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\hypersetup{pdftitle={Hilbert functions of chopped ideals of general plane points},pdfsubject={Chopped ideals of plane points, Hilbert functions, and Steiner bundles}}
\title[Chopped ideals of general plane points]{Hilbert functions of chopped ideals\\of general plane points}
\date{}
\begin{document}
\begin{abstract}
Let $Z$ be a finite set of general points in the projective plane over a
field of characteristic zero. We determine the Hilbert function of the
ideal generated by the degree-$d$ equations of $Z$, for every $d$.
This proves the plane case of the Expected Syzygy Conjecture.
The proof realizes the quotient of the point ideal by its degree-$d$
equations as the cokernel of a linear block in a Hilbert--Burch matrix,
and applies natural cohomology for general plane Steiner bundles.
\end{abstract}
\maketitle

\section{Introduction}
\label{plane:section}

Let $k$ be an algebraically closed field of characteristic zero, and put
\[
 S=k[x,y,z],\qquad
 N_t=\begin{cases}\binom{t+2}{2}&t\geq0,\\0&t<0.\end{cases}
\]
For a finite set $Z\subset\mathbb P^2$, write $I(Z)$ for its saturated
homogeneous ideal. Following Gesmundo, Kayser, and Telen~\cite{gkt2025},
the ideal
\[
 J=(I(Z)_d)
\]
is called the \emph{chopped ideal in degree $d$}. Its Hilbert function
measures which higher-degree equations of the points are generated by
the equations of degree $d$.

The Expected Syzygy Conjecture of Gesmundo, Kayser, and Telen predicts
this Hilbert function for general points. We prove its plane case.
The main input is the natural-cohomology theorem for general plane
Steiner bundles, in the form recorded by Coskun, Huizenga, and
Smith~\cite[Remark~1.5]{chs2025}. To apply it, we construct a family of
Hilbert--Burch matrices with independent quadratic and linear blocks,
and show that this family contains reduced point ideals. The quotient
$I(Z)/J$ is then presented by the linear block itself.

Here \emph{general} means that the assertion holds on a nonempty
Zariski open subset of the space of ordered, pairwise distinct points.
All degrees in the following theorem are covered simultaneously on
one such open subset. The descent to arbitrary characteristic-zero
fields is explained at the end of the proof.


\begin{theorem}\label{plane:main}
Let $d\geq2$, let $Z\subset\mathbb P^2$ consist of $p\geq1$ general
points, and set $I=I(Z)$ and $J=(I_d)$.  In degrees $t<d$ one has
$\dim_k(S/J)_t=N_t$.  The remaining Hilbert function is as follows.
\begin{enumerate}
\item If $p\geq N_d$, then $J=0$.
\item If $p\leq N_{d-1}$, then $J_t=I_t$ for every $t\geq d$;
in particular, $\dim_k(S/J)_t=p$ in those degrees.
\item Suppose that $N_{d-1}<p<N_d$, and put $r=N_d-p$, so that
$1\leq r\leq d$.  For $r=1$ or $r=2$, the ideal $J$ is generated by
a regular sequence of $r$ forms of degree $d$, and
\begin{equation}\label{plane:small-r}
 \operatorname{Hilb}(S/J;u)=\frac{(1-u^d)^r}{(1-u)^3}.
\end{equation}
For $r\geq3$ and every $e\geq0$,
\begin{align}
 \dim_k(S/J)_{d+e}
 &=\max\{p,N_{d+e}-rN_e\}\label{plane:hf}\\
 &=p+\max\left\{
 \frac e2\bigl(2d+3-3r-(r-1)e\bigr),0\right\}.
 \notag
\end{align}
\end{enumerate}
In the last case, the least positive integer $e$ for which
$J_{d+e}=I_{d+e}$ is
\begin{equation}\label{plane:cutoff}
 e_0=\max\left\{1,
 \left\lceil\frac{2d+3-3r}{r-1}\right\rceil\right\}<d,
\end{equation}
and this equality of graded pieces persists for every $e\geq e_0$.
\end{theorem}

The main case is proved by constructing point ideals through the
Hilbert--Burch theorem.  In the range where generators of degree $d+1$
are necessary, the quotient $I/(I_d)$ is presented by a general matrix
of linear forms.  Its Hilbert function is then determined by the
natural cohomology of general Steiner bundles on $\mathbb P^2$.

\section{Preliminaries}

\begin{lemma}\label{plane:generation}
General $p$ points have Hilbert function
$\dim_k(S/I(Z))_t=\min\{N_t,p\}$.  More generally, if a reduced set
$Z$ of $p$ points satisfies $\dim_k(S/I(Z))_q=p$ for some $q\geq0$,
then $I(Z)$ is generated in degrees at most $q+1$.
\end{lemma}

\begin{proof}
For each fixed $t$, evaluation at points spans $S_t^*$: a form
annihilated by every evaluation is zero.  One can therefore choose
successive distinct points whose evaluation functionals are independent
until the rank is $\min\{N_t,p\}$.  This rank condition defines a
nonempty open subset of the irreducible configuration space.  If $q$
is the least integer with $N_q\geq p$, intersect the two opens giving
injectivity in degree $q-1$ and surjectivity in degree $q$ (omitting
the first condition if $q=0$).  Injectivity propagates to lower degrees.
Choose a linear form $\lambda$ nonzero at all the points.  Multiplication
by $\lambda$ makes evaluation surjective in every degree above $q$.
This proves the Hilbert function assertion.

For the second assertion, put $R=S/I(Z)$.  Multiplication by $\lambda$
is injective on $R$, and surjectivity of evaluation in degree $q$
again implies $\dim_kR_t=p$ for all $t\geq q$.  Consequently,
\[
 \dim_k(R/\lambda R)_t=\dim_kR_t-\dim_kR_{t-1}=0
 \qquad(t\geq q+1).
\]
The image $\overline I$ of $I(Z)$ in
$S/(\lambda)\simeq k[u,v]$ thus contains every form of degree $q+1$
and is generated in degrees at most $q+1$.  Lift homogeneous generators
of $\overline I$ to $I(Z)$.  These lifts generate $I(Z)$: after
subtracting their multiples from an element of $I(Z)$, the remainder
is $\lambda h$, and $\lambda h\in I(Z)$ implies $h\in I(Z)$.
Induction on degree completes the argument.
\end{proof}

We use the following established result in its matrix formulation.

\begin{proposition}[Plane Steiner bundles]\label{plane:steiner}
Let $a>0$ and $c-a\geq2$.  For a general $a\times c$ matrix $L$
of linear forms in $S$, the map
\begin{equation}\label{plane:linear-map}
 L:S_t^c\longrightarrow S_{t+1}^a
\end{equation}
has maximal rank for every $t\geq0$.
\end{proposition}

\begin{proof}[Deduction from natural cohomology]
The general matrix defines a surjection of vector bundles
\[
 0\longrightarrow K\longrightarrow\mathcal O_{\mathbb P^2}^{c}
 \xrightarrow{L}\mathcal O_{\mathbb P^2}(1)^a\longrightarrow0.
\]
Indeed, the locus of scalar matrices of rank below $a$ has codimension
$c-a+1\geq3$, and general evaluation avoids this locus on
$\mathbb P^2$.  Its dual is the defining sequence of a Steiner bundle.
The general Steiner bundle on $\mathbb P^2$ has natural cohomology,
meaning that each twist has at most one nonzero cohomology group;
the result and this dual formulation are recorded in
\cite[Remark~1.5]{chs2025}.  Serre duality gives the same property
for $K$.  Twisting the displayed sequence by $t\geq0$ identifies the
kernel and cokernel of \eqref{plane:linear-map} with $H^0(K(t))$ and
$H^1(K(t))$, respectively.  They cannot both be nonzero, which is
precisely maximal rank.  Only finitely many of these open rank
conditions will be needed below.
\end{proof}

\section{Reduced point schemes from matrices}

We give the geometric justification that permits the linear matrix in
Proposition~\ref{plane:steiner} to occur inside a point ideal.  All
matrix entries in the next lemma vary independently.

\begin{lemma}\label{plane:reduced}
Let $\mathcal P$ be the affine space of homogeneous matrices defining
maps between fixed direct sums of line bundles on $\mathbb P^2$, of
ranks $c$ and $c+1$.  Suppose that every matrix entry has positive
degree, and that one matrix has a finite projective maximal-minor
scheme.  Then a general matrix has a finite reduced maximal-minor
scheme.  Its maximal minors generate a saturated ideal of height two,
and its graded resolution is the Hilbert--Burch resolution.
\end{lemma}

\begin{proof}
The universal maximal-minor scheme is closed in
$\mathcal P\times\mathbb P^2$ and is proper over $\mathcal P$.
The condition that its fibers have dimension at most zero is open.
For a matrix in this nonempty open, the homogeneous maximal-minor
ideal has height at least two.  The determinantal height bound gives
height at most two, so the height is exactly two.  In particular, the
projective scheme cannot be empty: an empty scheme would give an
irrelevant-primary ideal of height three.  These standard determinantal
facts and the Hilbert--Burch theorem may be found in
\cite[Chapter~20]{eisenbud1995}.

At any fixed point of $\mathbb P^2$, evaluation of the coefficients
surjects onto all scalar $(c+1)\times c$ matrices, since the line
bundles of the entries are globally generated.  Locally on
$\mathbb P^2$, this evaluation is a smooth morphism from
$\mathcal P\times\mathbb P^2$ to the corresponding matrix bundle.
The scalar rank-$(c-1)$ stratum is smooth of codimension two.  Its
inverse image is therefore smooth, of dimension $\dim\mathcal P$.
The locus of rank at most $c-2$ has codimension six (and is empty
when $c=1$).  Its universal inverse image has dimension at most
$\dim\mathcal P-4$; its proper projection cannot dominate
$\mathcal P$.  A general fiber consequently meets only the smooth
rank-$(c-1)$ stratum.  Generic smoothness in characteristic zero,
applied to its dominating components, now shows that a general
zero-dimensional fiber is smooth, hence reduced.  Components that
do not dominate are removed by shrinking the parameter open.

Finally, Hilbert--Burch gives a free resolution of the maximal-minor
ideal of length one.  Thus its quotient has depth one and dimension
one.  It has no elements annihilated by a power of the irrelevant
ideal, which is exactly saturation of the homogeneous ideal.
\end{proof}

\section{The quotient presented by the linear block}

Assume $N_{d-1}<p<N_d$, $r=N_d-p\geq3$, and
\begin{equation}\label{plane:ac}
 a=d+2-2r>0,\qquad c=d+1-r.
\end{equation}
Then $r+a=c+1$ and $c-a=r-1\geq2$.  Consider the full affine space
\begin{equation}\label{plane:parameter-space}
 \mathcal P=\operatorname{Mat}_{r\times c}(S_2)
 \times\operatorname{Mat}_{a\times c}(S_1),\qquad
 B=\begin{pmatrix}Q\\L\end{pmatrix}.
\end{equation}
Deleting a row of $Q$ gives a maximal minor of degree
$2(r-1)+a=d$; deleting a row of $L$ gives one of degree
$2r+a-1=d+1$.

To check the hypothesis of Lemma~\ref{plane:reduced}, number the rows
$0,\ldots,c$ and give row $i$ degree $\delta_i$, equal to $2$ in
the first $r$ rows and $1$ in the last $a$ rows.  Take the bidiagonal
matrix with entries
\[
 B_{i,i}=x^{\delta_i},\qquad
 B_{i+1,i}=y^{\delta_{i+1}}\qquad(0\leq i<c),
\]
and all other entries zero.  Two of its maximal minors are
$x^{d+1}$ and $y^d$, up to sign.  Its projective maximal-minor scheme
is therefore finite.  The lemma shows that a general $B$ defines
a reduced point ideal $I_B$ with resolution
\begin{equation}\label{plane:hb}
 0\longrightarrow S(-d-2)^c\xrightarrow{B}
 S(-d)^r\oplus S(-d-1)^a\longrightarrow I_B\longrightarrow0.
\end{equation}
In particular,
\begin{equation}\label{plane:point-series}
 \operatorname{Hilb}(S/I_B;u)
 =\frac{1-ru^d-au^{d+1}+cu^{d+2}}{(1-u)^3}.
\end{equation}
Taking coefficients, using \eqref{plane:ac}, gives $N_t$ for $t<d$
and the constant value $N_d-r=p$ for $t\geq d$.  Thus the reduced
scheme consists of exactly $p$ points and has the required Hilbert
function.

Let $J_B$ be generated by the $r$ minors of degree $d$.  Since
$I_B$ has no elements of lower degree, $J_B=((I_B)_d)$.  Quotienting
the middle term of \eqref{plane:hb} by $S(-d)^r$ gives an actual
presentation of the quotient module:
\begin{equation}\label{plane:block-quotient}
 I_B/J_B\simeq\operatorname{coker}\left(
 L:S(-d-2)^c\longrightarrow S(-d-1)^a\right).
\end{equation}
Consequently,
\begin{equation}\label{plane:residual}
 \dim_k(I_B/J_B)_{d+e}
 =\max\{aN_{e-1}-cN_{e-2},0\}
 =\max\left\{\frac e2
 \bigl(2d+3-3r-(r-1)e\bigr),0\right\}
\end{equation}
whenever $L:S_{e-2}^c\to S_{e-1}^a$ has maximal rank.  For $e=0$
both sides are zero; for $e=1$ both sides equal $a$; and for $e\geq2$
the first equality follows from Proposition~\ref{plane:steiner}.
The second follows by expanding the binomial coefficients.

It remains to choose one reduced point scheme for all the needed
ranks at once.  Let $e_0$ be as in \eqref{plane:cutoff}.  Maximal
rank for $2\leq e\leq e_0$ defines a nonempty open in the linear
factor of \eqref{plane:parameter-space}.  Its inverse image and the
nonempty reduced-point open intersect because $\mathcal P$ is
irreducible.  For a matrix in their intersection,
\eqref{plane:residual} holds through $e=e_0$.  The module
$I_B/J_B$ is generated in degree $d+1$, by
\eqref{plane:block-quotient}; its vanishing in degree $d+e_0$
therefore implies vanishing in all higher degrees.  This proves
\eqref{plane:residual} for every $e\geq0$ for one reduced set
of $p$ points.

\section{Transfer to general points and the remaining cases}

On the open configuration space of $p$ points with the Hilbert
function in Lemma~\ref{plane:generation}, the kernels $I_d$ of
evaluation form a vector bundle of rank $r$.  For each fixed $e$,
multiplication gives a morphism of vector bundles
\[
 I_d\otimes_k S_e\longrightarrow S_{d+e}.
\]
Its rank is lower semicontinuous.  On this space one always has
\begin{equation}\label{plane:lower-bounds}
 \dim_k(S/(I_d))_{d+e}\geq
 \max\{p,N_{d+e}-rN_e\}:
\end{equation}
the first bound follows from the quotient onto $S/I$, and the
second from the dimension of the multiplication source.  The
constructed point scheme attains both bounds in their maximum,
by \eqref{plane:residual} and \eqref{plane:point-series}.  Hence
general points attain this maximum for $0\leq e\leq e_0$.
Lemma~\ref{plane:generation}, with $q=d$, says that their ideal is
generated in degrees at most $d+1$; it has no elements below degree
$d$.  The quotient $I/(I_d)$ is thus generated in degree $d+1$.
Vanishing at $d+e_0$ again propagates to all higher degrees.
Only finitely many open conditions have been imposed, proving the
simultaneous assertion of the theorem when $a>0$.

Now assume $a=d+2-2r\leq0$, still with $r\geq3$.  Put
\[
 \ell=2r-d-2,\qquad q=d+1-r,
 \qquad \ell+q=r-1,\quad \ell+2q=d.
\]
Take a general $r\times(r-1)$ matrix having $\ell$ columns of
linear forms and $q$ columns of quadrics.  If column $j$ has degree
$\epsilon_j\in\{1,2\}$, the bidiagonal matrix with entries
$x^{\epsilon_j}$ and $y^{\epsilon_j}$ in rows $j$ and $j+1$
has maximal minors $x^d$ and $y^d$.  Lemma~\ref{plane:reduced}
therefore applies.  It gives a reduced point ideal with resolution
\begin{equation}\label{plane:saturated-resolution}
 0\longrightarrow S(-d-1)^\ell\oplus S(-d-2)^q
 \longrightarrow S(-d)^r\longrightarrow I_B\longrightarrow0.
\end{equation}
The Hilbert series has numerator
$1-ru^d+\ell u^{d+1}+qu^{d+2}$; its coefficients give $N_t$ for
$t<d$ and $p=N_d-r$ for $t\geq d$.  All its generators have degree
$d$, so $(I_B)_d$ generates $I_B$.  Semicontinuity transfers the
equality in degree $d+1$ to general points, and
Lemma~\ref{plane:generation} propagates it.  Finally,
\[
 \frac e2\bigl(2d+3-3r-(r-1)e\bigr)\leq0\qquad(e\geq1),
\]
since its inner linear factor is decreasing and its value at $e=1$
is $2a\leq0$.  This proves \eqref{plane:hf} in this range.

For either range, the residual expression is positive at a positive
integer $e$ precisely when $e<(2d+3-3r)/(r-1)$.  This gives
\eqref{plane:cutoff}.  Since $r\geq3$,
\[
 \frac{2d+3-3r}{r-1}\leq d-3,
 \qquad
 (r-1)(d-3)-(2d+3-3r)=(r-3)d\geq0.
\]
Thus $e_0\leq\max\{1,d-3\}<d$, as claimed.

For completeness we prove the two small-generator cases.  When
$r=1$, a nonzero generator is a regular sequence of length one.
When $r=2$, a reduced complete intersection supplies the required
configuration.  Choose pairwise distinct scalars $\alpha_i$ and
$\beta_j$ and set
\[
 f=\prod_{i=1}^d(x-\alpha_i z),\qquad
 g=\prod_{j=1}^d(y-\beta_j z).
\]
Their common zeros are the $d^2$ distinct affine grid points
$[\alpha_i:\beta_j:1]$, and they have no common zero at infinity.
The pair $f,g$ is regular, and its ideal is saturated, by its
complete-intersection resolution and the depth argument in
Lemma~\ref{plane:reduced}.  Consequently evaluation of $S_d/(f,g)_d$
on the grid is injective, and this space has dimension
$p=N_d-2\leq d^2$.  Choose $p$ grid points whose evaluation
coordinates form a basis of its dual.  For this subset $Z$, one
has $I(Z)_d=\langle f,g\rangle$.  Moreover $I(Z)_{d-1}=0$:
any nonzero form of degree $d-1$ would give three independent
multiples in the two-dimensional space $I(Z)_d$.

The condition that a basis $f,g$ of $I_d$ be a regular pair is open.
Indeed, it is equivalent to injectivity of
\[
 S_{d-1}^{\oplus2}\longrightarrow S_{2d-1},\qquad
 (u,v)\longmapsto uf+vg.
\]
If $f,g$ are coprime, every syzygy is a multiple of $(g,-f)$ and
none occurs in this degree.  If they have a common factor of
positive degree, the pair obtained by dividing $(g,-f)$ by that
factor gives a syzygy of coefficient degree at most $d-1$, which
can be multiplied to coefficient degree $d-1$.  The grid subset
therefore proves that the open condition is nonempty.  This proves
the regular-sequence assertion for general points and
\eqref{plane:small-r} follows from the Koszul resolution.

Finally, if $p\geq N_d$, general evaluation is injective in degree
$d$, so $J=0$.  If $p\leq N_{d-1}$, evaluation is surjective in
degree $d-1$ and Lemma~\ref{plane:generation} shows that $I$ is
generated in degrees at most $d$.  Multiplying these generators
shows $I_t=S_{t-d}I_d$ for all $t\geq d$.  This includes the boundary
$p=N_{d-1}$ and every choice of a higher generating degree, and
finishes the proof of Theorem~\ref{plane:main}.

\begin{corollary}\label{plane:expected}
The Expected Syzygy Conjecture for chopped ideals of general points
holds in $\mathbb P^2$.
\end{corollary}

\begin{proof}
Use the formulation in \cite[Conjecture~1]{gkt2025}, which takes
$d=\min\{t:N_t\geq p\}$. If $p=N_d$, there are no degree-$d$
equations and $J=0$, as required. Otherwise $N_{d-1}<p<N_d$.
When the predicted quotient dimension never reaches the point count,
the cutoff is interpreted as $+\infty$. For $r=1,2$ the alternating expression
is the regular-sequence series \eqref{plane:small-r}.  For $r\geq3$,
before degree $2d$ its value is $N_{d+e}-rN_e$; terms from the
second and higher Koszul modules first occur in degree $2d$.
By \eqref{plane:cutoff}, this expression first reaches or falls
below the point count $p$ at degree $d+e_0<2d$.  The conjecture
retains the earlier values and the constant value $p$ thereafter,
which is exactly \eqref{plane:hf}. The cases $d\leq1$ follow by
direct computation. The boundary $p=N_{d-1}$ in
Theorem~\ref{plane:main} also covers chopping one degree later than
the convention of the conjecture.
\end{proof}

\begin{remark}\label{plane:fields-and-linear}
The conclusions hold over any characteristic-zero field in the
generic sense.  All evaluation and multiplication rank conditions
used above are defined over the prime field, and only finitely many
are imposed.  Their nonemptiness can be tested after algebraic
closure; their intersection thus defines a nonempty open over the
original field.  The degree-one cases are immediate: one point has
an ideal generated by two independent linear forms, two general
points have a one-dimensional space of linear equations, and at
least three general points have no nonzero linear equation.
If $p=0$, with $I(\varnothing)=S$, then
$(I(\varnothing)_d)=(x,y,z)^d$.
\end{remark}


\begin{thebibliography}{99}
\raggedright

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I.~Coskun, J.~Huizenga, and G.~Smith,
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\href{https://doi.org/10.1307/mmj/20226238}{doi:10.1307/mmj/20226238}.

\bibitem{eisenbud1995}
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\bibitem{gkt2025}
F.~Gesmundo, L.~Kayser, and S.~Telen,
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\end{thebibliography}
\end{document}