\section{Hyperplane specialization}
\label{sec:hyperplane}

We write $\operatorname{Inj}(n,D,r)$, respectively
$\operatorname{Surj}(n,D,r)$, when multiplication in degrees $D,D+1$
is injective, respectively surjective, for general points whose
degree-$D$ ideal has dimension $r$. In this notation a point-count
cell is
\[
             1\le r\le N_D(n-1).
\]
All point configurations below are ordered and have distinct points.
Conditions are called independent when evaluation onto their stated
condition space is surjective. Their parameter spaces, as well as
the open subsets obtained by
requiring points to avoid a fixed hyperplane, are irreducible.

\begin{lemma}[Ordinary hyperplane specialization]
\label{lem:ordinary-hyperplane}
Let $n\ge3$, $D\ge2$, and suppose the counts below belong to their
point-count cells.
\begin{enumerate}
\item If $\operatorname{Inj}(n-1,D,r_2)$ and
$\operatorname{Inj}(n,D-1,r_1)$ hold, and
\[
                  (n-1)r_1+r_2\le N_D(n-1),
\]
then $\operatorname{Inj}(n,D,r_1+r_2)$ holds. The case $r_1=0$
is allowed, with no residual injectivity assumption.
\item If $\operatorname{Surj}(n-1,D,r_2)$ and
$\operatorname{Inj}(n,D-1,r_1)$ hold, where $r_1\ge1$, and
\[
                  (n-1)r_1+r_2\ge N_D(n-1),
\]
then $\operatorname{Surj}(n,D,r_1+r_2)$ holds.
\end{enumerate}
\end{lemma}

\begin{proof}
Put $H=(h=0)$ and $R=S/(h)$. Choose points $Z_1$ off $H$,
independent in degree $D-1$, and points $Z_2$ on $H$, independent
in degree $D$, so that
\[
 W=I_{Z_1}(D-1),\qquad V_H=I_{Z_2,H}(D),\qquad
 \dim W=r_1,\quad \dim V_H=r_2.
\]
Let $\rho:S\to R$ be restriction, $V=I_{Z_1\cup Z_2}(D)$, and
$U=\rho(WS_1)$. A lift of a polynomial on $H$ can be corrected at
$Z_1$ by a multiple of $h$, since degree-$(D-1)$ evaluation on
$Z_1$ is onto. Consequently
\[
 0\longrightarrow hW\longrightarrow V
      \xrightarrow{\rho}V_H\longrightarrow0,\qquad
 I_{Z_1}(D)\xrightarrow{\rho}R_D\longrightarrow0.
\]
The second map has kernel $hW$, and
$WS_1\cap hS_{D-1}=hW$: if $hG$ vanishes on $Z_1$, then
$G$ vanishes there because $h$ is nonzero at those points.
Thus residual injectivity gives $\dim U=(n-1)r_1$.

Choose $Z_1$ first. Evaluation of the fixed space $U$ at general
points of $H$ has maximal rank. Indeed, unless the current kernel
is zero, a further point can be chosen where one of its nonzero
polynomials does not vanish. This constructs a maximal-rank
evaluation, and a nonzero minor makes the condition open.
It follows that, simultaneously with the asserted trace
multiplication property, we may arrange
$U\cap V_H=0$ in the first case and $U+V_H=R_D$ in the second.

Choose a basis $w_j$ of $W$ and lifts $\widetilde g_i$ of a basis
$g_i$ of $V_H$. In the injective case, a relation
\[
          \sum_j hw_j a_j+\sum_i\widetilde g_i b_i=0,
          \qquad a_j,b_i\in S_1,
\]
restricts to a trace relation. Trace injectivity gives
$b_i=c_i h$, with $c_i\in k$. Divide by $h$ and restrict once
more. The resulting combination $\sum_i c_i g_i$ belongs to
$U\cap V_H$, so every $c_i$ vanishes. Residual injectivity now
gives $a_j=0$.

In the surjective case, let $\widetilde G$ denote the span of the
lifts. Since $WS_1+\widetilde G$ maps onto $R_D$ and contains
$hW$, it equals $I_{Z_1}(D)$. For
$F\in I_{Z_1\cup Z_2}(D+1)$, trace surjectivity allows us to
subtract an element of $\widetilde G S_1$ with the same trace.
The remainder is $hG$ with
$G\in I_{Z_1}(D)=WS_1+\widetilde G$, and therefore belongs to
$VS_1$. Finally, the exact sequence for $V$ gives the required
scalar postulation; openness passes the conclusion to general
points.
\end{proof}

\begin{corollary}[Integral trace]
\label{prop:integral-transfer}
Suppose $E(n-1,D)$ is an integer, the ideal generation statement
holds in degree $D$ in $n-1$ variables, and
\[
 \operatorname{Inj}\left(n,D-1,
           \left\lceil\frac{N_{D-1}(n)}{D+1}\right\rceil\right)
\]
holds. Then both critical degree-$D$ counts in $n$ variables
have maximal rank.
\end{corollary}

\begin{proof}
Set $r_2=E(n-1,D)$ and $x=N_{D-1}(n)/(D+1)$. The identities
\[
 E(n,D)=E(n-1,D)+x,\qquad
 (n-1)x+E(n-1,D)=N_D(n-1)
\]
show that the two cases of Lemma~\ref{lem:ordinary-hyperplane},
with $r_1=\lfloor x\rfloor$ and $r_1=\lceil x\rceil$, give
the injective and surjective critical counts.
\end{proof}

\begin{lemma}[The projective plane]
\label{lem:plane-base}
For every $D\ge2$, the ideal generation statement holds for
general points of $\mathbf P^2$.
\end{lemma}
\begin{proof}
On $\mathbf P^1$, the ideal of $D$ distinct points is generated
by their product of linear equations. Its multiplication map
from degree $D$ to degree $D+1$ is an isomorphism.
For $\mathbf P^2$ we prove the critical assertions with
\[
 r_{\rm inj}=\left\lfloor\frac{D+2}{2}\right\rfloor,\qquad
 r_{\rm surj}=\left\lceil\frac{D+2}{2}\right\rceil.
\]
In degree one, one linear form gives injectivity, and the two
linear equations of a point give surjectivity. For $D\ge2$,
apply Lemma~\ref{lem:ordinary-hyperplane} with $r_2=1$.
On the injective side take $r_1=\lfloor D/2\rfloor$, and on
the surjective side take $r_1=\lceil D/2\rceil$. Both residual
counts are on the preceding degree's injective side, and
\[
 2\lfloor D/2\rfloor+1\le D+1
       \le2\lceil D/2\rceil+1.
\]
These are precisely the two required inequalities.
\end{proof}

\subsection{A differential hyperplane step}
\label{sec:differential}

Write $H=(x_n=0)$, $R=k[x_1,\ldots,x_{n-1}]$, and
\[
 \Sigma_t=\left\{(f_1,\ldots,f_n)\in S_t^n:
                    \sum_i x_i f_i=0\right\},\qquad
 \mathcal E_{t-1}=S_{t-1}^{\,n-1}.
\]
Use $\Sigma'_t$ for the corresponding space over $R$. Thus
$\Sigma_t=H^0(\mathbf P^{n-1},\Omega_{\mathbf P^{n-1}}(t+1))$.
For $g=(g_i)\in\mathcal E_{t-1}$ put
$u_g=\sum_{i<n}x_i g_i$. There are two exact sequences
\begin{align}
 0\longrightarrow\mathcal E_{t-1}
 &\longrightarrow\Sigma_t\longrightarrow\Sigma'_t
 \longrightarrow0,
 &g&\longmapsto(x_ng_1,\ldots,x_ng_{n-1},-u_g),
 \label{eq:first-sequence}\\
 0\longrightarrow\Sigma_{t-1}
 &\longrightarrow\mathcal E_{t-1}
 \longrightarrow R_t\longrightarrow0,
 &&g\longmapsto\overline{u_g}.
 \label{eq:second-sequence}
\end{align}
In the first sequence the last map restricts the first $n-1$
components to $H$; a trace relation extends simply by using the
same polynomials in $S$ and adding a zero last component.
In the second sequence, $\overline{u_g}=0$ means
$u_g=x_n\phi$, so $(g,-\phi)\in\Sigma_{t-1}$; surjectivity is
the equality $R_1R_{t-1}=R_t$.

At $y\in H$ write
$T_y=\{v\in k^{n-1}:\sum_{i<n}y_iv_i=0\}$.
A \emph{trace condition} at $y$ asks that the first $n-1$
components of a section vanish there. It has $n-2$ scalar
components. More generally a \emph{partial trace condition}
specified by $L\subset T_y$ asks that those components lie in
$L$. These conditions will only be used as intermediate linear
conditions; the final configurations consist of ordinary points.

\begin{lemma}[Moving one point off the hyperplane]
\label{lem:moving-point}
Impose on $\Sigma_t$ full point conditions at $Z_{\rm off}$
off $H$ and $Z_{\rm on}$ on $H$, together with trace conditions
at $T\subset H$ and, if desired, a partial trace condition at a
further point. Let $B\subset\Sigma'_t$ be the space satisfying
the resulting trace conditions. Choose a further $y_0\in H$
such that evaluation $B\to T_{y_0}$ is injective, and put
$L=\operatorname{ev}_{y_0}(B)$. On $\mathcal E_{t-1}$ consider
the residual conditions
\[
 g(Z_{\rm off})=0,\qquad
 u_g(Z_{\rm on})=0,\qquad g(y_0)\in L.                 \tag{*}
\]
If these conditions have zero kernel, then adjoining a general
point on the chosen line with a general nonzero parameter gives zero kernel on
$\Sigma_t$. If the trace conditions and the residual conditions
\textup{(*)} are both independent, then all the resulting
conditions on $\Sigma_t$ are independent.
\end{lemma}

\begin{proof}
Take a vector $v$ with $v_n\ne0$ and move the point along
$y_s=[y_0+sv]$. In the injective case, a nonzero generic kernel
would yield, after clearing denominators and dividing by a
power of $s$, a polynomial relation
$f(s)=f_0+sf_1+\cdots$ with $f_0\ne0$ and $f(s)(y_s)=0$.
The trace of $f_0$ belongs to $B$ and vanishes at $y_0$, so it
is zero. Sequence~\eqref{eq:first-sequence} gives
$f_0=(x_ng,-u_g)$, with $g$ satisfying the first two residual
conditions. The coefficient of $s$ in the first $n-1$
components at $y_s$ gives
\[
                 v_n g(y_0)+\overline{f'_1}(y_0)=0.
\]
Since the trace of $f_1$ belongs to $B$, this says $g(y_0)\in L$.
The residual hypothesis gives $g=0$, contrary to $f_0\ne0$.

For independence, let $A$ be the space satisfying the fixed
conditions, before adjoining $y_s$, and let $M(s)$ be evaluation
at $y_s$ on $A$. The exact evaluation diagram for
\eqref{eq:first-sequence} shows that the fixed conditions are
independent and $A\to B$ is onto. Put $b=\dim B=\dim L$.
Residual independence implies that
\[
 K=\{g:g(Z_{\rm off})=0,\ u_g(Z_{\rm on})=0\}
       \longrightarrow k^{n-1}/L
\]
is onto. Therefore $M(0)$ has image of dimension $b+1$:
the $b$ trace components in $L$, and the last component
$-u_g(y_0)$. Its kernel consists of $(x_ng,-u_g)$ with
$g\in K$ and $u_g(y_0)=0$. The first derivative, modulo
$\operatorname{im}M(0)$, is
$g\mapsto v_ng(y_0)\bmod L$, and maps onto $T_{y_0}/L$.
It has rank $n-2-b$.

For a matrix $M(s)=M_0+sM_1+\cdots$, the rank for general
$s$ is at least
$\operatorname{rank}M_0+
\operatorname{rank}(\ker M_0\to\operatorname{coker}M_0)$,
where the second map is induced by $M_1$. To see this, put
$M_0$ in block form with an identity block; the relevant larger
minor has a nonzero leading coefficient. Here the lower bound
is $(b+1)+(n-2-b)=n-1$, the full fiber dimension.
\end{proof}

\begin{lemma}[Removing the scalar trace]
\label{lem:scalar-trace}
In a problem on $\mathcal E_{t-1}$, allow full vector
conditions, scalar conditions $u_g(y)=0$ on $H$, and partial
conditions $g(y)\in L\subset T_y$ on $H$. Let $Y$ be the set of
points of $H$ occurring in any of these conditions. If
$|Y|=N_t(n-1)$ and evaluation $R_t\to k^Y$ is an isomorphism,
then the problem has the same kernel and cokernel as the
following problem on $\Sigma_{t-1}$: keep full conditions off
$H$, turn full vector conditions on $H$ into trace conditions,
retain partial trace conditions, and discard scalar conditions.
\end{lemma}
\begin{proof}
Every condition at $y\in Y$ implies $\overline{u_g}(y)=0$.
In the evaluation diagram for \eqref{eq:second-sequence}, the
quotient condition space is thus exactly $k^Y$. Its evaluation
map from $R_t$ is an isomorphism. On the kernel, the conditions
are precisely those listed: off $H$, the first $n-1$ components
determine the last component of a syzygy, whereas on $H$ they
give its trace. The kernel-cokernel exact sequence of this
diagram proves the assertion.
\end{proof}

\begin{lemma}[A fixed partial trace condition]
\label{lem:fixed-partial}
Suppose ideal generation holds in degree $t$ in $n-1$
variables. Fix $y\in H$ and $L\subset T_y$, with $\dim L=c$.
For $k$ general points $P\subset H$, chosen after $(y,L)$,
the space
\[
       \{\beta\in\Sigma'_t:\beta(P)=0,\ \beta(y)\in L\}
\]
has dimension
$\max\{0,\dim\Sigma'_t-(n-2)k-(n-2-c)\}$.
The same assertion holds after one further general point is
added, provided $k,k+1,k+2$ all lie in the degree-$t$ point-count
cell.
\end{lemma}
\begin{proof}
The ideal generation statements at $k$ and $k+1$ determine
the rank of evaluation at $y$ on $\Sigma'_t(P)$. Its image
has dimension $\min\{\dim\Sigma'_t(P),n-2\}$. For a general
$c$-plane $L$ in $T_y$, intersection with that image therefore
has the expected dimension. The asserted dimension is the
smallest possible one, so it defines a nonempty open condition.
The stabilizer of $y$ in $\operatorname{PGL}(H)$ acts on
$T_y$ through the full general linear group, up to scalar, and
acts transitively on its $c$-planes. Consequently the good open
has a nonempty fiber over every fixed $L$. Repeat the same
argument with $k+1$ points, using the assertion at $k+2$.
\end{proof}

\begin{lemma}[Selection of scalar conditions]
\label{lem:scalar-selection}
Let $t\ge2$, let $Z$ be general points off $H$ with
$|Z|\ge N_{t-2}(n)$, and put $W=I_Z(t-1)$, $\dim W=\rho\ge1$.
Suppose multiplication $W\otimes S_1\to S_t$ is injective.
Choose general points $P\subset H$ after $Z$, with
$|P|\ge(n-1)\rho$. Let $y_0\in H$ satisfy $W(y_0)\ne0$,
and let $L\subset T_{y_0}$ be any $c$-dimensional subspace;
$L$ may depend on $P$. Set
\[
 K^\infty=\{g\in W^{n-1}:g(y_0)\in L\},\qquad
 k_0=(n-1)\rho-(n-1-c).
\]
Then $\dim K^\infty=k_0$, and scalar evaluation
$g\mapsto(u_g(z))_{z\in P}$ is injective on $K^\infty$.
For $0\le q\le|P|$, one may select $q$ points of $P$ whose scalar conditions
kill $K^\infty$ if $q\ge k_0$, or are independent on it if
$q\le k_0$.
\end{lemma}
\begin{proof}
Restriction $W\to\overline W$ is injective because its kernel is
$x_nI_Z(t-2)=0$. Moreover,
$\overline W\otimes R_1\to R_t$ is injective. Indeed, a trace
relation lifts to
$\sum_jw_j\ell_j=x_nG$ with $G\in W$. Expanding $G$ in a
basis $w_j$ gives a linear relation on $W$; its injectivity
forces every $\ell_j$ to be zero. Hence
$U=\overline W R_1$ has dimension $(n-1)\rho$, and evaluation
on general $P$ is injective on all of $U$.
It is therefore injective on the scalar polynomials $u_g$
arising from $K^\infty$. Since some member of $W$ is nonzero
at $y_0$, evaluation $W^{n-1}\to k^{n-1}$ is onto, which
gives the stated dimension. Selecting independent rows of
this injective evaluation proves both conclusions.
\end{proof}

\subsection{The two-level transfer}
\label{sec:two-level}

\begin{proposition}[Fractional trace]
\label{prop:fractional-trace}
Let $n\ge4$, $D\ge3$. Suppose ideal generation holds at
$(n-1,D)$ and $(n-1,D-1)$, and $E(n-1,D)$ is not integral.
Define
\begin{align*}
 q_0&=N_D(n-1)-\lceil E(n-1,D)\rceil,&
 b_0&=(n-2)(\lceil E(n-1,D)\rceil-E(n-1,D)),\\
 \tau&=\lfloor E(n-1,D)\rfloor,&
 D_1&=(n-1)N_{D-1}(n-1)-N_{D+1}(n-1),\\
 q_1&=\lfloor D_1/(n-2)\rfloor,& b_1&=D_1-(n-2)q_1.
\end{align*}
Let $a$ and $a'$ be the least nonnegative residues of
$N_{D+1}(n-1)$ and $-N_{D+1}(n-1)$ modulo $n-1$, and put
\[
 \rho_I=\frac{q_1+n-1-a-b_1}{n-1},\qquad
 \rho_S=\frac{q_1+n-1+a'-b_1}{n-1}.
\]
Assume
\[
 D_1\ge0,\quad
 1\le\rho_I\le\rho_S\le N_{D-2}(n-1),\quad
 \tau+q_1\ge(n-1)\rho_S,\quad
 \operatorname{Inj}(n,D-2,\rho_S).
\]
Assume also that $k,k+1,k+2$, where $k=\tau+q_1$, belong to
the degree-$(D-1)$ point-count cell on $H$; equivalently,
\[
       N_{D-2}(n-1)\le k,\qquad k+2<N_{D-1}(n-1).
\]
Then both critical degree-$D$ counts in $n$ variables have
maximal rank.
\end{proposition}

\begin{proof}
Let $r=\lfloor E(n,D)\rfloor$ or $\lceil E(n,D)\rceil$ and
$p=N_D(n)-r$. First fix $q_0$ general points $A\subset H$
and one further general $y^*\in H$. The degree-$D$ trace
space
$B_D=\{\beta\in\Sigma'_D:\beta(A)=0\}$
has dimension $b_0$, and evaluation at $y^*$ is injective.
These are exactly the surjective and injective critical
assertions at $(n-1,D)$. Set $L^*=\operatorname{ev}_{y^*}(B_D)$.
Lemma~\ref{lem:moving-point} reduces the problem to the
residual vector conditions
\[
         g(Z_1)=0,\qquad u_g(A)=0,\qquad g(y^*)\in L^*,
         \qquad |Z_1|=p-q_0-1,
\]
on $\mathcal E_{D-1}$.

Specialize $\tau$ of the points of $Z_1$ to a set
$T\subset H$, leaving a set $Z$ of
$p-N_D(n-1)$ full points. Since
$q_0+1+\tau=N_D(n-1)$, the scalar evaluation on
$A\cup\{y^*\}\cup T$ can be made an isomorphism.
Lemma~\ref{lem:scalar-trace} then reduces the residual problem
to $\Sigma_{D-1}$ with full points $Z$, trace conditions at
$T$, and the fixed partial trace condition $(y^*,L^*)$.

We now construct this latter problem. Choose general points
$Z_{\rm off}$ off $H$ with
\[
 |Z_{\rm off}|=p-N_D(n-1)-q_1-1,
\]
then general $P\subset H$ with $|P|=\tau+q_1$, and finally
a general $y_0\in H$. Put $W=I_{Z_{\rm off}}(D-2)$. Direct
substitution gives
\[
 \dim W=r+q_1+1-N_{D-1}(n-1)
       =\begin{cases}\rho_I,&r=\lfloor E(n,D)\rfloor,\\
                       \rho_S,&r=\lceil E(n,D)\rceil.
         \end{cases}
\]
In particular these numbers are integers. The assumed cell
bound gives $|Z_{\rm off}|\ge N_{D-3}(n)$, and the required
injectivity of $W$ follows from the hypothesis and point
addition.

Write $\sigma'_t=\dim\Sigma'_t$. The identity
\[
 \sigma'_{D-1}-(n-2)\tau-(n-2-b_0)=D_1
\]
and Lemma~\ref{lem:fixed-partial} show that
\[
 B=\{\beta\in\Sigma'_{D-1}:
                    \beta(P)=0,\ \beta(y^*)\in L^*\}
\]
has dimension $b_1$, and evaluation at $y_0$ is injective.
Let $L=\operatorname{ev}_{y_0}(B)$. The crucial point is that
$B$ and $L$ depend on the whole set $P$, before it is
partitioned. Lemma~\ref{lem:scalar-selection}, applied with
$t=D-1$, is valid for this $L$ even though it depends on $P$.
Its remaining dimension is
\[
 k_0=(n-1)\dim W-(n-1-b_1)
       =\begin{cases}q_1-a,&\text{injective side},\\
                       q_1+a',&\text{surjective side}.
         \end{cases}
\]
Thus $q_1$ scalar conditions selected from $P$ kill the
residual kernel in the first case, and are independent in
the second case. Call the selected set $Z_{\rm on}$ and put
$T=P\setminus Z_{\rm on}$. Lemma~\ref{lem:moving-point}
at degree $D-1$ gives the required rank after adjoining a
point moving off $y_0$. This supplies $Z$ with exactly the
prescribed number of full points.

It remains to ensure that the two specializations use the
same choices. Keep $(A,y^*,L^*)$ fixed. The construction
just given proves that the required rank is a nonempty open
condition on the actual parameter space
\[
                   H^\tau\times(\mathbf P^{n-1})^{|Z|}.
\]
That space is irreducible. The condition that $Z$ avoid $H$
is dense open. Independently, the condition that
$A\cup\{y^*\}\cup T$ interpolate $R_D$ is nonempty open:
the fixed points $A\cup\{y^*\}$ already impose independent
conditions and can be completed by general points.
Intersect these opens. Lemma~\ref{lem:scalar-trace} now
applies to this same configuration, after which openness
allows the points $T$ in the vector problem to move off
$H$. Finally apply Lemma~\ref{lem:moving-point} in degree
$D$ at $y^*$.

Scalar degree-$D$ postulation can also be imposed in this
intersection. The points off $H$ are general and no more
numerous than $N_{D-1}(n)$; the trace points
$A\cup\{y^*\}$ impose independent degree-$D$ conditions.
Lifting a trace polynomial and correcting the off-hyperplane
values by $x_n$ times a degree-$(D-1)$ polynomial proves
onto scalar evaluation at the specialized configuration.
It remains onto after the final movement. The Euler
evaluation criterion (Lemma~\ref{lem:euler-evaluation}) identifies the resulting
maximal ranks with the required multiplication ranks.
\end{proof}