonicity proves
Theorem~\ref{thm:cubic}.