text:
\section{Introduction}

    It is famously untrue that a deformation of a zero-dimensional projective
    scheme $\Gamma \subseteq \mathbb{P}^{n}$ induces a deformation of its
    homogeneous coordinate ring $S/I(\Gamma)$, where
    $I(\Gamma)\subseteq S =\kk[\alpha_0, \ldots,
    \alpha_n]$ is the saturated ideal. The standard example is given by the
    varying three points
    \[
        \Gamma_t = \{[1:0:0], [1:t:1], [0:0:1]\}\subseteq \mathbb{P}^2,
    \]
    which form a flat family over the affine line with parameter $t$. The ideal
    $I(\Gamma_t|_{t=\lambda})$
    contains a linear form exactly when $\lambda = 0$, so
    if the family $S[t]/I(\Gamma_t)$ existed, it had to be \emph{lower} semi-continuous
    in degree one; an absurd. What happens instead is that as $\Gamma_t$
    converges towards $\Gamma_0$, the family of
    ideals $I(\Gamma_t|_{t=\lambda})$ converges degreewise to a \emph{nonsaturated} ideal
    \[
        I(\Gamma_0)' = \left(\alpha_0\alpha_2(\alpha_0 -
        \alpha_2)\right) + \alpha_1\cdot\left(\alpha_0, \alpha_1, \alpha_2
        \right).
    \]
    For a homogeneous ideal $I$ we say that it is \emph{a limit of saturated
    ideals} or \emph{saturable} if there is a family of saturated ideals
    that converges to $I$. Hence $I(\Gamma_0)'$ above is saturable, yet not
    saturated. In this article we tackle the following.
    \begin{problem}\label{ref:mainProblem}
        Given a homogeneous ideal, decide whether it is saturable.
    \end{problem}
    There are numerical obstructions to being saturable: an ideal has to have
    an admissible Hilbert function.
    In the case of a standard graded polynomial ring such functions are
    classified by Macaulay~\cite[\S4.2]{BrunsHerzog}. In general we will
    consider only ideals having admissible Hilbert functions.

    \subsection{Motivation} Problem~\ref{ref:mainProblem} is important outside algebraic geometry for areas such as
    quantum physics and computational complexity. The connection is via the
    theory of tensors, and more specifically \emph{border apolarity} and
    \emph{varieties of sums of powers} which both lie in the intersection of projective
    algebraic geometry and its applications. Let us discuss them in
    a baby case of projective space. The case of most interest for
    applications is a product of projective spaces, however it does not differ
    much apart from notation, hence we restrict to a single projective space.

    Let $x_0, \ldots ,x_n$ be the dual basis for $\alpha_0, \ldots
    ,\alpha_n$ and $S^* = \mathbb{C}[x_0, \ldots ,x_n]$ be the polynomial
    ring.
    Let $0\neq F\in S^*_d$ and let $\nu_d\colon \mathbb{P}^n\to \mathbb{P}^{\binom{n+d}{n}-1}$ be
    the $d$-th Veronese embedding, where $\mathbb{P}^n$ has coordinates
    $\alpha_0, \ldots,\alpha_n$.
    \emph{Apolarity Lemma} says that for any linear forms $\ell_1$, \ldots ,
    $\ell_R$, the following are equivalent:
    \begin{itemize}
        \item the polynomial $F$ is a linear combination of $\ell_1^d$, \ldots
            ,$\ell_R^d$,
        \item $[F]\in \spann{\nu_d(\Gamma)}$ where $\Gamma = \left\{ [\ell_1],
            \ldots ,[\ell_R] \right\}$,
        \item $F\in I(\Gamma)_d^{\perp}$.
    \end{itemize}
    For a given $F$ the smallest $R$ for which linear forms as
    above exist is called the \emph{Waring rank} of $F$. Little is known about
    it; even the maximal possible Waring rank is known only for few $(d,
    n)$~\cite{blekherman_teitler_on_maximal_typical_and_generic_ranks}. Apolarity Lemma is usually used to mine
    the rich geometry of possible $\Gamma$ for information about the Waring
    rank: for example, for a given $F$ one attempts to construct $\Gamma$ with small $R$ or
    disprove its existence.

    The set of forms with given Waring rank is constructible, but usually not
    closed in Euclidean topology, and not a variety. Therefore it is
    more natural to consider the \emph{border rank of $F$}, denoted $\brr(F)$, which is the
    smallest $R$ such that $F$ is a limit of forms of Waring rank at most $R$.
    The forms of border rank at most $R$ form the \emph{$R$-th secant variety} of
    $\nu_d(\mathbb{P}^n)$. There has been a vast amount of study of this
    variety for varying $d, n$, both classical~\cite{AlexHirsch,
    Zak__tangents_and_secants, Geramita, Landsberg_survey,
iarrobino_kanev_book_Gorenstein_algebras} and more
    recent for example (the list is highly incomplete!)~\cite{Abo_brambilla, LO, bubu2010, Carlini_Grieve_Oeding,
    Abo_Vannieuwenhoven, Ein_Niu_Park, Buczyska_Buczynski__border, Choe_Kwak, Landsberg_complexity}.

    The main problem that arises in this context is that there is no obvious analogue of the Apolarity
    Lemma for border rank. Indeed, suppose that $\brr(F)\leq R$ and take
    $F_t\to F$, where $F_t$ has Waring rank at most $R$. Each $F_t$ has an
    associated tuple of $R$ points $\Gamma_t$ with $[F_t]\in
    \spann{\nu_d(\Gamma_t)}$. It is possible to take the limit of $\Gamma_t$
    inside the Hilbert scheme. It would be natural to expect that if
    $\Gamma_t\to \Gamma$, then $[F]\in \spann{\nu_d(\Gamma)}$, but this is not
    true!
    In fact, there are forms $F$ of border rank $R$ for
    which \emph{no} degree $R$ finite subscheme $\Gamma$ exists with
    $[F]\in\spann{\nu_d(\Gamma)}$.
    Such forms
    $F$ are examples of \emph{wild polynomials}. One concrete example is $F
    = x_0^2x_2 + x_0x_1x_3 + x_1^2x_4$ which has border rank $5$,
    see~\cite{Bucz_Bucz_smoothable} and
    also~\cite{ranestad_schreyer_on_the_rank_of_a_symmetric_form}.
    The problem in the expectation above is that
    taking $\spann{\nu_d(-)}$ is not even upper semicontinuous, already for
    $d=1$ and for the above
    example of three points. In that example $\spann{\Gamma_0}$ is
    is a line while
    $\spann{\Gamma_t|_{t=\lambda}} =\mathbb{P}^2$ for
    every $\lambda\neq 0$.

    The main idea of \emph{border apolarity}~\cite{Buczyska_Buczynski__border}
    is to take the limit in the \emph{multigraded Hilbert scheme} instead of
    the usual one. For a given Hilbert function $H$, the multigraded Hilbert
    scheme $\Hilb^H$ parameterizes homogeneous ideals (possibly nonsaturated)
    with this Hilbert function. Let $S = \mathbb{C}[\alpha_0, \ldots
    ,\alpha_n]$, as above. An ideal $I\subseteq S$ is encoded as a sequence
    $(I_e)_e$ of linear spaces which satisfy $S_1\cdot I_e \subseteq
    I_{e+1}$. Hence, the scheme $\Hilb^H$
    is a closed subscheme of the infinite product $\prod_e \Gr(\dim S_e - H(e), S_e)$.
    Fix $R$ and let $H$ be the Hilbert function of a general $R$-tuple in
    $\mathbb{P}^n$. If $F$ has border rank $\brr(F)\leq R$, then from $F_t\to
    F$ we obtain saturated ideals $I(\Gamma_t)\in \Hilb^H$ and, as their
    limit, a possibly nonsaturated ideal $I'$ such that $F\in (I'_d)^{\perp}$. This observation is named
    \emph{Weak border apolarity}
    in~\cite[Theorem~1.2]{Buczyska_Buczynski__border} and is analogous to
    Apolarity Lemma: border rank replaces rank while the possibly nonsaturated ideal $I'$ replaces $I(\Gamma)$.

    The introduction of border apolarity led to significant improvements of
    several lower bounds for the border rank of important
    forms~\cite{Conner_Huang_Landsberg}.
    These improvements were obtained because the set of possible $I'$ was
    empty due to numerical constraints and using the group
    action~\cite[Theorem~4.3]{Buczyska_Buczynski__border}. However once a
    candidate ideal $I'$ is constructed, the question is whether it arises from a
    limit of $I(\Gamma_t)$. For small $R$ this is effectively equivalent to
    asking the weaker question ``Is $I'$ saturable?'' which brings us to
    Problem~\ref{ref:mainProblem}.

    The \emph{Varieties of Sums of Powers} (VSP) are a classical source of examples
    of interesting projective varieties~\cite{Ranestad_Iliev__VSPcubic,
    Ranestad_Voisin__VSP}. They are defined for given $F$ and $R$ as the closure in the Hilbert
    scheme $\Hilb^R(\mathbb{P}^n)$ of the set of all length $R$ subschemes $\Gamma$ with
    $F\in \spann{\nu_d(\Gamma)}$. It is tedious to understand the closure,
    especially when one needs to prove the smoothness of VSP.
    The \emph{border VSP}
    is defined in~\cite[\S4]{Buczyska_Buczynski__border} as the set of all ideals
    $[I]\in \Hilb^H$ which are limits of ideals of points and are such that $F\in
    (I_d)^{\perp}$. For example, for $F = x_0^2x_2 + x_0x_1x_3 + x_1^2x_4$ and
    $(n,R) = (4,5)$, we see that the border VSP is nonempty yet contains only
    nonsaturated ideals, see also~\cite[\S7]{Huang_Michalek_Ventura}. Again,
    asking whether a given ideal $I$ with $F\in (I_d)^{\perp}$ is in the
    border VSP
    brings us to Problem~\ref{ref:mainProblem}.

    \subsection{Results}

        There is one case where the answer to Problem~\ref{ref:mainProblem}
        depends only on the Hilbert function: an ideal $I$ in
        $\mathbb{C}[\alpha_0,\alpha_1]$ is saturable if and only if it has an admissible
        Hilbert function $H$. This is because $\Hilb^H$ is
        irreducible~\cite{Maclagan_Smith__multigraded} and the locus of
        saturated ideals is open (Theorem~\ref{ref:opennessOfSaturation:thm})
        and nonempty since $H$ is admissible. This result holds for any
        grading. Unfortunately, already for three variables the situation
        drastically complicates.

        Before we present our results, we give an informal observation which
        underlies our criteria. Let $S$ be an $\mathbb{N}^r$-graded polynomial
        ring. Let $I$ be a homogeneous ideal. We say that
        an ideal $J$ with $I \subsetneq J\subseteq I^{\sat}$ \emph{sticks with}
        $I$ (or is a \emph{sticky ideal} for $I$) if every deformation of $I$ induces a deformation of
        $J\supseteq I$.  If $I$ admits a sticky ideal $J$, then it is not a limit of saturable
        ideals. Indeed, suppose $I_t\to I$ with $I_t|_{t=\lambda}$ saturated
        for $\lambda\neq 0$.
        Then we have a corresponding family $J_t\supsetneq I_t$. Due to
        equal Hilbert functions, the ideals $J_t|_{t=\lambda}$ and $I_t|_{t=\lambda}$ agree in large
        degrees, so $(J_t|_{t=\lambda})^{\sat} = (I_{t}|_{t=\lambda})^{\sat} = I_t|_{t=\lambda}$, in particular
        $J_t|_{t=\lambda}\subseteq I_t|_{t=\lambda}$, which is a contradiction.

        In this paper, we obtain a toolbox of cohomological criteria for proving that a given ideal
        $J$ sticks with $I$. The \emph{fiber obstruction group} for the pair
        $(I, J)$ is
        \begin{equation}\label{eq:introOb}
            \ObFib(I,J) := \Ext^1_S\left( \frac{J}{I},\ \frac{S}{J}
            \right)_{\mathbf{0}}.
        \end{equation}
        The most succinct version of our criterion is.
        \begin{theorem}[Theorem~\ref{ref:deformsWhenObFibVanishes:thm}]\label{ref:introObVanishes:thm}
            If $\ObFib(I, J) = 0$, then $J$ sticks with $I$, hence $I$ is not
            saturable.
        \end{theorem}
        This is effective already for the ideal $J :=
        I^{\sat}$, see~\S\ref{ssec:four}.
        A slightly changed version is the following.
        \begin{theorem}[Theorem~\ref{ref:deformWhenTangentSurjective:thm}]\label{ref:introSmoothAndTangentSpace:thm}
            If the natural map $\Hom_S(J, S/J)_{\mathbf{0}}\to\Hom_S(I,
            S/J)_{\mathbf{0}}$ is onto
            and $[J]\in \Hilb^{H_{S/J}}$ is smooth, then $J$ sticks with $I$, hence $I$ is not
            saturable.
        \end{theorem}

        We say that $I$ is \emph{entirely nonsaturable} if it is nonsaturated and
        there is an open neighbourhood of $[I]\in \Hilb^{H_{S/I}}$ such that any ideal
        $I'$ from this neighbourhood has $S/(I')^{\sat}$ with the same Hilbert
        function as $S/I^{\sat}$, so ``every near enough ideal is as nonsaturated
        as $I$''. If the conditions of
        Theorem~\ref{ref:introObVanishes:thm} or
        Theorem~\ref{ref:introSmoothAndTangentSpace:thm} are satisfied for $J :=
        I^{\sat}$, then $I$ is entirely nonsaturable (see
        Theorems~\ref{ref:saturationDeformsWhenObFibVanishes:thm}-\ref{ref:saturationDeformWhenTangentSurjective:thm}).

        The necessary conditions for both theorems above are easily computable
        with computer systems such as \emph{Macaulay2}~\cite{M2}. However,
        the presentation of fiber obstruction~\eqref{eq:introOb} is opaque.
        There is a more transparent version for a one-dimensional
        $J = I^{\sat}$ for $S$ being
        $\mathbb{N}$-graded.
        Namely, the vector space $\ObFib$ is dual to $(J/I\otimes_{S}
        \omega_{S/I^{\sat}})_0$, where $\omega_{S/I^{\sat}}$ is the canonical
        module, see Proposition~\ref{ref:CIAndcanonicalModule:prop}.
        This reinterpretation is especially transparent when we assume
        additionally that $S/I^{\sat}$ is Gorenstein (which holds for example if
        $I^{\sat}$ is a complete intersection). In this case, we have
        \begin{equation}\label{eq:introObEasy}
            \ObFib(I, I^{\sat})  \simeq
            \left(\frac{I^{\sat}}{I+(I^{\sat})^2}\right)_{a}^{\vee}
        \end{equation}
        where $a$ is the largest degree for which
        $H_{S/I^{\sat}}(a)
        <H_{S/I^{\sat}}(a+1)$ see~Corollary~\ref{ref:fiberObstructionForGorenstein:cor}. The obstruction group can be more
        generally computed for non-Gorenstein $J=I^{\sat}$ if additional
        conditions on the Hilbert functions are satisfied, see
        Example~\ref{ex:lineAndPoints}.

        Above we discussed \emph{necessary} conditions for being saturable. It
        is natural to ask what happens if those are not satisfied.
        In this case, interestingly, we do get information about possible
        families showing saturability of $I$. For example, in the context
        of Theorem~\ref{ref:introSmoothAndTangentSpace:thm} for
            $J=I^{\sat}$, if $\Hom_S(J,
        S/J)_0\to \Hom_S(I, S/J)_0$ is not surjective and yet $\Hom_S(I,
        S/I)_0\to \Hom_S(I, S/J)_0$ is surjective, then we obtain a tangent
        vector (see~\S\ref{ssec:maindiagram}) at $[I]$, which
        does not come from a tangent vector at $[I\subseteq I^{\sat}]$. If we are able to
        integrate along it (for example, if $[I]$ is smooth), then we obtain a
        one-dimensional family whose
        general member $I'$ has $H_{S/(I')^{\sat}} > H_{S/I^{\sat}}$, where
        the inequality means argument-wise inequality with at least one of
        them strict. Repeating this process we arrive at the case where
        $H_{S/(I')^{\sat}} = H_{S/I} = H_{S/I'}$, hence $I'$ is saturated.
        This procedure seems effective on examples even when $[I]$ is not
        smooth.

        \subsection{Examples} To illustrate the effectiveness of our ideas, we
        decide the saturability for all ideals with Hilbert function $(1,d,d,d,
        \ldots )$ for $d\leq 5$, see~\S\ref{sec:examples}. For $d = 4$ and
        all but one cases for $d = 5$ an ideal $I$ is saturable  if and only
        if the obstruction space~\eqref{eq:introObEasy} is nonzero.
        The unique case for $d = 5$ is when $H_{S/I^{\sat}} = (1,2,3,4,5,5,
        \ldots )$. In this case the nonvanishing of the obstruction boils down
        to $(I^{\sat})^2_3\subseteq I$. It is still
        necessary, yet not sufficient: there is a second necessary condition
        of the form $(I^{\sat}\cdot J)_2\subseteq I$ for a certain
        ideal $I\subseteq J \subseteq I^{\sat}$ determined by $I$.
        The same argument would apply to other Hilbert functions
        eventually equal to $d\leq
        5$; we have restricted ourselves to $(1,d,d, \ldots )$ in order to
        better exhibit the main ideas.

    \subsection{Previous work}

        To our best knowledge, there are almost no papers concerned with
        Problem~\ref{ref:mainProblem}. This problem is mentioned
        in~\cite{Buczyska_Buczynski__border, Conner_Huang_Landsberg}. One may also ask
        which ideals are limits of saturated and radical ideals.
        The latter problem was investigated in~\cite{Mandziuk} in the $\mathbb{N}$-graded case.
        It gives two necessary conditions: a semicontinuity of tangent spaces type of statement
        \cite[Theorem~1.1]{Mandziuk} and a necessary condition in the case that $V(I)$ is contained in the
        line~\cite[Theorem~2.7]{Mandziuk}. The latter statement is an immediate corollary of
        Theorem~\ref{ref:introObVanishes:thm} above. We stress that while
        formulating the
        precise necessary condition from \cite[($\dagger$) above
        Theorem~2.7]{Mandziuk} required insight from the author,
        here it was deduced automatically from the general principle of
        searching for sticky
        ideals. Ma{\'n}dziuk's paper contains also a sufficient condition in the case
        $n=2$~\cite[Theorem~1.3]{Mandziuk} which we plan to extend in future
        work.

    \subsection{Applications to nonexistence of wild polynomials}

        We apply the above results also to the theory of border rank in the plane case.
        It is an open question whether wild
        forms in three variables exist. By~\cite[Proposition~3.4]{Mandziuk} degree $d$
        forms of border rank at most $d+2$ are not wild. Here we improve the
        bound by one.
        \begin{proposition}[Proposition~\ref{prop:no_wild_polynomials}]
            Let $F$ be a degree $d$ form in three variables. If the border
            rank of $F$ is at most $d+3$, then the smoothable rank of $F$ is
            equal to its border rank, so $F$ is not wild.
        \end{proposition}
        There are lower bounds for the smoothable rank such
        as~\cite{ranestad_schreyer_on_the_rank_of_a_symmetric_form}.
        Coupled with the above they yield constraints on the possible forms of
        low border rank. To give a concrete example, a septic with Hilbert
        function $(1,3,6,9,9,6,3,1)$ and border rank $9$ has, by the above,
        smoothable rank $9$ and so,
        by~\cite[Proposition~1]{ranestad_schreyer_on_the_rank_of_a_symmetric_form},
        its apolar ideal cannot be generated in degrees at
        most $4$. This also follows from border apolarity \cite[Theorem~3.15]{Buczyska_Buczynski__border}.

    \subsection{Additional results} While our focus is on nonsaturated ideals,
    we obtain two results about saturated ideals. They are of
    independent interest.

    The first result concerns the deformations of a saturated ideal in the
    multigraded setting.
    For an $\mathbb{N}^r$ grading on a polynomial ring $S$, the
    \emph{irrelevant ideal} $S_+$ is spanned by monomials $m$ such that
    every coordinate of $\deg(m)$ is strictly positive. Our first result is as
    follows.
    \begin{theorem}[openness of saturation, Theorem~\ref{ref:opennessOfSaturation:thm}]
        For any $\mathbb{N}^r$-grading on $S$ and any Hilbert function
        $H\colon \mathbb{N}^r\to \mathbb{N}$, the subset of $[I]\in \Hilb^H$
        which are saturated (with respect to $S_+$) is open.
    \end{theorem}
    The above seems not found in the literature even in the case of standard
    grading. Although possibly known to experts in commutative algebra, it was
    definitely not known to experts in the tensor world, where it is important as it
    allows to define the Slip component(s), see
    Definition~\ref{ref:Slip:def}.
    A preliminary proof based on the
    ideas of the first-named author is contained in an unpublished
    note~\cite{jabuSaturationOpen} by Buczy{\'n}ska-Buczy{\'n}ski (who
    considered slightly more general grading: with possible torsion in the
    grading group). Here we
    give a more direct proof.

    Our second result gives a functorial description of the set of all
    zero-dimensional ideals with given Hilbert function and a criterion for
    being smooth there.
    \begin{theorem}[Theorems~\ref{ref:constantHilbertFunction:prop}, \ref{ref:smoothPoints:thm},
        \ref{ref:usualClassesSmooth:cor}]\label{ref:intro:saturated}
        For any $n$ and a function $H\colon \mathbb{N}\to \mathbb{N}$ with $d
        = H(i)$ for $i\gg 0$,
        the locus of $[\Gamma]\in \Hilb^d (\mathbb{P}^{n-1})$ with
        $H_{S/I(\Gamma)} = H$ is locally closed in $\Hilb^d
        (\mathbb{P}^{n-1})$ and admits
        a scheme structure with which it is isomorphic to $\Hilb^{H,\sat}$, the saturated locus
        of $\Hilb^{H}$.
        A point $[\Gamma]$ in this locus is smooth whenever the Artin
    reduction of $S/I(\Gamma)$ is unobstructed. In particular, $[\Gamma]$
        is smooth when $n \leq 3$ or when $n \leq 4$ and $S/I(\Gamma)$ Gorenstein.
    \end{theorem}
    In the case $n = 3$ and for a field of characteristic zero Gotzmann proved
    that the above locus is smooth~\cite{gotzmann_fixed_Hilbert_function}.
    Kleppe proved smoothness in the Gorenstein $n=4$
    case~\cite{kleppe__smoothness}. The
    interpretation as a functor seems new even in these special cases.

\section{Preliminaries}

Throughout the paper, $\kk$ is a field. We do not assume it to be
algebraically closed or characteristic zero. Any willing reader can assume
$\kk = \mathbb{C}$ without losing much content.

\subsection{Conventions and notation}\label{ssec:setup}

    Let $S$ be a
    polynomial ring. We assume that it has a grading by an abelian group $A$.
    We will eventually be only interested in $\mathbb{N}^{r}$-gradings;
    the first time reader may safely assume that the grading is the standard
    $\mathbb{N}$-grading.
    We assume that for every $a\in A$ the number $\dim_{\kk} S_a$ is finite.
    A \emph{homogeneous ideal} in $S$ means an ideal
    homogeneous with respect to $A$-grading. For any homogeneous ideal $J$ we have Hilbert
    function $H_{S/J}\colon A\to \mathbb{N}$ given by
    $H_{S/J}(a) := \dim_{\kk} (S/J)_a$.

    By $\Hilb^H$ we denote the multigraded
    Hilbert scheme parameterizing ideals with Hilbert function $H$, defined
    in~\cite{Haiman_Sturmfels__multigraded}. For an ideal $J$ we define the
    shorthand notation $\Hilb_{J} := \Hilb^{H_{S/J}}$.
    Since we are mostly concerned with multigraded Hilbert schemes, we will
    refer to the classical Hilbert scheme of $d$ points on a variety $X$ as
    \emph{Grothendieck Hilbert scheme} and denote it by $\Hilb^d(X)$.

    There is a natural morphism from the multigraded Hilbert scheme to the
    Grothendieck one. We restrict
    to the case when $S$ is a standard graded polynomial ring of dimension $n$
    and fix an $H\colon \mathbb{N}\to \mathbb{N}$ which is an admissible
    Hilbert function for a zero-dimensional subscheme of
    $\mathbb{P}^{n-1}$. Then $H$ is nondecreasing and eventually constant, let
    $d$ be its maximum. For clarity of notation fix one ideal $I$ with
    $H = H_{S/I}$. By~\cite[Lemma~4.1]{Haiman_Sturmfels__multigraded} for
    all $e\gg 0$, we have $\Hilb^d(X) \simeq \Hilb_{I_{\geq e}}$.
    A homogeneous deformation of $I$ induces by definition a deformation of
    $I_{\geq e}$, hence $\pr_I\colon \Hilb_{I_{\geq e}\subseteq I} \to
    \Hilb_I$ is an isomorphism. We obtain a natural map
    \[
        \begin{tikzcd}
            \Hilb_I\ar[r, "\simeq", "\pr_I^{-1}"'] & \Hilb_{I_{\geq e}
        \subseteq I} \ar[r, "\pr_{I_{\geq e}}"] & \Hilb_{I_{\geq e}}  \simeq
        \Hilb^d(\mathbb{P}^{n-1})
        \end{tikzcd}
    \]
    that sends a point $[I]$ to $\Proj(S/I)\subseteq \mathbb{P}^{n-1}$.

    \subsection{Macaulay's inverse
    systems}\label{ssec:macaulaysInverseSystems}

    Throughout the paper, we will frequently consider surjections such as
    $S/I\twoheadrightarrow S/I^{\sat}$ and analyse the possible choices of the subideal $I$ for fixed
    $I^{\sat}$. It seems cleaner to do this dually: consider superobjects
    $I^{\perp}\supseteq (I^{\sat})^{\perp}$ instead. Hence we employ
    Macaulay's inverse systems. They are a standard tool in the Artinian case and
    introduced in the higher-dimensional case at least
    in~\cite{Elias_Rossi__Gorenstein_for_higher_dims}. Here we loosely
    summarize exactly as much of the theory as we need.

    The idea is quite simple yet requires some notation. Given $S = \kk[\alpha_0,  \ldots ,\alpha_n]$ we
    define the dual graded vector space $S^* = \kkdp[x_0, \ldots ,x_n]$ and
    make it into a graded $S$-module by the \emph{contraction} action $S\,\lrcorner\,
    S^*\to S^*$ defined on a monomial $m\in S^*$ by
    \[
        \alpha_i \,\lrcorner\, m =
        \begin{cases}
            0 & \mbox{if } x_i\mbox{ does not divide } m,\\
            m/x_i & \mbox{otherwise}.
        \end{cases}
    \]
    For an ideal $I$ we define $I^{\perp}\subseteq S^*$ as $I^{\perp} =
    \left\{ f\in S^*\ |\ I\,\lrcorner\, f = 0 \right\}$. The subspace $I^{\perp}$ is
    an $S$-submodule of $S^*$. Similarly, for a submodule $M\subseteq S^*$ we
    define an ideal $M^{\perp} \subseteq S$ as the annihilator of $M$, namely
    $M^{\perp} = \left\{ \sigma\in S\ |\ \sigma\,\lrcorner\, M = 0 \right\}$.
    We have $(-)^{\perp\perp} = \id$, hence $(-)^{\perp}$ gives a bijection
    between graded ideals of $S$ and graded $S$-submodules of $S^*$. This
    bijection reverses inclusions and is equivariant with respect to the usual
    $\GL_{n+1}$-actions.

	\begin{remark}\label{ref:contraction_and_differentiation:rmk}
        For experts: the subscript in $\kkdp$ underlies the fact that $S^*$ can
        be made into a divided powers ring on which $S_1$ acts as differential
        operators (we will not use this structure).
    \end{remark}

    In characteristic zero (or large enough) a slightly different,
    equivalent, action by \emph{partial differentiation}
    is usually employed.  In this setup, we assume that $\kk$ has
    characteristic zero, take the polynomial ring $\kk[x_0, \ldots ,x_n]$ (note the lack of
    subscript $dp$) and make it into an $S$ module where $\alpha_i$ acts as
    $\frac{\partial}{\partial x_i}$. We denote this action by $\circ$.
    There is an isomorphism $\kk[x_0, \ldots ,x_n]\to S^*$ of $S$-modules that
    sends a monomial $x^{\aa}$ to $\aa!x^{\aa}$, and so all the duality
    claims for $S, S^*$ hold --- still in characteristic zero --- also for
    the pair $S, \kk[x_0, \ldots ,x_n]$. See~\cite{Jel_classifying} for more details.

	We use the following example in \S\ref{ssec:five} where we assume
	that $\kk$ has characteristic zero.
	Therefore, we present it under the same assumption and we consider the action
	$\circ$ of $S$ on $\kk[x_0,\ldots, x_n]$.

    \begin{example}\label{ex:points}
		Assume that $\kk$ has characteristic zero.
        Let $L\in \spann{x_0, \ldots ,x_n}$ be a nonzero linear form and let $I = (1, L, L^2, L^3,
        \ldots , )^{\perp}$. Then $I$ is the homogeneous ideal of the point $[L]\in
        \Proj \kk[\alpha_0, \ldots ,\alpha_n] \simeq  \mathbb{P}^{n}$, or in
        symbols
        \[
            I(\{L\}) = \left\{ L^i\ |\ i\geq 0 \right\}^{\perp}.
        \]
        Similarly, if $L_1, \ldots ,L_r$ are pairwise different linear forms,
        then $I(\{L_1, \ldots ,L_r\}) = \left\{ L^i_j\ |\ i\geq 0, 1\leq j\leq
        r\right\}^{\perp}$.
    \end{example}

    For further use, we note the following result:
    \begin{proposition}\label{ref:HilbertFunctionFromLimits:prop}
        Let $F_1, \ldots ,F_s$ be homogeneous degree $\ee$ elements of $S[t]$, where
        $\deg(t) = \mathbf{0}$. Assume that there are homogeneous degree $\ee$ elements $G_1$, \ldots
        ,$G_s\in S[t]$
        such that
        \begin{enumerate}
            \item for every $i$ there exists $d_i\geq 0$ with $t^{d_i}G_i\in
                \spann{F_1, \ldots ,F_s}S[t]$,
            \item\label{it:indep} $G_1|_{t=0}$, \ldots ,$G_{s}|_{t=0}$ are linearly
                independent. (we call them \emph{limit forms}).
        \end{enumerate}
        Then $F_1|_{t=\lambda}$, \ldots ,$F_{s}|_{t=\lambda}$ are linearly
        independent for general $\lambda$ and the limit of the subspaces
        $\spann{F_1, \ldots ,F_s}$ at $t\to 0$ is $\spann{G_1|_{t=0}, \ldots ,G_{s}|_{t=0}}$.
    \end{proposition}
    \begin{proof}
        Let $M := \kk[t]^{\oplus s}$ and let $\iota\colon M\to S_{\ee}[t]$ the
        $\kk[t]$-linear map that sends the $i$-th generator to $G_i$.
        Using~\ref{it:indep} and the classification of $\kk[t]$-modules
        and~\cite[Lemma~A.7]{Sernesi__Deformations} we conclude that after
        localizing at some $f\in \kk[t]\setminus (t)$ the homomorphism $\iota$ is
        injective and $(S_{\ee}[t]/\im \iota)_{f}$ is a free
        $\kk[t]_{f}$ module.
        By this freeness it follows that $\iota|_{t=\lambda}$ is injective
        for every $\lambda$ with $f(\lambda)\neq 0$.
        This shows that for every nonzero such $\lambda$, the forms
        $(G_j|_{t=\lambda})_{j=1}^s$ are linearly independent. They are linear
        combinations of $(F_j|_{t=\lambda})_{j=1}^s$, which are thus linearly
        independent as well. The last statement is purely formal.
    \end{proof}

    \subsection{Main diagram}\label{ssec:maindiagram}

        For ideals $K\subseteq J \subseteq S$
        the two long exact sequences coming from $\Hom(K, -)$ and
        $\Hom(-,\frac{S}{J})$ give the following \emph{main diagram}
        \[
            \begin{tikzcd}[column sep=small]
                &&&0\ar[d]  \\
                &&&\Hom\left( K, \frac{J}{K} \right)_{\mathbf{0}}\ar[d]\\
                && \TgFlag\ar[r]\ar[d]\ar[dr, phantom, "\usebox\pullback" ,
                very near start, color=black] &\Hom\left( K, \frac{S}{K}
                \right)_{\mathbf{0}}\ar[d]\\
                0\ar[r] & \Hom\left( \frac{J}{K}, \frac{S}{J} \right)_{\mathbf{0}}\ar[r] &
                \Hom\left( J, \frac{S}{J} \right)_{\mathbf{0}} \ar[r] & \Hom \left( K,
                \frac{S}{J}
                \right)_{\mathbf{0}} \ar[r]\ar[d] & \Ext^1_S\left( \frac{J}{K},
                \frac{S}{J} \right)_{\mathbf{0}} \ar[rd, bend left=30]\\
                &&& \Ext^1_S\left(K, \frac{J}{K}\right)_{\mathbf{0}}\ar[rd, bend
                right=30] & \ObFlag\ar[r]\ar[d]\arrow[dr, phantom,
                "\usebox\pullback" , very near start, color=black] &
                \Ext^1_S\left( J,
                \frac{S}{J} \right)_{\mathbf{0}}\ar[d]\\
                &&&& \Ext^1_S\left(K, \frac{S}{K}\right)_{\mathbf{0}}\ar[r] & \Ext^1_S\left( K,
                \frac{S}{J} \right)_{\mathbf{0}}.
            \end{tikzcd}
        \]
        We define the spaces $\TgFlag$ and $\ObFlag$ as pullbacks of vector
        spaces: their elements are pairs of elements which map to the same
        element. Their names are justified by the following theorem. Recall
        that the flag multigraded Hilbert scheme $\Hilb_{K\subseteq J}$
        parameterizes pairs of ideals $K'\subseteq J'$ with Hilbert functions
        $H_{K'} = H_K$ and $H_{J'} = H_J$, respectively. The tangent space to
        $\Hilb_{K\subseteq J}$ at $[K\subseteq J]$ is $\TgFlag$, see, for
        example~\cite[Theorem~4.10]{Jelisiejew__Elementary}.
        The flag Hilbert scheme is
        a closed subscheme of $\Hilb_K \times \Hilb_J$, so we obtain
        projections
        \begin{equation}\label{eq:Hilbs}
            \begin{tikzcd}
                \Hilb_K & \Hilb_{K\subseteq J} \ar[r, "\pr_J"]\ar[l, "\pr_K"'] & \Hilb_J.
            \end{tikzcd}
        \end{equation}
        Let
        \[
            \psi\colon \Hom_S(K, S/K) \oplus \Hom_S(J, S/J)\to \Hom_S(K, S/J).
        \]
        be the sum of the natural maps $\Hom_S(K, S/K), \Hom_S(J, S/J)\to
        \Hom_S(K, S/J)$. We prove that if $\psi_{\mathbf{0}}$ is surjective,
        then $\Hilb_{K\subseteq J}$ admits an obstruction theory at
        $[K\subseteq J]$ with obstruction space $\ObFlag$, see
        Theorem~\ref{ref:obstructionAtFlag:thm}. Since the proof is
        technical, we delegate it to an appendix.

        Going back to the main diagram, we see that the ``linking'' group
        $\Ext^1_S(K,
        J/K)_{\mathbf{0}}$ is the obstruction group for equivariantly deforming
        $K$ inside $J$, in other words, it is the obstruction group of the
        fiber
        $(\pr_J^{-1}([J]), [K])$, while $\Ext^1_S(J/K, S/J)_{\mathbf{0}}$ is the obstruction
        group for deforming $J/K$ in $S/K$, hence the obstruction group of
        $(\pr_K^{-1}([K]), [J])$. We will be principally interested in the map
        $\pr_K$, hence we define the \emph{fiber obstruction group}
        \begin{equation}\label{eq:obfib}\tag{ObFib}
            \ObFib(K, J) := \Ext^1_S\left( \frac{J}{K}, \frac{S}{J}
            \right)_{\mathbf{0}}.
        \end{equation}
        \stepcounter{equation}

        The Main Theorem of obstruction calculus asserts that a map of pointed schemes
        which induces a surjection of tangent spaces and an injection of
        obstruction spaces (at the base point) is smooth (at the base point).
        It follows from the main diagram that the vanishing of $\ObFib$
        implies \emph{both} these conditions for $\pr_K$ (when $\psi_{\mathbf{0}}$ is onto).
        We will exploit this observation
        in~\S\ref{ssec:smoothnessOfProjections}.

        \subsection{Saturations}

        In this section we assume that $S$ is $\mathbb{N}^r$-graded, so
        that the group $A$ defined in~\S\ref{ssec:setup} is
        $\mathbb{Z}^r$.
        For degrees $\aa, \aa'\in \mathbb{N}^r$ we define $\aa \geq \aa'$ if
        $\aa - \aa'$ has nonnegative entries and $\aa > \aa'$ if $\aa - \aa'$
        has strictly positive entries.
        We define the \emph{irrelevant ideal} $S_+ :=
        \bigoplus_{m\in \mathbb{Z}_{>0}^r} S_m$. Important special cases
        covered by this definition are
        \begin{itemize}
            \item $r = 1$ and $S$ the standard graded polynomial ring.
                Geometrically this is the case of a projective space.
            \item $S = S^{1}\otimes_{\kk}  \ldots \otimes S^{r}$, where
                $S^{i}$ is a standard graded polynomial rings.
                Geometrically this is the case of a product of projective spaces.
        \end{itemize}
        We stress that $S_+$ need not be a maximal ideal of $S$.\\
        For a homogeneous ideal $I$ in a Noetherian $S$-algebra $S'$, its \emph{saturation} is the ideal
        \[
            I^{\sat} := \bigcup_{i\geq 0} (I : S_+^i).
        \]
        The ideal $I$ is \emph{saturated} if $I = I^{\sat}$. The saturation of any
        ideal is saturated. Let $S_+' := S_+\cdot S'$. By prime avoidance, an ideal $I$ is saturated if
        an only if there exists a homogeneous $\ell\in S_+'$ such that the multiplication
        by $\ell$ is injective on $S'/I$. For a homogeneous ideal $I$ a
        \emph{transverse element} is any homogeneous
        $\ell\in S_+'$ which is a nonzerodivisor on $S'/I^{\sat}$. For every
        such $\ell$, the ideal
        $I^{\sat}/I$ is the kernel of the natural map $S'/I\to (S'/I)_{\ell}$.
        The only $S'$ we consider in this article are of the form
        $S\otimes_{\kk} A$ for $A$ a $\kk$-algebra.

        Proposition~\ref{ref:transverseElementDeforms:prop} is the technical part of the proof that saturation is an open
        property, see Theorem~\ref{ref:opennessOfSaturation:thm} below for
        motivation.
        \begin{lemma}\label{ref:Gotzmann:lem}
            Let $S$ be $\mathbb{N}$-graded, let $U$ be a Noetherian $\kk$-scheme and $Z \subset
            \Spec(S) \times U$ be a closed subscheme given by a
            homogeneous ideal. Denote by $p\colon Z\to U$ the projection. Then
            the set of Hilbert functions of fibers of $p$ is finite.
        \end{lemma}
        \begin{proof}
            We do induction on the dimension of $U$. For a given $U$, we
            replace it with $U_{\redd}$ and then by the disjoint union of the
            irreducible components of its
            affine open subschemes. In this manner, we reduce to
            the case $U$ affine and integral, say $U = \Spec(A)$ and $Z =
            \Spec(B)$. By Generic Flatness
            (\cite[\href{https://stacks.math.columbia.edu/tag/051R}{Tag
            051R}]{stacks-project} or \cite[Theorem~14.4]{Eisenbud}) there is
            a nonzero element $a\in A$ such that $B_a$ is a free $A_a$-module.
            Every degree of $B_a$ is a direct summand of $B_a$ hence a locally
            free $A_a$-module.
            This means that the Hilbert function of $p^{-1}(u)$ is
            independent of the choice of $u\in (a\neq0)\subseteq
            U$.
            By induction, the set of Hilbert
            functions for the fibers over $U\setminus (a\neq 0)$ is finite.
            This
            concludes the proof.
        \end{proof}
        \begin{proposition}\label{ref:transverseElementDeforms:prop}
            Let $f\in S$ be a homogeneous element of positive degree. Let $U$ be a
            locally Noetherian $\kk$-scheme and $Z \subset \Spec (S) \times U$ be a closed
            subscheme given by a homogeneous ideal.
            Denote by $p\colon Z\to U$ the projection and assume
            it is flat. Then the subset
            \begin{equation}\label{eq:goodfibers}
                U' := \left\{ u\in U\ |\ f \mbox{ is a nonzerodivisor on
                }\OO_{p^{-1}(u)} \right\} \subseteq U
            \end{equation}
            is open. The multiplication by $f$ on
            $\OO_{p^{-1}(U')}$ is injective and $V(f)\cap p^{-1}(U')$ is flat
            over $U'$.
        \end{proposition}
        \begin{proof}
            \def\HilbFunc{\mathcal{H}}%
            We replace the $\mathbb{N}^r$-grading on $S$ by
            $\mathbb{N}$-grading, where the new degree of a homogeneous
            element is the sum of its previous degree, so we assume that $S$
            is $\mathbb{N}$-graded.
            Everything is local on $U$, so we assume $U$ is connected, $U = \Spec(A)$ and $Z =
            \Spec(B)$. Every degree $B_e$ is a finitely generated flat module over
            a Noetherian ring $A$, hence it is locally free.
            Pick a point $u_0\in U$, let $\mm \subset A$ be the maximal ideal
            corresponding to $u_0$ and $\kappa_0 = A/\mm$. Suppose that $f$ is a
            nonzerodivisor on $p^{-1}(u_0) = \Spec(B/\mm B)$. Let $T = A\setminus
            \mm$ be the multiplicatively closed subset, so that $T^{-1}A =
            A_{\mm}$.
            Let $d = \deg(f) > 0$ and let $\mu_f\colon B\to B$ be the multiplication
            by $f$.
            Consider the map $(\mu_f)_e\colon B_e\to B_{d+e}$. By assumption, the
            map $(\mu_f)_e\colon B_e \otimes_A \kappa_0 \to B_{d+e} \otimes_A
            \kappa_0$ is injective. By~\cite[Lemma~A.7]{Sernesi__Deformations} the map
            $T^{-1}((\mu_f)_e)\colon T^{-1}B_{e}\to T^{-1}B_{d+e}$
            is split injective; in particular
            \[
                T^{-1}\left( \frac{B_{d+e}}{f B_e} \right)  \simeq
                \frac{T^{-1}B_{d+e}}{f\cdot  T^{-1}B_{e}}
            \]
            is a finitely
            generated free $A_{\mm}$-module.
            Thus $T^{-1}(B/fB)$ is a free $A_{\mm}$-module.
            We now prepare to ``smear out'' the freeness of $T^{-1}(B/fB)$ to
            the freeness of some
            $(B/fB)_s$, $s\in T$. The issue here is that $T^{-1}(B/fB)$ is a
            \emph{countable} direct sum of finitely generated free $A_{\mm}$-modules.
            Let $H$ be the
            Hilbert function of the fiber $B\otimes_A
            \kappa_0$ and $H'$ be the Hilbert function of $(B/fB)\otimes_A \kappa_0$,
            so that $H'(i) = H(i) - H(i-d)$ for every $i\in \mathbb{Z}$ under the convention
            that $H'(i) = H(i) = 0$ for $i < 0$.
            By Lemma~\ref{ref:Gotzmann:lem} the set $\HilbFunc$ of possible Hilbert
            functions for the fibers of $B/fB$ is finite. Fix $i_0$ such that
            every Hilbert function $H''\in \HilbFunc\setminus \{H'\}$ differs from
            $H'$ in some position less than or equal to $i_0$.

            Fix $s\in T$ such that $(B_e/fB_{e-d})_s$ is a free $A_s$-module for every
            $e \leq i_0$. For every $u\in \Spec(A_s)$ with residue field $\kappa
            = \kappa(u)$, the graded ring
            $B\otimes_{A} \kappa$ has Hilbert function $H$.
            The Hilbert function $\tilde{H'}$ of
            $(B/fB)\otimes_A \kappa$ is an element of $\HilbFunc$ and agrees
            with $H'$ for all arguments $i\leq i_0$, hence is equal to $H'$.
            In the right-exact sequence
            \[
                \begin{tikzcd}
                    B_{i-d}\otimes_A \kappa\ar[r, "\mu_{f}\otimes_A \kappa"] & B_i\otimes_A
                    \kappa \ar[r] & (B/fB)_i\otimes_A \kappa\ar[r] & 0
                \end{tikzcd}
            \]
            the dimensions of the $\kappa$-vector spaces are $H(i-d)$, $H(i)$,
            $\tilde{H}'(i)$. Since $\tilde{H}'(i) = H'(i) = H(i) - H(i-d)$,
            this sequence is left-exact. This proves that $u\in
            U'$. Since $u$ was chosen arbitrarily, we have
            $\Spec(A_s)\subseteq U'$. The injectivity of $(\mu_f)_s$ and
            flatness of $(B/fB)_s$
            can be checked after localizing to every maximal
            ideal of a point $u\in \Spec(A_s)$ and then follow from the argument given
            above for $u_0$.
        \end{proof}

        \begin{theorem}\label{ref:opennessOfSaturation:thm}
            For any Hilbert function $H$, the set of points of $\Hilb^H$ which
            correspond to saturated ideals is open; we denote the resulting
            open subscheme by $\Hilb^{H, \sat}$. Moreover, for every
            morphism $\varphi\colon T\to \Hilb^H$ the following are equivalent
            \begin{enumerate}
                \item the (set-theoretic) image of $\varphi$ is contained in $\Hilb^{H,
                    \sat}$,
                \item the fibers of the universal ideal sheaf pulled back to
                    $T$ are saturated ideals.
            \end{enumerate}
        \end{theorem}
        \begin{proof}
            \def\barL{\overline{L}}%
            If $[I]\in \Hilb^{H, \sat}$, then there exists an $f$ such that
            multiplication by $f$ on $S/I\to S/I$ is injective. By
            Proposition~\ref{ref:transverseElementDeforms:prop} this
            multiplication is injective for every point of some open
            neighbourhood $U'$ of $[I]$, so $U'\subseteq \Hilb^{H, \sat}$.
            This proves openness.

            The claim for
            $T$ reduces immediately to the case of $T=\Spec(L)$, where
            $\kk\subseteq L$ is a field extension, so actually the claim is
            that for an ideal $I$ and an ideal $I'$ obtained by field
            extension, $I$ is saturated iff so is $I'$. If $I'$ is saturated,
            then so is $I$ since the quotient by $I$ injects into the quotient
            by $I'$. If $I$ is saturated, then a transverse element $\ell\in
            S_+$ is a nonzerodivisor in this quotient, hence also in the
            quotient by $I'$, hence $I'$ is saturated.
        \end{proof}

        \newcommand{\Satbar}{\overline{\Sat}}%
        \begin{definition}
            We define the \emph{saturable locus} $\Satbar^H$ of $\Hilb^H$ as the closure
            of $\Hilb^{H, \sat}$. It is a union of irreducible components of
            $\Hilb^H$.
        \end{definition}

        In view of the discussion in the introduction, it is interesting to
        understand the locus where a deformation of $I$ induces a deformation
        of $I^{\sat}$. The following is our most general result in this
        direction (see~\S\ref{ssec:smoothnessOfProjections} for more fleshy
        results under additional assumptions).
        \begin{proposition}\label{ref:projectionFromSatLocallyClosed:prop}
            Let $I$ be a homogeneous ideal of $S$ and $I^{\sat}$ be its
            saturation. Let
            \[
                p := \pr_I\colon \Hilb_{I\subseteq I^{\sat}}\to \Hilb_I,
            \] then $p$ is a locally closed immersion near $[I\subseteq
            I^{\sat}]$.
            More precisely, let $U$ be the open neighborhood of $[I\subseteq
            I^{\sat}]\in \Hilb_{I\subseteq I^{\sat}}$ that consists of pairs
            $[K\subseteq J]$ with $J$ saturated and let $Z = \Hilb_{I\subseteq
            I^{\sat}} \setminus\; U$.
            Then $p|_U\colon U\to \Hilb_I\setminus\; p(Z)$ is a closed
            immersion.
        \end{proposition}
        \begin{proof}
            First, the subset $U$ is equal to
            $\pr_{I^{\sat}}^{-1}(\Hilb_{I^{\sat}}^{\sat})$, hence indeed open
            by Theorem~\ref{ref:opennessOfSaturation:thm}.
            Observe that
            $p^{-1}(p(Z)) = Z$ as closed subsets. Suppose not, then for
            some $[I'\subseteq (I')^{\sat}]\in \Hilb_{I\subseteq I^{\sat}}$
            there exist also $[I' \subseteq J]\in Z$ with $J$ nonsaturated.
            The ideals $I'$, $(I')^{\sat}$ and $J$ all have the same Hilbert function
            for indices $\ee$ with $\min \ee \gg 0$, hence $I'_{\ee} =
            J_{\ee}$ for such $\ee$. It follows that $(I')^{\sat} = J^{\sat}$,
            but then again from equality of Hilbert functions
            we get that the inclusion $J \subseteq J^{\sat}$ is an equality.
            The proof of $p^{-1}(p(Z)) = Z$ is completed.
            The projective morphism $p$ restricts to a \emph{projective}
            morphism
        \[
            p|_U\colon U = \Hilb_{I\subseteq I^{\sat}}\setminus\; p^{-1}(p(Z))
            \to \Hilb_I \setminus\; p(Z).
        \]
        We claim that this morphism is a closed immersion. By~
        \cite[\href{https://stacks.math.columbia.edu/tag/04XV}{Tag 04XV},
        \href{https://stacks.math.columbia.edu/tag/05VH}{Tag
        05VH}]{stacks-project} it is enough to prove that for every point
        $[I']\in \Hilb_I \setminus\; p(Z)$ the fiber
        $p^{-1}([I'])$
        is either empty or maps isomorphically to the point $[I']$.
        It is enough to prove this claim after field
        extension~\cite[\href{https://stacks.math.columbia.edu/tag/02L4}{Tag
        02L4}]{stacks-project}. For a point $[I']$, we base change by
        replacing $\kk$ with the field
        $\overline{\kappa([I'])}$, so we get that
        the base field is algebraically closed and $[I']$ is closed.
        Consider the
        fiber $p^{-1}([I'])$.
        Arguing as above, we get that \emph{as a set} the fiber is
        \begin{itemize}
            \item a singleton $[I'\subseteq (I')^{\sat}]$ if
                $H_{S/(I')^{\sat}} = H_{S/I^{\sat}}$,
            \item empty if $H_{S/(I')^{\sat}} \neq H_{S/I^{\sat}}$.
        \end{itemize}
        Assume that the fiber is nonempty.
        Since $\Hom_S((I')^{\sat}/I', S/(I')^{\sat}) = 0$, by the main
        diagram~\ref{ssec:maindiagram} we get that the tangent map to $p$
        at $[I'\subseteq (I')^{\sat}]$ is injective. Hence the fiber has zero
        tangent space, so \emph{as a scheme} it is a point.
        \end{proof}

\subsection{Miscellaneous results}
In this section we gather two rather technical yet general results, which form
important parts of the arguments in~\S\ref{sec:examples}
and~\S\ref{ssec:smoothnessOfSaturated}, respectively.
\begin{lemma}[small tangent space implies large
        intersection, {\cite[Lemma~2.6]{Mandziuk}}]\label{ref:boundOnIntersectionFromTangentSpace:lem}
    Let $X$ be a scheme locally of finite type over $\kk$ and $Z_1,
    Z_2\subseteq X$ be irreducible closed subschemes. Let $x\in Z_1\cap Z_2$
    be a $\kk$-point. Then every irreducible component of $Z_1\cap Z_2$ has
    dimension at least
    \[
        \dim Z_1 + \dim Z_2 - \dim T_x X.
    \]
\end{lemma}

Let $X$ be a $\kk$-scheme and $\varphi\colon \Spec(A)\to X$ be a morphism of
schemes. The $A$-module
\[
    \Hom_A(H^0(\varphi^*\Omega_{X/\kk}), A)
\]
is called
the \emph{module of (pulled back) vector fields on $\Spec(A)$}. We have the
following description.
\begin{lemma}[vector fields and cotangent module]\label{ref:vectorfields:lem}
    There is a bijection functorial in $A$ between the elements of the module
    of vector fields on $\Spec(A)$ and the set of morphisms of $\kk$-schemes
    $\varphi'\colon \Spec(A[\varepsilon]/\varepsilon^2)\to X$ which are equal
    to $\varphi$
    when restricted to $\Spec(A)$.
\end{lemma}
The proof is standard, yet we could not find a ready reference. Likely the
most natural way of arguing is gluing on affine parts, but this seems messy to
write down. As a result, the proof is quite hermetic, the reader might like to
follow it for affine $X$ first.
\begin{proof}
    \def\mathcal{A}{\mathcal{A}}
    Let $\mathcal{A} := \varphi_*\OO_{\Spec(A)}$. To give a morphism $\varphi'$ as in the statement is
    to give a sheaf-of-$\kk$-algebras homomorphism $(\varphi')^{\#}\colon \OO_X\to
    \mathcal{A}[\varepsilon]/\varepsilon^2$ that restricts to $\varphi^{\#}$ when
    composed with $\mathcal{A}[\varepsilon]/\varepsilon^2\twoheadrightarrow \mathcal{A}$. Write $(\varphi')^{\#}
    = \varphi^{\#} + \varepsilon \delta$ for $\delta\colon \OO_X\to \mathcal{A}$.
    The fact that $(\varphi')^{\#}$ is a homomorphism is equivalent to the
    fact that $\delta$ is a $\kk$-linear derivation of the $\OO_X$-module
    $\mathcal{A}$. By the universal property of $\Omega_{X/\kk}$, such derivations are
    in bijection with $\Hom_{\OO_X}(\Omega_{X/\kk}, \mathcal{A})$, which by adjunction
    is isomorphic $\Hom_{\OO_{\Spec(A)}}(\varphi^* \Omega_{X/\kk},
    \OO_A)$, which is in bijection with the module of vector fields on
    $\Spec(A)$.
\end{proof}

        \section{General results}

        \subsection{Fiber obstruction group for saturation}
        \newcommand{\cokern}[1]{C(#1)}
        Let $I$ be a homogeneous ideal in $S$.
        For the pair $(K, J) := (I, I^{\sat})$, the fiber obstruction group
        from~\eqref{eq:obfib}
        can be ``underived'' as follows. Choose a transverse element $\ell$
        and let $C(I)$ be defined by the exact sequence
        \begin{equation}\label{eq:cokernel}
            0\to \frac{S}{I^{\sat}}\to \left( \frac{S}{I} \right)_{\ell}\to
            \cokern{I}\to 0.
        \end{equation}
        Applying $\Hom(I^{\sat}/I, -)$ we obtain a long exact sequence.
        The multiplication by a large enough power of $\ell$ annihilates
        $I^{\sat}/I$ and is an isomorphism on $(S/I)_{\ell}$, hence
        $\Ext^{\bullet}_S(I^{\sat}/I, (S/I)_{\ell}) = 0$ and the long exact
        sequence induces a homogeneous isomorphism
        \begin{equation}\label{eq:fiberObstructionConcrete}
            \Ext^1_S\left( \frac{I^{\sat}}{I}, \frac{S}{I^{\sat}}
            \right)  \simeq \Hom\left( \frac{I^{\sat}}{I},
            \cokern{I} \right).
        \end{equation}
        For further use, we refine it taking into account degrees.
        Fix degrees $\midx \leq \ee$.
        Applying $\Hom(I^{\sat}_{\geq \mm}/I_{\geq \ee}, -)$ to $0\to
        (S/I^{\sat})_{\geq \midx} \to ((S/I)_{\ell})_{\geq \midx} \to
        \cokern{I}_{\geq \midx}\to 0$ we arrive
        at the homogeneous isomorphism
        \begin{equation}\label{eq:gradedfiberObstructionConcrete}
            \Ext^1_S\left( \frac{I^{\sat}_{\geq \midx}}{I_{\geq \ee}},
            \left(\frac{S}{I^{\sat}}\right)_{\geq \midx} \right) \simeq \Hom\left( \frac{I^{\sat}_{\geq
             \midx}}{I_{\geq \ee}}, \cokern{I}_{\geq \midx} \right).
        \end{equation}
        By construction, the module $M := I^{\sat}_{\geq \midx}/I_{\geq \ee}$ is
        generated in degrees $\geq \midx$. The same is true for modules in the minimal resolution
        of $M$ since the ring $S$ is positively
        graded. The Ext groups are computed from this resolution, so if we
        restrict to homogeneous maps, we can
        drop the subscripts in~\eqref{eq:gradedfiberObstructionConcrete} to obtain
        \begin{equation}\label{eq:fiberObstructionConcreteRestricted}
            \Ext^1_S\left( \frac{I^{\sat}_{\geq \midx}}{I_{\geq \ee}},
            \frac{S}{I^{\sat}_{\geq \midx}} \right)_{\mathbf{0}} \simeq \Hom\left( \frac{I^{\sat}_{\geq
            \midx}}{I_{\geq \ee}}, \cokern{I_{\geq \ee}} \right)_{\mathbf{0}}.
        \end{equation}

        \newcommand{\Homstar}{{}^*\!\Hom}%
        The $S/I^{\sat}$-module $\cokern{I}$ is pleasantly explicit. If $\dim
        S/I^{\sat} = 1$ it admits connections to the canonical module
        $\omega_{S/I^{\sat}}$~\cite[\S13.1]{24hours}, which
        are useful since the latter is sometimes even easier to describe. We discuss
        this connection now.

        Recall the graded dual operator: for a graded vector space $M =
        \bigoplus_{\ee} M_{\ee}$ with every $M_{\ee}$ finite-dimensional we define $\Homstar(M, \kk) :=
        \bigoplus_{\ee} \Hom(M_{-\ee}, \kk) \subseteq \Hom(M, \kk)$. The
        inclusion is an equality if $M$ has finite dimension but not in
        general. If $M$ is an $S$-module, then $\Homstar(M, \kk)$ is an
        $S$-module as well by $(s\cdot \varphi)(m) := \varphi(sm)$ and the
        canonical map $M\to \Homstar(\Homstar(M, \kk), \kk)$ is an isomorphism
        of $S$-modules.

        \begin{proposition}\label{ref:CIAndcanonicalModule:prop}
            Suppose that $S$ is standard $\mathbb{N}$-graded and $\dim
            S/I^{\sat} = 1$. Then the module $\cokern{I}$ is isomorphic to
            $\Homstar(\omega_{S/I^{\sat}}, \kk)$ and to the top local cohomology
            of $S/I^{\sat}$.
        \end{proposition}
        \begin{proof}
            Let $\ell_1$ be a transversal element for $I$.
            Since $S/I^{\sat}$ is one-dimensional, we may pick $\dim S - 1$
            elements $\ell_2, \ldots ,\ell_{\dim S}$ of $S_+$ that annihilate
            $S/I^{\sat}$ and such that the radical of $(\ell_1, \ell_2, \ldots
            ,\ell_{\dim S})$ is $S_+$.
            Using properties of local cohomology~\cite[Proposition~7.3(b),
            Theorem~7.13]{24hours}, we conclude that $\cokern{I} \simeq
            H^1_{S_+}(S/I^{\sat})$. Since $S/I^{\sat}$ is one-dimensional,
            by~\cite[Theorem~13.5]{24hours}
            we obtain $\Homstar(\cokern{I}, \kk) \simeq \omega_{S/I^{\sat}}$.
            Applying $\Homstar(-, \kk)$ to this isomorphism we get the claim.
        \end{proof}
        The above Proposition~\ref{ref:CIAndcanonicalModule:prop} becomes
        easier in the Gorenstein case. Recall that $S/I^{\sat}$ is
        \emph{Gorenstein} if for some (or every) transverse element $\ell$ the
        socle $S/(I^{\sat}+(\ell))$ is
        one-dimensional~\cite[Proposition~21.5]{Eisenbud}.
        \begin{corollary}[Gorenstein case]\label{ref:CIForGorenstein:prop}
            Suppose that $S$ is standard $\mathbb{N}$-graded and $
                S/I^{\sat}$ is one-dimensional and Gorenstein. Then the module $\cokern{I}$ is isomorphic to
            $\Homstar(S/I^{\sat}, \kk)(-a)$ where $a$ is the largest index
            such that $H_{S/I^{\sat}}(a) \neq H_{S/I^{\sat}}(a+1)$.
        \end{corollary}
        \begin{proof}
            It follows from Grothendieck's local
            duality~\cite[Theorem~18.7]{24hours} and from
            Proposition~\ref{ref:CIAndcanonicalModule:prop} that $\cokern{I}
            \simeq \Homstar(S/I^{\sat}, \kk)(-a)$ for a certain $a$. It
            follows from this isomorphism that $a$ is the largest degree in
            which $\cokern{I}$ is nonzero. From~\eqref{eq:cokernel} it follows
            that this agrees with the $a$ in the statement.
        \end{proof}
        \begin{corollary}\label{ref:fiberObstructionForGorenstein:cor}
            Let $S$ be standard $\mathbb{N}$-graded.
            Let $I$ be a homogeneous ideal such that $I^{\sat}$ satisfies the
            assumptions of Corollary~\ref{ref:CIForGorenstein:prop}. Then
            \[
                \ObFib(I, I^{\sat}) \simeq
                \Hom_{\kk}\left(\left(\frac{I^{\sat}}{I+(I^{\sat})^2}\right)_a,
                \kk \right).
            \]
        \end{corollary}
        \begin{proof}
            Let $R = S/I^{\sat}$. Using Corollary~\ref{ref:CIForGorenstein:prop}
            and~\eqref{eq:fiberObstructionConcrete} we obtain that
            \begin{align*}
                &\ObFib(I, I^{\sat})
                \simeq \Hom_{R}\left( \frac{I^{\sat}}{I+(I^{\sat})^2},
                \cokern{I} \right)_{0}
                \simeq \Hom_{R}\left( \frac{I^{\sat}}{I+(I^{\sat})^2},
                \Homstar(R, \kk)(-a) \right)_{0}\\
                & \simeq \Hom_{R}\left( \frac{I^{\sat}}{I+(I^{\sat})^2},
                \Homstar(R, \kk) \right)_{-a}\stackrel{(\star)}{\simeq}
                \Hom_{\kk}\left(\frac{I^{\sat}}{I+(I^{\sat})^2}, \kk
                \right)_{-a}  \simeq
                \Hom_{\kk}\left(\left(\frac{I^{\sat}}{I+(I^{\sat})^2}\right)_a, \kk
                \right),
            \end{align*}
            where $(\star)$ follows from the graded version of the tensor-Hom
            adjunction.
        \end{proof}

    \subsection{Smoothness of projection maps in the case of
    saturation}\label{ssec:smoothnessOfProjections}

    In this section we give cohomological criteria for the existence of sticky
    ideals (see introduction). The only known to us deformation theoretic way of proving
    that a given map is surjective around a given point is to prove that it is
    smooth. We give criteria for that.

        \begin{theorem}\label{ref:deformsWhenObFibVanishes:thm}
            Suppose that we have $\ObFib(K, J) = 0$. Then
            $\pr_K\colon \Hilb_{K\subseteq J}\to \Hilb_K$ is smooth at $[K\subseteq J]$.
            In particular, if $K\subsetneq J\subseteq K^{\sat}$ then
            $K$ is nonsaturable.
        \end{theorem}
        \begin{proof}
            Since $\ObFib = 0$, the assumptions of
            Theorem~\ref{ref:obstructionAtFlag:thm} are satisfied.
            By the main diagram~\S\ref{ssec:maindiagram}, the tangent map to $\pr_K\colon \Hilb_{K\subseteq
                J}\to \Hilb_K$ is surjective, while the obstruction map
                is injective. This proves that $\pr_K$ is smooth at $[K]$. In
                particular its image contains an open neighborhood of $[K]$.
                Suppose that $K$ is saturable and let $K_t$ be a family with
                special fiber $K$ and general fiber $K_{\eta}$ such that
                $K_{\eta}$ is saturated. Then, after perhaps shrinking the
                base of the family, we would also get a family $K_t\subseteq
                J_t$ with special fiber $J$.
                Since $J\subseteq K^{\sat}$, the ideal $J/K$ is annihilated by
                a large enough power of $S_+$, hence its Hilbert function satisfies $H_{J/K}(\ee) = 0$ for
                all $\ee$ with
                $\min \ee \gg 0$. The same holds for $J_{\eta}/K_{\eta}$, hence
                $S_+^{e}(J_{\eta}/K_{\eta}) = 0$ for $e\gg 0$. But this is a
                contradiction with $K_{\eta}$ being saturated.
        \end{proof}

        It may happen that the obstruction group is nonzero, yet the map
        $\pr_K$ is smooth, so we provide the following refined criterion.
        \begin{theorem}\label{ref:deformWhenTangentSurjective:thm}
            Suppose that $(K,J)$ are such that the map $\Hom_S(J,
            S/J)_{\mathbf{0}}\to \Hom_S(K, S/J)_{\mathbf{0}}$ is surjective
            and that $[J]\in \Hilb_J$ is a smooth point. Then $\pr_K$ is
            smooth at $[K\subseteq J]$.
            In particular, if $K\subsetneq J\subseteq K^{\sat}$ then
        $K$ is nonsaturable.
        \end{theorem}
        \begin{proof}
            By assumption, the map $\psi_{\mathbf{0}}$ is surjective, so we can apply
            Theorem~\ref{ref:obstructionAtFlag:thm}.
            By pullback, the map $d\pr_K\colon \TgFlag\to
            \Hom_S(K,S/K)_{\mathbf{0}}$ is surjective.
            The map $\ObFlag\to
            \Ext^1_{S}(J, S/J)_{\mathbf{0}}$ is a map of obstruction theories,
            so the image of any obstruction is the obstruction to deforming
            $[J]$, which is zero by assumption. Thus the obstructions actually
            live in a smaller space $O := \ker(\ObFlag\to \Ext^1_{S}(J,
            S/J)_{\mathbf{0}})$ which is thus by definition another obstruction space at
            $[K\subseteq J]\in \Hilb_{K\subseteq J}$. Since $O\to \Ext^1_S(K,
            S/K)_{\mathbf{0}}$ is injective, it follows from the Main Theorem of
            obstruction calculus that $\pr_K$ is smooth.
        \end{proof}

        In the special case $(K, J) = (I, I^{\sat})$ we can be a little more
        precise.
        \begin{theorem}\label{ref:saturationDeformsWhenObFibVanishes:thm}
            Let $I$ be a homogeneous ideal and
            suppose that $\ObFib(I, I^{\sat}) = 0$. Then $\pr_{I}$ is an open immersion
            near $[I \subseteq I^{\sat}]$. In particular, if $I$ is non
            saturated, it is entirely nonsaturable.
        \end{theorem}
        \begin{proof}
            Since $\ObFib = 0$, the assumptions of
            Theorem~\ref{ref:obstructionAtFlag:thm} are satisfied.
            The multiplication by a transverse element $\ell$ for $I$ is an
            injective map on $S/I^{\sat}$ and a nilpotent one on $I^{\sat}/I$,
            hence $\Hom_S(I^{\sat}/I, S/I^{\sat}) = 0$.
            By the main diagram, the tangent map to $\pr_I\colon \Hilb_{I\subseteq
                I^{\sat}}\to \Hilb_I$ is bijective, while the obstruction map
                is injective. This proves that $\pr_I$ is \'etale at $[I]$. In
                particular its image contains an open neighborhood of $[I]$,
                whence if $I^{\sat}\neq I$, then $I$ is entirely nonsaturable.
                Consider  $U := \pr^{-1}_{I^{\sat}}(\Hilb_{I^{\sat}}^{\sat})$
                which is an open
                (Theorem~\ref{ref:opennessOfSaturation:thm}) neighbourhood of
                $[I\subseteq I^{\sat}]$. For every point $[I'
                \subseteq I'']\in U$ we have $I'' = (I')^{\sat}$. Thus the map
                $(\pr_I)|_U\colon U\to \Hilb_I$ is universally
                injective~\cite[\href{https://stacks.math.columbia.edu/tag/01S4}{Tag
                01S4}]{stacks-project}. By shrinking $U$ to a neighborhood of
                $[I\subseteq I^{\sat}]$, we assume that it is \'etale on $U$
                as well. By~\cite[\href{https://stacks.math.columbia.edu/tag/025G}{Tag
                025G}]{stacks-project} the morphism $(\pr_I)|_U$ is an open
                immersion. This immediately implies that $I$ is entirely
                nonsaturable.
        \end{proof}

        Again, the vanishing of $\ObFib(I, I^{\sat})$ is not
        necessary for $I$ to be nonsaturable, see~\S\ref{ssec:nonvanishingGroup}
        for an explicit example. We state a slightly more
        general version of the theorem.
        \begin{theorem}\label{ref:saturationDeformWhenTangentSurjective:thm}
            Suppose that the map $\Hom_S(I^{\sat},
            S/I^{\sat})_{\mathbf{0}}\to\Hom_S(I, S/I^{\sat})_{\mathbf{0}}$ is
            surjective and that $[I^{\sat}]\in \Hilb_{I^{\sat}}$
            is a smooth point. Then $\pr_I$ is an open immersion near $[I\subseteq
            I^{\sat}]$. In particular, if $I$ is not saturated, it is entirely
            nonsaturable.
        \end{theorem}
        \begin{proof}
            By Theorem~\ref{ref:deformWhenTangentSurjective:thm} the map is
            smooth and arguing as in the proof of
            Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} we see
            that $\pr_I$ is an open immersion near $[I\subseteq I^{\sat}]$.
        \end{proof}

        \begin{remark}\label{ref:canonicalModule:remark}
            Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} is
            inspired by the works of Cid-Ruiz and
            Ramkumar~\cite{CidRuiz_Ramkumar__Fiber_full,
            CidRuiz_Ramkumar__Fiber_full_local} which in turn were inspired
            by~\cite{Conca_Varbaro__squareFree}. Roughly speaking,
            the difference is that Cid-Ruiz and Ramkumar employ local cohomology and
            impose stronger conditions that guarantee that also the higher local
            cohomology groups $H^i$ deform with the ideal, while we only investigate
            whether $H^0_{S_+}(S/I) = I^{\sat}/I$ deforms with $I$.
        \end{remark}

    \subsection{Smoothness of the saturated locus in the multigraded Hilbert
    schemes}\label{ssec:smoothnessOfSaturated}

    In this section we slightly change focus to provide classes of
    \emph{saturated} ideals which yield smooth
    points of the multigraded Hilbert schemes.
    Throughout this section, the ring $S$ is $\mathbb{N}$-graded in the
    standard way.
    Likely the results of this section generalize (certainly
    Corollary~\ref{ref:usualClassesSmooth:cor} does, under the assumption of
    Cohen-Macaulayness), but we restrict to the $\mathbb{N}$-grading for clarity
    of presentation. We first identify the saturated locus with a locally
    closed subscheme of the usual Hilbert scheme.
    \begin{proposition}\label{ref:constantHilbertFunction:prop}
        Fix a Hilbert function $H$ admissible for a saturated one-dimensional ideal in
        $S$. Let $n = \dim S$ and $d = H(e)$ for $e\gg 0$. The natural map
        \begin{equation}\label{eq:toUsualHilb}
            \Hilb^{H,\sat}\to \Hilb^d(\mathbb{P}^{n-1})
        \end{equation}
        is a locally closed immersion whose image consists of subschemes
        $\Gamma\subseteq \mathbb{P}^{n-1}$ with $S/I_{\Gamma}$ having Hilbert
        function $H$.
    \end{proposition}
    \begin{proof}
        Let $I^{\sat}$ be a saturated ideal with Hilbert function $H$, so that
        $\Hilb_{I^{\sat}} = \Hilb^H$, and let $e\gg
        0$.
        As discussed in~\S\ref{ssec:setup}, the projective map $\Hilb^{H}\to
        \Hilb^d(\mathbb{P}^{n-1})$ is the composition
        \[
            \begin{tikzcd}
                \Hilb_{I^{\sat}} \ar[r, "\simeq"] & \Hilb_{I^{\sat}_{\geq e}\subseteq I^{\sat}}
                \ar[r, "\pr_{I^{\sat}_{\geq e}}"] & \Hilb_{I^{\sat}_{\geq e}}
            \end{tikzcd}
        \]
        so the claim follows directly from
        Proposition~\ref{ref:projectionFromSatLocallyClosed:prop} applied to
        the ideal $I^{\sat}_{\geq e}$.
    \end{proof}

    Recall that $\Hilb^d(\mathbb{P}^{n-1})$ has a distinguished component $\Hilb^{d, \sm}(\mathbb{P}^{n-1})$,
    called the \emph{smoothable component}. The general point $[\Gamma]$ of
    this component corresponds to a reduced $\Gamma$, that is, to a tuple of $d$ points of $\mathbb{P}^{n-1}$.
    Even more, the set of points $[\Gamma]\in \Hilb^d(\mathbb{P}^{n-1})$ such
    that $\Gamma$ is smooth, is open. For the purposes of this discussion, we
    call it the \emph{locus of smooth subschemes} of $\Hilb^d(\mathbb{P}^{n-1})$.
    \begin{definition}\label{ref:Slip:def}
        The \emph{Scheme of Limits of Ideals of Points} $\Slip^H\subseteq
        \Hilb^H$ is the closure of the open subscheme: the intersection of
        $\Hilb^{H, \sat}$ and the preimage of the locus of smooth subschemes.
        For $[I]\in \Hilb^H$, we also denote $\Slip^H$ by $\Slip_I$.
    \end{definition}
    By definition, a general point of $\Slip^H$ corresponds to a saturated
    ideal $I^{\sat}$ such that $V(I^{\sat})$ is a tuple of points with Hilbert function $H$.
    The map $\Slip^H\to \Hilb^{d, \sm}(\mathbb{P}^{n-1})$ can be described a
    bit more precisely when $H$ is the Hilbert function of a general $d$-tuple
    of points in $\mathbb{P}^{n-1}$, that is, $H(i) = \min(d, \binom{n-1+i}{i})$
for every $i$.
    For a zero-dimensional subscheme $\Gamma$ in a smooth quasi-projective scheme
    $X$, we say that
    $\Gamma$ is \emph{unobstructed} if $[\Gamma]\in
    \Hilb^d(X)$ is a smooth point, where $d = \dim_{\kk} H^0(\OO_{\Gamma})$ and
    $\Hilb^d(X)$ is a Grothendieck Hilbert scheme. Unobstructedness is an
    intrinsic property of $\Gamma$, it does not depend on the embedding.
    \begin{corollary}\label{ref:exceptionalIsDivisor:cor}
        Let $H$ be the Hilbert function of a general $d$-tuple of points in
        $\mathbb{P}^{n-1}$. Let $[I]\in \Slip^H$ be such that $V(I^{\sat})\subseteq
        \mathbb{P}^{n-1}$ is unobstructed and the fiber of the
        map $\Slip^H\to \Hilb^d(\mathbb{P}^{n-1})$ over $V(I^{\sat})$ is not a point. Then there is a divisor $E\subseteq
        \Slip^H$ with $[I]\in E$ such that the image of $E$ in
        $\Hilb^d(\mathbb{P}^{n-1})$ has codimension at least two.
    \end{corollary}
    \begin{proof}
        The locally closed immersion from
        Proposition~\ref{ref:constantHilbertFunction:prop} defines a birational
        morphism $\Slip^H \to \Hilb^{d, \sm}(\mathbb{P}^{n-1})$. Indeed, its restriction to
        the inverse image of the locus $\mathcal{V}$ of all smooth subschemes with Hilbert
        function $H$ is a surjective locally closed immersion onto the smooth
        scheme $\mathcal{V}$ and thus an isomorphism.
        The claim follows from applying~\cite[Theorem~2, p.120]{Shafarevich_1}
        to this map.
    \end{proof}

    We now proceed to the question of when a saturated ideal corresponds to a
    smooth point of the multigraded Hilbert scheme.

    For a saturated homogeneous one-dimensional ideal $I^{\sat}$ with a transverse element
    $\ell$ which is a linear form we fix a decomposition $S = S'[\ell]$,
    where $S'$ is a polynomial subring. We define the one-parameter family for
    $(I^{\sat},\ell, S')$ as
    \[
        \begin{tikzcd}
            \Spec(S/I^{\sat})\ar[rd, "\pi"']\ar[r, hook]  & \Spec(S)  \simeq  \Spec(S')\times
            \mathbb{A}^1\ar[d, "\pr_2"]\\
            & \mathbb{A}^1 = \Spec(\kk[\ell])
        \end{tikzcd}
    \]
    where $\pi$ is induced by the inclusion $\kk[\ell]\to S/I^{\sat}$.
    In the projective space $\Proj(S)$, the quotient $\Proj(S/I^{\sat})$ corresponds
    to a zero-dimensional subscheme disjoint from the hyperplane $(\ell = 0)$. It follows
    that for every nonzero $\lambda\in \kk$ the scheme
    $\pi^{-1}(\lambda) = \Spec(S/(I^{\sat}+(\ell-\lambda)))$ is isomorphic to $\Proj(S/I^{\sat})$.
    \begin{lemma}\label{ref:family:lem}
        For every $(I^{\sat}, \ell, S')$ as above, the map $\pi$ is finite and flat.
    \end{lemma}
    \begin{proof}
        The $\kk$-algebra $S/(I^{\sat}+(\ell))$ is finitely generated and zero-dimensional, hence a
        finite-dimensional $\kk$-vector space.
        Fix its basis $\mathcal{B}$. Since $S/I^{\sat}$ is $\mathbb{N}$-graded,
        induction by degree proves that $S/I^{\sat}$ is spanned by $\mathcal{B}$ as a
        $\kk[\ell]$-module, so $\pi$ is finite. Suppose that there is a nonzero element of $S/I^{\sat}$ annihilated
        by some nonzero polynomial in $\kk[\ell]$. Taking leading forms, we
        find a nonzero element of $S/I^{\sat}$ annihilated by a power of $\ell$, a
        contradiction with transversality of $\ell$. Hence, the $\kk[\ell]$-module $S/I^{\sat}$ is torsion-free, so
        $\pi$ is flat.
    \end{proof}
    \begin{proposition}\label{ref:extendingTangentVector:prop}
        Suppose that $\ell$ is a linear form transverse to a saturated
        one-dimensional ideal
        $I^{\sat}$ and that $\Spec(S/(I^{\sat}+(\ell))\subseteq \Spec(S)$ is unobstructed.
        Then the natural map
        \[
            \frac{\Hom_S\left( I^{\sat}, \frac{S}{I^{\sat}} \right)}{\ell\cdot \Hom_S\left( I^{\sat}, \frac{S}{I^{\sat}} \right)}\to \Hom_{S}\left( I^{\sat},
            \frac{S}{I^{\sat}+(\ell)}
            \right)
        \]
        is an isomorphism.
    \end{proposition}
    The theorem does not hold under the slightly weaker assumption that
    $\Proj(S/I^{\sat}) \simeq \Spec(S/(I^{\sat}+(\ell-1))\subseteq \Proj(S)$ is
    unobstructed; one counterexample is a general tuple of $8$ points on
    $\mathbb{P}^3$.
    \begin{proof}
        By left-exactness of Hom, it is enough to prove that $\Hom_S(I^{\sat},
        S/I^{\sat})\to \Hom_S(I^{\sat}, S/(I^{\sat}+(\ell))$ is surjective.
        The family $\pi$ above provides a map $\widetilde{\pi}\colon \mathbb{A}^1\to
        \Hilb^d(\Spec(S'))$, where
        $d = \dim_{\kk} \frac{S}{I^{\sat}+(\ell)}$. By assumption, the point
        $\widetilde{\pi}(0)\in \Hilb^d(\Spec(S'))$ is smooth, hence a general
        point of the image is smooth as well. But all $\kk$-points of
        $\widetilde{\pi}|_{\mathbb{A}^1\setminus \{0\}}$ correspond to different
        embeddings of the same abstract scheme $\Proj(S/I^{\sat})$, hence we conclude that every point in the
        image of $\widetilde{\pi}$ is smooth. In particular, the cotangent sheaf
        of $\Hilb^d(\Spec(S'))$ pulled back to $\mathbb{A}^1$ is a
        \emph{free} sheaf of $\mathbb{A}^1$. As a result, we obtain
        surjectivity of the natural
        restriction map from
        the module of pulled back vector fields on $\Spec(\kk[\ell])$ to such
        a module on $\Spec(\kk[\ell]/(\ell))$.

        By Lemma~\ref{ref:vectorfields:lem} and a generalization
        of~\cite[Theorem~10.1]{Stromme_Hilbert}, we conclude that the
        $\kk[\ell]$-module
        of (pulled back) vector fields for $\widetilde{\pi}$ is
        $\Hom_S(I^{\sat}, S/I^{\sat})$ and that this is a free $\kk[\ell]$-module. Applying
        the same Lemma and reference
        to $\Spec(\kk) = (\ell=0)\to \Hilb^d(\Spec(S'))$ we obtain
        $\Hom_{S'}(I^{\sat}/(I^{\sat}\cap (\ell), S/(I^{\sat}+(\ell))$.
        As mentioned above,
        the restriction map of modules of vector fields
        \[
            \Hom_S\left( I^{\sat}, \frac{S}{I^{\sat}} \right)\to
            \Hom_{S'}\left(\frac{I^{\sat}}{I^{\sat}\cap(\ell)},\; \frac{S}{I^{\sat}+(\ell)}
            \right)
        \]
        is surjective. Since $\ell$ is a nonzerodivisor in $S/I^{\sat}$, we have
        $I^{\sat}\cap (\ell) = I^{\sat}\cdot (\ell)$, so the natural map
        \[
            \Hom_S\left( I^{\sat}, \frac{S}{I^{\sat}+(\ell)} \right)\to \Hom_{S'}\left(
            \frac{I^{\sat}}{I^{\sat}\cap (\ell)},\; \frac{S}{I^{\sat}+(\ell)}
            \right)
        \]
        is bijective. It follows that the restriction $\Hom_S(I^{\sat}, S/I^{\sat})\to
        \Hom_{S}\left(I^{\sat}, S/(I^{\sat}+(\ell))\right)$ is surjective as well.
    \end{proof}

    \begin{corollary}\label{ref:zeroMap:cor}
        Under the assumptions of
        Proposition~\ref{ref:extendingTangentVector:prop}, the multiplication
        by $\ell$ on $\Ext^1_S(I^{\sat}, S/I^{\sat})$ is injective. Consequently, for every
        subideal $K \subseteq I^{\sat}$ such that $I^{\sat}\cdot \ell^{\gg 0} \subseteq K$ the natural map
        $\Ext^1_S(I^{\sat}/K, S/I^{\sat})\to \Ext^1_S(I^{\sat}, S/I^{\sat})$
        is zero.
    \end{corollary}

    \begin{proof}
        Applying $\Hom_{S}(I^{\sat}, -)$ to the short exact sequence
        \[
            0\to \frac{S}{I^{\sat}}(-1)\to \frac{S}{I^{\sat}}\to \frac{S}{I^{\sat}+(\ell)}\to 0
        \]
        and using
        Proposition~\ref{ref:extendingTangentVector:prop} we obtain that the
        multiplication by $\ell$ on $\Ext_S^1(I^{\sat}, S/I^{\sat})$ is injective. The
        multiplication by $\ell$ on $I^{\sat}/K$ is nilpotent, so it is
        nilpotent on
        $\Ext^1_S(I^{\sat}/K, S/I^{\sat})$ as well, whence the claim.
    \end{proof}

    \begin{theorem}[unobstructedness of Artinian reduction implies smoothness
        in multigraded setting]\label{ref:smoothPoints:thm}
        Let $S$ be a standard graded polynomial ring and $I^{\sat}$ be a saturated one-dimensional
        homogeneous ideal with transverse element $\ell$ which is a linear
        form. If $\Spec(S/(I^{\sat}+(\ell))\subseteq \Spec(S)$ is unobstructed, then
        $[I^{\sat}]\in \Hilb_{I^{\sat}}$ is a smooth point.
    \end{theorem}
    \begin{proof}
        Let $d = \dim_{\kk} S/(I^{\sat}+(\ell))$ and fix $e\gg 0$.
        Consider the canonical map defined in~\S\ref{ssec:setup}:
        \[
            \begin{tikzcd}
                \Hilb_{I^{\sat}} \ar[r, "\simeq", "\pr_{I^{\sat}}^{-1}"'] &
                \Hilb_{I^{\sat}_{\geq e}\subseteq I^{\sat}} \ar[r, "\pr_{I^{\sat}_{\geq e}}"] &
                \Hilb_{I^{\sat}_{\geq e}} \simeq \Hilb^d(\mathbb{P}^{\dim S-1}).
            \end{tikzcd}
        \]
        The natural map $\Hom_{S}(I^{\sat}_{\geq e}, S/I^{\sat}_{\geq e})_{\mathbf{0}}\to
        \Hom_S(I^{\sat}_{\geq e}, S/I^{\sat})_{\mathbf{0}}$ is surjective, hence
        Theorem~\ref{ref:obstructionAtFlag:thm} applies and shows that
        $[I^{\sat}_{\geq e}\subseteq I^{\sat}]$ has an obstruction group $\ObFlag$ given by
        \[
            \begin{tikzcd}
                \ObFlag\ar[r]\ar[d]\arrow[dr, phantom,
                "\usebox\pullback" , very near start, color=black] &
                \Ext^1_S\left( I^{\sat},
                \frac{S}{I^{\sat}} \right)_{\mathbf{0}}\ar[d]\\
                \Ext^1_S\left(I^{\sat}_{\geq e}, \frac{S}{I^{\sat}_{\geq e}}\right)_{\mathbf{0}}\ar[r] & \Ext^1_S\left( I^{\sat}_{\geq e},
                \frac{S}{I^{\sat}} \right)_{\mathbf{0}}.
            \end{tikzcd}
        \]
        Using the one-parameter family $\pi$ and the fact that
        unobstructedness is open, we conclude that
        $\Spec(S/(I^{\sat}+(\ell-1))\subseteq \Spec(S)$ is unobstructed, hence also $\Proj(S/I^{\sat}) \subseteq
        \Proj(S)$ is unobstructed, so $[I^{\sat}_{\geq e}]\in \Hilb_{I^{\sat}_{\geq e}}$ is
        a smooth point. This means that for any obstruction
        at $[I^{\sat}_{\geq e}\subseteq I^{\sat}]$
        arising from any small extension, its image under the left vertical
        map is zero.
        By Corollary~\ref{ref:zeroMap:cor}, the right vertical map is
        injective.
        But then it follows that every such obstruction is zero,
        hence $[I^{\sat}]\in \Hilb_{I^{\sat}}$ is smooth.
    \end{proof}

    \begin{corollary}\label{ref:usualClassesSmooth:cor}
        Suppose that $I^{\sat}\subseteq S$ is a homogeneous
        saturated ideal and that one of the following holds
        \begin{enumerate}
            \item $S$ is at most three-dimensional (so that $\Proj(S) =
                \mathbb{P}^2$ or $\mathbb{P}^1$),
            \item $S$ is four-dimensional (so that $\Proj(S) \simeq
                \mathbb{P}^3$) and $S/I^{\sat}$ is Gorenstein,
            \item $I^{\sat}$ is a complete intersection.
        \end{enumerate}
        Then $I^{\sat}\in \Hilb_{I^{\sat}}$ is unobstructed.
    \end{corollary}
    This corollary is
    classical and proven, for example, in the references given in the proof.
    \begin{proof}
        To prove smoothness, we may enlarge the base field $\kk$, so we assume
        it is infinite. Then there exists a linear form $\ell$ transverse to $I^{\sat}$.
        The ideal $I^{\sat}/(I^{\sat}\cap (\ell))\subseteq S/(\ell)$ is respectively
        \begin{enumerate}
            \item an ideal in $S/(\ell) \simeq \kk[\alpha_1, \alpha_2]$ or in
                $\kk[\alpha_1]$,
            \item a Gorenstein ideal in $\kk[\alpha_1, \alpha_2, \alpha_3]$,
            \item a complete intersection ideal.
        \end{enumerate}
        In these cases unobstructedness of $S/(I^{\sat}+(\ell))$ follows from classical
        results: Hilbert-Burch Theorem~\cite[Theorem~1,
        Corollary~1]{Schaps__codimension_two}, Buchsbaum-Eisenbud
        theorem~\cite[Theorem~9.7]{HarDeform} and finally~\cite[Theorem~9.2]{HarDeform}, so
        the results follow from Theorem~\ref{ref:smoothPoints:thm}.
    \end{proof}

    \section{Examples}\label{sec:examples}

\subsection{The fiber obstruction group for cases with
$r=1$.}\label{ssec:ObFibGorenstein}

    Let $S$ be standard $\mathbb{N}$-graded, with $\kk$ infinite. Let $I\subseteq S$ be
    homogeneous and let $a$ be the largest degree such that $\dim_{\kk}
    (S/I^{\sat})_a \neq \deg(S/I^{\sat})$.
    As proven in Corollary~\ref{ref:fiberObstructionForGorenstein:cor} if
    $S/I^{\sat}$ is Gorenstein, then $\ObFib := \ObFib(I,I^{\sat})$ is
    dual to the degree $a$ part of $I^{\sat}/(I+(I^{\sat})^2)$.

    \begin{example}[Line]\label{ref:necessaryConditionOnLine:exa}
        In the special case when $I^{\sat}$ is a degree $d$ subscheme of a
        line, we have $a=d-2$. In this case $H_{S/I^{\sat}}(d-2) = d-1$,
        $H_{S/I^{\sat}}(d-1) = d$ and $\ObFib = 0$ if and only if the
        ideals $I^{\sat}$ and $I+(I^{\sat})^2$ agree in degree $d-2$. Assuming additionally that $H_{S/I}$ is
        nondecreasing, we get $\dim_{\kk} (I^{\sat}/I)_{d-2} \leq 1$, hence $\ObFib \neq
        0$ only if $(I^{\sat})^2_{d-2} \subseteq I_{d-2}\subsetneq
        (I^{\sat})_{d-2}$.
    \end{example}

    \begin{example}[Line and some general points]\label{ex:lineAndPoints}
        Generalizing Example~\ref{ref:necessaryConditionOnLine:exa}, consider
        the case where $S/I^{\sat}$ has Hilbert function $(1,e+2, e+3, e+4,
        \ldots , e + d-1, e+ d, e+d,
        \ldots )$ and there exists an $J\supseteq I$ with $S/J^{\sat}$ having Hilbert
        function $(1,2,3,4, \ldots , d-1, d, d,  \ldots , \ldots )$.
        A choice of an element $\ell$ transversal both for $I$ and $J$ yields
        a surjective map $\cokern{I}\to \cokern{J}$. The spaces $\cokern{I}$,
        $\cokern{J}$ have the same
        Hilbert functions in positive degrees, so this map is bijective in positive
        degrees. The quotient $S/J^{\sat}$ is Gorenstein, so we have that
        \[
            \ObFib(I)  \simeq \Hom_S\left( \frac{I^{\sat}}{I},
            \cokern{I}_{\geq 1} \right)_{0}  \simeq \Hom_S\left(
            \frac{I^{\sat}}{I}, \cokern{J}_{\geq 1}
            \right)_0  \simeq \Hom_{\kk}\left(  \left(\frac{I^{\sat}}{I + I^{\sat}\cdot
            J^{\sat}}\right)_{d-2}, \kk \right),
        \]
        which is nonzero if and only if $I^{\sat}$ and $I + I^{\sat}\cdot
        J^{\sat}$
        differ in degree $d-2$.
    \end{example}

    \subsection{Three points}\label{ssec:threePoints}
    Consider the example $I := I(\Gamma_0)'$ from the introduction. In this
    example $\ObFib$ is
    one-dimensional.

    We claim that all points from $\Hilb_I$ are saturable. The restriction of natural map
    $\Hilb_I \to \Hilb^3(\mathbb{P}^2)$ to the saturable locus is dominant and projective, hence
    onto $\Hilb^3(\mathbb{P}^2)$. If $[I']\in \Hilb_I$
    is not saturated, then $S/(I')^{\sat}$ has Hilbert function $(1,2,3,3, \ldots)$.
    Therefore, $[I']$ is the unique point of $\Hilb_{I}$ whose saturation is
    $(I')^{\sat}$.
    It follows that $I'$ is saturable.
    This claim follows also from \cite[Thm.~1.3]{Mandziuk}.

    \subsection{Four points}\label{ssec:four}

    Suppose that $I$ is a (not necessarily saturated) homogeneous ideal in a standard graded polynomial
    ring $S = \kk[\alpha_0, \ldots ,\alpha_3]$ such that the Hilbert function
    of $S/I$ is $(1,4,4,4, \ldots )$, where \emph{$\kk$ has characteristic
    zero}. The Hilbert function of $S/I^{\sat}$
    is $(1,3,4,4,\ldots)$ or $H=(1,2,3,4,4, \ldots)$.
    In the first case, $[I]$ is the unique point in $\Hilb_I$
    whose saturation is equal to $I^{\sat}$. Therefore, we get that $I$ is
    saturable arguing as in~\S\ref{ssec:threePoints}. We assume that $S/I^{\sat}$ has Hilbert function~$H$.
    By Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} and
    Example~\ref{ref:necessaryConditionOnLine:exa} a necessary condition for $I$ to be
    saturable is that $(I^{\sat})^2_2 \subseteq I$. We claim that it is also a sufficient
    condition.

    Let $Z$ in $\Hilb^4(\mathbb{P}^3)$ consist of subschemes with Hilbert
    function $H$. This is a closed, irreducible and $8$-dimensional locus
    parameterized by a line in $\mathbb{P}^3$ and a quartic on it.
    Let $Y$ in $\Hilb_I$ be the preimage of $Z$. The fiber of $Y\to Z$ over any $[R]$
    is $\Gr(6, I(R)_2) \cong \mathbb{P}^6$, so $Y$ is irreducible and
    $14$-dimensional.
    Let $Y'\subseteq Y$ be the locus consisting of $[I']$
    such that $((I')^{\sat})^2_2\subseteq I'$. The fiber of $Y'\to Z$
    over any $[R]$ is isomorphic with
	$\operatorname{Gr}(3, I(R)_2/I(R)^2_2) \cong \mathbb{P}^3$, so $Y'$
    is irreducible and $11$-dimensional.

    Let $X = \Slip^{(1,4,4, \ldots )}$ be the locus of saturable ideals in $\Hilb_I$. It is irreducible
    and $12$-dimensional.
    By
    Example~\ref{ref:necessaryConditionOnLine:exa} and
    Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm}, all saturable
    ideals from $Y$ lie in $Y'$, so $X \cap Y \subseteq Y'$. We now prove that
    this containment is an equality.
	Take any line and four points $\Gamma$ on it. Converging four general points
        to $\Gamma$ yields an ideal $[I']\in X\cap Y$ with $V(I'^{\sat}) =
        \Gamma$; in this way we obtain more than one $I'$ for given $\Gamma$. The
        scheme $V(I'^{\sat})$ is a complete intersection, hence
        unobstructed. Applying Corollary~\ref{ref:exceptionalIsDivisor:cor} we
        obtain a divisor $[I']\in E \subseteq X$, so $\dim E = 11$. Since
    $E$ gets contracted in $\Hilb^4(\mathbb{P}^3)$, the saturation of a point of $E$ cannot
    have Hilbert function $(1,3,4, \ldots )$, thus is has
    Hilbert function $H$, so $E \subseteq X \cap
    Y\subseteq Y'$. But $Y'$ is irreducible and $\dim Y' = 11 = \dim
    E$, so $Y' = E \subseteq X$.

    We have thus proven the following.
    \begin{proposition}
        Let $I\subseteq
        \kk[\alpha_0,  \ldots , \alpha_3]$ be a nonsaturated ideal such that $H_{S/I} =
        (1,4,4,4, \ldots )$. Then $I$ is a limit of saturated ideals if and
        only if $\ObFib(I, I^{\sat})\neq 0$ if and only if $(I^{\sat})^2_2
        \subseteq I$.
    \end{proposition}

    \subsection{Five points}\label{ssec:five}
    Suppose that $I$ is a (not necessarily saturated) homogeneous ideal in a standard graded polynomial
    ring $S = \kk[\alpha_0, \ldots ,\alpha_4]$ such that the Hilbert function
    of $S/I$ is $(1,5,5,5, \ldots )$. \emph{We assume that $\kk$ has
    characteristic zero.} There are a few possible Hilbert
    functions of $S/I^{\sat}$. We divide them according to the number of
    quadrics.

    \subsubsection{Case $H_{S/I^{\sat}}(2) = 5$.} In this case $[I]$ is the unique point
    of the fiber of the natural map $\Hilb_I\to \Hilb^5(\mathbb{P}^4)$ defined
    in~\S\ref{ssec:setup}, hence
    is saturable.

    \subsubsection{Case $H_{S/I^{\sat}}(2) = 4$.}\label{ssec:fivePts4Quadrics} Using Macaulay's Growth
    Theorem as in~\cite[Lemma~2.9]{cjn13} we conclude that $H_{S/I^{\sat}} =
    (1, 3, 4, 5, 5, \ldots )$.
    Geometrically, the scheme $V(I^{\sat}_{\leq 3})\subseteq \mathbb{P}^4$ is a line
    and a (possibly embedded) point. Let $J\supseteq I^{\sat}_{\leq 3}$ define this
    line. The locus $\mathcal{L}\subseteq \Hilb^5(\mathbb{P}^4)$ of
    possible $V(I^{\sat})$ is parameterized by the choice of the
    point in $\mathbb{P}^4$, the line and four points on it, hence it is
    irreducible and its dimension is $4+6+4=14$. Every $V(I^{\sat})$ in this
    locus is contained in $\mathbb{P}^2$, hence unobstructed.

    The ideal $I$ contains no elements of degree less than two, so the $i$-th
    element $F_i$ of its minimal free resolution $ \ldots \to F_1\to F_0\to I\to 0$ has generators of degree at least
    $2+i$. In particular, every generator of $F_1$ has degree at least three.
    Since $H_{I^{\sat}/I} = (0,2,1,0, \ldots )$ this shows that
    $\Ext^1_S(I, I^{\sat}/I)_0 = 0$ and so $\psi_0$ is surjective.

    From~Example~\ref{ex:lineAndPoints}, we see that $\dim_{\kk} \ObFib(I, I^{\sat})$
    is zero when $(I^{\sat})_1\cdot J_1\not\subset I_2$.
    By Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm}, the ideal $I$ is saturable only if
    $(I^{\sat})_1\cdot J_1\subseteq I_2$.
    Let $\mathcal{U}'$ be the preimage of $\mathcal{L}$ in $\Hilb_{I\subseteq I^{\sat}}$.
    For every $[I\subseteq I^{\sat}]\in \mathcal{U}'$ the space
    $I^{\sat}_1\cdot J_1$ is $5$-dimensional and it varies continuously with
    $I\subseteq I^{\sat}$, hence we obtain a closed subvariety $\mathcal{V}
    \subseteq \mathcal{U}'$ whose fiber over $[I^{\sat}]$ is
    given by $[I]$ with $I^{\sat}_1\cdot J_1\subseteq I_2$; this fiber is
    isomorphic to $\Gr(5, I^{\sat}_2/I^{\sat}_1\cdot J_1)$, hence is
    $5$-dimensional.
    By~\cite[11.4.C, 25.2.E]{Vakil_foag} the variety $\mathcal{V}$ is smooth,
    irreducible and $19$-dimensional, hence also
        $\pr_I(\mathcal{V})$ is irreducible and $19$-dimensional.

    The locus $\mathcal{V}$ contains a point $[I'\subseteq (I')^{\sat}]$ with $I'$ saturable: when
    a general $5$-tuple of points $\Gamma_t$ deforms to $\Gamma$ a $4$-tuple on a line and fifth
    point is ``static'' outside it, the limit $I'$ of $I(\Gamma_t)$ is one such
    ideal; for given $\Gamma$ one obtains more than $I'$
        (compare~\S\ref{ssec:four}).
    Let $X = \Slip^{(1,5,5, \ldots)}$. By Corollary~\ref{ref:exceptionalIsDivisor:cor} there is a
    $19$-dimensional family $[I']\in E \subseteq X$,
    which gets contracted in $\Hilb^5(\mathbb{P}^4)$. The saturation of a
    general element $[I'']\in E$ satisfies $H_{S/I''}(2) \geq 4$ by
    semicontinuity and $H_{S/I''}(2) \neq 5$ since $E$ gets contracted.
    Therefore, $E \subseteq \pr_I(\mathcal{U'})\cap X \subseteq
    \pr_I(\mathcal{V})$. Since $\pr_I(\mathcal{V})$ is irreducible and
    $19$-dimensional, we conclude that $E = \pr_I(\mathcal{V})$.

    \subsubsection{Case $H_{S/I^{\sat}}(2) = 3$.} Using Macaulay's Growth
    Theorem as in~\cite[Lemma~2.9]{cjn13} we conclude that $H_{S/I^{\sat}} =
    (1, 2, 3, 4, 5, 5, \ldots )$. When $(I^{\sat})^2_3\not\subseteq I$, we
    obtain by Example~\ref{ref:necessaryConditionOnLine:exa} that $\ObFib = 0$, hence $I$ is entirely nonsaturable.

    The space of possible $I^{\sat}$ has dimension $\dim\Gr(3, S_1) + \dim
    \mathbb{P}(\kk[\alpha_3, \alpha_4]_5) = 11$.
    Constructing $I$ from given $I^{\sat}$ is a bit more tricky in the current
    subcase. Each such ideal requires a choice of a $10$-dimensional space of
    quadrics inside a $12$-dimensional space $(I^{\sat})_2$, but a general
    choice $\mathcal{Q}$ of such quadrics will yield $H_{S/(\mathcal{Q})}(3)
    <5$, so does not give rise to $I$.
    Instead of dealing with ideals directly, we employ Macaulay's inverse
    systems~\S\ref{ssec:macaulaysInverseSystems} in their partial
    differentiation flavour.

    Up to coordinate change, we have $I^{\sat}_1 = (\alpha_0, \alpha_1,
    \alpha_2)$, so in terms of inverse systems, for every $e\leq 4$ the space
    $(I_e^{\sat})^{\perp}$ is $\kk[x_3, x_4]_e$.  To construct $I$, we need
    to add a one-dimensional space of cubics $\spann{c}$ and a 2-dimensional
    space of quadrics $\spann{q_1, q_2}$, taking into account that the
    derivatives of the cubic lie in the $5$-dimensional space $\spann{x_3^2,
    x_3x_4, x_4^2, q_1, q_2}$.

    We construct two families. First, the \emph{cubic only} family is obtained
    by taking a cubic $c$ whose space of partial derivatives $S_1\circ c$
    is two-dimensional modulo $\kk[x_3, x_4]_2$. The condition $(I^{\sat})^2_3
    \subseteq I$ is equivalent to $(\alpha_0, \alpha_1, \alpha_2)^2\circ c = 0$,
    which in turn implies that
    \[
        c = x_0q_0 + x_1q_1 + x_2q_2 + c_0,
    \]
    where $q_0, q_1, q_2\in \kk[x_3, x_4]_2$ and $c_0\in \kk[x_3, x_4]_3$.
    The choice of $c_0$ does not affect the ideal $I$, hence we have a $(3\cdot
    3 - 1)$-dimensional choice of $c$. Altogether, we obtain a
    $19$-dimensional irreducible family.
    We now prove that it lies in $\Slip_I$. It is enough to prove this for a
    general point of this family. Passing to the algebraic closure if
    necessary, we assume that
    $V(I^{\sat}) = \{[\mu_1], \ldots ,[\mu_5]\} \subseteq \mathbb{P}\spann{x_3,
    x_4}$, where $[\mu_i]\neq [x_3]$ for $i=1,2 \ldots ,5$.
    Since $H_{S/I^{\sat}}(2) = H_{\kk[x_3,x_4]}(2)$, we have $\spann{\mu_1^2, \ldots
    ,\mu_5^2} = \kk[x_3,
    x_4]_2$. Rescaling $\mu_i$ if necessary, we assume $\sum_{j=1}^5 \mu_j^3 =
    0$. Fix $c_1, c_2, c_3\in \kk^5$ such that $\sum_{j=1}^5 (c_{i})_j
    \mu_j^2 = q_i$ for $i=0,1,2$.

	For $j=1,\ldots, 5$ let $p_j = \mu_j + t\sum_{i=0}^2 (c_i)_j x_i$ and consider $
	\Gamma_t = \left\{ p_j \ |\ j=1, \ldots ,5 \right\}\subseteq\mathbb{P}^4.$
	By Example~\ref{ex:points}, for every $e > 0$ and nonzero $t$ we have
	$(I(\Gamma_t)_e)^{\perp} = \spann{p_j^e\ |\ j=1,2, \ldots ,5}$.
    We now use Proposition~\ref{ref:HilbertFunctionFromLimits:prop} to show that for every $e>0$ and general
    $\lambda$, the vector space $(I(\Gamma_\lambda)_e)^{\perp}$ is five-dimensional and that the limit of those
    spaces is $I_e^\perp$.
	In the notation of that Proposition, for every $e>0$ and $i\in \{1,\ldots, 5\}$ we take $F_i = p_i^e$.
	If $e>3$, then $\mu_1^e, \ldots, \mu_5^e$ are linearly independent so we
    may take $G_i=F_i$ for all $i$ and conclude using that Proposition.
	Let
	\[
	H = \frac{1}{3t}\sum_{j=1}^5\left( \mu_j + t\sum_{i=0}^2 (c_i)_j x_i\right)^3.
	\]
	Observe that for every homogeneous partial
	derivative operator $D$ of order $k\leq 3$  we have
	\begin{equation}\label{eq:5points_in_the_span}
        DH \in \frac{1}{t}\spann{(p_1)^{3-k}, \ldots, (p_5)^{3-k}}\kk[t]
	\end{equation}
	and
	\begin{equation}\label{eq:derivatives_agree}
		\lim_{t\to 0 } (DH) = D(x_0 q_0 + x_1 q_1 + x_2 q_2) \equiv
		Dc \mod \kk[x_3, x_4]_{3-k},
	\end{equation}
    if $D(x_0 q_0 + x_1 q_1 + x_2 q_2)\neq 0$.
	Assume that $e\in \{1,2,3\}$. Let $G_1, \ldots, G_{4-e}$ be $D_1 H, \ldots, D_{4-e} H$, where
	$D_i$ are homogeneous partial derivative operators in $x_3$ and $x_4$ of order $3-e$ such that
	$D_1 (x_0 q_0 + x_1 q_1 + x_2 q_2), \ldots, D_{4-e} (x_0 q_0 + x_1 q_1 + x_2 q_2)$ are linearly independent.
	Let $G_{5-e}, \ldots, G_5$ be some of $F_1, \ldots, F_5$ such that the
	corresponding $\mu_j^e$ are linearly independent.
	The above claims about $(I(\Gamma_t)_e)^\perp$ follow from Proposition~\ref{ref:HilbertFunctionFromLimits:prop},
	\eqref{eq:5points_in_the_span} and \eqref{eq:derivatives_agree}.
	We proved that for general $\lambda$ the Hilbert function of
    $S/I(\Gamma_{\lambda})$ is $(1,5,5, \ldots )$ and the limit of
    $[I(\Gamma_{t})]\in\Hilb^{(1,5,5 \ldots )}$ with $t\to 0$ is the ideal
    $I$, so $[I]\in \Slip^{(1,5,5 \ldots )}$.

    Next, the \emph{cubic and quadric} family is obtained by taking the cubic
    $c$ to be a cube of a linear form and all possible limits of $c$.
    The choice of $c$ is $4$-dimensional. The only nontrivial possible limit
    is $\ell_1\cdot (\ell_2)^2$, where $\ell_1$ is a linear form and $\ell_2\in
    \spann{x_3, x_4}$. We call such cubics \emph{impure}.
    As for the choice of quadrics, we fix the quadric $q_1$ as any basis element of the
    one-dimensional space $S_1\circ c$. The second quadric can be chosen
    arbitrarily modulo $\spann{x_3^2, x_3x_4, x_4^2, q_1}$, hence yields a
    $10$-dimensional choice.
    It follows that the whole family has dimension $11+10+4 = 25$.
    In particular, its general element does not lie in $\Slip_I$.
    Moreover, from the condition $(I^{\sat})^2_3\subseteq I$ we see that the
    only elements that are possibly in $\Slip_I$ have impure
    $c$. Hence we assume that $c=x_0x_3^2$.

    We consider the subcase when $q_2(x_0, x_1, x_2, 0, 0) = \nu x_0^2$
    for some $\nu\in \kk$, hence
    $q_2 = \nu x_0^2 + x_3\ell_1 + x_4\ell_2$ for some $\ell_1, \ell_2\in \spann{x_0,
    \ldots ,x_4}$.
    In this case we claim that the ideal $I$ is in $\Slip_I$ for every choice
    of $\nu$ and $q_2$ and the quintic in $I^{\sat}$. Since $\Slip_I$ is
    closed, it is enough to
    prove this for a general choice and over an algebraically closed field
    $\kk$, thus we assume that the quintic is completely decomposable and
    $V(I^{\sat}) = \{[\mu_1], \ldots ,[\mu_5]\} \subseteq \mathbb{P}\spann{x_3,
    x_4}$, where $[\mu_i]\neq [x_3]$ for $i=1,2 \ldots ,5$.
    The five cubes $\mu_1^3$, \ldots ,$\mu_5^3$ are linearly dependent as they
    lie in a four-dimensional space $\kk[x_3, x_4]_3$. Rescaling the forms $\mu_i$, we assume
    $\sum_{i=1}^5 \mu_i^3 = 0$. Moreover, since $H_{S/I^{\sat}}(2) = 3$, the squares $\mu_1^2$, \ldots ,$\mu_5^2$
    span $\kk[x_3, x_4]_2$.

    Take $c_1, c_2, c_3, c_4\in \kk^5$ such that
    \[
        \sum_{i=1}^5 c_{1i}\mu_i^2 = x_3^2,\qquad \sum_{i=1}^5 \left(
        \frac{\partial\mu_i}{\partial x_4}
        \right)c_{1i}^2 \neq 0
	\]
	and
	\[
		\sum_{i=1}^5 \mu_i \left(
        \frac{\partial \mu_i}{\partial x_4} \right)(c_{2i}x_0 + c_{3i}x_1 + c_{4i}x_2) \equiv
        \frac{\sum_{i=1}^5 \left(\frac{\partial \mu_i}{\partial x_4}\right) c_{1i}^2}{2\nu}(\ell_1x_3+\ell_2x_4) \mod 			\kk[x_3, x_4]_{2}.
    \]
    To prove the existence of $c_1$, it is enough to prove it on a single
    example, for example with $[\mu_{\bullet}] = ([x_4 - x_3], [x_4+x_3],
    [x_4-2x_3], [x_4 + 2x_3], [x_4])$ and $c_1 = \frac{1}{8}(0, 0, 1, 1, -2)$.
	For $i=1,\ldots, 5$ let $p_i= \mu_i + tc_{1i}x_0 + t^2(c_{2i}x_0 + c_{3i}x_1 + c_{4i}x_2)$
	and consider the family $\Gamma_t =  \{p_i\ |\ i=1,2, \ldots ,5\}\subseteq \mathbb{P}^4$.
   	We show that $S/I(\Gamma_\lambda)$ has Hilbert function $(1,5,5,\ldots)$ for general $\lambda$ and
   	the limit of ideals $I(\Gamma_t)$ taken degree-by-degree (hence, in the
    multigraded Hilbert scheme) is $I$. As above, we use
    Proposition~\ref{ref:HilbertFunctionFromLimits:prop} and for each degree
    $e>0$ we take $F_i = p_i^e$ for each $i=1,\ldots, 5$.
	For $e>3$ we may take $G_i=F_i$ for every $i$ and conclude using that Proposition. Let
	\[
	H = \frac{1}{3t} \sum_{i=1}^5 \left(\mu_i + tc_{1i}x_0 + t^2(c_{2i}x_0 + c_{3i}x_1 + c_{4i}x_2)\right)^3.
	\]
	For $e\in \{1,2,3\}$ we take $G_1,\ldots, G_{e+1}$ to be any $e+1$ of $F_1,\ldots, F_5$ such that the
	corresponding $\mu_i^e$ are linearly independent.
	For $e=3$ we take $G_5 = H$ and conclude using Proposition~\ref{ref:HilbertFunctionFromLimits:prop} since
	\[
	\lim_{t\to 0} H = \sum_{i=1}^5 \mu_i^2c_{1i}x_0 = x_0x_3^2.
	\]
	Before proceeding with cases $e=1,2$ observe that
	\[
	\lim_{t \to 0} \frac{1}{2t} \frac{\partial H}{\partial x_4} \equiv
	\left(\frac{1}{2}\sum_{i=1}^5 \frac{\partial \mu_i}{\partial x_4} c_{1i}^2 \right)
	\cdot \left(x_0^2 + \frac{\ell_1x_3}{\nu} + \frac{\ell_2 x_4}{\nu}\right) \mod \kk[x_3,x_4]_2.
	\]
	By assumption, the first factor is nonzero. Replacing $H$ by $H$ divided by that
	scalar and multiplied by $\nu$ we get
	\[
	\lim_{t \to 0} \frac{1}{2t} \frac{\partial H}{\partial x_4} \equiv q_2 \mod \kk[x_3,x_4]_2.
	\]
	Therefore, for $e=2$ we may take $G_4 = \frac{\partial H}{\partial x_3}$ and
	 $G_5 = \frac{1}{2t} \frac{\partial H}{\partial x_4}$.
	Finally, for $e=1$ we take
	$G_3 = \frac{1}{2t} \frac{\partial^2 H}{\partial x_4^2}$, $G_4 =
	\frac{1}{2t} \frac{\partial^2 H}{\partial x_4 \partial x_3}$ and
	 $G_5 = \frac{1}{2t} \frac{\partial^2 H}{\partial x_4 \partial x_0}$.
    This finishes the proof of our claims about $I(\Gamma_t)$ and implies
     that $[I] \in \Slip^{(1,5,5,\ldots)}$.

	Finally, we consider the subcase when $q_2(x_0, x_1, x_2, 0, 0)$ is not a
	multiple of $x_0^2$. Consider the natural map
	$\pi\colon \Hilb_I \to \Hilb_{I+S_{\geq 4}}$.
	The images of $\Slip_I$ and the cubic and quadric family are both $20$-dimensional.
	The tangent space at the image of any ideal corresponding to
	$(q_2,c) = (x_0x_1, x_0x_3^2)$ is $20$-dimensional, so no such ideal is in $\Slip_I$.
	If the quadric $q_2(x_0, x_1, x_2, 0, 0)$ has rank at least
	two, then $q_2$ degenerates to $x_0x_1$ so the corresponding ideal is not in $\Slip_I$.
	We are left with the rank one case, so $q_2(x_0, x_1, x_2, 0, 0) = x_1^2$.
	It is enough to show that no ideal with $(q_2, c) = (x_1^2, x_0x_3^2)$ is in $\Slip_I$.
	Let $K$ be the image under $\pi$ of any such ideal. A computation using the \emph{VersalDeformations}
    package~\cite{Ilten_Versal_Deformations} verifies that $[K]$ lies on the intersection of two irreducible components:
    $20$-dimensional and $19$-dimensional. It follows that $[K]\notin \pi(\Slip_I)$.

    Summing up the five points case, we have the following:
    \begin{proposition}
        Let $I\subseteq
        \kk[\alpha_0,  \ldots , \alpha_4]$ be an ideal such that $H_{S/I} =
        (1,5,5,5, \ldots )$.
        \begin{enumerate}
            \item Suppose $H_{S/I^{\sat}}(2) = 5$. Then $I$ is a limit of
                saturated ideals.
            \item Suppose $H_{S/I^{\sat}}(2) = 4$.
                Then $I$ is a limit of saturated ideals if and
                only if $\ObFib(I, I^{\sat})\neq 0$ if and only if
                $(I^{\sat})_1\cdot J_1
                \subseteq I$, where $J$ was defined
                in~\ref{ssec:fivePts4Quadrics}.
            \item Suppose $H_{S/I^{\sat}}(2) = 3$. Assume that $\kk$ has characteristic zero.
                Then $I$ is a limit of saturated ideals if and only if both
                $(I^{\sat})^2_3 \subseteq I$ and $(I:S_+^2)_1 \cdot
                I_1^{\sat} \subseteq I$.
        \end{enumerate}
    \end{proposition}

\subsection{A partially saturable ideal with nonzero fiber obstruction
group}\label{ssec:nonvanishingGroup}

    Let $S = \kk[\alpha_0, \alpha_1, \alpha_2]$ with standard grading and
    consider the degree $9$ ideal
    \[
        I = (\alpha_0^2\alpha_2,\, \alpha_0\alpha_1^3,\,
        \alpha_0^2\alpha_1^2,\, \alpha_0^3\alpha_1,\, \alpha_0^5,
        \alpha_1^6).
    \]
    Its Hilbert function is $(1,3,6,9, \ldots )$ and it has $I^{\sat} =
    (\alpha_{0}^{2},\,\alpha_{0}\alpha_{1}^{3},\,\alpha_{1}^{6})$, whose Hilbert function is
    $(1,3,5,7,8,9, \ldots )$.
    In this example we check that the map $\Hom_S(I^{\sat}, S/I^{\sat})_0\to
    \Hom_S(I, S/I^{\sat})_0$ has one-dimensional cokernel and the same holds for $\TgFlag\to
    \Hom_S(I, S/I)_0$. Choosing a complementary tangent vector we discover the
    deformation
    \[
        I_t = (t\alpha_0\alpha_1^2 + \alpha_0^2\alpha_2,\, \alpha_0\alpha_1^3,\,
        \alpha_0^2\alpha_1^2,\, \alpha_0^3\alpha_1,\, \alpha_0^5, \alpha_1^6).
    \]
    The saturation of the ideal $I_1$ is $(\alpha_0\alpha_1^2 +
    \alpha_0^2\alpha_2, \alpha_0^2\alpha_1, \alpha_0^3, \alpha_1^6)$ which has
    Hilbert series $(1,3,6,7,8,9, \ldots )$. This shows that $I = I_0$ is not
    entirely nonsaturable. In fact, with the help of \emph{VersalDeformations}
    package~\cite{Ilten_Versal_Deformations} we check that $[I_1]\in
    \Hilb_{I_1}$ is smooth and has $\ObFib = 0$, so it is entirely
    nonsaturable. We check directly that the tangent space dimensions at $I_0$
    and $I_1$ are equal, so it also follows that $I$ is a smooth point on the
    component of $I_1$, so $I$ is nonsaturable as well.

\subsection{An entirely nonsaturable ideal with nonzero fiber obstruction
    group using Bia{\l{}}ynicki-Birula{} decomposition}

    Let $S = \kk[\alpha_0, \alpha_1, \alpha_2]$ with standard grading and
    consider the ideal
    \[
        I = (\alpha_0^3, \alpha_0\alpha_1^2, \alpha_0^2\alpha_2,
        \alpha_0\alpha_1\alpha_2, \alpha_0\alpha_2^4, \alpha_2^6).
    \]
    In this case $\ObFib$ is one-dimensional, so
    Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} does not apply. We will
    nevertheless prove that $I$ is entirely nonsaturable.
    Refine the standard grading on $S$ to an $\mathbb{N}^2$-grading, where
    $\deg(\alpha_0) = \deg(\alpha_2) = (1, 0)$ and $\deg(\alpha_1) = (1,1)$.
    With the help of \emph{Macaulay2} we check that for the pair $(I,
    I^{\sat})$ we have $\ObFib_{0, *} =
    \ObFib_{0, -1}$, so in particular $\ObFib_{0, \geq 0} = 0$.
    The second coordinate of the grading gives a $\kk^*$-action on $\Hilb_I$
    and $\Hilb_{I\subseteq I^{\sat}}$.
    We now use the theory of Bia{\l{}}ynicki-Birula{} decompositions. Introducing this theory
    would take too much place, so we only sketch the main results.
    Repeating the proofs of
    Theorem~\ref{ref:obstructionAtFlag:thm} and
    Theorem~\ref{ref:deformsWhenObFibVanishes:thm} for degrees $(0, \geq 0)$,
    we conclude that the map $\Hilb_{I\subseteq I^{\sat}}^+\to \Hilb_I^{+}$ on
    Bia{\l{}}ynicki-Birula{} decompositions is smooth, hence also the map
    \[
        \id_{\mathbb{A}^2} \times \pr_{I}\colon \mathbb{A}^2 \times \Hilb_{I\subseteq I^{\sat}}^+\to
        \mathbb{A}^2 \times \Hilb_I^{+}
    \]
    is smooth.
    Consider the embedding
    \[
        \mathbb{A}^2\ni (v_1, v_2)\to \begin{pmatrix}
            1 & v_1 & 0\\
            0 & 1 & 0\\
            0 & v_2 & 1
        \end{pmatrix}\in \GL_3
    \]
    The scheme $\mathbb{A}^2 \times \Hilb_I^{+}$ maps to $\Hilb_I$
    by the forgetful map $\Hilb_I^+\to \Hilb_I$ followed by the
    $\mathbb{A}^2$-action via $\mathbb{A}^2\hookrightarrow \GL_3$.
    Under this map, the tangent space to $\mathbb{A}^2$ maps to a
    $2$-dimensional subspace of $\Hom_{S}(I, S/I)_{0,-1}$.
    A final \emph{Macaulay2} computation
    shows that this subspace is the degree $(0, <0)$ part of the
    tangent space, hence the above tangent map is surjective in degrees
    $(0, *)$. Arguing as in~\cite[Theorem~4.5]{Jelisiejew__Elementary},
    we conclude that $\mathbb{A}^2 \times \Hilb_I^+\to \Hilb_I$ is an open
    immersion at
    $(0, [I])$, hence we conclude that the image of the natural morphism
    \[
        \mathbb{A}^2 \times \Hilb_{I\subseteq I^{\sat}}^+ \to \Hilb_I
    \]
    contains an open neighbourhood of $[I]$. Then the same is true for the
    morphism
    \[
        \mathbb{A}^2 \times \Hilb_{I\subseteq I^{\sat}}\to \Hilb_I,
    \]
    which proves that $[I]$ is entirely nonsaturable.
    This proof is interesting also because while the $\TgFlag\to \Hom_S(I,
    S/I)_{0}$ is surjective, the map $\psi_{0}$ is not; only the
    map $\psi_{0, \geq 0}$ is surjective.

    We provide the \emph{Macaulay2} code below.

{\small\begin{verbatim}
S = QQ[a_0 .. a_2, Degrees=>{{1, 0}, {1, 1}, {1, 0}}];
I = ideal mingens ideal(a_0^3, a_0*a_1^2, a_0^2*a_2, a_0*a_1*a_2, a_0*a_2^4, a_2^6);
Isat = ideal mingens saturate I;
obfib = Ext^1(Isat/I, S^1/Isat);
hs = hilbertSeries(obfib, Order=>1);
T1 := (gens ring hs)_1;
assert(select(terms hs, el -> first degree el == 0) == {T1^(-1)});
assert(sum select(terms hilbertSeries(Hom(I, S^1/I), Order=>1),
    el -> first degree el == 0) == T1^6 + T1^5 + T1^4 + T1^3 + T1^2 + 4*T1 + 3 + 2*T1^(-1));
    \end{verbatim}}

Since $H_{S/I^2}(9) = 17$ the fact that $I$ is not saturable could also be deduced from \cite[Theorem~1.1]{Mandziuk}.
Here, by a completely different argument, we obtain a stronger result, that $I$ is entirely nonsaturable.

\section{Wild polynomials}

For positive integers $r$ and $n$ we denote by $\genHF{r}{n}$ the Hilbert
function of $r$ points in general position on $\mathbb{P}^n$, that is, we have
$\genHF{r}{n}(i) = \min\{r, \binom{n+i}{i}\}$ for every $i$.
By $\genMHS{r}{n}$ we denote the multigraded Hilbert scheme
$\Hilb^{\genHF{r}{n}}$
and by  $\genSat{r}{n}$ we denote
$\Satbar^{\genHF{r}{n}} \subseteq \genMHS{r}{n}$.

\begin{remark}
If $r$ and $n$ are such that $\Hilb^r(\mathbb{P}^n)$ is irreducible, then $\genSat{r}{n}$ coincides with
the closure of the locus of saturated and radical ideals. That, is we have
$\genSat{r}{n} = \operatorname{Slip}_{r, \mathbb{P}^n}$ where $\operatorname{Slip}_{r, \mathbb{P}^n}$ is as defined in \cite{Buczyska_Buczynski__border}.
\end{remark}

In this section $\kk$ is an algebraically closed field and $S=\kk[\alpha_0,\alpha_1,\alpha_2]$
is a polynomial ring with graded dual ring $S^* = \kkdp[x_0,x_1,x_2]$.

\begin{definition}
Let $F \in S^*_d$ be a nonzero form of positive degree. Given a subscheme $\Gamma$ of $\mathbb{P}S^*_d$,
by $\langle \Gamma \rangle$ we denote the projective linear span of $\Gamma$.
If $[F]\in \langle \Gamma \rangle$, we say that $\Gamma$ is \emph{apolar to
$F$}.
The \emph{cactus rank} of $F$ is the smallest $r$ such that there exists
    $\Gamma\subseteq \Proj(S) \simeq
    \mathbb{P}^{2}$ apolar to $F$ and of degree $r$.
    The \emph{smoothable rank} of $F$ is defined in the same way, but
        only considering $\Gamma$ from the smoothable component
        (see~\S\ref{ssec:smoothnessOfSaturated}).
\end{definition}

We assume that the characteristic of $\kk$ is zero or larger than the degree of any form whose rank we want to compute.
Therefore, the reader may think of $S^*$ as an ordinary polynomial ring,
    compare~\S\ref{ssec:macaulaysInverseSystems}.
The main result of this section is
Proposition~\ref{prop:no_wild_polynomials}, in which we show that if a divided
power polynomial $F\in S^*_d$ satisfies  $\brr(F)\leq d+3$ then we have
$\srr(F) = \brr(F)$.

\subsection{Apolarity in three variables}

This subsection gathers some technical lemmas necessary for the
proof of Proposition~\ref{prop:no_wild_polynomials}.
Throughout we assume that $F\in S^*_d$ is a nonzero degree $d$ form,
for $d > 0$.

\begin{lemma}\label{lem:apolar_to_wild_is_nonsaturated}
Assume that $\brr(F) \leq r < \crr(F)$ for
some positive integer $r$. Then there exists an ideal $I \subseteq \Ann(F)$ such that $[I] \in \genSat{r}{2}$ and $I^{\sat}_d \neq I_d$.
\end{lemma}
\begin{proof}
By the border apolarity lemma
\cite[Theorem~3.15]{Buczyska_Buczynski__border}, there is an ideal $[I] \in \genSat{r}{2}$ such that $I \subseteq \Ann(F)$. If $I^{\sat}_d = I_d$,
then $I^{\sat} \subseteq \Ann(F)$. Thus, $\crr(F) \leq r$ follows from the cactus apolarity lemma \cite[Theorem~4.7]{teitler14}.
\end{proof}

\begin{lemma}\label{lem:middle_generator}
Let $e=\lceil \frac{d+1}{2} \rceil$. For every linear form $\ell\in S_1$,
there is an element of $\Ann(F)_e$ that is not divisible by $\ell$.
\end{lemma}
\begin{proof}
    Let $G = \ell\,\lrcorner\, F$. Suppose that $\Ann(F)_e \subseteq (\ell)$, then also $\Ann(F)_{\leq e}
    \subseteq (\ell)$, so the multiplication by $\ell$ is a bijection from
    $\Ann(G)_a$ to $\Ann(F)_{a+1}$ for every $a\leq e-1$.
    Let $f$, $g$ be the Hilbert functions of
$S/\Ann(F)$, $S/\Ann(G)$, respectively. For $a\leq e-1$ we have
\begin{equation}\label{eq:relation_between_annihilators}
f(a+1) = \dim_{\kk} S_{a+1} - \dim_\kk \Ann(F)_{a+1} = a+2 + \dim_\kk S_{a} - \dim_\kk \Ann(G)_a = a+2 + g(a).
\end{equation}

Assume that $d$ is even. Then $d=2e-2$ and we have
\begin{align*}
f(e-2) &= f((2e-2)-(e-2)) = f(e) \stackrel{\eqref{eq:relation_between_annihilators}}{=}  g(e-1)+e+1 \\
&= g((2e-3)-(e-1))+e+1 = g(e-2) + e+1 \stackrel{\eqref{eq:relation_between_annihilators}}{=} f(e-1)+1.
\end{align*}
This contradicts the unimodality of the Hilbert function of $S/\Ann(F)$,
see~\cite[Theorem~4.2]{stanley_hilbert_functions_of_graded_algebras}, \cite[Theorem~5.25]{iakanev}.

Assume that $d$ is odd. Then $d=2e-1$ and we have
\begin{align*}
g(e) &= g((2e-2)-e) = g(e-2) \stackrel{\eqref{eq:relation_between_annihilators}}{=} f(e-1) - e \\
&= f((2e-1)-(e-1)) -e = f(e) - e  \stackrel{\eqref{eq:relation_between_annihilators}}{=} g(e-1)+1.
\end{align*}
This contradicts the unimodality of the Hilbert function of $S/\Ann(G)$.
\end{proof}

\begin{lemma}\label{lem:square_annihilates}
If there is a linear form $\ell$ in $S_1$ such that $(\ell^2)\subseteq \Ann(F)$, then $\crr(F) \leq 2\cdot \lceil \frac{d+1}{2} \rceil$.
\end{lemma}
\begin{proof}
    Let $e=\lceil\frac{d+1}{2}\rceil$. By Lemma~\ref{lem:middle_generator} there exists a $\sigma\in \Ann(F)_{e}$
    not divisible by $\ell$. The ideal $(\ell^2, \sigma)$ is a complete
    intersection, hence is saturated and proves that $\crr(F)\leq 2e$.
\end{proof}

\begin{lemma}\label{lem:apolar_contained_in_line_has_special_form}
Let $k\geq 2$ be an integer. Assume that there exists $[I]\in \genSat{d+k}{2}$ such that
$I^{\sat}_1\neq 0$ and $I\subseteq \Ann(F)$. Then
\begin{enumerate}[label = (\arabic*)]
    \item\label{it:special_form_part_one} Up to a linear change of variables, we have
\[
I_{d+k-2} = \langle x_0G, x_1^{d+k-2}, x_1^{d+k-3}x_2, \ldots, x_2^{d+k-2}\rangle^\perp
\]
for some $G\in \kkdp[x_1,x_2]_{d+k-3}$.
\item\label{it:special_form_part_two} If $H_{\kk[\alpha_1, \alpha_2]/\Ann(G)} (k-2) = k-1$, then $\alpha_0^2 \in \Ann(F)$.
\end{enumerate}
\end{lemma}
\begin{proof}
    The ideal $I^{\sat}$ defines a degree $d+k$ subscheme on a line, hence
    $(I^{\sat})_{\leq d+k-2} \subseteq (\ell)$ where $\ell\in S_1$ defines the
    line. We change coordinates so that $\ell = \alpha_0$. Comparing Hilbert functions, we obtain that $\dim_{\kk} (I^{\sat}/I)_{d+k-2}
    = 1$, so $I_{d+k-2} = \langle x_0G, x_1^{d+k-2}, x_1^{d+k-3}x_2, \ldots,
    x_2^{d+k-2}\rangle^\perp$ for some $G\in \kkdp[x_0, x_1,
    x_2]_{d+k-3}$.
Since  $[I]\in \genSat{d+k}{2}$ we get from Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} that
$\ObFib  = \Ext^1_S\left(\frac{I^{\sat}}{I}, \frac{S}{I^{\sat}}\right)_0\neq 0$.
It follows from Example~\ref{ref:necessaryConditionOnLine:exa} that $(\alpha_0^2)_{d+k-2} \subseteq I_{d+k-2}$.
As a result, $I_{d+k-2}$ is as claimed in
part~\ref{it:special_form_part_one}.
If $H_{\kk[\alpha_1, \alpha_2]/\Ann(G)} (k-2) = k-1$, then
the space $\mathcal{L} = x_0\cdot (\kk[\alpha_1, \alpha_2]_{k-2}\,\lrcorner\, G)
+ \langle x_1^d, x_1^{d-1}x_2, \ldots, x_2^d \rangle$ is $(d+k)$-dimensional.
It is contained in $I_d^{\perp}$, so comparing dimensions we get
$I_d = \mathcal{L}^{\perp}$, so $(\alpha_0^2)_d \subseteq (F)^{\perp}$
and part~\ref{it:special_form_part_two} follows.
\end{proof}

For $r\in \mathbb{Z}_{>0}$ let $f_r$ be the Hilbert function $(1,3,4,5, \ldots , r-1, r, r, r,
 \ldots )$.

\begin{lemma}\label{lem:apolar_containded_in_f}
    Assume that $d \geq 3 $
and that there exists $[I]\in \genSat{d+3}{2}$ such that $S/I^{\sat}$ has Hilbert function $f_{d+3}$ and $I\subseteq \Ann(F)$. Up to a linear change of variables we have
$(\alpha_0^2\alpha_1, \alpha_0^2\alpha_2)\subseteq \Ann(F)$  or $(\alpha_0^3, \alpha_0^2\alpha_1)\subseteq \Ann(F)$.
\end{lemma}
\begin{proof}
We may and do assume that $I^{\sat}_2 = (\alpha_0\alpha_1, \alpha_0\alpha_2)_2$ or $I^{\sat}_2 = (\alpha_0^2,\alpha_0\alpha_1)_2$. Let $J=(\alpha_0, \theta_{d+2})$,
where $\theta_{d+2}\in S_{d+2}$ is a minimal generator of $I^{\sat}$. It
follows from Theorem~\ref{ref:saturationDeformsWhenObFibVanishes:thm} and
Example~\ref{ex:lineAndPoints} that we have $(\alpha_0\cdot I^{\sat})_d \subseteq
I_d\subseteq \Ann(F)_d$.
From this we deduce that $\alpha_0\cdot I^{\sat}_2 \subseteq \Ann(F)$.
\end{proof}

\begin{lemma}\label{lem:generators_in_degree_3}
Assume that $d\geq 3$ and that there exists $[I]\in \genSat{d+3}{2}$ such that $S/I^{\sat}$ has Hilbert function $\genHF{d+3}{1}$ and $I\subseteq \Ann(F)$. Up to a linear change of variables we have $(\alpha_0^2\ell, \alpha_0\alpha_1\ell) \subseteq \Ann(F)$ for a linear form  $\ell\in S_1$.
\end{lemma}
\begin{proof}
By
Lemma~\ref{lem:apolar_contained_in_line_has_special_form}\ref{it:special_form_part_one} applied with
$k=3$, we obtain
$I_{d+1} = \langle x_0G_d, x_1^{d+1}, x_1^dx_2, \ldots,
x_2^{d+1}\rangle^\perp$ for some nonzero $G_d\in \kkdp[x_1,x_2]_d$.
Consider $\langle \alpha_1 \,\lrcorner\, G_d, \alpha_2 \,\lrcorner\, G_d\rangle $. If
this space is two-dimensional, then applying
Lemma~\ref{lem:apolar_contained_in_line_has_special_form}\ref{it:special_form_part_two}, we get
$\alpha_0^2\in \Ann(F)$. We thus assume that
$\dim_\kk \langle \alpha_1 \,\lrcorner\, G_d, \alpha_2 \,\lrcorner\, G_d \rangle =
1$. We may and do assume that $G_d = x_2^d$. As a result, we have
\begin{equation}\label{eqn:no_terms_with_alpha0alpha2n-2}
I_{d-1} \subseteq  \langle x_0x_2^{d-2} \rangle^\perp.
\end{equation}

If $H_{S/\Ann(\alpha_0\,\lrcorner\, F)}(1) \leq 2$, then there is a linear form $\ell
\in S_1$ such that $\alpha_0\ell \in \Ann(F)$ and the proof is complete. Thus, we assume that  $H_{S/\Ann(\alpha_0\,\lrcorner\, F)}(1) = 3$. We have
\begin{align*}
H_{S/\left(\Ann(F)\cap (\alpha_0)\right)}(d-1) &= \dim_\kk S_{d-1} - \big(\dim_\kk S_{d-2} - H_{S/\Ann(\alpha_0\,\lrcorner\, F)}(d-2)\big) \\
&= d + H_{S/\Ann(\alpha_0\,\lrcorner\, F)}\big((d-1)-(d-2)\big) = d+3.
\end{align*}
It follows that $\big(\Ann(F)\cap(\alpha_0)\big)_{d-1} = I_{d-1}$.
Thus, from Equation \eqref{eqn:no_terms_with_alpha0alpha2n-2} we get
\[
\big(\kk\cdot (\alpha_0\alpha_2^{d-2}\,\lrcorner\, F)\big) \cap \big((\alpha_0^2, \alpha_0\alpha_1)_{d-1}\,\lrcorner\, F\big) = 0.
\]
In particular, $\dim_\kk \big((\alpha_0^2, \alpha_0\alpha_1)_{d-1}\,\lrcorner\, F\big) \leq 2$, so there exists a linear form $\ell\in S_1$ such that
\[
\big((\alpha_0^2, \alpha_0\alpha_1)_{d-1}\,\lrcorner\, F\big) \subseteq \ell^\perp.
\]
We get $(\alpha_0^2\ell, \alpha_0\alpha_1\ell)_d \subseteq \Ann(F)$ and we conclude that $\alpha_0^2\ell, \alpha_0\alpha_1\ell$ are in $\Ann(F)$.
\end{proof}

\subsection{Nonexistence of wild polynomials of small border rank}

A form $F\in S^*_d$ is called \emph{wild} if its border rank is smaller than its smoothable rank.

\begin{lemma}[{\cite[Proposition~3.4]{Mandziuk}}]\label{lem:d+2_not_wild}
If $\brr(F) \leq d+2$ for some $F \in S^*_d$, then $\crr(F) = \srr(F) = \brr(F)$. In particular, $F$ is not wild.
\end{lemma}
\begin{proof}
By \cite[Proposition~2.5]{bubu2010} we may and do assume that $\brr(F) = d+2$. Suppose that $\crr(F) > d+2$. By Lemma~\ref{lem:apolar_to_wild_is_nonsaturated} there exists $[I]\in \genSat{d+2}{2}$ such that $I\subseteq \Ann(F)$ and $I^{\sat}_d\neq I_{d}$. Then necessarily, $S/I^{\sat}$ has Hilbert function $\genHF{d+2}{1}$ and it follows from Lemma~\ref{lem:apolar_contained_in_line_has_special_form}(2) that the square of a linear form annihilates $F$. From Lemma~\ref{lem:square_annihilates} we get $\crr(F) \leq 2\lceil\frac{d+1}{2}\rceil \leq d+2$.
\end{proof}

\begin{lemma}\label{lem:special_case_3}
Let $d\geq 4$ and $F_d\in S^*_d$ be of the form $F_d = ax_0x_1x_2^{d-2} +
G_d(x_0,x_2) + H_d(x_1,x_2)$ for some $a\in \kk$ and homogeneous, degree $d$
divided power polynomials $G_d$ and $H_d$.
The cactus rank of $F_d$ is at most $d+3$.
\end{lemma}
\begin{proof}
We have $\crr(G_d) \leq \lceil \frac{d+1}{2} \rceil$ so we may and do assume that $a=1$.
Assume first that $d$ is even and let $d=2k$. We consider two cases:
\begin{enumerate}[label=(\arabic*)]
\item\label{it:special_case_3_part_one} $\Ann(\alpha_0^2\,\lrcorner\, G_d)_{k-1} \neq 0$ and $\Ann(\alpha_1^2\,\lrcorner\, H_d)_{k-1} \neq 0$;
\item\label{it:special_case_3_part_two} $\Ann(\alpha_0^2\,\lrcorner\, G_d)_{k-1} = 0$ or $\Ann(\alpha_1^2\,\lrcorner\, H_d)_{k-1} = 0$.
\end{enumerate}
In the first case, there are $\theta_{k-1}(\alpha_0, \alpha_2)$ and $\eta_{k-1}(\alpha_1, \alpha_2)$ such that $I=(\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\theta_{k-1}, \alpha_1^2\eta_{k-1}) \subseteq \Ann(F)$. Its initial ideal with respect to any term order is of the form $J=(\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\theta'_{k-1}, \alpha_1^2\eta'_{k-1})$ for some monomials $\theta'\in \kk[\alpha_0,\alpha_2]_{k-1}$ and $\eta' \in \kk[\alpha_1, \alpha_2]_{k-1}$. The ideal $J$ is saturated and the Hilbert function of $S/J$ is $2k+2 = d+2$. Therefore, $\crr(F) \leq d+2$ follows.

If \ref{it:special_case_3_part_two} holds, we may and do assume that $\Ann(\alpha_0^2\,\lrcorner\, G_d)_{k-1} = 0$. The map
\[
\kk[\alpha_0,\alpha_2]_{k-1} \to \kkdp[x_0,x_2]_{d-k-1} = \kkdp[x_0,x_2]_{k-1} \text{ given by }\theta \mapsto (\alpha_0^2\theta) \,\lrcorner\, G_d
\]
is injective and thus bijective.
It follows that for some $\theta_{k-1}\in \kk[\alpha_0,\alpha_2]_{k-1}$, $\eta_{k-1}\in \kk[\alpha_1, \alpha_2]_{k-1}$ and $c\in \{0,1\}$, the ideal
\[
I = (\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\theta_{k-1} - \alpha_0\alpha_1\alpha_2^{k-1}, \alpha_1^2\eta_{k-1} - c\alpha_0\alpha_1\alpha_2^{k-1})
\]
is contained in $\Ann(F)$. If $c=0$ or $\theta_{k-1}\notin \kk\cdot\alpha_2^{k-1}$ or $\eta_{k-1}\notin \kk\cdot \alpha_2^{k-1}$, then these generators form a Gr\"obner basis with respect to the grevlex order with $\alpha_0>\alpha_1>\alpha_2$ or $\alpha_1 > \alpha_0 > \alpha_2$ and the proof is as in case~\ref{it:special_case_3_part_one}. Therefore, we are left with the case that
$
I = (\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\alpha_2^{k-1} -\alpha_0\alpha_1\alpha_2^{k-1}, c'\alpha_1^2\alpha_2^{k-1}-\alpha_0\alpha_1\alpha_2^{k-1}) \subseteq \Ann(F)
$ for some nonzero $c'$. This ideal contains $I'=(\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\alpha_2^{k-1} -c'\alpha_1^2\alpha_2^{k-1}) = (\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\alpha_2^{k-1} -c'\alpha_1^2\alpha_2^{k-1}, \alpha_1^3\alpha_2^{k-1})$. By taking an initial ideal, one may verify that $I'$ is saturated and $S/I'$ has Hilbert polynomial $2k+3 = d+3$.

Now assume that $d$ is odd and let $d=2k+1$. Then for dimension reasons, there are $\theta_{k}\in \kk[\alpha_0,\alpha_2]_k$ and $\eta_{k}\in \kk[\alpha_1, \alpha_2]_k$ such that $I=(\alpha_0^2\alpha_1, \alpha_0\alpha_1^2, \alpha_0^2\theta_{k}, \alpha_1^2\eta_k)\subseteq \Ann(F)$. By taking an initial ideal, one may verify that $I$ is saturated and $S/I$ has Hilbert polynomial $2k+4=d+3$.
\end{proof}

\begin{lemma}\label{lem:special_case_4}
Let $d$ be a positive integer and $F_d\in S^*_d$ be of the form $F_d = ax_0x_2^{d-1} + G_d(x_0,x_1) + H_d(x_1,x_2)$
for some $a\in \kk$ and homogeneous, degree $d$ divided power polynomials $G_d$ and $H_d$.
The cactus rank of $F_d$ is at most $d+3$.
\end{lemma}
\begin{proof}
    Let $e = \lceil \frac{d+1}{2} \rceil$.
The cactus rank of $G_d$ is at most $e$, so it suffices to show that the
cactus rank of $x_0x_2^{d-1}+H_d(x_1,x_2)$ is at most $e+1$.
For some $\theta \in \kk[\alpha_1,\alpha_2]_{e-1}$ and $a\in \{0,1\}$, we
have $I=(\alpha_0^2, \alpha_0\alpha_1, \alpha_1\theta-a\alpha_0\alpha_2^{e-1})
\subseteq \Ann(x_0x_2^{d-1}+H_d(x_1,x_2))$. The ideal $I$ is saturated and
$S/I$ has Hilbert polynomial $e+1$.
\end{proof}

\begin{proposition}\label{prop:no_wild_polynomials}
Assume that $\brr(F) \leq d+3$ for some $F\in S^*_d$. If the characteristic of $\kk$ is zero or larger than $d$, then $\crr(F) = \srr(F) = \brr(F)$. In particular, $F$ is not wild.
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:d+2_not_wild} we may and do assume that $\brr(F) = d+3$.
By the Alexander-Hirschowitz theorem, see \cite{AlexHirsch} or
\cite[Theorem~1.61]{iakanev}, for $d\leq 5$ such a high border rank does not
occur,
hence we obtain $d\geq 6$.
Suppose that $\crr(F) > d+3$. By Lemma~\ref{lem:apolar_to_wild_is_nonsaturated} there  is an ideal $I\subseteq \Ann(F)$ such that $I^{\sat}_d\neq I_d$ and $[I]\in \genSat{d+3}{2}$. There are two possibilities for the Hilbert function of $S/I^{\sat}$. It is either $\genHF{d+3}{1}$ or $f_{d+3}$. It follows from Lemmas~\ref{lem:apolar_containded_in_f} and \ref{lem:generators_in_degree_3} that we may assume that one of the following holds
\begin{enumerate}[label=(\arabic*)]
\item\label{it:wild_polynomials_part_one} $(\alpha_0^2\alpha_1, \alpha_0^2\alpha_2)\subseteq \Ann(F)$;
\item\label{it:wild_polynomials_part_two} $(\alpha_0^3, \alpha_0^2\alpha_1)\subseteq \Ann(F)$;
\item\label{it:wild_polynomials_part_three} $(\alpha_0^2\alpha_1, \alpha_0\alpha_1^2) \subseteq \Ann(F)$;
\item\label{it:wild_polynomials_part_four} $(\alpha_0^2\alpha_2, \alpha_0\alpha_1\alpha_2)\subseteq \Ann(F)$.
\end{enumerate}
Let $e = \lceil\frac{d+1}{2}\rceil$.  Let $<$ be the grevlex monomial order with $\alpha_0<\alpha_1<\alpha_2$.

Assume that \ref{it:wild_polynomials_part_one} holds. Let $\eta_e\in \Ann(F)_e$ be not divisible by $\alpha_0$ (see Lemma~\ref{lem:middle_generator}). We choose one of the form  $\eta_e = \zeta_e + a\alpha_0^e + \alpha_0\alpha_1\zeta_{e-2} + \alpha_0\alpha_2\zeta'_{e-2}$, where $a\in \kk$, $\zeta_e \in \kk[\alpha_1,\alpha_2]_{e}\setminus \{0\}$ and $\zeta_{e-2}, \zeta'_{e-2}\in \kk[\alpha_1,\alpha_2]_{e-2}$. Furthermore, if $a\neq 0$, then $\alpha_0^{e+2} \in \Ann(F)$. It follows that $\alpha_0^2\in \Ann(F)$, so by Lemma~\ref{lem:square_annihilates} the cactus rank of $F$ is at most $d+2$.
Let $J = \operatorname{in}_<(\alpha_0^2\alpha_1, \alpha_0^2\alpha_2, \eta_e)$. Then $J= (\alpha_0^2\alpha_1, \alpha_0^2\alpha_2, M)$ where $M$ is a monomial in $\kk[\alpha_1,\alpha_2]_e$. It follows that $S/J$ has Hilbert polynomial $2e+1\leq d+3$ and $J$ is saturated.

Assume that \ref{it:wild_polynomials_part_two} holds. Let $\eta_e\in \Ann(F)_e$ be not divisible by $\alpha_0$ (see Lemma~\ref{lem:middle_generator}). We choose one of the form $\eta_e = a\alpha_2^e + \alpha_1\zeta_{e-1} + b\alpha_0^2\alpha_2^{e-2} + \alpha_0\zeta'_{e-1}$, where $a,b\in \kk$, $\zeta_{e-1}, \zeta_{e-1}' \in \kk[\alpha_1,\alpha_2]_{e-1}$ and $a\alpha_2^2+\alpha_1\zeta_{e-1} \neq 0$. If $a\neq 0$, then $\alpha_0^2\alpha_2^e \in \Ann(F)$ and it follows that $\alpha_0^2\in \Ann(F)$. Then by Lemma~\ref{lem:square_annihilates} we get $\crr(F) \leq d+2$. Therefore, $J=\operatorname{in}_<(\alpha_0^3, \alpha_0^2\alpha_1, \eta_e) = (\alpha_0^3, \alpha_0^2\alpha_1, \alpha_1M)$ for a monomial $M\in \kk[\alpha_1,\alpha_2]_{e-1}$. It follows that $J$ is saturated and $S/J$ has Hilbert polynomial $2e+1 \leq d+3$.

If \ref{it:wild_polynomials_part_three} or \ref{it:wild_polynomials_part_four} hold, then $\crr(F) \leq d+3$ by Lemma~\ref{lem:special_case_3} or \ref{lem:special_case_4}, respectively.
\end{proof}

redpajama_set_name: RedPajamaArXiv